Thermodynamics Theory + Questions

Basic Concepts S K Mondal’s

1.

Chapter 1

Basic Concepts

Theory at a Glance (For GATE, IES & PSUs) Intensive and Extensive Properties Intensive property: Whose value is independent of the size or extent i.e. mass of the system. These are, e.g., pressure p and temperature T. Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case letters as the symbols). e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases. e.g., volume V, internal energy U, enthalpy H, entropy S, etc. Specific property: It is a special case of an intensive property. It is the value of an extensive property per unit mass of system. (Lower case letters as symbols) eg: specific volume, density (v, ρ).

Thermodynamic System and Control Volume • In our study of thermodynamics, we will choose a small part of the universe to which we will apply the laws of thermodynamics. We call this subset a SYSTEM. • The thermodynamic system is analogous to the free body diagram to which we apply the laws of mechanics, (i.e. Newton’s Laws of Motion). • The system is a macroscopically identifiable collection of matter on which we focus our attention (e.g., the water kettle or the aircraft engine).

System Definition • System: A quantity of matter in space which is analyzed during a problem. • Surroundings: Everything external to the system. • System Boundary: A separation present between system and surrounding. Classification of the system boundary:• Real solid boundary • Imaginary boundary

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B Basic C Conce epts S K Mondal’s

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Th he system bou undary may be b further cla assified as:: • Contrrol Mass Systtem. • Contrrol Volume Sy ystem. Th he choice of booundary depends on the problem p beiing analyzed d.

Ty ypes off Syste em

Closed System S (Contro ( ol Mass System m) 1. It’s a system of fixed masss with fixed identity. 2. Thiss type of system s is ussually referred d to as “closed system”.. 3. Therre is no masss transfer across a the systtem boundarry. 4. Energy transferr may take place into or out of the system. Fig. A Co ontrol Masss System or Closed C Systtem

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O Open Sy ystem (Contr rol Vollume System S ) 1. 2.

3. 4. 5.

6.

7. 8. 9. 10.

Its a system of fixed f volum me. Thiis type of system s is ussually refeerred to as "open system m” or a "c control volu ume" Mass transfer can take place acrooss a control volume. Eneergy transfer may also occur intoo or out of the system. A control volum me can be seen n as a fixeed region accross which mass and d energy tran nsfers are stu udied. Con ntrol Surfa ace – Its the bou undary of a control voolume Fig. A Co ontrol Volum me System acrooss which th he transfer off both or r Open System mass and energ gy takes placee. Thee mass of a co ontrol volum me (open systeem) may or may m not be fix xed. Wh hen the net in nflux of mass across the control surfa ace equals zeero then the mass of the systtem is fixed and a vice-verssa. Thee identity of mass in a control c volum me always ch hanges unlik ke the case for f a control mass system (cllosed system)). Most of the engiineering deviices, in generral, represent an open sysstem or contrrol volume.

Ex xample: • Heat exchanger - Fluid enters an nd leaves thee system con ntinuously with w the transfer of heat across the sy ystem bound dary. • Pump - A continuous c fllow of fluid takes place th hrough the system s with a transfer off mechanical energy from m the surroun ndings to the system.

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Issolated d Syste em 1. 2.

3.

It is a sysstem of fixed d mass with h same identity and fixed energy. e No intera action of masss or energy takes place beetween the system and d the surround dings. In more informal words w an isoolated system iss like a closed shop amiidst a busy marrket.

Fig. An n Isolated System S

Q Quasi-Sttatic Prrocess Th he processees un nrestrained

can

be e

restrained d

or

Wee need restra ained processses in practice. Aq quasi – staticc process is one o in which • The dev viation from m thermodyn namic equilibriu um is infinite esimal. • All statess of the syste em passes th hrough are equillibrium state es. • If we rem move the weig ghts slowly one o by Fig. A quasi q – static process one the pressure of the gass will displace the piston gradually. It is quasista atic. • On the other hand if we remove all a the weights at once th he piston will be kicked up p by the gas pressure. (This is un nrestrained expansion) but b we don’tt consider th hat the work k is done – because it i is not in a sustained manner. • In both cases c the systtems have un ndergone a ch hange of statte. • Another e.g., if a person climbs down d a ladder from roof to t ground, it is a quasista atic process. On the otther hand if he jumps theen it is not a quasistatic process. p

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La aws of Thermodynam mics

h Law deals with therrmal equilib brium and provides a means for measuring • The Zeroth temperaturees. L deals with w the consservation of energy and introduces the concept of internal • The First Law energy. modynamics provides witth the guideliines on the coonversion heeat energy of • The Second Law of therm matter into work. It also o introduces the t concept of o entropy. L of therm modynamics defines the absolute zerro of entropy y. The entrop py of a pure • The Third Law crystalline substance s at absolute zeroo temperaturre is zero.

Summattion of 3 Laws s • Firstly, there isn’t a meaningful m teemperature of the sourcce from whiich we can get g the full conversion of o heat to wo ork. Only at infinite tem mperature onee can dream of getting th he full 1 kW work outputt. • Secondly, more m interestiingly, there isn’t i enough work availab ble to producce 0 K. In oth her words, 0 K is unattaiinable. This is i precisely th he Third law w. • Because, wee don’t know what 0 K loooks like, we haven’t got a starting pooint for the temperature t scale!! That is why all te emperature scales are at best b empirica al. Yo ou can’t get something for nothing g: To get work k output you must m give som me thermal energy. e Yo ou can’t get something for very litttle: To get some e work outputt there is a minimum m amount of therm mal energy th hat needs to be given. 5

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Yo ou can’t get every thing g: However mu uch work you u are willing to give 0 K can’t c be reach hed. Vio olation of all a 3 laws: Try to get ev verything forr nothing.

eroth Law L of Thermo T odynam mics Ze • If two systeems (say A an nd B) are in thermal equ uilibrium witth a third sy ystem (say C)) separately (that is A an nd C are in th hermal equillibrium; B an nd C are in th hermal equillibrium) then n they are in thermal equ uilibrium themselves (tha at is A and B will be in thermal equilib brium).

•

All tem mperature measuremen m nts are base ed on Zeroth h law of the ermodynam mics

In nternatiional Te empera ature Sc cale To provide a standard s forr temperaturre measurem ment taking g into accoun nt both theooretical and pra actical consid derations, th he Internation nal Tempera ature Scale (IITS) was adoopted in 1927 7. This scale has been refin ned and exte ended in several revisioons, most reecently in 1990. The In nternational Temperature Scale S of 1990 (ITS-90) is defined in su uch a way th hat the temp perature mea asured on it con nforms with the thermod dynamic temp perature, thee unit of which is the kellvin, to withiin the limits of accuracy of measuremen nt obtainablle in 1990. The T ITS–90 is based on the assigneed values of tem mperature off a number of o reproducib ble fixed poin nts (Table). In nterpolation between thee fixed-point tem mperatures is i accomplisshed by form mulas that give g the rela ation between readings of o standard insstruments an nd values of the ITS. In the t range froom 0.65 to 5.0 K, ITS-90 is defined by b equations giv ving the temp perature as functions f of the t vapor prressures of pa articular helium isotopess. The range froom 3.0 to 24.5561 K is ba ased on meassurements ussing a helium m constant-voolume gas th hermometer. In the range frrom 13.8033 to 1234.93 K, K ITS-90 is defined by means m of certtain platinum m resistance theermometers. Above 1234 4.9 K the tem mperature iss defined usiing Planck’s equation for blackbody rad diation and measuremen m nts of the inteensity of visib ble-spectrum m radiation th he absolute temperature t

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ale. The absoolute temperrature scale is also know wn as Kelvin n temperaturre scale. In defining d the sca Keelvin tempera ature scale also, the triple point of wa ater is taken as the stand dard referencce point. For aC Carnot engin ne operating between reseervoirs at tem mperature θ and θtp, θtp being the triiple point of wa ater arbitrariily assigned the t value of 273.16 K. Tim me Constan nts: The time constant iss the amountt of time req quired for a thermocouple t e to indicate 63.2% of step change c in te emperature of o a surround ding media. Some of thee factors influ uencing the meeasured timee constant are sheath walll thickness, degree of inssulation com mpaction, and d distance of jun nction from the t welded ca an on an ung grounded theermocouple. In I addition, the t velocity of o a gas past thee thermocoup ple probe gre eatly influencces the time constant c mea asurement. In general, tim me constants for f measurem ment of gas can c be estima ated to be ten times as loong as those forr measuremeent of liquid. The time con nstant also varies v inverseely proportional to the sq quare root of thee velocity of the t media.

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W Work a path p fun nction Woork is one of the basic mo odes of energy y transfer. Th he work donee by a system m is a path fu unction, and nott a point fun nction. There efore, work is not a pr roperty of th he system, and it cann not be said tha at the work k is posses ssed by the system. It is an intera action acrosss the bounda ary. What is stoored in the system is ene ergy, but nott work. A decrease in energy of the system s appea ars as work don ne. Therefore, work is energy in transit and d it can be identified i o only when the t system un ndergoes a process. p

Free Exp pansion n with Zero Z W Work Tra ansfer Fr ree Expansio on Let us con nsider an insu ulated containeer (Figure) wh hich is divided d into two com mpartments A and d B by a thin n diaphragm. Compartmen nt A contains a mass of ga as, while com mpartment B is i completely eva acuated. If thee diaphragm is punctured, the gas in A will expand into B until the t pressures in A and B beccome equal. This T is known n as free or unrestrained expansion.

T The process of free ex xpansion is

irr reversible.. Also work done is zeroo during freee expansion n.

Free Expa ansion

pd dV-worrk or Displacement Work W Let the gas in the t cylinder (Figure show wn in below) be a system having initia ally the pressure p1 and vollume V1. Th he system is in thermody ynamic equilibrium, the state of wh hich is described by the cooordinates p1 , V1. The pistton is the on nly boundary which moves due to gas pressure. Leet the piston moove out to a new final po osition 2, wh hich is also a thermodyn namic equilib brium state specified by preessure p2 an nd volume V2. At any inteermediate poiint in the tra avel of the piiston, let the pressure be pa and the volu ume V. This must m also bee an equilibrium state, siince macrosccopic propertties p and V 8

B Basic C Conce epts S K Mondal’s

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sig gnificant only.

m states. Whe en the piston n moves an in nfinitesimal distance dl, and if ‘a' bee the area of forr equilibrium thee piston, the force F actin ng on the pistton F = p.a. and a the infinitesimal amoount of work done by the gass on the pistoon.

dW W=

F ⋅ dl d = pad dl = pdV V

wh here dV = ad dl = infinitesimal displa acement volu ume. The diffferential sig gn in dW with the line dra awn at the toop of it will be explained later. l Wh hen the pistoon moves out from positioon 1 to positioon 2 with thee volume cha anging from V1 to V2, the am mount of work k W done by the t system will w be

W1−2 =

∫

V2

V1

9

pd dV

B Basic C Conce epts S K Mondal’s

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Th he magnitudee of the worrk done is given g by thee area underr the path 1--2, as shown n in Fig. Sin nce p is att all times a thermod dynamic cooordinate, all the states pa assed through h by the sysstem as the volume cha anges from V1 to V2 mu ust be equiliibrium statess, and the path p 1-2 mu ust be qua asi-static. The T piston moves inffinitely slow wly so that every state passed thrrough is an equilibrium e state. Th he integration n

∫ pdV

can n be perform med only

on a quasi-sta atic path.

Fig. Quasi-Static pdV Work k

Heat Tra ansfer-A A Path Functio on Heeat transfer is i a path fu unction, thatt is, the amoount of heat transferred when a systtem changes froom state 1 to state 2 depe ends on the intermediate i e states throu ugh which th he system pa asses, i.e. its patth. Thereforee dQ is an ine exact differen ntial, and wee write

∫

2

1

dQ Q = Q1−2 or

1

Q2 ≠ Q2 − Q1

Th he displacemeent work is given g by W1−2 =

∫

2

1

dW =

∫

2

1

PR ROBLEM MS & SO OLUTION NS

pdV ≠ W2 − W1

Ex xample 1 In a closed systtem, volume changes from m 1.5m3 to 4.5 m3 and hea at addition iss 2000 kJ. Ca alculate the cha ange in interrnal energy given g the presssure volumee relation as 10 ⎞ ⎛ W Where p is in n kPa and V is in m3. p = ⎜V 2 + V ⎟⎠ ⎝ So olution:

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Basic Concepts S K Mondal’s Work done =

V2

∫

p.dV =

V1

Chapter 1

V2

⎛

∫ ⎜⎝ V

V1

2

+

10 ⎞ dV V ⎟⎠

⎡1 V ⎤ = ⎢ V23 − V13 + 10ln 2 ⎥ V1 ⎦ ⎣3 4.5 ⎤ ⎡1 = ⎢ 4.53 − 1.53 + 10ln 1.5 ⎥⎦ ⎣3

(

(

)

)

⎡1 ⎤ = ⎢ ( 91.125 − 3.375 ) + 10ln3⎥ 3 ⎣ ⎦ = [29.250 + 10.986] = 40.236 kJ

First Law of Thermodynamics:Q = W + ΔU 2000 = 40.236 + ΔU ∴ ΔU = 2000 – 40.236 = 1959.764 kJ Example 2. A fluid is contained in a cylinder piston arrangement that has a paddle that imparts work to the fluid. The atmospheric pressure is 760 mm of Hg. The paddle makes 10,000 revolutions during which the piston moves out 0.8m. The fluid exerts a torque of 1.275 N-m one the paddle. What is net work transfer, if the diameter of the piston is 0.6m? Solution: Work done by the stirring device upon the system W1 = 2πTN = 2π × 1.275 × 10000 N-m = 80kJ This is negative work for the system.

(Fig.) Work done by the system upon the surroundings. W2 = p.dV = p.(A × L)

π

(0.6)2 × 0.80 = 22.9kJ 4 This is positive work for the system. Hence the net work transfer for the system. W = W1 + W2 = - 80 + 22.9 = - 57.l kJ. = 101.325 ×

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Basic Concepts S K Mondal’s

Chapter 1

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions GATE-1.

List-I A. Heat to work B. Heat to lift weight C. Heat to strain energy D. Heat to electromagnetic energy 5. Thermal radiation 6. Bimetallic strips Codes: A B C D (a) 3 4 6 5 (c) 3 6 4 2

1. 2. 3. 4.

(b) (d)

List II [GATE-1998] Nozzle Endothermic chemical reaction Heat engine Hot air balloon/evaporation A 3 1

B 4 2

C 5 3

D 6 4

Open and Closed systems GATE-2.

An isolated thermodynamic system executes a process, choose the correct statement(s) form the following [GATE-1999] (a) No heat is transferred (b) No work is done (c) No mass flows across the boundary of the system (d) No chemical reaction takes place within the system

GATE-2a. Heat and work are (a) intensive properties (c) point functions

[GATE-2011]

(b) extensive properties (d) path functions

Quasi-Static Process GATE-3.

A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ/kg) during this process will be: [GATE-2009] (a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00

Free Expansion with Zero Work Transfer GATE-4.

A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process? (a) The internal energy of the gas decreases from its initial value, but the enthalpy remains constant [GATE-2008] (b) The internal energy of the gas increases from its initial value, but the enthalpy remains constant (c) Both internal energy and enthalpy of the gas remain constant (d) Both internal energy and enthalpy of the gas increase

12

Basic Concepts S K Mondal’s GATE-5.

Chapter 1

Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by: [GATE-1996, IAS-2000] (a) –∫Pdv (b) +∫Pdv (c) –∫vdp (d) +∫vdp

pdV-work or Displacement Work GATE-6.

In a steady state steady flow process taking place in a device with a single inlet and a single outlet, the work done per unit mass flow rate is given by

ω=−

outlet

∫

vdp , where v is the specific volume and p is the pressure. The

inlet

expression for w given above: (a) Is valid only if the process is both reversible and adiabatic (b) Is valid only if the process is both reversible and isothermal (c) Is valid for any reversible process

(d) Is incorrect; it must be ω = −

[GATE-2008]

outlet

∫

vdp

inlet

GATE-7.

A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work that could be utilized from the above process is: [GATE-2008] (a) 0kJ (b) 1kJ (c) 2kJ (d) 3kJ

GATE-8.

For reversible adiabatic compression in a steady flow process, the work transfer per unit mass is: [GATE-1996]

(a) ∫ pdv

(b) ∫ vdp

(c) ∫ Tds

(d ) ∫ sdT

Previous 20-Years IES Questions IES-1.

Which of the following are intensive properties? [IES-2005] 1. Kinetic Energy 2. Specific Enthalpy 3. Pressure 4. Entropy Select the correct answer using the code given below: (a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4

IES-2.

Consider the following properties: [IES-2009] 1. Temperature 2. Viscosity 3. Specific entropy 4. Thermal conductivity Which of the above properties of a system is/are intensive? (a) 1 only (b) 2 and 3 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4

IES-2a.

Consider the following: 1. Kinetic energy

2. Entropy

13

[IES-2007, 2010]

Basic Concepts S K Mondal’s

Chapter 1

3. Thermal conductivity 4. Pressure Which of these are intensive properties? (a) 1, 2 and 3 only (b) 2 and 4 only (c) 3 and 4 only (d) 1, 2, 3 and 4 IES-3.

Which one of the following is the extensive property of a thermodynamic system? [IES-1999] (a) Volume (b) Pressure (c) Temperature (d) Density

IES-4.

Consider the following properties: [IES-2009] 1. Entropy 2. Viscosity 3. Temperature 4. Specific heat at constant volume Which of the above properties of a system is/are extensive? (a) 1 only (b) 1 and 2 only (c) 2, 3 and 4 (d) 1, 2 and 4

IES-4a

Consider the following: 1. Temperature 2. Viscosity 3. Internal energy 4. Entropy Which of these are extensive properties? (a) 1, 2, 3 and 4 (b) 2 and 4 only (c) 2 and 3 only (d) 3 and 4 only.

[IES-2010]

Thermodynamic System and Control Volume IES-5.

Assertion (A): A thermodynamic system may be considered as a quantity of working substance with which interactions of heat and work are studied. Reason (R): Energy in the form of work and heat are mutually convertible. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false [IES-2000] (d) A is false but R is true

IES-5a

A control volume is [IES-2010] (a) An isolated system (b) A closed system but heat and work can cross the boundary (c) A specific amount of mass in space (d) A fixed region in space where mass, heat and work can cross the boundary of that region

Open and Closed systems IES-6.

A closed thermodynamic system is one in which [IES-1999, 2010, 2011] (a) There is no energy or mass transfer across the boundary (b) There is no mass transfer, but energy transfer exists (c) There is no energy transfer, but mass transfer exists (d) Both energy and mass transfer take place across the boundary, but the mass transfer is controlled by valves

IES-7

Isothermal compression of air in a Stirling engine is an example of

14

Basic Concepts S K Mondal’s (a) (b) (c) (d)

Chapter 1

Open system Steady flow diabatic system Closed system with a movable boundary Closed system with fixed boundary

[IES-2010]

IES-8.

Which of the following is/are reversible process(es)? [IES-2005] 1. Isentropic expansion 2. Slow heating of water from a hot source 3. Constant pressure heating of an ideal gas from a constant temperature source 4. Evaporation of a liquid at constant temperature Select the correct answer using the code given below: (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4

IES-9.

Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a reversible process is the most efficient process. [IES-2001] Reason (R): The energy transfer as heat and work during the forward process as always identically equal to the energy transfer is heat and work during the reversal or the process. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-9a

Which one of the following represents open thermodynamic system? (a) Manual ice cream freezer (b) Centrifugal pump (c) Pressure cooker (d) Bomb calorimeter

[IES-2011]

IES-10.

Ice kept in a well insulated thermo flask is an example of which system? (a) Closed system (b) Isolated systems [IES-2009] (c) Open system (d) Non-flow adiabatic system

IES-10a

Hot coffee stored in a well insulated thermos flask is an example of (a) Isolated system (b) Closed system (c) Open system (d) Non-flow diabatic system

[IES-2010]

IES10b

A thermodynamic system is considered to be an isolated one if (a) Mass transfer and entropy change are zero (b) Entropy change and energy transfer are zero (c) Energy transfer and mass transfer are zero (d) Mass transfer and volume change are zero

[IES-2011]

IES-10c.

Match List I with List II and select the correct answer using the code given below the lists: [IES-2011] List I List II A. Interchange of matter is not possible in a 1. Open system B. Any processes in which the system returns to 2. System its original condition or state is called C. Interchange of matter is possible in a 15

Basic Concepts S K Mondal’s

Chapter 1

D. The quantity of matter under consideration in thermodynamics is called

Code: (a) (c)

A 2 2

B 1 4

C 4 1

D 3 3

(b) (d)

A 3 3

B 1 4

3. Closed system 4. Cycle

C 4 1

D 2 2

Zeroth Law of Thermodynamics IES-11.

Measurement of temperature is based on which law of thermodynamics? [IES-2009] (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics

IES-12.

Consider the following statements: [IES-2003] 1. Zeroth law of thermodynamics is related to temperature 2. Entropy is related to first law of thermodynamics 3. Internal energy of an ideal gas is a function of temperature and pressure 4. Van der Waals' equation is related to an ideal gas Which of the above statements is/are correct? (a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4

IES-13.

Zeroth Law of thermodynamics states that [IES-1996, 2010] (a) Two thermodynamic systems are always in thermal equilibrium with each other. (b) If two systems are in thermal equilibrium, then the third system will also be in thermal equilibrium with each other. (c) Two systems not in thermal equilibrium with a third system are also not in thermal equilibrium with each other. (d) When two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

International Temperature Scale IES-14.

Which one of the following correctly defines 1 K, as per the internationally accepted definition of temperature scale? [IES-2004] (a) 1/100th of the difference between normal boiling point and normal freezing point of water (b) 1/273.15th of the normal freezing point of water (c) 100 times the difference between the triple point of water and the normal freezing point of water (d) 1/273.15th of the triple point of water

IES-15.

In a new temperature scale say °ρ, the boiling and freezing points of water at one atmosphere are 100°ρ and 300°ρ respectively. Correlate this scale with the Centigrade scale. The reading of 0°ρ on the Centigrade scale is: [IES-2001] (a) 0°C (b) 50°C (c) 100°C (d) 150°C

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Basic Concepts S K Mondal’s

Chapter 1

IES-16.

Assertion (a): If an alcohol and a mercury thermometer read exactly 0°C at the ice point and 100°C at the steam point and the distance between the two points is divided into 100 equal parts in both thermometers, the two thermometers will give exactly the same reading at 50°C. [IES-1995] Reason (R): Temperature scales are arbitrary. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-17.

Match List-I (Type of Thermometer) with List-II (Thermometric select the correct answer using the code given below the List-I List-II 1. Pressure A. Mercury-in-glass B. Thermocouple 2. Electrical resistant C. Thermistor 3. Volume D. Constant volume gas 4. Induced electric voltage Codes: A B C D A B (a) 1 4 2 3 (b) 3 2 (c) 1 2 4 3 (d) 3 4

Property) and [IES 2007]

C 4 2

D 1 1

IES-18.

Pressure reaches a value of absolute zero (a) At a temperature of – 273 K (b) Under vacuum condition (c) At the earth's centre (d) When molecular momentum of system becomes zero

[IES-2002]

IES-19.

The time constant of a thermocouple is the time taken to attain: (a) The final value to he measured [IES-1997, 2010] (b) 50% of the value of the initial temperature difference (c) 63.2% of the value of the initial temperature difference (d) 98.8% of the value of the initial temperature difference

Work a Path Function IES-20.

Assertion (A): Thermodynamic work is path-dependent except for an adiabatic process. [IES-2005] Reason(R): It is always possible to take a system from a given initial state to any final state by performing adiabatic work only. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-20a

Work transfer between the system and the surroundings (a) Is a point function (b) Is always given by ∫ pdv

(c) Is a function of pressure only

[IES-2011]

(d) Depends on the path followed by the system 17

Basic Concepts S K Mondal’s

Chapter 1

Free Expansion with Zero Work Transfer IES-21.

Match items in List-I (Process) with those in List-II (Characteristic) and select the correct answer using the codes given below the lists: List-I List-II [IES-2001] A. Throttling process 1. No work done B. Isentropic process 2. No change in entropy C. Free expansion 3. Constant internal energy D. Isothermal process 4. Constant enthalpy Codes: A B C D A B C D (a) 4 2 1 3 (b) 1 2 4 3 (c) 4 3 1 2 (d) 1 3 4 2

IES-22.

The heat transfer, Q, the work done W and the change in internal energy U are all zero in the case of [IES-1996] (a) A rigid vessel containing steam at 150°C left in the atmosphere which is at 25°C. (b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly outwards. (c) A rigid vessel containing ammonia gas connected through a valve to an evacuated rigid vessel, the vessel, the valve and the connecting pipes being well insulated and the valve being opened and after a time, conditions through the two vessels becoming uniform. (d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated bottle.

pdV-work or Displacement Work IES-23.

One kg of ice at 0°C is completely melted into water at 0°C at 1 bar pressure. The latent heat of fusion of water is 333 kJ/kg and the densities of water and ice at 0°C are 999.0 kg/m3 and 916.0 kg/m3, respectively. What are the approximate values of the work done and energy transferred as heat for the process, respectively? [IES-2007] (a) –9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ (c) –333.0 kJ and –9.4 J (d) None of the above

IES-24.

Which one of the following is the correct sequence of the three processes A, B and C in the increasing order of the amount of work done by a gas following idealgas expansions by these processes?

18

Basic Concepts S K Mondal’s IES-25.

Chapter 1

(a) A – B – C (b) B – A – C An ideal gas undergoes an isothermal expansion from state R to state S in a turbine as shown in the diagram given below:

[IES-2006] (d) C – A – B

(c) A – C – B

The area of shaded region is 1000 Nm. What is the amount is turbine work done during the process? (a) 14,000 Nm (b) 12,000 Nm (c) 11,000 Nm (d) 10,000 Nm IES-26.

[IES-2004]

Assertion (A): The area 'under' curve on pv plane,

∫ pdv represents the work of

reversible non-flow process.

[IES-1992]

Reason (R): The area 'under' the curve T–s plane

∫ Tds represents heat of any

reversible process. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-27.

If

∫ pdv

∫

and − vdp for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process undergone by the system is: [IES-2003]

(a) Isomeric IES-28.

(b) Isentropic

(c) Isobaric

(d) Isothermal

Which one of the following expresses the reversible work done by the system (steady flow) between states 1 and 2? [IES-2008] 2

(a)

∫ pdv 1

2

2

(b) − ∫ vdp

(c) − ∫ pdv

1

1

2

(d)

∫ vdp 1

Heat Transfer-A Path Function IES-29.

Assertion (A): The change in heat and work cannot be expressed as difference between the end states. [IES-1999] Reason (R): Heat and work are both exact differentials. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

19

Basic Concepts S K Mondal’s

Chapter 1

Previous 20-Years IAS Questions Thermodynamic System and Control Volume IAS-1.

The following are examples of some intensive and extensive properties: 1. Pressure 2. Temperature [IAS-1995] 3. Volume 4. Velocity 5. Electric charge 6. Magnetisation 7. Viscosity 8. Potential energy Which one of the following sets gives the correct combination of intensive and extensive properties? Intensive Extensive (a) 1, 2, 3, 4 5, 6, 7, 8 (b) 1, 3, 5, 7 2, 4, 6, 8 (c) 1, 2, 4, 7 3, 5, 6, 8 (d) 2, 3, 6, 8 1, 4, 5, 7

Zeroth Law of Thermodynamics IAS-2.

Match List-I with List-II and select the correct answer using the codes given below the lists: [IAS-2004] List-I List-II A. Reversible cycle 1. Measurement of temperature B. Mechanical work 2. Clapeyron equation C. Zeroth Law 3. Clausius Theorem D. Heat 4. High grade energy 5. 3rd law of thermodynamics 6. Inexact differential Codes: A B C D A B C D (a) 3 4 1 6 (b) 2 6 1 3 (c) 3 1 5 6 (d) 1 4 5 2

IAS-3.

Match List-I with List-II and select the correct answer: [IAS-2000] List-I List-II A. The entropy of a pure crystalline 1. First law of thermodynamics substance is zero at absolute zero temperature B. Spontaneous processes occur 2. Second law of thermodynamics in a certain direction C. If two bodies are in thermal 3. Third law of thermodynamics equilibrium with a third body, then they are also in thermal equilibrium with each other D. The law of conservation of energy 4. Zeroth law of thermodynamics. Codes: A B C D A B C D (a) 2 3 4 1 (b) 3 2 1 4 20

Basic Concepts S K Mondal’s (c)

Chapter 1 3

2

4

1

(d)

2

3

1

4

International Temperature Scale IAS-4.

A new temperature scale in degrees N is to be defined. The boiling and freezing on this scale are 400°N and 100°N respectively. What will be the reading on new scale corresponding to 60°C? [IAS-1995] (c) 220°N (d) 280°N (a) 120°N (b) 180°N

Free Expansion with Zero Work Transfer IAS-5.

In free expansion of a gas between two equilibrium states, the work transfer involved [IAS-2001] (a) Can be calculated by joining the two states on p-v coordinates by any path and estimating the area below (b) Can be calculated by joining the two states by a quasi-static path and then finding the area below (c) Is zero (d) Is equal to heat generated by friction during expansion.

IAS-6.

Work done in a free expansion process is: (a) Positive (b) Negative (c) Zero

IAS-7.

In the temperature-entropy diagram of a vapour shown in the given figure, the thermodynamic process shown by the dotted line AB represents (a) Hyperbolic expansion (b) Free expansion (c) Constant volume expansion (d) Polytropic expansion

[IAS-2002] (d) Maximum

[IAS-1995] IAS-8.

If

∫ pdv

∫

and − vdp for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process undergone by the system is: [IAS-1997, IES-2003] (a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal IAS-9.

For the expression

∫ pdv

to represent the work, which of the following

conditions should apply? [IAS-2002] (a) The system is closed one and process takes place in non-flow system (b) The process is non-quasi static (c) The boundary of the system should not move in order that work may be transferred (d) If the system is open one, it should be non-reversible

21

Basic Concepts S K Mondal’s IAS-10.

IAS-11.

Chapter 1

Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by: [IAS-2000, GATE-1996] (a) –∫pdv (b) +∫pdv (c) –∫vdp (d) +∫vdp Match List-I with List-II and select the correct answer using the codes given below the lists: [IAS-2004] List-I List-II A. Bottle filling of gas 1. Absolute Zero Temperature B. Nernst simon Statement 2. Variable flow C. Joule Thomson Effect 3. Quasi-Static Path D. ∫pdv 4. Isentropic Process 5. Dissipative Effect 6. Low grade energy 7. Process and temperature during phase change. Codes: A B C D A B C D (a) 6 5 4 3 (b) 2 1 4 3 (c) 2 5 7 4 (d) 6 1 7 4

pdV-work or Displacement Work IAS-13.

Thermodynamic work is the product of (a) Two intensive properties (b) Two extensive properties (c) An intensive property and change in an extensive property (d) An extensive property and change in an intensive property

[IAS-1998]

Heat Transfer-A Path Function IAS-14.

Match List-I (Parameter) with List-II using the codes given below the lists: List-I A. Volume B. Density C. Pressure D. Work Codes: A B C D (a) 3 2 4 1 (c) 2 3 4 1

22

(Property) and select the correct answer List-II 1. Path function 2. Intensive property 3. Extensive property 4. Point function A B C (b) 3 2 1 (d) 2 3 1

[IAS-1999]

D 4 4

Basic Concepts S K Mondal’s

Chapter 1

Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. Ans. (a) GATE-2. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system.

dQ = 0,

dW = 0,

∴ dE = 0

or E = Constant

GATE-2a. Ans. (d)

⎛ V2 ⎞ ⎟ ⎝ V1 ⎠ ⎛V ⎞ = P1V1 ln ⎜ 2 ⎟ ⎝ V1 ⎠

GATE-3. Ans. (a) Iso-thermal work done (W) = RT1 ln ⎜

⎛ 0.030 ⎞ = 800 × 0.015 × ln ⎜ ⎟ ⎝ 0.015 ⎠ = 8.32 kJ/kg GATE-4. Ans. (c) It is free expansion. Since vacuum does not offer any resistance, there is no work transfer involved in free expansion. 2

Here,

∫ δω = 0

and Q1-2=0 therefore Q1-2 = ΔU + W1-2 so, ΔU = 0

1

GATE-5. Ans. (c) For closed system W =

+ ∫ pdv , for steady flow W = − ∫ vdp

GATE-6. (c) GATE-7. Ans. (b) W = Resistance pressure. Δ V = 1 × Δ V = 100 × 0.1 kJ = 1kJ GATE-8. Ans. (b) W = − ∫ vdp

Previous 20-Years IES Answers IES-1. Ans. (b) IES-2. Ans. (d) Intensive property: Whose value is independent of the size or extent i.e. mass of the system. Specific property: It is a special case of an intensive property. It is the value of an extensive property per unit mass of system (Lower case letters as symbols) e.g., specific volume, density (v, ρ). IES-2a.

Ans. (c) Kinetic energy

1 mv 2 depends on mass, Entropy kJ/k depends on mass so 2

Entropy is extensive property but specific entropy kJ/kg K is an intensive property. 23

Basic Concepts S K Mondal’s

Chapter 1

IES-3. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive property. IES-4. Ans. (a) Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case letters as the symbols) e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases. IES-4a Ans. (d) The properties like temperature, viscosity which are Independent of the MASS of the system are called Intensive property IES-5. Ans. (d) • But remember 100% heat can’t be convertible to work but 100% work can be converted to heat. It depends on second law of thermodynamics. • A thermodynamic system is defined as a definite quantity of matter or a region in space upon which attention is focused in the analysis of a problem. • The system is a macroscopically identifiable collection of matter on which we focus our attention IES-5a Ans. (d) IES-6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transfer exists. IES-7. Ans. (c) IES-8. Ans. (d) Isentropic means reversible adiabatic. Heat transfer in any finite temp difference is irreversible. IES-9. Ans. (a) The energy transfer as heat and work during the forward process as always identically equal to the energy transfer is heat and work during the reversal or the process is the correct reason for maximum efficiency because it is conservative system. IES-9a. Ans. (b) IES-10. Ans. (b) Isolated System - in which there is no interaction between system and the surroundings. It is of fixed mass and energy, and hence there is no mass and energy transfer across the system boundary. IES-10a Ans. (a) IES-10b. Ans. (c) IES-10c. Ans. (d) IES-11. Ans. (a) All temperature measurements are based on Zeroth law of thermodynamics. IES-12. Ans. (a) Entropy - related to second law of thermodynamics. Internal Energy (u) = f (T) only (for an ideal gas) Van der Wall's equation related to => real gas. IES-13. Ans. (d) IES-14. Ans. (d) IES-15.Ans. (d)

C −0 0 − 300 = ⇒ C = 150°C 100 − 300 100 − 0

24

Basic Concepts S K Mondal’s

Chapter 1

IES-16. Ans. (b) Both A and R are correct but R is not correct explanation for A. Temperature is independent of thermometric property of fluid. IES-17. Ans. (d) IES-18. Ans. (d) But it will occur at absolute zero temperature. IES-19. Ans. (c) Time Constants: The time constant is the amount of time required for a thermocouple to indicated 63.2% of step change in temperature of a surrounding media. Some of the factors influencing the measured time constant are sheath wall thickness, degree of insulation compaction, and distance of junction from the welded cap on an ungrounded thermocouple. In addition, the velocity of a gas past the thermocouple probe greatly influences the time constant measurement. In general, time constants for measurement of gas can be estimated to be ten times as long as those for measurement of liquid. The time constant also varies inversely proportional to the square root of the velocity of the media. IES-20. Ans. (c) IES-20a Ans. (d) IES-21. Ans. (a) IES-22. Ans. (c) In example of (c), it is a case of free expansion heat transfer, work done, and changes in internal energy are all zero.

⎛m ⎝ ρ2

IES-23. Ans. (a) Work done (W) = P Δ V = 100 × (V2 – V1) = 100 × ⎜

−

m⎞ ⎟ ρ1 ⎠

1 ⎞ ⎛ 1 = 100 kPa × ⎜ − ⎟ = –9.1 J 999 916 ⎝ ⎠

IES-24. Ans. (d)

WA = ∫ pdV = 4 × (2 − 1) = 4 kJ

1 WB = ∫ pdV = × 3 × (7 − 4) = 4.5 kJ 2 WC = ∫ pdV = 1× (12 − 9) = 3kJ IES-25. Ans. (c) Turbine work = area under curve R–S

= ∫ pdv = 1 bar × ( 0.2 − 0.1) m3 + 1000 Nm

= 105 × ( 0.2 − 0.1) Nm + 1000Nm = 11000 Nm IES-26. Ans. (b) IES-27. Ans. (d) Isothermal work is minimum of any process.

pv = mRT

pdv + vdp = 0[ ∵ T is constant]

∫ pdv = − ∫ vdp 2

∫

IES-28. Ans. (b) For steady flow process, reversible work given by − vdp . 1

IES-29. Ans. (c) A is true because change in heat and work are path functions and thus can't be expressed simply as difference between the end states. R is false because both work and heat are inexact differentials.

25

Basic Concepts S K Mondal’s

Chapter 1

Previous 20-Years IAS Answers IAS-1. Ans. (c) Intensive properties, i.e. independent of mass are pressure, temperature, velocity and viscosity. Extensive properties, i.e. dependent on mass of system are volume, electric charge, magnetisation, and potential energy. Thus correct choice is (c). IAS-2. Ans. (a) IAS-3. Ans. (c) IAS-4. Ans. (d) The boiling and freezing points on new scale are 400° N and 100°N i.e. range is 300°N corresponding to 100°C. Thus conversion equation is °N = 100 + 3 × °C = 100+ 3 × 60 = 100 + 180 = 280 °N IAS-5. Ans. (c) IAS-6. Ans. (c) Since vacuum does not offer any resistance, there is no work transfer involved in free expansion. IAS-7. Ans. (b) IAS-8. Ans. (d) Isothermal work is minimum of any process. IAS-9. Ans. (a) IAS-10. Ans. (c) For closed system W =

+ ∫ pdv , for steady flow W = − ∫ vdp

IAS-12. Ans. (b) Start with D. ∫PdV only valid for quasi-static path so choice (c) & (d) out. Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2. IAS-13. Ans. (c) W = ∫ pdv where pressure (p) is an intensive property and volume (v) is an

extensive property IAS-14. Ans. (a) Pressure is intensive property but such option is not there.

26

Fir rst Law w of Therm T odyna amics S K Mondal’s

2.

Cha apter 2

Firstt Law w of T herm modyn namic cs

Theo ory at a Glance (Fo or GAT TE, IES S & PS SUs) First Law w of Th hermody ynamic cs Sttatement: •

When a closed syste em executes a complete cy ycle the sum of heat interacttions is equa al to the sum of work interacttions. Mathem matically

• (Σ Q)cyclle = (Σ Σ W)cycle The sum mmations be eing over the entire cycle.

Allternate sttatement:: Wh hen a closed system undergoes a cyclle the cyclic integral of heat h is equal to the cyclicc integral of woork. Ma athematicallly

∫ δQ

=

∫ δW

In other word ds for a two process cycle

Q A1 A − 2 + QB2 −1 = WA1− 2 + WB 2 −1 In nternal Energy y – A Prroperty y of Sys stem

W Which can be e written ass

28

Fir rst Law w of Therm T odyna amics S K Mondal’s

•

• •

Cha apter 2

Since A and B are arbitrarily a ch hosen, the con nclusion is, as a far as a prrocess is conccerned (A or B) the difference δQ – δW rema ains a consta ant as long as a the initial and the fina al states are the sam me. The diffe erence depends only on th he end points of the proccess. Note that Q and W themseelves depend on the path followed. f Butt their differeence does nott. This im mplies that th he difference between thee heat and woork interactioons during a process is a propertty of the system. This prroperty is called the energ gy of the systtem. It is dessignated as E and is equa al to some of all the energies e at a given state.

Wee enunciate the t FIRST LA AW for a proccess as

δQ δ – δW W = dE

• • • • •

An isollated system m which do oes not inter ract with th he surround dings Q = 0 and W = 0. Thereffore, E rema ains constan nt for such a system. Let us reconsider r th he cycle 1-2 along a path A and 2-1 alon ng path B as shown s in fig. Work done d during the path A = Area A under 1-A-2-3-4 1 Work done d during the path B = Area A under 1-B-2-3-4 1 Since these two are eas are not equal, e the neet work interraction is tha at shown by the shaded area.

29

Fir rst Law w of Therm T odyna amics S K Mondal’s • • •

• • • •

• • • •

Cha apter 2

The nett area is 1A2B1. Therefoore some worrk is derived by the cycle. First la aw compels that this iss possible only o when there t is also o heat interaction betweeen the systtem and the surrounding gs. In oth her words, if i you have e to get wo ork out, yo ou must give e heat in. Thus, the t first law can be consttrued to be a statement of o conservatioon of energy - in a broad sense. In the example e show wn the area under u curve A < that und der B The cyccle shown ha as negative work outputt or it will receive work from the su urroundings. Obviously, the net heat h interacttion is also neegative. Thiss implies that this cycle will w heat the nment. (as pe er the sign coonvention). environ For a process we can n have Q = 0 or W = 0 We can n extract worrk without su upplying heat (during a process) bu ut sacrificing g the energy of the system. s We can n add heat to the system without doin ng work (in process) p wh hich will go too increasing the eneergy of the sy ystem. Energy y of a system m is an exte ensive prop perty

he internal en nergy depend ds only upon the initial an nd final statees of the systtem. Internall energy of a Th sub bstance does not include any energy that t it may possess p as a result r of its macroscopic m c position or moovement. Tha at so why in SFEE S (Stead dy flow energ gy equation) C2/2 and gz iss there. Reecognize that h and similarly

= u + pv p from whicch u2 + p2 v2 = h2

u1 + p1v1 = h1

Q Q–W = [(h2 + C22/2 + gZ2) - (h1+C C12/2 + gZ1)] = [(h2 - h1) + (C22/2 - C12/2) + g(Z2 - Z1)]

or

[W Where C = Vellocity (m/s), h = Specific enthalpy e (J/k kg), z = elevattion (m)

Bu ut Reme ember: Miicroscopic vieew of a gas is a collection of particles in random mootion. Energy y of a particle consists of tra anslational energy, rottational ene ergy, vibrattional energ gy and speciific electron nic energy. Alll these energ gies summed d over all thee particles off the gas, form m the specifiic internal en nergy, e , of thee gas.

30

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

Perpetua al Motio on Mac chine off the First Kind-PMM M1 Th he first law sttates the gen neral principlle of the consservation of energy. e Energ gy is neither created nor desstroyed, but only gets tra ansformed froom one form to t another. here can be no n machine which would d continuoussly supply mechanical m woork without some other Th forrm of energy disappearing g simultaneoously (Fig. sh hown in below w). Such a fictitious mach hine is called ap perpetual mootion machine e of the first kind, or in brrief, PMM1. A PMM1 is thus t impossible. he converse of the above statementt is also tru ue, i.e. theree can be no machine which w would Th con ntinuously coonsume work k without som me other form m of energy appearing a sim multaneously y (Fig.).

A PMM1 P

The e Converse of o PMM1

Enthalpy y Th he enthalpy of a substance e H is defined d as

H = U + PV P

It iis an extenssive propertty of a system m and its uniit is kJ.

h = u + pv p

Specific Enthallpy

m and its unitt is kJ/kg. It iis an intensiive propertty of a system Intternal energy y change is equal to the heat transfeerred in a coonstant volum me process involving i no woork other tha an pdv work. It is possiblle to derive an a expression n for the hea at transfer in n a constant preessure process involving no work oth her than pdv v work. In su uch a processs in a closed d stationary sysstem of unit mass m of a pure substancee. d Q = du + pdv At constan nt pressure pdv = d(pv) Therefoore ( dQ ) = du + d ( pv ) P

or ( dQ ) or

P

= d(u+pv)

( dQ ) P =

dh

W Where H = U + PV is the enthalpy, a property of system. s Specific enth halpy h = H/m m, kJ/kg and also h = u + pdv Where h = sp pecific enthalpy, kJ/kg 31

First Law of Thermodynamics S K Mondal’s

Chapter 2

u = specific internal energy, kJ/kg dv = change in specific volume, m3/kg.

Specific heat at constant volume

The specific heat of a substance at constant volume Cv is defined as the rate of change of specific internal energy with respect to temperature when the volume is held constant, i.e., ⎛ ∂u ⎞ Cv = ⎜ ⎟ ⎝ ∂T ⎠v For a constant volume process

( Δu )v =

T2

∫ C .dT v

T1

The first law may be written for a closed stationary system composed of a unit mass of a pure substance. Q = Δu + W or d Q = du + d W For a process in the absence of work other than pdv work d W = pdv Therefore d Q = du + pdv Therefore, when the volume is held constant (Q )v  = ( Δu )v T2

(Q )v  = ∫ Cv .dT T1

⋅

Since u, T and v are properties, Cv is a property of the system. The product m Cv is called the heat capacity at constant volume (J/K).

Specific heat at constant pressure The specific heat at constant pressure Cp is defined as the rate of change of specific enthalpy with respect to temperature when the pressure is held constant. ⎛ ∂h ⎞ CP = ⎜ ⎟ ⎝ ∂T ⎠ P For a constant pressure process

( Δh )P

=

T2

∫C

P

.dT

T1

The first law for a closed stationary system of unit mass dQ = du + pdv Again, h = u + pv Therefore dh = du + pdv + vdp = d Q + vdp Therefore

dQ = dh – vdp

Therefore ( dQ )P = dh 32

First Law of Thermodynamics S K Mondal’s or

(Q ) p

Chapter 2 = ( Δh) p

∴ Form abow equations

( Q )P Cp

=

T2

∫C

P

.dT

T1

is a property of the system, just like Cv. The heat capacity at constant pressure is equal to m C p (J/K).

Application of First Law to Steady Flow Process S.F.E.E S.F.E.E. per unit mass basis

(i)

C 12 C 22 dQ dW + gz1 + = h2 + + gz2 + h1 + 2 dm 2 dm

[h, W, Q should be in J/Kg and C in m/s and g in m/s2]

(ii)

C12 dQ C22 dW gZ1 gZ 2 h1 + + + = h2 + + + 2000 1000 dm 2000 1000 dm

[h, W, Q should be in KJ/Kg and C in m/s and g in m/s2]

S.F.E.E. per unit time basis

⎛ ⎞ dQ C12 w1 ⎜ h1 + + z1 g ⎟ + 2 ⎝ ⎠ dτ ⎛ ⎞ dWx C22 = w2 ⎜ h2 + + z2 g ⎟ + dτ 2 ⎝ ⎠ Where, w = mass flow rate (kg/s) Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of the Control Volume

33

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

Ma ass balance

w A 1C v1

1

+

1

A 2C v2

+ w 2

2

=

= w A 3C v3

3 3

+ w +

4

A 4C v4

4

here v = speccific volume (m ( 3/kg) Wh En nergy balan nce

⎛ ⎞ ⎛ ⎞ dQ C12 C22 w1 ⎜ h1 + + Z1 g ⎟ + w2 ⎜ h2 + + Z2 g ⎟ + 2 2 ⎝ ⎠ ⎝ ⎠ dτ ⎛ ⎞ ⎛ ⎞ dWx C2 C2 = w3 ⎜ h3 + 3 + Z3 g ⎟ + w4 ⎜ h4 + 4 + Z4 g ⎟ + 2 2 dτ ⎝ ⎠ ⎠ ⎝ ome examplle of steady y flow proce esses:So Th he following examples e illu ustrate the applications of o the steady flow energy equation in some of the eng gineering sysstems. No ozzle and Diiffuser: An nozzle is a device d which increases th he velocity or o K.E. of a fluid at thee expense of its pressure droop, whereas a diffuser in ncreases the pressure of a fluid at thee expense of its K.E. Figu ure show in bellow a nozzle which is insu ulated. The steady s flow en nergy equation of the con ntrol surface gives dWx C2 C2 dQ = h2 + 2 + Z 2 g + h1 + 1 + Z1 g + dm dm 2 2

F Fig. 34

First Law of Thermodynamics S K Mondal’s

Chapter 2

dWx dQ = 0; = 0, and the change in potential energy is zero. The equation reduces to dm dm C2 C2 h1 + 1 = h2 + 2 (a) 2 2 The continuity equation gives AC AC (b) w= 1 1 = 2 2 v1 v2 When the inlet velocity or the ‘velocity of approach’ V1 is small compared to the exit velocity V2, Equation (a) becomes Here

C22 h1 = h2 + 2 or

C2 = 2(h1 − h2 )m / s

where (h1 – h2) is in J/kg. Equations (a) and (b) hold good for a diffuser as well. Throttling Device: When a fluid flows through a constricted passage, like a partially opened value, an orifice, or a porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled. Figure shown in below, the process of throttling by a prettily opened value on a fluid flowing in an insulated pipe. In the steady-flow energy equationdWx dQ = 0, =0 dm dm And the changes in P. E. are very small and ignored. Thus, the S.F.E.E. reduces to C2 C2 h1 + 1 = h2 + 2 2 2

(Fig.- Flow Through a Valve) Often the pipe velocities in throttling are so low that the K. E. terms are also negligible. So

h1 = h2

or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling. Turbine and Compressor: Turbines and engines give positive power output, whereas compressors and pumps require power input. For a turbine (Fig. below) which is well insulated, the flow velocities are often small, and the K.E. terms can be neglected. The S.F.E.E. then becomes

35

First Law of Thermodynamics S K Mondal’s

Chapter 2

(Fig.-. Flow through a Turbine) h1 = h2 +

dWx dm

Wx = h1 − h2 m The enthalpy of the fluid increase by the amount of work input. or

Heat Exchanger: A heat exchanger is a device in which heat is transferred from one fluid to another, Figure shown in below a steam condenser where steam condense outside the tubes and cooling water flows through the tubes. The S.F.E.E for the C.S. gives wc h1 + ws h2 = w c h3 + ws h4 or , ws ( h2 − h4 ) = w c (h3 − h1 )

Here the K.E. and P.E. terms are considered small, there is no external work done, and energy exchange in the form of heat is confined only between the two fluids, i.e. there is no external heat interaction or heat loss.

Fig. Figure (shows in below) a steam desuperheater where the temperature of the superheated steam is reduced by spraying water. If w1, w2, and w3 are the mass flow rates of the injected water, of the steam entering, and of the steam leaving, respectively, and h1, h2, and h3 are the corresponding enthalpies, and if K.E. and P.E. terms are neglected as before, the S.F.E.E. becomes w1h1 + w2 h2 = w3 h3 and the mass balance gives w1 + w2 = w3

36

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

Th he above law w is also callled as stead dy flow ene ergy equatiion. This can n be applied d to various pra actical situattions as work k developing system and work absorp ption system. Let the ma ass flow rate un nity. (1)) Work deve eloping systtems (a) Water turbines In this casse Q = 0 and ΔU = 0 and equation e becomes 2 C p1 v1 +z1g + 1 = z2g + p2 v2  + W 2 (b) Steam or gas turbin nes In this casse generally ΔZ can be assumed to be zero and thee equation becomes ⎛ C 2 − C22 ⎞ W = ( h1  – h2 ) + ⎜ 1 ⎟ + ΔQ 2 ⎝ ⎠ orbing syste ems (2)) Work abso (a) Centrifuggal water pum mp The system m is shown in the Figure below b

Fig. he energy equ uation now becomes, b In this sysstem Q = 0 and ΔU = 0; th C22 p1 v1 +z1g + W = z2g + p2 v2  + 2 al compress sor – In this system s Δz = 0 and the equ uation becom mes, (b) Centrifuga 2 2 C1 C  + h1 + W – Q = 2 + h2   2 2 (c) Blowers – In this case we have Δ z = 0, p1 v1 = p2 v 2 and Q = 0; now thee energy simp plifies to 37

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

C2 u1 +W = u2  +   2 ass C2 C1 2 f the tem mperature rise is very small and heat loss is negleected (i.e.) Δh h = 0, q = 0 (d)) Fans – In fans and hence th he energy equ uation for fan ns becomes, C2 W= 2 2 essor – In a reciprocating g compressorr ΔKE and ΔP PE are neglig gibly energy (e) Reciprocatting compre equation app plied to a reciprocating coompressor is h1 – Q = h 2 – W or

W = Q +  ( h2 – h1 )

g and absorb bing system ms (3)) Non-work developing (a) Steam boiiler – In thiss system we neglect ΔZ, ΔKE Δ and W (i.e.) ΔZ = ΔK KE = W = 0;; the energy equation for a boiler beco omes Q = (h2 – h1) (b) Steam con ndenser – In n this system m the work doone is zero and a we can also a assume ΔZ Δ and ΔKE are very sma all. Under ste eady conditioons the chang ge in enthalp py is equal too heat lost by y steam. Q = (h1 – h2) and d this heat is also equal to the changee in enthalpy y of cooling water w circulatted (i.e.) the heat lost by steam will be e equal to heat gained by the cooling water. w (c) Steam nozz zle:

In this systeem we can assume ΔZ an nd W to be zero z and hea at transfer which w is nothing but any possible heatt loss also zero. The eneergy equation n for this casse becomes. C2 C2 h1 + 1 = h2 + 2 2 2 or

C2 = C12 + 2(h1 − h2 )

(viii) Unsteady y Flow Anallysis:Ma any flow proccesses, such as a filling up and a evacuatiing gas cylind ders, are not steady, Such h processes can n be analyzed d by the conttrol volume teechnique. Coonsider a dev vice through which w a fluid d is flowing un nder non-stea ady state cond ditions (Figu ure-shown in below). The rate r at which h the mass off fluid witthin the conttrol volume iss accumulateed is equal too the net ratee of mass flow w across the control c surrface, as giveen below:

38

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

F Fig. dmv dm1 dm d 2 = w1 − w2 = − dτ dτ dτ here mv is th he mass of flu uid within th he control volume at any instant. i Wh Over an ny finite periiod of time Δmv = Δm1  – Δm2 Th he rate of acccumulation off energy with hin the contrrol volume iss equal to thee net rate of energy flow acrross the control surface. If I Ev is the en nergy of fluid d within the control c volum me at any instant, Ra ate of energy increase = Rate R of energy y inflow – Ra ate of energy outflow. ⎞ dQ ⎛ ⎛ ⎞ dW dEv C2 C2 − w2 ⎜ h 2 + 2 +Z2g ⎟ − = w1 ⎜ h1 + 1 +Z Z1g ⎟ + equatiion... A dτ 2 2 ⎠ dτ ⎝ ⎝ ⎠ dτ ⎞ ⎛ mC2 Ev = ⎜ U + + mgZ m ⎟ 2 ⎠v ⎝ Wh here m is thee mass of fluiid in the conttrol volume at a any instantt 2 ⎞ dE d ⎛ mC C ∴ v = + mgZ ⎟ = ⎜U + dτ dτ ⎝ 2 ⎠v ⎞ dm1 dQ ⎛ ⎛ ⎞ dm 2 dW C12 C2 + − ⎜ h 2 + 2 +Z2g ⎟ − +Z1g ⎟ ⎜ h1 + 2 2 dτ dτ dτ ⎝ ⎠ dτ ⎠ ⎝

( equaation........B )

Following Figu ure shows all these energy y flux quantitties. For any y time interva al, equation (B) ( becomes 2 2 ⎛ ⎞ ⎛ ⎞ C C ΔEv = Q − W + ∫ ⎜ h1 + 1 +Z1g ⎟dm m1 − ∫ ⎜ h2 + 2 +Z2g ⎟dm2 2 2 ⎝ ⎠ ⎝ ⎠

Fig. Equatioon (A) is gene eral energy equation. e Forr steady flow, dEv =0 dτ and the equatioon reduces For F a closed system s w1 = 0, 0 w2 = 0, theen from equattion (A), dEv dQ dW = − dτ dτ dτ Orr dEv = dQ d – dW or dQ = dE + dW d 39

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

Floow Processess

Ex xample of a variable flo ow problem m: Va ariable flow processes p may y be analyzed d either by th he system tecchnique or th he control vollume tecchnique, as illlustrated below. Consideer a process in i which a ga as bottle is filled from a pipeline p (Figu ure shown in below). In thee beginning the t bottle con ntains gas of mass m1 at state s p1, T1, v1, h1 and u1. The valve iss opened and gas flows in nto the bottle e till the masss of gas in th he bottle is m2 at state p2, T2, v2, h2 an nd u2. The sup pply to the piipeline is verry large so th hat the state of gas in the pipeline is constant at p p ,TP , v P , h P , u P and v P .

Sy ystem Techn nique: Assum me an enveloope (which is extensible) of o gas in the pipeline p and the tube wh hich would ev ventually entter the bottlee, as shown in n Figure abov ve. Energy y of the gas be efore filling. ⎛ C2 ⎞ E1  = m1  u1  + ( m2  – m1 ) ⎜ P + uP ⎟ 2 ⎝ ⎠ Wh here ( m2  – m1 ) is the ma ass of gas in the t pipeline and a tube whiich would enter the bottlee. E2 = m2 u2 ⎛ C2 ⎞  ΔE = E2 – E1 = m2 u2 – m1 u1 ( m1 – u1 ) ⎜ P + uP ⎟ ⎠ ⎝ 2 The P.E. terms t are neglected. The gas in the boottle is not in n motion, and d so the K.E. terms have beeen omitted. Now, therre is a chang ge in the volume of gas because of the t collapse of the envellope to zero vollume. Then the t work done    W = p (V2 – V1 ) p

        = p p ⎡⎣0 – ( m2 – m1 ) v P ⎤⎦ = – ( m2 – m1 ) p p v P Ussing the firstt for the process Q = ΔE + W

⎡C2 ⎤ = m2 u2  – m1 u1 – ( m2  – m1 ) ⎢ P + uP ⎥ − ( m2  – m1 ) pP vP ⎣ 2 ⎦ 40

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

⎛ C2 ⎞ = m2 u2 – m1 u1 – ( m2 – m1 ) ⎜ P + h P ⎟ ⎝ 2 ⎠ hich gives thee energy bala ance for the process. p wh Co ontrol Volum me Techniq que: Assume a control vollume bounded by a controol surface as shown in Fig gure above, Applying A the energy equa ation in this case, the folllowing energy y balance ma ay be wrritten on a tim me rate basiss dEv dQ ⎛ C 2 ⎞ dm m = + ⎜ hP + P ⎟ dτ dτ ⎝ 2 ⎠ dττ Sin nce hP and CP are constan nt, the equatiion is integra ated to give for f the Total process 2 ⎛ C ⎞ ΔEv = Q + ⎜ h P + P ⎟ ( m2 − m1 ) 2 ⎠ ⎝ Noow ΔEv = U 2  – U1  = m2 u2  – m1 u1  

⎛ C2 ⎞ Q = m2 u2  – m1 u1  – ⎜ h P + P ⎟ ( m2 − m1 ) 2 ⎠ ⎝

D Discharg ging an nd Charrging a Tank Let us considerr a tank disccharging a flu uid into a su upply line (Fiigure). Sincee dWx = 0 an nd dmin = 0, app plying first la aw to the con ntrol volume,,

⎛ ⎞ C2 + gz ⎟ dmout dUV = dQ Q + ⎜h + 2 ⎝ ⎠ out Assuming K.E. and P.E. of the t fluid to be b small and dQ = 0 d(mu) = hdm mdu+ udm = udm+ pv dm d

Ag gain or

dm du = m pv onst. V = vm = co vdm + mdv = 0 dm dv =− m v du dv =− pv v d ( u + pv ) = 0

or

dQ = 0

wh hich shows that t the process is adiab batic and quasi-static. For cha arging the tan nk 41

Charging and Disc charging a Tank

First Law of Thermodynamics S K Mondal’s

Chapter 2 ∫ ( hdm )

in

= ΔUV = m2u2 − m1u1

mphp = m2u2 − m1u1

where the subscript p refers to the constant state of the fluid in the pipeline. If the tank is initially empty, m1 = 0. m p h p = m2 u2 Since

mp = m2 hp = u2 If the fluid is an ideal gas, the temperature of the gas in the tank after it is charged is given by

c pTp = cvT2 T2 = γTp

or

PROBLEMS & SOLUTIONS

Example 1 The work and heat transfer per degree of temperature change for a closed system is given by dW 1 dQ 1 = kJ / ° C; = kJ / ° C dT 30 dT 10 Calculate the change in internal energy as its temperature increases from 125ºC to 245ºC. Solution: dT dW = 30 W =

T2

dT 1 1 = (T2 − T1 ) = ( 245 − 125 ) 30 30 T1 30

∫

dQ = Q=

T2

dT 10 dT

∫ 10

T1

=

1 ( 245 − 125 ) = 12 kJ 10

Applying First Law of Thermodynamics Q = W + ΔU ΔU = Q – W = 12 – 4 = 8kJ. Example 2 Air expands from 3 bar to 1 bar in a nozzle. The initial velocity is 90 m/s. the initial temperature is 150ºC. Calculate the velocity of air at the exit of the nozzle. Solution: The system in question is an open one. First Law of Thermodynamics for an open system gives ⎡ ⎤ ⎡ ⎤ C2 C2 w1 ⎢h1 + 1 + Z1 g ⎥ + Q = w2 ⎢h2 + 2 + Z2 g ⎥ + W 2 2 ⎣ ⎦ ⎣ ⎦ Since the flow is assumed to be steady. w1 = w2 Flow in a nozzle is adiabatic flow. 42

First Law of Thermodynamics S K Mondal’s

Chapter 2

Hence Q = 0 Also W = 0 The datum can be selected to pass through axis; then Z1 = Z2. Hence C2 C2 h1 + 1 = h2 + 2 2 2 C22 C2 ( or ) = ( h1 − h2 ) + 1 2 2 and γ −1

⎛p ⎞γ T2 = T1 ⎜ 2 ⎟ ⎝ p1 ⎠ γ for air = 1.4

T1 = 150 + 273 = 423 0.4/1.4

⎛1 ⎞ ∴ T2 = 423 ⎜ ⎟ = 309 K ⎝3⎠ For air Cp = 1.005 kJ/kgºC Cv = 0.718 kJ/kgºC. R = 287 J/kg K = 0.287 kJ/kg K S.F.E.E. : - We have (h1 – h2) = Cp(T1 - T2) C22 C2 902 = ( h1 − h2 ) + 1 = 1.005 × 103 × (423 − 309) + 2 2 2 or, C2 = 487 m / s.

Example 3 An evacuated cylinder fitted with a valve through which air from atmosphere at 760 mm Hg and 25°C is allow to fill it slowly. If no heat interaction is involved, what will be the temperature of air in the bottle when the pressure reaches 760 mm Hg? Use the following: (1) Internal energy of air u = u0 + 0.718T kJ/kg where T is temperature in °C. (2) R = 0.287 kJ/kg K. Solution: Applying first law, ignoring potential and kinetic energy terms, to the vessel as control volume.

43

First Law of Thermodynamics S K Mondal’s

Chapter 2

Q + m i h i = m e h e + m2 u2 – m1 u1 + W

Here Q = 0, W = 0, m e = 0 ( no mass leaving from control vol.)

m1 = 0 ( evacuated ) ∴ m2 = mi ∴ hi = u2

hi = ui + pv = u0 + 0.718Ti + 0.287 (Ti + 273 ) = u0 + 0.718 × 25 + 0.287 × 298 = u0 + 103.48kJ / kg = u2 or u2 − u0 = 103.48kJ / kg u2 = u0 + 0.718T2 T2 =

u2 − u0 103.48 = = 144.2°C 0.718 0.718

Example 4 A system whose mass is 4.5 kg undergoes a process and the temperature changes from 50° C to 100°C. Assume that the specific heat of the system is a function of temperature only. Calculate the heat transfer during the process for the following relation ship. 80 cn = 1.25 + kJ / kg °C [t is in oC] t + 160 Solution: 100 100 80 ⎞ ⎛ = = Q mc dt 4.5 1 2 ∫50 n ∫50 ⎜⎝1.25 + t + 160 ⎟⎠ dt 100 ⎧⎪100 ⎫⎪ dt = 4.5 ⎨ ∫ 1.25dt + ∫ ⎬ 0.0125t + 2.0 ⎭⎪ 50 ⎩⎪ 50 100 100 ⎪⎧ ⎡ 1 ⎤ ⎪⎫ = 4.5 ⎨[1.25t ]50 + ⎢ ln ( 0.0125t + 2.0 )⎥ ⎬ ⎣ 0.0125 ⎦50 ⎭⎪ ⎩⎪

⎧ ⎫ ⎡ 1 = 4.5 ⎨[1.25 × 50] + ⎢ (ln (1.25 + 2.0 ) − ln (0.625 + 2 )) ⎤⎥ ⎬ 0.0125 ⎣ ⎦⎭ ⎩ 1 3.25 ⎧ ⎫ = 4.5 ⎨62.5 + ln ⎬ = 358 kJ 0.0125 2.625 ⎭ ⎩

44

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

ASKED D OBJEC CTIVE QUESTIONS (G GATE, IES, IAS) P Previou us 20-Y Years GATE E Ques stions A Applicatiion of First F Law w to Ste eady Flo ow Proc cess S.F F.E.E Co ommon Data for Qu uestions Q1 Q and Q2 2:

[GA ATE-2009]

Th he inlet and d the outle et conditions of

ste eam for an n adiabatic c steam tur rbine are as indic cated in the t figure. The otations are as usually followed. no

ATE-1. GA

If mass m flow rate r of steam through the turbine is 20 kg/ss the power r output of the e turbine (in n MW) is: (a) 12.157

GA ATE-2.

[GA ATE-2009] (b) 12.941

(cc) 168.001

(d) 168.785

Asssume the ab bove turbin ne to be par rt of a simp ple Rankine e cycle. The density of water at the inlet to the e pump is 1000 kg/m3. Ignoring kinetic k and d potential ene ergy effects s, the specifi fic work (in kJ/kg) supp plied to the pump is: [G GATE-2009] (a) 0.293 (b) 0.35 1 (c)) 2.930 (d) 3.510 3

GA ATE-3. The e following g four figures have been dra awn to re epresent a fictitious the ermodynam mic cycle, on n the p-v and d T-s planess. [G GATE-2005]

Acc cording to the t first law w of thermodynamics, equal e areas are enclose ed by (a) Figures 1and d 2 (b) Fig gures 1and 3 (c) Figuress 1and 4 (d) Figures 2 and 3

45

First Law of Thermodynamics S K Mondal’s

Chapter 2

Internal Energy – A Property of System GATE-4.

A gas contained in a cylinder is compressed, the work required for compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to the heated. The change in internal energy of the gas during the process is: [GATE-2004] (a) – 7000 kJ (b) – 3000 kJ (c) + 3000 kJ (d) + 7000 kJ

GATE-4a. The contents of a well-insulated tank are heated by a resistor of 23 Ω in

which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy ( ΔU) during the process in kW are [GATE-2011] (a) Q = 0, W = –2.3, Δ U = +2.3 (b) Q = +2.3, W = 0, Δ U = +2.3 (c) Q = –2.3, W = 0, Δ U = –2.3 (d) Q = 0, W = +2.3, Δ U = –2.3

Discharging and Charging a Tank GATE-5.

A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at I MPa, 350°C. A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches I MPa. The final temperature inside the tank (A) Is greater than 350°C (B) Is less than 350°C (C) Is equal to 350°C (D) May be greater than, less than, or equal to 350°C, depending on the volume of the tank

Previous 20-Years IES Questions First Law of Thermodynamics IES-1.

Which one of the following sets of thermodynamic laws/relations is directly involved in determining the final properties during an adiabatic mixing process? [IES-2000] (a) The first and second laws of thermodynamics (b) The second law of thermodynamics and steady flow relations (c) Perfect gas relationship and steady flow relations (d) The first law of thermodynamics and perfect gas relationship

46

First Law of Thermodynamics S K Mondal’s

Chapter 2

IES-2.

Two blocks which are at different states are brought into contact with each other and allowed to reach a final state of thermal equilibrium. The final temperature attained is specified by the [IES-1998] (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics

IES-3.

For a closed system, the difference between the heat added to the system and the work done by the system is equal to the change in [IES-1992] (a) Enthalpy (b) Entropy (c) Temperature (d) Internal energy

IES-4.

An ideal cycle is shown in the figure. Its thermal efficiency is given by

IES-5.

⎛ v3 ⎞ ⎜ − 1⎟ v (a)1 − ⎝ 1 ⎠ ⎛ p2 ⎞ ⎜ − 1⎟ ⎝ p1 ⎠

⎛ v3 ⎞ ⎜ − 1⎟ 1 ⎝ v1 ⎠ (b) 1 − γ ⎛ p2 ⎞ ⎜ − 1⎟ ⎝ p1 ⎠

(c)1 − γ

(b) 1 −

( v3 − v1 ) p1 ( p2 − p1 ) v1

1 ( v3 − v1 ) p1 γ ( p2 − p1 ) v1

[IES-1998]

Which one of the following is correct? The cyclic integral of (δQ − δW ) for a process is:

(a) Positive

(b) Negative

(c) Zero

[IES-2007]

(d) Unpredictable

IES-6.

A closed system undergoes a process 1-2 for which the values of Q1-2 and W1-2 are +20 kJ and +50 kJ, respectively. If the system is returned to state, 1, and Q2-1 is [IES-2005] 10 kJ, what is the value of the work W2-1? (a) + 20 kJ (b) –40 kJ (c) –80 kJ (d) +40 kJ

IES-7.

A gas is compressed in a cylinder by a movable piston to a volume one-half of its original volume. During the process, 300 kJ heat left the gas and the internal energy remained same. What is the work done on the gas? [IES-2005] (a) 100kNm (b) 150 kNm (c) 200 kNm (d) 300 kNm

IES-8.

In a steady-flow adiabatic turbine, the changes in the internal energy, enthalpy, kinetic energy and potential energy of the working fluid, from inlet to exit, are -100 kJ/kg, -140 kJ/kg, -10 kJ/kg and 0 kJ/kg respectively. Which one of the following gives the amount of work developed by the turbine? [IES-2004] (a) 100 kJ/kg (b) 110 kJ/kg (c) 140 kJ/kg (d) 150 kJ/kg

47

First Law of Thermodynamics S K Mondal’s

Chapter 2

IES-9.

Gas contained in a closed system consisting of piston cylinder arrangement is expanded. Work done by the gas during expansion is 50 kJ. Decrease in internal energy of the gas during expansion is 30 kJ. Heat transfer during the process is equal to: [IES-2003] (a) –20 kJ (b) +20 kJ (c) –80 kJ (d) +80 kJ

IES-10.

A system while undergoing a cycle [IES-2001] A – B – C – D – A has the values of heat and work transfers as given in the Table: Process

Q kJ/min

W kJ/min

A–B B–C C–D D–A

+687 -269 -199 +75

+474 0 -180 0

The power developed in kW is, nearly, (a) 4.9 (b) 24.5 (c) 49 IES-11.

(d) 98

The values of heat transfer and work transfer for four processes of a thermodynamic cycle are given below: [IES-1994] Process 1 2 3 4

Heat Transfer (kJ)

Work Transfer (kJ)

300 Zero -100 Zero

300 250 -100 -250

The thermal efficiency and work ratio for the cycle will be respectively. (a) 33% and 0.66 (b) 66% and 0.36. (c) 36% and 0.66 (d) 33% and 0.36. IES-12.

A tank containing air is stirred by a paddle wheel. The work input to the paddle wheel is 9000 kJ and the heat transferred to the surroundings from the tank is 3000 kJ. The external work done by the system is: [IES-1999] (a) Zero (b) 3000 kJ (c) 6000 kJ (d) 9000 kJ

Internal Energy – A Property of System IES-13.

For a simple closed system of constant composition, the difference between the net heat and work interactions is identifiable as the change in [IES-2003] (a) Enthalpy (b) Entropy (c) Flow energy (d) Internal energy

IES-14.

Assertion (A): The internal energy depends on the internal state of a body, as determined by its temperature, pressure and composition. [IES-2006] Reason (R): Internal energy of a substance does not include any energy that it may possess as a result of its macroscopic position or movement. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

48

First Law of Thermodynamics S K Mondal’s

Chapter 2

IES-15.

Change in internal energy in a reversible process occurring in a closed system is equal to the heat transferred if the process occurs at constant: [IES-2005] (a) Pressure (b) Volume (c) Temperature (d) Enthalpy

IES-16.

170 kJ of heat is supplied to a system at constant volume. Then the system rejects 180 kJ of heat at constant pressure and 40 kJ of work is done on it. The system is finally brought to its original state by adiabatic process. If the initial value of internal energy is 100 kJ, then which one of the following statements is correct? [IES-2004] (a) The highest value of internal energy occurs at the end of the constant volume process (b) The highest value of internal energy occurs at the end of constant pressure process. (c) The highest value of internal energy occurs after adiabatic expansion (d) Internal energy is equal at all points

IES-17.

85 kJ of heat is supplied to a closed system at constant volume. During the next process, the system rejects 90 kJ of heat at constant pressure while 20 kJ of work is done on it. The system is brought to the original state by an adiabatic process. The initial internal energy is 100 kJ. Then what is the quantity of work transfer during the process? [IES-2009] (a) 30 kJ (b) 25 kJ (c) 20 kJ (d) 15 kJ

IES-17a

A closed system receives 60 kJ heat but its internal energy decreases by 30 kJ. Then the [IES-2010] work done by the system is (a) 90 kJ (b) 30 kJ (c) –30 kJ (d) – 90 kJ

IES-18.

A system undergoes a process during which the heat transfer to the system per degree increase in temperature is given by the equation: [IES-2004] dQ/dT = 2 kJ/°C The work done by the system per degree increase in temperature is given by the equation dW/dT = 2 – 0.1 T, where T is in °C. If during the process, the temperature of water varies from 100°C to 150°C, what will be the change in internal energy? (a) 125 kJ (b) –250 kJ (c) 625 kJ (d) –1250 kJ

IES-19.

When a system is taken from state A to state B along the path A-C-B, 180 kJ of heat flows into the system and it does 130 kJ of work (see figure given): How much heat will flow into the system along the path A-D-B if the work done by it along the path is 40 kJ? (a) 40 kJ (b) 60 kJ (c) 90 kJ (d) 135 kJ

IES-20.

[IES-1997]

The internal energy of a certain system is a function of temperature alone and is given by the formula E = 25 + 0.25t kJ. If this system executes a process for which the work done by it per degree temperature increase is 0.75 kJ/K, then the heat interaction per degree temperature increase, in kJ, is: [IES-1995] (a) –1.00 (b) –0.50 (c) 0.50 (d ) 1.00

49

First Law of Thermodynamics S K Mondal’s IES-21.

Chapter 2

When a gas is heated at constant pressure, the percentage of the energy supplied, which goes as the internal energy of the gas is: [IES-1992] (a) More for a diatomic gas than for triatomic gas (b) Same for monatomic, diatomic and triatomic gases but less than 100% (c) 100% for all gases (d) Less for triatomic gas than for a diatomic gas

Perpetual Motion Machine of the First Kind-PMM1 IES-22.

Consider the following statements: [IES-2000] 1. The first law of thermodynamics is a law of conservation of energy. 2. Perpetual motion machine of the first kind converts energy into equivalent work. 3. A closed system does not exchange work or energy with its surroundings. 4. The second law of thermodynamics stipulates the law of conservation of energy and entropy. Which of the statements are correct? (a) 1 and 2 (b) 2 and 4 (c) 2, 3 and 4 (d) 1, 2 and 3

Enthalpy IES-23.

Assertion (A): If the enthalpy of a closed system decreases by 25 kJ while the system receives 30 kJ of energy by heat transfer, the work done by the system is 55 kJ. [IES-2001] Reason (R): The first law energy balance for a closed system is (notations have their usual meaning) ΔE = Q − W

(a) (b) (c) (d)

Both A and R are individually true and R is the correct explanation of A Both A and R are individually true but R is NOT the correct explanation of A A is true but R is false A is false but R is true

Application of First Law to Steady Flow Process S.F.E.E IES-24.

IES-25.

Which one of the following is the steady flow energy equation for a boiler?

(a) h1 +

v12 v2 = h2 + 2 2 gJ 2 gJ

(b) Q = ( h2 − h1 )

(c) h1 +

v12 v2 + Q = h2 + 2 2 gJ 2 gJ

(d) Ws = ( h2 − h1 ) + Q

[IES-2005]

A 4 kW, 20 litre water heater is switched on for 10 minutes. The heat capacity Cp for water is 4 kJ/kg K. Assuming all the electrical energy has gone into heating the water, what is the increase of the water temperature? [IES-2008] (a) 15°C (b) 20°C (c) 26°C (d) 30°C

Discharging and Charging a Tank 50

First Law of Thermodynamics S K Mondal’s IES-26.

Chapter 2

An insulated tank initially contains 0.25 kg of a gas with an internal energy of 200 kJ/kg .Additional gas with an internal energy of 300 kJ/kg and an enthalpy of 400 kJ/kg enters the tank until the total mass of gas contained is 1 kg. What is the final internal energy(in kJ/kg) of the gas in the tank? [IES-2007] (a) 250 (b) 275 (c) 350 (d) None of the above

Previous 20-Years IAS Questions IAS-1.

A system executes a cycle during which there are four heat transfers: Q12 = 220 kJ, Q23 = -25kJ, Q34 = -180 kJ, Q41 = 50 kJ. The work during three of the processes is W12 = 15kJ, W23 = -10 kJ, W34 = 60kJ. The work during the process 4 1 is: [IAS-2003] (a) - 230 kJ (b) 0 kJ (c) 230 kJ (d) 130 kJ

IAS-2.

Two ideal heat engine cycles are represented in the given figure. Assume VQ = QR, PQ = QS and UP =PR =RT. If the work interaction for the rectangular cycle (WVUR) is 48 Nm, then the work interaction for the other cycle PST is: (a) 12Nm (b) 18 Nm (c) 24 Nm (d) 36 Nm

IAS-2001]

IAS-3.

A reversible heat engine operating between hot and cold reservoirs delivers a work output of 54 kJ while it rejects a heat of 66 kJ. The efficiency of this engine is: [IAS-1998] (a) 0.45 (b) 0.66 (c) 0.75 (d) 0.82

IAS-4.

If a heat engine gives an output of 3 kW when the input is 10,000 J/s, then the thermal efficiency of the engine will be: [IAS-1995] (a) 20% (b) 30% (c) 70% (d) 76.7%

IAS-5.

In an adiabatic process, 5000J of work is performed on a system. The system returns to its original state while 1000J of heat is added. The work done during the non-adiabatic process is: [IAS-1997] (a) + 4000J (b) - 4000J (c) + 6000J (d) - 6000J

IAS-6.

In a thermodynamic cycle consisting of four processes, the heat and work are as follows: [IAS-1996] Q: + 30, - 10, -20, + 5 W: + 3, 10, - 8, 0 The thermal efficiency of the cycle will be: (a) Zero (b) 7.15% (c) 14.33% (d) 28.6%

51

First Law of Thermodynamics S K Mondal’s

Chapter 2

IAS-7.

Match List-I (Devices) with List-II (Thermodynamic equations) and select the correct answer using the codes below the lists: [IAS-1996] List-I List-II A. Turbine 1. W = h2 – h1 B. Nozzle 2. h1 = h2 C. Valve 3. h1 = h2 + V2/2 D. Compressor 4. W = h1 – h2 Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 3 1 4 (c) 4 3 1 2 (d) 3 2 4 1

IAS-8.

Given that the path 1-2-3, a system absorbs 100kJ as heat and does 60kJ work while along the path 14-3 it does 20kJ work (see figure given). The heat absorbed during the cycle 1-4-3 is: (a) - 140 Kj (b) - 80 kJ (c) - 40kJ (d) + 60 kJ

IAS-9.

The given figure shows the variation of force in an elementary system which undergoes a process during which the plunger position changes from 0 to 3 m. lf the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed during the process is: (a) 15 J (b) 20 J

IAS-10.

[IAS 1994]

(c) 25 J

[IAS-1994] (d) 30 J

The efficiency of a reversible cyclic process undergone by a substance as shown in the given diagram is: (a) 0.40 (b) 0.55 (c) 0.60 (d) 0.80

[IAS 1994]

Internal Energy – A Property of System IAS-11.

Which one of the following is the correct expression for change in the internal energy for a small temperature change Δ T for an ideal gas? [IAS-2007] (a) ΔU = Cv × ΔT (b) ΔU = C p × ΔT

52

First Law of Thermodynamics S K Mondal’s (c)

ΔU =

Cp Cv

Chapter 2

(

× ΔT

)

(d) ΔU = C p − Cv × ΔT

IAS-12.

The heat transferred in a thermodynamic cycle of a system consisting of four processes is successively 0, 8, 6 and -4 units. The net change in the internal energy of the system will be: [IAS-1999] (a) – 8 (b) Zero (c) 10 (d) –10

IAS-13.

During a process with heat and work interactions, the internal energy of a system increases by 30 kJ. The amounts of heat and work interactions are respectively [IAS-1999] (a) - 50 kJ and - 80 kJ (b) -50 kJ and 80 kJ (c) 50 kJ and 80 kJ (d) 50 kJ and - 80 kJ

IAS-14.

A mixture of gases expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is: [IAS 1994] (a) 30 kJ (b) 54 kJ (c) 84 kJ (d) 114 kJ

IAS-15.

In an adiabatic process 6000 J of work is performed on a system. In the nonadiabatic process by which the system returns to its original state 1000J of heat is added to the system. What is the work done during non-adiabatic process? [IAS-2004] (a) + 7000 J (b) - 7000 J (c) + 5000 J (d) - 5000 J

Enthalpy IAS-16.

The fundamental unit of enthalpy is: (a) MLT-2 (b) ML-2T-1 (c) ML2T-2

[IAS 1994]

(d) ML3T-2

Application of First Law to Steady Flow Process S.F.E.E IAS-17.

In a test of a water-jacketed compressor, the shaft work required is 90 kN-m/kg of air compressed. During compression, increase in enthalpy of air is 30 kJ/kg of air and increase in enthalpy of circulating cooling water is 40 kJ/ kg of air. The change is velocity is negligible. The amount of heat lost to the atmosphere from the compressor per kg of air is: [IAS-2000] (a) 20kJ (b) 60kJ (c) 80 kJ (d) 120kJ

IAS-18.

When air is compressed, the enthalpy is increased from 100 to 200 kJ/kg. Heat lost during this compression is 50 kJ/kg. Neglecting kinetic and potential energies, the power required for a mass flow of 2 kg/s of air through the compressor will be: [IAS-1997] (a) 300 kW (b) 200 kW (c) 100 kW (d) 50 kW

53

First Law of Thermodynamics S K Mondal’s

Chapter 2

Variable Flow Processes IAS-19.

Match List-I with List-II and select the correct answer using the codes given below Lists: [IAS-2004] List-I List-II A. Bottle filling of gas 1. Absolute zero temperature B. Nernst Simon statement 2. Variable flow C. Joule Thomson effect 3. Quasistatic path D.

∫ pdv

Codes: (a) (c) IAS-20.

4. Isenthalpic process

A 6 2

B 5 5

C 4 7

5. Dissipative effect 6. Low grade energy 7. Process and temperature change A B C (b) 2 1 4 (d) 6 1 7

D 3 4

during phase D 3 4

A gas chamber is divided into two parts by means of a partition wall. On one side, nitrogen gas at 2 bar pressure and 20°C is present. On the other side, nitrogen gas at 3.5 bar pressure and 35°C is present. The chamber is rigid and thermally insulated from the surroundings. Now, if the partition is removed, (a) High pressure nitrogen will get throttled [IAS-1997] (b) Mechanical work, will be done at the expense of internal energy (c) Work will be done on low pressure nitrogen (d) Internal energy of nitrogen will be conserved

54

First Law of Thermodynamics S K Mondal’s

Chapter 2

Answers with Explanation (Objective) Previous 20-Years GATE Answers C12 C2 gZ1 gZ2 dQ dW + + = h2 + 1 + + 2000 1000 dm 2000 1000 dm 2 160 9.81× 10 100 2 9.81× 6 dW 3200 + + = 2600 + + + 2000 1000 2000 1000 dm dW 600 + 7.8 + 0.04 = + dm

GATE-1. Ans. (a)

h1 +

GATE-2. Ans. (c)

W = ν (P2 − P1 ) =

1 ( 3000 − 70 ) × kJ/kg = 2.93 1000

GATE-3. Ans. (a) Fig-1 & 2 both are power cycle, so equal areas but fig-3 & 4 are reverse power cycle, so area is not meant something. GATE-4. Ans. (c) dQ = du + dw Q = u2 − u1 + W or − 2000 = u2 − u1 − 5000 or u2 − u1 = 3000kJ GATE-4a. Ans. (a) Q = 0, W = –2.3, ΔU =

I

+2.3 Tank is well insulated so Q = 0 Work is given to the system in the form of electric current.

R

I

So, W = − I2 R = − 102 × 23 = –2300 W = –2.3 kW By 1st Law of Thermodynamics Q1 − 2 = U2 − U1 + W1 − 2 0 = U 2 − U1 − 2.3 ΔU = 2.3 kW GATE-5. Ans (a) The final Temp. (T2)=

γT1

Previous 20-Years IES Answers IES-1. Ans. (a) If we adiabatically mix two liquid then perfect gas law is not necessary. But entropy change in the universe must be calculated by Second law of thermodynamics. Final entropy of then system is also a property. That so why we need second law. IES-2. Ans. (b) Using conservation of energy law we may find final temperature. IES-3. Ans. (d) From First law of thermodynamics, for a closed system the net energy transferred as heat Q and as work W is equal to the change in internal energy, U, i.e. Q – W = U

55

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

IES-4. Ans. (c)) Total heat addition a is coonstant volum me heat addiition, Q12 = cv (T2 − T1 ) Tottal heat rejection is consta ant pressure heat rejectioon, Q31 = c p (T3 − T1) Now w from equattion of state

P P1 P2 = (∵ v = const.) or T2 = 2 × T1 P1 T1 T2 and d

v v1 v 3 = (∵ p = const.)) or T3 = 3 × T1 v1 T1 T3

c (T − T ) Q31 (T − T ) = 1− p 3 1 = 1− γ 3 1 cv (T2 − T1) Q12 (T2 − T1)

Effiiciency, η = 1 −

⎛ v3 ⎞ ⎜ × T1 − T1 ⎟ v ⎠ = 1 − γ (v 3 − v1) p1 or η = 1 − γ ⎝ 1 ( p2 − p1) v1 ⎛ P2 ⎞ ⎜ × T1 − T1 ⎟ ⎝ P1 ⎠ IES-5. Ans. (c c) It is du = đQ đ – đW, as u is a thermoodynamic prooperty and itts cyclic integ gral must be zeroo. W or Q1− 2 + Q2 −1 = W1− 2 + W2 −1 IES-6. Ans. (b) ΣdQ = ΣdW

or 20 + ( −10 ) = 50 + W2−1

or W2−1 = −40kJ

IES-7. Ans. (d) dQ = du + dw as u = const. Theerefore du = 0 or dQ = dw d = 300kNm IES-8. Ans. (d) ⎞ ⎛ V2 Q − Wx = Δ ⎜ h + + gz ⎟ 2 ⎝ ⎠

O − Wx = −140 − 10 + 0 or Wx = 150 kJ / kg Cha ange of internal energy = -100 kJ/kg is superfluous data. IES-9. Ans. (b) Q = Δ E+ Δ W Δ E = –30 kJ (ddecrease in in nternal energ gy) ( done by the system m) Δ W = + 50 kJ (work Q = –30 + 50 = + 20 kJ IES-10. Ans. (a a) Net work = 74 – 180 kJ/m min = 294 kJ J/min = 294/6 60 kJ/s = 4.9 kW dW =47

∑

IES-11. Ans. (b b)

ηth =

Work done 3000 − 100 = 0..66 = 300 heaat added

Woork ratio =

∑ ( + w) − ∑ ( − w) = 550 − 350 = 0.366 5 550 ∑ ( + w)

IES-12. Ans. (a) This iss a case of constant voolume process or an n is isochooric process.. By perrforming worrk on the sysstem tempera ature can n be raised. In an irrev versible constant volu ume process, the system m doesn't perrform worrk on the surrrounding at the expense of its inteernal energy. 56

Fir rst Law w of Therm T odyna amics S K Mondal’s

Cha apter 2

IES-13. Ans. (d d) IES-14. Ans. (a) ( The interrnal energy depends d only y upon the in nitial and fin nal states of the system. Inteernal energy y of a substa ance does noot include an ny energy th hat it may possess p as a resu ult of its ma acroscopic position or movement. m T That so why in SFEE v2/2 / and gz is there.

If in nternal energ gy include poosition or movement then n why this v2/2 / and gz term ms is there. Bur Remembe er: Miccroscopic view of a gas is a collectioon of particlles in randoom motion. Energy E of a parrticle consists of transla ational energy, rotatio onal energy y, vibration nal energy and d specific ellectronic en nergy. All th hese energies summed oveer all the parrticles of the gass, form the sp pecific interna al energy, e , of the gas. IES-15. Ans. (b b) dQ = dU + pdV if V is con s tan t ( dQ )v = ( dU)v

IES-16. Ans. (a a) Q2 = 180 0kJ = Δu + ΔW = Δu + ( −40)) ∴U1 = 100kJ, U2 = 100 0 + 170 = 270 0 kJ, U3 = 270 − 180 + 40 = 130 kJ

IES-17. Ans. (d d)

IES-17a.

For the process p 1-2 dQ = +85 , dW = 0 For the process p 2-3 dQ = -90 kJ, dW = -20kJ For the process p 3-1 dQ = 0, dW = ? For a cycclic process ∑ dQ = ∑ dW ⇒ 85-90+ +0 = 0-20+ dW W ⇒ -5 = -20+ dW ⇒ 5 = +15kJ dW = -20+5

An ns. (a) dQ = dU + dW d or 60 = – 30 + dW or dW = 90 KJ 57

First Law of Thermodynamics S K Mondal’s

Chapter 2

IES-18. Ans. (c) dQ = du + dw

2.dt = du + ( 2 − 0.1T ) dT 150 0.1 0.1 ⎡1502 − 1002 ⎤⎦ = 625kJ × ⎡⎣T 2 ⎤⎦ = 100 2 2 ⎣ IES-19. Ans. (c) Change of internal energy from A to B along path ACB = 180 - 130 = 50 kJ. It will be same even along path ADB. ∴ Heat flow along ADB = 40 + 50 = 90 kJ.

or

∫ du = ∫ 0.1TdT =

IES-20. Ans. (d) dQ = dE + dW

Given: E = 25 + 0.25t kJ and

or

dQ dE dW = + dt dt dt

dW = 0.75 kJ / k dt

dE = 0.25 kJ / K dt dQ dE dW Therefore = + = 0.25 + 0.75 kJ / K = 1.00 kJ / K dt dt dt then

IES-21. Ans. (a) IES-22. Ans. (a) A closed system does exchange work or energy with its surroundings. option ‘3’ is wrong. 4. “The law of conservation of entropy” is imaginary so option ‘4’ is also wrong. IES-23. Ans. (a) v2 v2 dQ dw IES-24. Ans. (b) h1 + 1 + gz1 + = h2 + 2 + gz 2 + =0 2 dm 2 dm dw =0 For boiler v1, v2 is negligible and z1 = z2 and dm dQ or = ( h2 − h1 ) dm IES-25. Ans. (d) mCP ΔT = ( 4 × 10 × 60 )

⇒ 20 × 4 × ΔT = 2400 ⇒ ΔT = 30C° IES-26. Ans. (c) Enthalpy of additional gas will be converted to internal energy. Uf= miui+(mf-mi)hp = 0.25x200+(1-0.25)x400 = 350 kJ As total mass = 1kg, uf=350 kJ/kg Note: You cannot simply use adiabatic mixing law here because it is not closed system. This is a problem of variable flow process. If you calculate in following way it will be wrong. Final internal energy of gas(mixture) is m u + m2 u2 u= 1 1 m1 + m2 kJ ⎞ kJ ⎞ ⎛ ⎛ (0.25kg) ⎜ 200 ⎟ + (0.75kg) ⎜ 300 kg ⎟ kg ⎝ ⎠ ⎝ ⎠ u= (0.25 + 0.75)kG

u = 275

kJ kg

58

First Law of Thermodynamics S K Mondal’s

Chapter 2

It is valid for closed system only.

Previous 20-Years IAS Answers IAS-1. Ans. (b)

∑ dQ = ∑ dW

or 220 -25 -180 +50 = 15 -10 +60 +W4-1

IAS-2. Ans. (c) Area under p-v diagram is represent work.

Areas Δ PTS= IAS-3. Ans. (a) η =

1 1 Area (WVUR) ∴ Work PTS= × 48 =24 Nm 2 2

work output work out put 54 = = = 0.45 Heat input work output + heat rejection 54 + 66

IAS-4. Ans. (b) Thermal efficiency =

W 3 × 103 watts = = 0.3 = 30% Q 10.000 J/s

IAS-5. Ans. (c) Q1− 2 = (U2 − U1 ) + W1−2 or 0 = (U2 − U1 ) + ( −5000 ) or (U2 − U1 ) = 5000 J Q2−1 = (U1 − U2 ) + W2−1 or W2−1 = Q2−1 − (U1 − U2 ) = Q2−1 + (U2 − U1 ) = 1000 + 5000 = 6000 J

IAS-6. Ans. (c) Net work output = 3 + 10 – 8 = 5 unit

Therefore efficiency, η =

and Heat added = 30 + 5 = 35 unit

5 × 100% = 14.33% 35

IAS-7. Ans. (a) IAS-8. Ans. (d) Q123 = U13 + W123 or, 100 = U13 + 60 or, U13 = 40 kJ And Q143 = U13 + W143 = 40+20 = 60 kJ IAS-9.Ans. (b) Total work = 5 × 3 + IAS-10. Ans. (c) Efficiency =

=

1 × 5 × 1 = 17.5 J or δW = du + δW = 2.5 + 17.5 = 20 J 2

Area under 500 and 1500 Area under 0 and 1500

1 × {(5 − 1) + (4 − 2)} × (1500 − 500) 2

1 × {(5 − 1) + (4 − 2)} × (1500 − 500) + (5 − 1) × 500 2

=

3000 = 0.6 5000

IAS-11. Ans. (a) IAS-12. Ans. (b) Internal energy is a property of a system so

∫ du = 0

IAS-13. Ans. (a) dQ = du + dW if du = +30kJ then dQ = −50kJ and dW = −80kJ IAS-14. Ans. (b) δW = du + δW = du + pdV

or 84 × 103J = du + 1 × 106 × (0.06 – 0.03) = du +30 kJ or du = 83 – 30 = 54 kJ 59

First Law of Thermodynamics S K Mondal’s

Chapter 2

IAS-15. Ans. (a) Q1-2 = U2 –U1 +W1-2 Or 0 = U2 –U1 - 6000 or U2 –U1 = +6000 Q2-1 = U1-U2+W2-1 or W2-1 = Q2-1 - (U1-U2) =1000+6000=7000J

IAS-16. Ans. (c) IAS-17. Ans. (a) Energy balance gives as dW dQ = ( Δh )air + ( Δh ) water + dm dm dQ or = 90 − 30 − 40 dm = 20kJ / kg of air compressed. IAS-18. Ans. (a) dQ dw = m ( h2 ) + dt dt dw dQ or = m ( h1 − h2 ) + = 2 × (100 − 200 ) − 50 × 2 = −300kW dt dt i.e. 300kW work have to given to the system.

m ( h1 ) +

IAS-19. Ans. (b) IAS-20. Ans. (a)

60

Second Law of Thermodynamics S K Mondal’s

3.

Chapter 3

Second Law of Thermodynamics

Theory at a Glance (For GATE, IES & PSUs) The first law of thermodynamics states that a certain energy balance will hold when a system undergoes a change of state or a thermodynamic process. But it does not give any information on whether that change of state or the process is at all feasible or not. The first law cannot indicate whether a metallic bar of uniform temperature can spontaneously become warmer at one end and cooler at the other. All that the law can state is that if this process did occur, the energy gained by one end would be exactly equal to that lost by the other. It is the second law of thermodynamics which provides the criterion as to the probability of various processes.

Regarding Heat Transfer and Work Transfer (a)

Heat transfer and work transfer are the energy interactions. A closed system and its surroundings can interact in two ways: by heat transfer and by work transfer. Thermodynamics studies how these interactions bring about property changes in a system.

(b)

The same effect in a closed system can be brought about either by heat transfer or by work transfer. Whether heat transfer or work transfer has taken place depends on what constitutes the system.

(c)

Both heat transfer and work transfer are boundary phenomena. Both are observed at

the boundaries of the system, and both represent energy crossing the boundaries of the system. (d)

It is wrong to say 'total heat' or 'heat content' of a closed system, because heat or work is not a property of the system. Heat, like work, cannot be stored by the system. Both heat and

work are the energy is transit. (e)

Heat transfer is the energy interaction due to temperature difference only. All other energy interactions may be termed as work transfer.

(f)

Both heat and work are path functions and inexact differentials. The magnitude of

heat transfer or work transfer depends upon the path the system follows during the change of state. (g)

Heat transfer takes place according to second law of thermodynamics as it tells about the direction and amount of heat flow that is possible between two reservoirs.

62

Seco ond La aw of Therm modyn namicss S K Mondal’s

Cha apter 3

Q Qualitatiive Diffference e between Hea at and Work W •

•

Thermoodynamic deffinition of woork: Positivee work is done by a system when thee sole effec ct external to t the system m could be reduced d to the rise of o a weight. Thermoodynamic deffinition of heeat: It is th he energy in n transition n between th he system an nd the surrou undings by virtue v of the differen nce in temperrature.

Siign Conv ventions s • • • •

Work done d BY the system s is +ve e Obviously work don ne ON the sysstem is –ve Heat giiven TO the system s is +v ve Obviously Heat reje ected by the system s is –v ve

Heeat and work are not comp pletely intercchangeable forms f of energy. When woork is convertted into hea at, we alway ys have

W =Q but when heat is converted into work in n a complete closed c cycle process p

Q >W Th he arrow indicates the direction of energy transform mation.

Wo ork is said too be a high grade g energ gy and heat is i low grade e energy. Th he complete conversion c of low grade en nergy into high grade eneergy in a cycle is impossib ble.

HEAT T and WORK K are NOT pr roperties beecause they depend d on thee 63

Second Law of Thermodynamics S K Mondal’s

Chapter 3

path and end states. HEAT and WORK are not properties because their net change in a cycle is not zero.

Heat and work are inexact differentials. Their change cannot be written as differences between their end states.

Thus Similarly

∫ ∫

2

1 2

1

δQ ≠ Q2 − Q1 and is shown as 1 Q2 or Q1−2 δW ≠ W2 − W1 and is shown as 1W2 or W1−2

Note. The operator δ is used to denote inexact differentials and operator d is used to denote exact differentials.

Kelvin-Planck Statement of Second Law There are two statements of the second law of thermodynamics, the Kelvin-Planck statement, and the Clausius statement. The Kelvin-Planck statement pertains to heat engines. The Clausius statement pertains to refrigerators/heat pumps .

Kelvin-Planck statement of second law It is impossible to construct a device (engine) operating in a cycle that will produce no effect other than extraction of heat from a single reservoir and convert all of it into work. Mathematically, Kelvin-Planck statement can be written as:

Wcycle ≤ 0 (for a single reservoir) Clausius’ statement of second law It is impossible to transfer heat in a cyclic process from low temperature to high temperature without work from external source.

Reversible and Irreversible Processes A process is reversible with respect to the system and surroundings if the system and the surroundings can be restored to their respective initial states by reversing the direction of the process, that is, by reversing the heat transfer and work transfer. The process is irreversible if it cannot fulfill this criterion.

64

Seco ond La aw of Therm modyn namicss S K Mondal’s

Cha apter 3

C Clausius s' Theo orem Let a system be b taken from m an equilib brium state i to another equilibrium state f by foollowing the rev versible path h i-f(Figure). Let a reverssible adiabatiic i-a be draw wn through i and anotheer reversible adiiabatic b-f bee drawn thro ough f. Then a reversiblee isothermal a-b is drawn n in such a way w that the areea under i-a-b-f is equal to t the area under u i-f. App plying the firsst law for

Processs i − f Qi − f = U f − U i + Wif Processs i − a − b − f Qiabf = U f − U i + Wiabf i Since Wif = Wiabf ∴

Qi f = Qiabf = Qia + Qab + Qbf

Since

Qia = 0 and d Qbf = 0 Qif = Qab

Fig. Rev versible Patth Substitutted by Two o Reversible e Adiabatics and a Reversible e Isothermall

Heeat transferreed in the process i-f is equ ual to the hea at transferreed in the isoth hermal proceess a-b. Th hus any reveersible path may m be subsstituted by a reversible zigzag path, between th he same end sta ates, consistiing of a rev versible adia abatic follow wed by a rev versible isotthermal and then by a rev versible adia abatic, such that t the hea at transferred during thee isothermal process is the t same as tha at transferred during the original proccess. Let a smooth closed c curve representing g a reversiblee cycle (Fig below.) b be considered. Leet the closed cyccle be divided d into a larg ge number off strips by means m of reversible adiaba atics. Each strip may be cloosed at the toop and bottom m by reversib ble isotherma als. The origiinal closed cy ycle is thus reeplaced by a zig gzag closed path p consistiing of altern nate adiabatiic and isotheermal processses, such th hat the heat tra ansferred durring all the isothermal prrocesses is eq qual to the heat h transferrred in the orriginal cycle. Th hus the origin nal cycle is replaced by a large number of Carnot cycles. If th he adiabatics are close to onee another an nd the numbe er of Carnot cycles c is larg ge, the saw-tooothed zig-za ag line will cooincide with thee original cyccle.

65

Seco ond La aw of Therm modyn namicss S K Mondal’s

Cha apter 3

F Fig. A Reverssible Cycle Split S into a Large Num mber of Carn not Cycles a dQ1 heat is absorb bed reversiblly at T1, an nd dQ2 heat is rejected For the elemeental cycle abcd versibly at T2 rev dQ1 dQ2 = T1 T2

If h heat supplied d is taken as positive and d heat rejecteed as negative dQ1 dQ2 + =0 T2 T1

Sim milarly, for th he elementall cycle efgh dQ3 dQ4 + =0 T3 T4

If ssimilar equattions are wriitten for all th he elementall Carnot cycles, then for the t whole original cycle dQ3 dQ4 d Q1 dQ2 + + + + ... = 0 T1 T2 T3 T4

or

∫

R

dQ d =0 T

Th he cyclic inttegral of d Q/T for a re eversible cy ycle is equall to zero. Th his is known as a Clausius' theeorem. The leetter R emph hasizes the fa act that the equation is va alid only for a reversible cycle. c

66

Second Law of Thermodynamics S K Mondal’s

Chapter 3

Thus the original cycle is replaced by a large number of Carnot cycles. If the adiabatics are close to one another and the number of Carnot cycles is large, the saw-toothed zig-zag line will coincide with the original cycle.

Refrigerator and Heat Pump [with RAC] Equivalence of Kelvin-Planck and Clausius Statements II Law basically a negative statement (like most laws in society). The two statements look distinct. We shall prove that violation of one makes the other statement violation too. Let us suspect the Clausius statement-it may be possible to transfer heat from a body at colder

to a body at hotter temperature without supply of work

Let us have a heat engine operating between T1 as source and T2 as a sink. Let this heat engine reject exactly the same Q2 (as the pseudo-Clausius device) to the reservoir at T2. To do this an amount of Q1 needs to be drawn from the reservoir at T1. There will also be a W = Q1 – Q2. Combine the two. The reservoir at T2 has not undergone any change (Q2 was taken out and by pseudo-Clausius device and put back by the engine). Reservoir 1 has given out a net Q1-Q2. We got work output of W. Q1-Q2 is converted to W with no net heat rejection. This is violation of KelvinPlanck statement.

67

Second Law of Thermodynamics S K Mondal’s •

Chapter 3

Let us assume that Clausius statement is true and suspect Kelvin-Planck statement

Pseudo Kelvin Planck engine requires only Q1–Q2 as the heat interaction to give out W (because it does not reject any heat) which drives the Clausius heat pump Combining the two yields: • The reservoir at T1 receives Q1 but gives out Q1–Q2 implying a net delivery of Q2 to it. • Q2 has been transferred from T2 to T1 without the supply of any work!! • A violation of Clausius statement

Moral: If an engine/refrigerator violates one version of II Law, it violates the other one too.

All reversible engine operating between the same two fixed temperatures will have the same η and COP. If there exists a reversible engine/ or a refrigerator which can do better than that, it will violate the Clausius statement. Let us presume that the HP is super efficient!! For the same work given out by the engine E, it can pick up an extra Δ Q from the low temperature source and deliver over to reservoir at T1 . The net effect is this extra has Δ Q been transferred from T2 to T1 with no external work expenditure. Clearly, a violation of Clausius statement!!

SUM UP • Heat supplied = Q1; Source temperature = T1 ; Sink temperature = T2 • Maximum possible efficiency = W/Q1= (T1 – T2)/T1 • Work done = W = Q1(T1 – T2)/T1

68

Seco ond La aw of Therm modyn namicss S K Mondal’s

Cha apter 3

Carnot Engine E with w sam me effic ciency or o same e work output o T1 Q1 = T2 Q2

Sin nce,

T1 − T2 Q1 − Q2 = T2 Q2 T1 − T2 = (Q1 − Q2 )

T2 Q2

T2 − T3 = (Q2 − Q3 )

T3 T = (Q2 − Q3 ) 2 Q2 Q3

• For sam me work ou utput W1 = W2 Q1 – Q 2 = Q2 – Q3

T1 – T2 = T2 – T3 •

For sam me efficiency

η1 = η2 orr 1 −

T T2 = 1− 3 T1 T2

o T2 = T1 × T3 or • Whic ch is the t mo ore effe ective way w to increase the efficiency off a Carn not engiine: to increase e T1 keeping T2 stant; orr to decrrease T2 keeping g T1 con nstant? cons Th he efficiency of o a Carnot engine is giveen by

η = 1−

T2 T1

If T2 is constan nt

⎛ ∂η ⎞ T2 ⎜ ⎟ = 2 ⎝ ∂T1 ⎠T2 T1 ⎛ ∂η ⎞ AS S T1 increasess, η increasess, and the sloope ⎜ d (Fiigure). If T1 is i constant, ⎟ decreases ⎝ ∂T1 ⎠T2

69

Seco ond La aw of Therm modyn namicss S K Mondal’s

Cha apter 3 ⎛ ∂η ⎞ 1 ⎜ ⎟ = − T1 ⎝ ∂T2 ⎠T1

⎛ ∂η ⎞ As T2 decreasess, η increasess, but the slo ope ⎜ r consttant (Figure)). ⎟ remains ⎝ ∂T2 ⎠T1

Also Since

⎛ ∂η ⎞ ⎛ ∂η T2 T1 η⎞ ⎟ =− 2 ⎜ ⎟ = 2 and ⎜ T1 ⎝ ∂T1 ⎠T2 T1 ⎝ ∂T2 ⎠T1 ⎛ ∂η ⎞ ⎛ ∂∂η ⎞ T1 > T2 , ⎜ ⎟ >⎜ ⎟ ⎝ ∂T2 ⎠T1 ⎝ ∂T1 ⎠T2

So, the more effective e way y to increasse the efficieency is to deecrease T2. Alternatively A y, let T2 be deccreased by ΔT with T1 rem maining the same s

η1 = 1 −

T2 − ΔT T1

If T1 is increase ed by the sam me ΔT, T2 rem maining the same s

η2 = 1 −

T2 T1 + ΔT

The en η1 − η 2 =

T2 T − ΔT − 2 T1 + ΔT T1

(T =

1

Sin nce

− T2 ) ΔT + ( ΔT )

2

T1 (T1 + ΔT )

T1 > T2 , ( η1 − η 2 ) > 0

Th he moree effectiv ve way to increease thee cycle efficiency e y is

d decrea ase T2. 70

to

Second Law of Thermodynamics S K Mondal’s

Chapter 3

PROBLEMS & SOLUTIONS Example 1. An inventor claims that his petrol engine operating between temperatures of 2000°C and 600°C will produce 1 kWhr consuming 150g of petrol having 45000 kJ/kg calorific value. Check the validity of the claim. Solution: By Carnot's theorem, the thermal efficiency of a reversible cycle engine which cannot be exceeded is given by T − T2 2273 − 873 ηmax = 1 = = 0.616 or 61.6% 2273 T1 Actual thermal efficiency is given by 1 ×103 × 3600 = 0.53 or 53% ηt = 0.15 × 45000 ×103 Since actual efficiency is less than the maximum obtainable, the inventor's claim is feasible. Example 2. Two reversible engines takes 2400 kJ per minute from a reservoir at 750 K and develops 400 kJ of work per minute when executing complete cycles.The engines reject heat two reservoirs at 650 K and 550 K. Find the heat rejected to each sink. Solution: QA + QB = 2400 QA2 + QB2 = 2000 QA − QA2 QA

=

750 − 650 = 0.1333 750

QA = 1.1539QA 2

71

Second Law of Thermodynamics S K Mondal’s

Chapter 3

Similarly QB = 1.3636QB 2 i.e. 1.1539Q A 2 + 1.3636QB 2 = 2400kJ Q A 2 = 2000 − QB 2

(

)

i.e. 1.1539 × 2000 − QB 2 + 1.3636QB 2 = 2400kJ . 0.2091QB 2 = 92 ∴ QB 2 = 440kJ and QA 2 = 1560kJ

Example 3. A solar powerd heat pump receives heat from a solar collector at temperature Th, uses the entire energy for pumping heat from cold atmosphere at temperature ‘Tc’ to a room at temperature ‘Ta’. The three heat transfer rates are Qh, Qa and Qc respectively. Derive an expression for the minimum ratio Qh/Qc, in terms of the three temperatures. If Th = 400 K, Ta = 300 K, Tc = 200 K, Qc = 12 kW, what is the minimum Qh? If the collector captures 0.2 kW/m2, what is the minimum collector area required? Solution: Let Qh, Qa and Qc be the quantity of heat transferred from solar collector, room and atmosphere respectively.

Qa = Qh + Qc Qa − Qc = Qh

or,

COPHP =

Ta Qa Q + Qc = h and also Ta − Tc Qa − Qc Qh

Qh + Qc Ta = Qh Ta − Tc

∴

Qc Ta T − Ta + Tc Tc = −1 = a = Qh Ta − Tc Ta − Tc Ta − Tc ∴ Qh =

Qh (Ta − Tc ) = Qc Tc

Qc × (Ta − Tc )

Area =

Tc

=

12 × ( 300 − 200 ) 200

= 6kW

6 = 30m2 0.2

72

Second Law of Thermodynamics S K Mondal’s

Chapter 3

Example 4. A reversible engine works between 3 thermal reservoirs A, B and C. The engine absorbs an equal amount of heat from the reservoirs A and B, maintained at temperatures of T1 and T2 respectively and rejects heat to the thermal reservoir C maintained at T3. The efficiency of this engine is α times the efficiency of reversible engine operating between reservoirs A and C only. Show that T1 2T = ( 2α − 1) + 1 (1 − α ) T2 T3

Solution:

η1 =

W1 2Q1

η2 =

T1 − T3 T1 ⎛ T1 − T3 ⎞ ⎟ ⎝ T1 ⎠

η1 = αη2 = α ⎜

The cycle is reversible so ΔS = 0 Q Q Q ∴ 1+ 1 = 2 T1 T2 T3 Also, 2Q1 = W1 + Q2 Combining above equations we have Q1 Q1 2Q1 − W1 + = T1 T2 T3 ⎛ T − T3 ⎞ W1 =α⎜ 1 ⎟ 2Q1 ⎝ T1 ⎠

73

Second Law of Thermodynamics S K Mondal’s

Chapter 3

⎛ T − T3 ⎞ W1 = 2Q1α ⎜ 1 ⎟ ⎝ T1 ⎠ α ( T1 − T3 ) ⎞ Q1 Q1 2Q1 ⎛ + = ⎜⎜1 − ⎟⎟ T1 T2 T3 ⎝ T1 ⎠ 1 1 2 2α 2α + = − + T1 T2 T3 T3 T1 T1 2T1 = (1 − α ) + ( 2α − 1) T2 T3

Example 5. Ice is to be made from water supplied at 15°C by the process shown in figure. The final temperature of the ice is -10°C, and the final temperature of the water that is used as cooling water in the condenser is 30°C.

Determine the minimum work required to produce 1000 kg of ice. Solution: Quantity of ice produced = 1000 kg. Specific heat of ice = 2070 J/kg K Specific heat of water = 4198 J/kgK Latent heat of ice = 335 kJ/kg 15 + ( −10 ) Mean temperature of ice bath = = 2.5º C 2 15 + 30 = 22.5° C Mean temperature of condenser bath = 2 Q2 = 1000 × 4198(15 - 2.5) + 1000 × 335 × 103 +1000 × 2070(2.5 + 10) = 413.35MJ For a reversible refrigerator system T Q T Q 1− L = 1− 2 ; L = 2 ; TH Q1 TH Q1 Q1 = Q2

TH ⎛ 22.5 + 273 ⎞ = 413.35 × 106 ⎜ ⎟ = 443.36 MJ TL ⎝ 2.5 + 273 ⎠

Minimum work required = Q1 – Q2 = 443.36 - 413.35 = 30.01 MJ.

74

Second Law of Thermodynamics S K Mondal’s

Chapter 3

75

Second Law of Thermodynamics S K Mondal’s

Chapter 3

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years IES Questions IES-1.

Which one of the following is correct on basis of the second law of thermodynamics? [IES 2007] (a) For any spontaneous process, the entropy of the universe increases (b) ∆S =qrev/T at constant temperature (c) Efficiency of the Stirling cycle is more than that of a Carnot cycle (d) ∆E=q+w (The symbols have their usual meaning)

IES-2.

Assertion (A): Second law of thermodynamics is called the law of degradation of energy. [IES-1999] Reason (R): Energy does not degrade each time it flows through a finite temperature difference. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-3.

Heat transfer takes place according to (a) Zeroth Law of Thermodynamics (c) Second Law of Thermodynamics

IES-3a.

Consider the following statements: [IES-2010] 1. Slow heating of water from an electric heater. 2. Isentropic expansion of air. 3. Evaporation of a liquid from a heat source at the evaporation temperature. 4. Constant pressure heating of a gas by a constant temperature source. Which of these processes is/are reversible? (a) 3 only (b) 2 and 3 only (c) 2 and 4 only (d) 1, 2, 3 and 4

IES-3b.

Consider the following statements: [IES-2010] 1. Boiling of water from a heat source at the same boiling temperature. 2. Theoretical isothermal compression of a gas. 3. Theoretical polytropic compression process with heat rejection to atmosphere. 4. Diffusion of two ideal gases into each other at constant pressure and temperature. Which of these processes are irreversible? (a) 1, 2, 3 and 4 (b) 1 and 4 only (c) 2, 3 and 4 only (d) 3 and 4 only

76

[IES-1996] (b) First Law of Thermodynamics (d) Third Law of Thermodynamics.

Second Law of Thermodynamics S K Mondal’s

Chapter 3

Kelvin-Planck Statement of Second Law IES-4.

Consider the following statements: The definition of 1. Temperature is due to Zeroth Law of Thermodynamics. 2. Entropy is due to First Law of Thermodynamics. 3. Internal energy is due to Second Law of Thermodynamics. 4. Reversibility is due to Kelvin-Planck's statement. Of these statements (a) 1, 2 and 3 are correct (b) 1, 3 and 4 are correct (c) 1 alone is correct (d) 2 alone is correct

[IES-1993]

Clausius' Statement of the Second Law IES-5.

Assertion (A): Heat cannot spontaneously pass from a colder system to a hotter system without simultaneously producing other effects in the surroundings. Reason (R): External work must be put into heat pump so that heat can be transferred from a cold to a hot body. [IES-1999] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Clausius' Theorem IES-6.

A steam power plant is shown in figure,

(a) The

cycle

violates

first

and

second laws of thermodynamics. (b) The cycle does not satisfy the condition of Clausius inequality. (c) The cycle only violates the second laws of thermodynamics (d) The cycle satisfies the Clausius inequality [IES-1992] IES-7.

An inventor says that his new concept of an engine, while working between temperature limits of 27°C and 327°C rejects 45% of heat absorbed from the source. His engine is then equivalent to which one of the following engines? (a) Carnot engine (b) Diesel engine [IES-2009] (c) An impossible engine (d) Ericsson engine

77

Second Law of Thermodynamics S K Mondal’s IES-7a

Chapter 3

An inventor states that his new engine rejects to the sink 40% of heat absorbed from the source while the source and sink temperatures are 327º C and 27º C respectively. His engine is therefore equivalent to [IES-2010] (a) Joule engine (b) Stirling engine (c) Impossible engine (d) Carnot engine

Equivalence of Kelvin-Planck and Clausius Statements IES-8.

Assertion (A): Efficiency of a reversible engine operating between temperature [IES-2002] limits T1 and T2 is maximum. Reason (R): Efficiency of a reversible engine is greater than that of an irreversible engine. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Carnot Engine with same efficiency or same work output IES-9.

A reversible engine operates between temperatures T1, and T2, The energy rejected by this engine is received by a second reversible engine at temperature T2 and rejected to a reservoir at temperature T3. If the efficiencies of the engines are same then the relationship between T1, T2 and T3 is given by: [IES-2002]

(a) T2

=

(T1 + T3 ) 2

(b) T2

=

(T

2 1

+ T32

)

(c) T2

= T1T3

(d)

T2 =

(T1 + 2T3 ) 2

IES-10.

A reversible engine operates between temperatures 900 K & T2 (T2 < 900 K), & another reversible engine between T2 & 400 K (T2 > 400 K) in series. What is the value of T2 if work outputs of both the engines are equal? [IES-2005] (a) 600 K (b) 625 K (c) 650 K (d) 675 K

IES-10a.

An engine operates between temperature limits of 900 K and T2 and another between T2 and 400 K. For both to be equally efficient, the value of T2 will be (a) 700 K (b) 600 K (c) 750 K (d) 650 [IES-2010]

IES-11.

Two reversible engine operate between thermal reservoirs at 1200 K, T2K and 300 K such that 1st engine receives heat from 1200 K reservoir and rejects heat to thermal reservoir at T2K, while the 2nd engine receives heat from thermal reservoir at T2K and rejects heat to the thermal reservoir at 300 K. The efficiency of both the engines is equal. [IES-2004] What is the value of temperature T2? (a) 400 K (b) 500 K

IES-12.

(c) 600 K

(d) 700 K

A series combination of two Carnot’s engines operate between the temperatures of 180°C and 20°C. If the engines produce equal amount of work, then what is the intermediate temperature? [IES-2009]

78

Second Law of Thermodynamics S K Mondal’s (a) 80°C

Chapter 3 (b) 90°C

(c) 100°C

(d) 110°C

IES-13.

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at 27°C, then what is the source temperature? [IES-2009] (a) 750°C (b) 477°C (c) 203°C (d) 67.5°C

IES-13a

A Carnot engine rejects 30% of absorbed that to a sink at 30º C. The temperature of the heat source is [IES-2010] (a) 100º C (b) 433º C (c) 737º C (d) 1010º C

IES-14.

A reversible heat engine rejects 50 percent of the heat supplied during a cycle of operation. If this engine is reversed and operates as a heat pump, then what is its coefficient of performance? [IES-2009] (a) 1.0 (b) 1.5 (c) 2.0 (d) 2.5

IES-15. A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of 227°C; the heat is rejected at 27°C, the cycle is reversible, then what amount of heat is rejected? [IES-2009] (a) 250 kJ/s (b) 200 kJ/s (c) 180 kJ/s (d) 150 kJ/s IES-16.

One reversible heat engine operates between 1600 K and T2 K, and another reversible heat engine operates between T2K and 400 K. If both the engines have the same heat input and output, then the temperature T2 must be equal to: [IES-1993] (a) 1000 (b) 1200 (c) 1400 (d) 800

Previous 20-Years IAS Questions Kelvin-Planck Statement of Second Law IAS-1.

Assertion (A): No machine would continuously supply work without expenditure of some other form of energy. [IAS-2001] Reason (R): Energy can be neither created nor destroyed, but it can only be transformed from one form into another. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Equivalence of Kelvin-Planck and Clausius Statements IAS-2.

A heat engine is supplied with 250 KJ/s of heat at a constant fixed temperature of 227°C. The heat is rejected at 27°C. The cycle is reversible, if the amount of heat rejected is: [IAS-1995] (a) 273 KJ/s (b) 200 KJ/s (c) 180 KJ/s (d) 150 KJ/s.

79

Second Law of Thermodynamics S K Mondal’s IAS-3.

Chapter 3

A reversible engine En as shown in the given figure draws 300 kcal from 200 K reservoir and does 50 kcal of work during a cycle. The sum of heat interactions with the other two reservoir is given by: (a) Q1 + Q2 = + 250 kcal (b) Q1 + Q2 = –250 kcal (c) Q1 + Q2 = + 350 kcal (d) Q1 + Q2 = –350 kcal

[IAS-1996]

Carnot Engine with same efficiency or same work output IAS-4.

Consider the following statements: [IAS-2007] 1. Amount of work from cascaded Carnot engines corresponding to fixed temperature difference falls as one goes to lower absolute level of temperature. 2. On the enthalpy-entropy diagram, constant pressure lines diverge as the entropy increases. Which of the statements given above is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

IAS-5.

In a cyclic heat engine operating between a source temperature of 600°C and a sink temperature of 20°C, the least rate of heat rejection per kW net output of the engine is: [IAS 1994] (a) 0.460 kW (b) 0.505 kW (c) 0.588 kW (d) 0.650 kW

Answers with Explanation (Objective) Previous 20-Years IES Answers IES-1. Ans. (a) IES-2. Ans. (c) A is true but R is false. IES-3. Ans. (c) Heat transfer takes place according to second law of thermodynamics as it tells about the direction and amount of heat flow that is possible between two reservoirs. IES-3a. Ans. (b) All spontaneous processes are irreversible. Statement-1 and statement-4 heat is transferred with a finite temperature difference they are irreversible. IES-3b. Ans. (d) Any natural process carried out with a finite temperature gradient is an irreversible process. All spontaneous processes are irreversible. Statement -4 is a spontaneous process. IES-4. Ans. (c) Out of 4 definitions given, only first definition is correct and balance three are wrong. IES-5. Ans. (b) A and R are true. A is the Clausius statement of second law of thermodynamics. Spontaneously means without change in surroundings. If question comes like following then answer will be (a).

80

Second Law of Thermodynamics S K Mondal’s

Chapter 3

Assertion (A): External work must be put into heat pump so that heat can be transferred from a cold to a hot body. Reason (R): Heat cannot spontaneously pass from a colder system to a hotter system without simultaneously producing other effects in the surroundings. IES-6. Ans. (d) 300 T = 0.5 IES-7. Ans. (c) Carnot efficiency of engine = η = 1 − 2 = 1 − 600 T1

IES-7a

But according to the inventor’s Claim Efficiency of engine = 1-0.45 = 0.55 ∵ Efficiency of Actual Engine cannot be greater then Carnot efficiency. So this is an impossible engine. Ans. (c) We know Carnot efficiency ηcarnot = 1 −

T2 T1

⇒1−

300 1 = = 0.5 600 2

ηcarnot = 50% But inventor claims 60% efficiency (means 40% heat rejection). It is then impossible. IES-8. Ans. (a) IES-9. Ans. (c) IES-10. Ans. (c) Figure from another question W1 = W2

or Q1 − Q2 = Q2 − Q3 or T1 − T2 = T2 − T3 or T2 = IES-10a.

Ans. (b) When equally efficiency

1−

T T2 = 1− 3 T1 T2

or T2 = T1T3 = 900 × 400 = 600 K

IES-11. Ans. (c) η1 = η2

or 1 −

T2 300 = 1− 1200 T2

or T2 = 1200 × 300 = 600K

81

T1 + T3 900 + 400 = = 650K 2 2

Seco ond La aw of Therm modyn namicss S K Mondal’s

Cha apter 3

IES-12. Ans. (c c) Sou urce Tempera ature = T1, Inte ermediate Temperature = T Sin nk Temperatu ure = T2

W1 = W2

∵

⎛ T⎞ ⎛ T ⎞ Q1 ⎜ 1 − ⎟ = Q2 ⎜ 1 − 2 ⎟ T⎠ ⎝ ⎝ T1 ⎠

T1 ⎛ T1 T2 T ⎞ ⎛ T2 ⎞ ⎜1− ⎟ = ⎜1− ⎟ ⇒ − 1 = 1− T ⎝ T1 ⎠ ⎝ T⎠ T T T1 + T2 T + T2 180 + 20 0 =2 = = 100°C ⇒ ⇒T = 1 2 2 T IES-13. Ans. (b b) Sink temp perature = 27 7°C = 27 + 27 73 = 300K It is given that engine rejeccts 40% of ab bsorbed heat from the sou urce Q Q Forr a carnot cyccle engine 1 = 2 T2 T1 ⇒

Q 0.4 4Q 30 00 = ⇒T= = 750K = 477°C T 30 00 0..4

IES-13a

Ans. (c) Q1 Q2 = T1 T2 or T1 = T2 ×

Q1 Q = 30 03 × 1 = 1010 K = 737 o C Q2 0.3Q1

Q1 Q2 = T1 T2 Q 0.5 Q1 ⇒ 1 = T1 T2

IES-14. Ans. (c c)

⇒

T2 = 0.5 T1

If the engine is reversed an nd operated as a the Heat Pum mp. Theen COP coeffficient of perfformance

=

T1 = T1 − T2

1 1 = =2 T2 5 1 − 0.5 1− T1

IES-15. Ans. (d d) Heat supp plied by the Heat H Engine = Q1 = 250 kJ/sec k Sou urce tempera ature = 227°C = 500 K Sin nk temperatu ure = 27°C = 300K 3 82

Second Law of Thermodynamics S K Mondal’s

Chapter 3

250 Q2 = ⇒ Q2 = 250 × 0.6 = 150kJ / sec 500 300 IES-16. Ans. (d) Two reversible heat engines operate between limits of T2 and 400K 1600K and T2; Both have the same heat input and output,

T1 − T2 1600 − T2 T2 − 400 is same for both or = or T2 = 800 K T1 1600 T2

i.e.

Previous 20-Years IAS Answers IAS-1. Ans. (a) IAS-2. Ans. (d) IAS-3. Ans. (b)

Q1 Q2 = T1 T2

∑Q = ∑W

300 + Q1 + Q2 = 50 IAS-4. Ans. (b) For reversible cycle

T1 T2 T3 = = Q1 Q2 T3 or

T1 − T2 Q1 − Q2 = T2 Q2

or T1 − T2 = (Q1 − Q2 ) × Similarly

T2 Q2

T2 − T3 = ( Q2 − Q3 ) ×

T3 Q3

If T1 − T2 = T2 − T3 then Q1 − Q2 = Q2 − Q3 or W1 = W2 IAS-5. Ans. (b) Reversible engine has maximum efficiency where

Therefore least heat rejection per kW net output, W 1 Q2 = × T2 = × 293 = 0.505 kW T1 − T2 873 − 293

83

Q1 Q2 Q1 − Q2 W = = = T1 T2 T1 − T2 T1 − T2

Enttropy S K Mondal’s

4.

Cha apter 4

Entro opy

Theo ory at a Glance (Fo or GAT TE, IES S & PS SUs) Tw wo Reve ersible Adiabati A c Paths cannot Intersec ct Each Other Let it be assum med that two o reversible adiabatics a AC and BC in ntersect each h other at point C (Fig.). A be drawn in such a wa ay that it inteersects the reeversible adiabatics at A Let a reversiblee isotherm AB e processes AB, A BC, and CA C together constitute c a reversible cy ycle, and the and B. The three reversible r th he net work output in a cycle. But such s a cycle is impossiblle, since net areea included represents woork is being produced p in a cycle by a heat h engine by b exchangin ng heat with a single reseervoir in the proocess AB, which w violattes the Kelv vin-Planck statement s o the secon of nd law. Theerefore, the asssumption of the intersection of the reversible r ad diabatics is Fig. F wrong. Through T one point, there can n pass only one o reversiblee adiabatic. nce two consstant propertty lines can never intersect each otther, it is infferred that a reversible Sin adiiabatic path must represe ent some prooperty, which h is yet to be identified. i

83

Entropy S K Mondal’s

Chapter 4

The Property of Entropy

⎛ dQ ⎞ ds = ⎜ ⎟ ⎝ T ⎠ •

•

Reversible

dS is an exact differential because S is a point function and a property. The subscript R in dQ indicates that heat dQ is transferred reversibly.

S s = J / kg K m

•

It is an extensive property, and has the unit J/K. The specific entropy is an intensive property and has unit J/kgK

•

The change of entropy may be regarded as a measure of the rate of availability of heat for transformation into work.

If the system is taken from an initial equilibrium state i to a final equilibrium state f by an irreversible path, since entropy is a point or state function, and the entropy change is independent of the path followed, the non-reversible path is to be replaced by a reversible path to integrate for the evaluation of entropy change in the irreversible process

Sf − Si =

∫

i

f

dQrev = ( ΔS )irrev path T

Integration can be performed only on a reversible path.

84

Enttropy S K Mondal’s

Cha apter 4

Te empera ature-Entropy Plot

The Ineq quality of Clau usius Th hen for any cy ycle

∫

dQ ≤ T

∫

ds

Sin nce entropy is i a property and the cycliic integral off any property is zero

∫

dQ T

≤ 0

Th his equation is i known as the t inequalitty of Clausiu us. It providess the criterion of the reverrsibility of a cyccle.

If

∫ ∫

dQ = 0 , the cyclle is reversibble, T d dQ < 0, the cycle isi irreversibble and possible T 85

Entropy S K Mondal’s

∫

Chapter 4

dQ > 0 , the cycle is impossible, since it violates the second law. T

Entropy Change in an Irreversible Process

Flow of current through a resistance – when a battery discharges through a resistance heat is dissipated. You can’t recharge the battery by supplying heat back to the resistance element!! Pickpocket !!!Marriage!!!!.............................................are irreversible Process.

Applications of Entropy Principle (S1-S2)irreversible > (S1-S2)reversible An irreversible process generates more entropy than a reversible process. An irreversible engine can’t produce more work than a reversible one.

•

An irreversible heat pump will always need more work than a reversible heat pump.

•

An irreversible expansion will produce less work than a reversible expansion

•

An irreversible compression will need more work than a reversible compression.

86

Enttropy S K Mondal’s

Cha apter 4

Maximum Work W Obtainable fr rom two Finite F Bod dies at tem mperature es T1 and T2 Let us consideer two inden ntical finite bodies b of con nstant heat capacity at temperaturee T1 and T2 resspectively, T1 being high her than T2. If the two bodies b are meerely brough ht together in nto thermal con ntact, deliverring no work,, the final tem mperature Tf reached wou uld be the maximum T + T2 Tf = 1 2 ne is operated d between th he two bodies acting as thermal t enerrgy reservoirrs (shown in If a heat engin Fig g. below), parrt of the hea at withdrawn n from body 1 is converted to work W by the heat engine, and thee remainder is rejected to t body 2. Th he lowest atttainable fina al temperaturre Tf corresp ponds to the dellivery of the largest possiible amount of o work, and is associated d with a reverrsible process. A As work is deelivered by th he heat engin ne, the temp perature of body 1 will bee decreasing and that of bod dy 2 will be increasing. i When W both th he bodies atta ain the final temperaturee Tf, the heatt engine will stoop operating. Let the bodiies remain att constant pressure and undergo u no ch hange of phase.

Fig. Total heat with hdrawn from body 1 Q1  = Cp  ( T1  − Tf ) Wh here Cp is thee heat capaciity of the twoo bodies at coonstant presssure. Total heat rejeccted to body 2 Q2 = Cp (Tf – T2) ∴A Amount of tootal work dellivered by thee heat enginee W = Q1 – Q2 = Cp (T1 + T2 - 2T Tf) For given valuees of Cp, T1 and a T2, the magnitude m off work W dep pends on Tf. Work obtain nable will be ma aximum when n Tf is minim mum. Noow, for body 1, 1 entropy ch hange ΔS1 is given g by T Tf dT f ΔS1 = ∫ C p = C pln T1 T1 T For body 2, enttropy change ΔS2 would be Tf Tf dT Δ S2 = ∫ C p = C pln T2 T T2 nce the work king fluid ope erating in thee heat engine cycle does not undergo any entropy y change, ΔS Sin of tthe working fluid in heat engine = ∫ dS d = 0 . Applying the entrropy principlle

87

Enttropy S K Mondal’s

Cha apter 4 ΔSuniv ≥ 0

∴

C pln

Tf T1

C pln

+ C pln Tf2 T1T2

Tf T2

≥0

≥0

nimum From abovee Eq. for  Tf  to be a min C pln n

Tf2 T1 .T2

=0 Tf2

= 0 = ln 1

or

ln

∴

Tf = T1 .T2

T1T2

For W to be e  a maximum m, Tf  will be T1 .T2 . From m abow Equattion   Wmax = C p (T1 + T2 − 2 T1T2 ) = C p ( T1 − T2 )2 Th he final temp peratures of the t two bodiees, initially at a T1 and T2, can range from f (T1 + T2)/2 with no dellivery of work k to

T1 .T2 with w maximu um delivery of o work.

• Maximu um Work Obtainable from a Finite Bo ody and a TER :-

Let one of the bodies consiidered in thee previous seection be a thermal t enerrgy reservoirr. The finite bod dy has a therrmal capacity y Cp and is at a temperaturre T and the TER is at teemperature T0, such that T > T0. Let (Q - W). Then

F Fig. – Ma aximum Worrk Obtainable e When One of the Bodiess is a TER.

88

Enttropy S K Mondal’s

Cha apter 4 T0

ΔSBody = ∫ C p T

T dT = C pln n 0 T T

ΔSHE =

∫ dss = 0

ΔSTER =

Q −W To

T0 Q − W + T To By the entropy principle,.

∴

ΔSuniv = C pln l

ΔSuniv ≥ 0 C pln

T0 Q − W ≥0 + T To

or

C pln

T0 W − Q ≥ To T

or

T W −Q ≤ C pln l 0 T To

or

W ≤ Q + T0C pln

∴

Wmax = Q + T0C pln

or

⎡ T⎤ Wmax = C p ⎢(T − T0 ) − T0ln n ⎥ To ⎦ ⎣

To T To T

• Processses Exhibiiting Exte ernal Mec chanical Ir rreversibility :on of Work :: Let us con nsider the isoothermal disssipation of work w through [i] Isothermall Dissipatio a ssystem into the internall energy of a reservoir, as a in the flow of an elecctric current through a ressistor in conttact with a re eservoir (Fig.in below.) At A steady statte, the intern nal energy of the resistor and hence its teemperature is i constant. So, S by first la aw: W=Q he flow of currrent represents work trransfer. At steady s state the work is dissipated isothermally Th intto heat transsfer to the surroundings. Since the surrroundings absorb a Q unitt of heat at temperature t T,

89

Enttropy S K Mondal’s

Cha apter 4 (Fig.. )

Q W = T T At stea ady state, ΔSsys = 0 ΔSsurr =

W T he irreversible process is thus t accompa anied by an entropy e increease of the un niverse. Th ∴ ΔSuni u v = ΔSsys + ΔSsurr =

[ii]] Adiabatic c Dissipation of Work :: Let W be the t stirring work w supplied to a viscou us thermally inssulated liquid, which is dissipated adiabatically a into interna al energy inccrease of thee liquid, the tem mperature off which incre eases from Ti to Tf ( show wn in fig beloow). Since th here is no flow w of heat to or from the surrroundings.

( (Fig. ) To calculate th he entropy ch hange of the system, thee original irreversible patth (dotted lin ne) must be rep placed by a reversible r on ne between the t same end d states, i an nd f. Let us replace the irreversible perrformance off work by a reversible r isoobaric flow off heat from a series of resservoirs rang ging from Ti to Tf to cause th he same chan nge in the sta ate of the sysstem. The entropy changee of the systeem will be C p dT Tf dQ ΔSsys = ∫ f =∫f = C p ln T Ti Ri T Ri here Cp is thee heat capaccity of the liq quid. wh ΔSuniv = ΔSsys + ΔSsurr = C pln n

Tf Ti

Entropy Genera ation •

Irreversible Processses increase the t entropy of o the universse.

•

Reversiible Processe es do not effecct the entrop py of the univ verse.

•

Impossible Processe es decrease th he entropy off the universse.

ΔSuniverse = 0 90

Entropy S K Mondal’s

Chapter 4

Entropy Generation in the universe is a measure of lost of work.

ΔS Universe = ΔS System + ΔS Surroundings The losses will keep increasing. The sin keeps accumulating. Damage to environment keeps increasing. When the entropy of the universe goes so high, then some one has to come and set it right. HE SAYS HE WILL COME. Every religion confirms this. Let us all wait. Cheer up, things are not that bad yet!!

Entropy and Direction: The Second Law a Directional law of Nature The second law indicates the direction in which a process takes place. A process always occurs in such a direction as to cause an increase in the entropy of the universe.

91

Entropy S K Mondal’s

Chapter 4

Summary 1.

Clausius theorem:

2.

S f − Si =

f

∫ i

⎛ dQ ⎞ =0 T ⎟⎠ rev.

∫ ⎜⎝

dQ rev = ( Δ s ) irrev . path . T

Integration can be performed only on a reversible path.

4.

dQ ≤0 T At the equilibrium state. The system is at the peak of the entropy hill. (isolated)

5.

Tds = du + Pdv

6.

Tds = dh – Vdp

7.

Famous relation S = K lnW

3.

8.

Clausius Inequality:

∫

where K =Boltzmann constant W = thermodynamic probability General case of change of entropy of a Gas.

⎧ P V ⎫ S2 − S1 = m ⎨Cv ln 2 + CP ln 2 ⎬ P1 V1 ⎭ ⎩ Initial condition of gas P1, V1, T1, S1 and Final condition of gas P2, V2, T2, S2 9. Process and property change Table:

92

Entropy S K Mondal’s

Chapter 4

PROBLEMS & SOLUTIONS Example 1. 5 kg of air is compressed in a reversible polytrophic process from 1 bar and 40°C to 10 bar with an index of compression 1.25. Calculate the entropy change during the process. Solution:T2 ⎛ p2 ⎞ =⎜ ⎟ T1 ⎝ p1 ⎠

n −1 n

n −1

0.25

⎛p ⎞n ⎛ 10 ⎞ 1.25 = ( 273 + 40 ) ⎜ ⎟ T2 = T1 ⎜ 2 ⎟ p ⎝1 ⎠ ⎝ 1⎠ Therefore T2 = 496 K 2

2

2

2

dT vdp dT dp −∫ = Cp ∫ − R∫ T T T p 1 1 1 1

φ2 − φ1 = C p ∫

⎛T ⎞ ⎛p ⎞ =1.005ln ⎜ 2 ⎟ − R ln ⎜ 2 ⎟ T ⎝ 1⎠ ⎝ p1 ⎠ 496 = 1.005ln − 0.287 ln(10) 313 = 0.4627 − 0.6608 = − 0.1981 kJ / kg K . Total change in entropy = 5 × (-0.1981) = - 0.9905 kJ/K (Reduction). (Hence heat is rejected during the process).

Example 2.

Two compartments of an insulated vessel each of 3 m3 contain air at 0.7 MPa, 95°C and 0.35 MPa, 205°C. If the removed, find the change in entropy, if the two portions mix completely and adiabatically. Solution:

93

Entropy S K Mondal’s

Chapter 4

m1 =

p1V1 0.7 ×106 × 3 = = 19.883 kg RT1 287 × ( 273 + 95 )

m2 =

p2V2 0.35 ×106 × 3 = = 7.654 kg RT2 287 × ( 273 + 205 )

Assuming specific heat to be constant Tf =

Cv (m1T1 + m2T2 ) Cv (m1 + m2 )

ΔS1 = m1Cv ln

Tf T1

= 398.6 K = 125.6°C

= 19.883 × 1005 × ln

398.6 = 1595J / K 368

398.6 = − 1396.6J / K 478 ΔS = 1595 − 1397 = 198J / K ΔS2 = 7.654 ×1005 × ln

94

Entropy S K Mondal’s

Chapter 4

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions Applications of Entropy Principle GATE-1.

A 1500 W electrical heater is used to heat 20 kg of water (Cp = 4186 J/kg K) in an insulated bucket, from a temperature of 30°C to 80°C. If the heater temperature is only infinitesimally larger than the water temperature during the process, the change in entropy for heater is….. J/k and for water ............. J/K. [GATE-1994]

Entropy Generation in a Closed System GATE-1A An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ΔS of the gas is ⎛P ⎞ ⎛P⎞ ⎛P ⎞ Q (a) mR ln ⎜ 2 ⎟ (b) mR ln ⎜ 1 ⎟ (c) mR ln ⎜ 2 ⎟ − (d ) zero [GATE-2012] P P ⎝ 1⎠ ⎝ 2⎠ ⎝ P1 ⎠ T1

Data for Q2 and Q3 are given below. Solve the problems and choose correct answers. Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2m3. Heat exchange occurs with the atmosphere at 298 K during this process. GATE-2.

The work interaction for the Nitrogen gas is: (a) 200 kJ (b) 138.6 kJ (c) 2 kJ

[GATE-2003] (d) –200 kJ

GATE-3.

The entropy change for the Universe during the process in kJ/K is: [GATE-2003] (a) 0.4652 (b) 0.0067 (c) 0 (d) –0.6711

GATE-4.

If a closed system is undergoing an irreversible process, the entropy of the system [GATE-2009] (a) Must increase (b) Always remains constant (c) Must decrease (d) Can increase, decrease or remain constant

Entropy and Direction: The Second Law a Directional law of Nature GATE-5.

One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is: (a) Equal to entropy change of the reservoir [GATE-2010]

95

Entropy S K Mondal’s

Chapter 4

(b) Equal to entropy change of water (c) Equal to zero (d) Always positive

Common Data for Questions GATE-6 and GATE-7: In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air: Specific heat at constant pressure, cp = 1.005 kJ/kg K; Specific heat at constant volume, cv = 0.718 kJ/kg K; Characteristic gas constant, R = 0.287 kJ/kg K. Enthalpy, h = cpT Internal energy, u = cvT GATE-6.

If the pressure at station Q is 50 kPa, the change in entropy (sQ – sP ) in kJ/kg K is (a) – 0.155

GATE-7.

(b) 0

(c) 0.160

[GATE-2011] (d) 0.355

If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to [GATE-2011] (a) 50 (b) 87 (c) 128 (d) 150

Previous 20-Years IES Questions Two Reversible Adiabatic Paths cannot Intersect Each Other IES-1.

The relation ds =

dQ , where s represents entropy, Q represents heat and T T

represents temperature (absolute), holds good in which one of the following processes? [IES-2009] (a) Reversible processes only (b) Irreversible processes only (c) Both reversible and irreversible processes (d) All real processes IES-2.

IES-2a.

Which of the following statement is correct? [IES-2008] (a) The increase in entropy is obtained from a given quantity of heat transfer at a low temperature. (b) The change in entropy may be regarded as a measure of the rate of the availability of heat for transformation into work. (c) The entropy represents the maximum amount of work obtainable per degree drop in temperature (d) All of the above A heat engine receives 1000 kW of heat at a constant temperature of 285°C and rejects 492 kW of heat at 5°C. Consider the following thermodynamic cycles in this regard: [IES-2000] 1. Carnot cycle 2. Reversible cycle 3. Irreversible cycle 96

Entropy S K Mondal’s

Chapter 4

Which of these cycles could possible be executed by the engine? (a) 1 alone (b) 3 alone (c) 1 and 2 (d) None of 1, 2 and 3

The Property of Entropy IES-3.

Assigning the basic dimensions to mass, length, time and temperature respectively as M, L, T and θ (Temperature), what are the dimensions of entropy? [IES-2007] (a) M LT-2 θ (b) M L2 T-1 θ-1 (c) M L2 T-2θ-1 (d) M L3T-2 θ -1

IES-4.

A Carnot engine operates between 327°C and 27°C. If the engine produces 300 kJ of work, what is the entropy change during heat addition? [IES-2008] (a) 0.5 kJ/K (b) 1.0 kJ/K (c) 1.5 kJ/K (d) 2.0 kJ/K

Temperature-Entropy Plot IES-4a

Isentropic flow is

[IES-2011]

(a) Irreversible adiabatic flow (c) Ideal fluid flow

IES-5.

(b) Reversible adiabatic flow (d) Frictionless reversible flow

A system comprising of a pure substance executes reversibly a cycle 1 -2 -3 -4 -1 consisting of two isentropic and two isochoric processes as shown in the Fig. 1. Which one of the following is the correct representation of this cycle on the temperature – entropy coordinates? [IES-2002]

97

Enttropy S K Mondal’s

IES-6.

Cha apter 4

A cycle of pressure p – volume dia agram is sh hown in th he given Fig g. I, Same S cyc cle on tem mperature-e entropy diagram willl be represe ented by:

[IES-1995]

IES-7.

An ideal air standard cycle c is given sho own in the tem mperature-e entropy diagram.

[IES-1997] 98

Enttropy S K Mondal’s

Cha apter 4

The e same cyc cle, when re epresented on the pre essure-volum me coordin nates takes the e form

IES-8.

Ma atch figures s of Colum mn-I with th hose given in Column n-II and se elect given bellow the colu umns: [IES-1994] Collumn-I (p-v diagram) Colum mn-II (T-s diagram)

Cod des: (a) (c) IES-9.

A 1 3

B 2 1

C 3 2

(b) (d)

A 2 3

B 3 2

C 1 1

A cyclic c proce ess ABCD shown s in the e V-T diagra am perform med with a constant c ma ass of an id deal gas. The e process of o p-V diagr ram will be as shown in n [IES-1992]

99

Enttropy S K Mondal’s

IES-10.

Thr ree process ses are repr resented on n the p-v an nd T-s diagr rams in the e following figu ures. Match h processess in the tw wo diagramss and selec ct the corre ect answer usiing the code es given bellow the diag grams: [IES-1994]

Cod des: (a) (c) IES-11.

Cha apter 4

A 1 3

B 2 2

C 3 1

(b) (d)

A 2 1

B 3 3

C 1 2

Tw wo polytropiic processes undergon ne by a perffect gas are e shown be elow in the pre essure-volum me co-ordin nates. [IES-2008]

Wh hich represe entation sho ows correcttly the abov ve processess on the tem mperature– enttropy co-ord dinates?

100 0

Entropy S K Mondal’s

IES-12.

Chapter 4

Assertion (A): If a graph is plotted for absolute temperature as a function of entropy, the area under the curve would give the amount of heat supplied. Reason (R): Entropy represents the maximum fraction of work obtainable from heat per degree drop in temperature. [IES-1998] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

The Inequality of Clausius IES-13.

For real thermodynamic cycle:

(a) IES-14.

∫

∫

(c)

∫

dQ =0 T

δQ T

=0

(b)

∫

δQ T

0

For an irreversible cycle:

(a) IES-16.

(b)

[IES-2005]

dQ 0 but < ∞ T

∫

dQ ≤0 T

(b)

(d)

∫

dQ =∞ T

[IES-1998]

∫

δQ T

≥0

[IES-1994, 2011]

∫

dQ >0 T

(c)

∫

dQ 0 T dQ (c) ∫ > 0 and ΔS = 0 T (a)

dQ = 0 and ΔS = 0 T dQ (d) ∫ < 0 and ΔS < 0 T

∫

(b)

101

∫

Entropy S K Mondal’s

Chapter 4

Entropy Change in an Irreversible Process IES-17.

Consider the following statements: [IES-1998] In an irreversible process 1. Entropy always increases. 2. The sum of the entropy of all the bodies taking part in a process always increases. 3. Once created, entropy cannot be destroyed. Of these statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1, 2 and 3 are correct

IES-18.

Consider the following statements: [IES-1997] When a perfect gas enclosed in a cylinder piston device executes a reversible adiabatic expansion process, 1. Its entropy will increase 2. Its entropy change will be zero 3. The entropy change of the surroundings will be zero Of these statements (a) 1 and 3 are correct (b) 2 alone is correct (c) 2 and 3 are correct (d) 1 alone is correct

IES-19.

A system of 100 kg mass undergoes a process in which its specific entropy increases from 0.3 kJ/kg-K to 0.4 kJ/kg-K. At the same time, the entropy of the surroundings decreases from 80 kJ/K to 75 kJ/K. The process is: [IES-1997] (a) Reversible and isothermal (b) Irreversible (c) Reversible (d) Impossible

IES-20.

Which one of the following temperature entropy diagrams of steam shows the reversible and irreversible processes correctly? [IES-1996]

Applications of Entropy Principle IES-21.

A Carnot engine operates between 27°C and 327°C. If the engine produces 300 kJ of Work, What is the entropy change during heat addition? [IES-2005] 102

Entropy S K Mondal’s

Chapter 4

(a) 0.5 kJ/K

(b) 1.0 kJ/K

(c) 1.5 kJ/K

(d) 2.0 kJ/K

IES-22.

The entropy of a mixture of ideal gases is the sum of the entropies of constituents evaluated at: [IES-2005] (a) Temperature and pressure of the mixture (b) Temperature of the mixture and the partial pressure of the constituents (c) Temperature and volume of the mixture (d) Pressure and volume of the mixture

IES-23.

The heat added to a closed system during a reversible process is given by Q = αT + β T 2 , where α and β are constants. The entropy change of the system as its temperature changes from T1 to T2 is equal to: [IES-2000]

β ⎡ ⎤ (b) ⎢α (T2 − T1 ) + T22 − T12 ⎥ / T1 2 ⎣ ⎦

(

(a) α + β (T2 − T1 )

⎛ T2 ⎞ ⎟ + 2β (T2 − T1 ) ⎝ T1 ⎠

β ⎡α ⎤ (c) ⎢ T22 − T12 + T23 − T13 ⎥ / T12 2 ⎣2 ⎦

(

)

(

)

)

( d ) α ln ⎜

IES-24.

One kg of air is subjected to the following processes: [IES-2004] 1. Air expands isothermally from 6 bar to 3 bar. 2. Air is compressed to half the volume at constant pressure 3. Heat is supplied to air at constant volume till the pressure becomes three fold In which of the above processes, the change in entropy will be positive? (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3

IES-25.

A reversible heat engine receives 6 kJ of heat from thermal reservoir at temperature 800 K, and 8 kJ of heat from another thermal reservoir at temperature 600 K. If it rejects heat to a third thermal reservoir at temperature 100 K, then the thermal efficiency of the engine is approximately equal to: [IES-2002] (a) 65% (b) 75% (c) 80% (d) 85%

IES-26.

A reversible engine exceeding 630 cycles per minute drawn heat from two constant temperature reservoirs at 1200 K and 800 K rejects heat to constant temperature at 400 K. The engine develops work 100kW and rejects 3200 KJ heat per minute. The ratio of heat drawn from two reservoirs

(a) 1 IES-27.

(b) 1.5

(c) 3

Q1200 is nearly. Q800 [IES-1992] (d) 10.5

In which one of the following situations the entropy change will be negative (a) Air expands isothermally from 6 bars to 3 bars [IES-2000] (b) Air is compressed to half the volume at constant pressure (c) Heat is supplied to air at constant volume till the pressure becomes three folds (d) Air expands isentropically from 6 bars to 3 bars

103

Enttropy S K Mondal’s

Cha apter 4

Entropy and Direction n: The Second Law a Direc ctional aw of Nature la IES-28.

IES-29

A mass m M of a fluid at tem mperature T1 is mixed d with an eq qual mass of o the same fluiid at tempe erature T2. The T resultan nt change in n entropy of o the universe is: [IES-1992] (a) Zero (b) Negliigible (c) Alwayss negative (d) Always positive p Increase in n entropy off a system represents r (a) Increase in availabiliity of energy (b) Increasse in tempera ature e in pressure (d) Degrad dation of energy (c) Decrease

[IES-2011]

Previo ous 20 0-Years s IAS Quest Q ions Tw wo Reve ersible Adiabati A c Paths cannot Intersec ct Each Other IA AS-1.

Wh hich one of the t followin ng is the cor rrect statem ment? Tw wo adiabatic c will: (a) Intersect at absolute zerro temperature (b) Never interssect (c) Become orth hogonal at ab bsolute zero temperature t (d) Become parallel at absollute zero tem mperature

[IAS-2007]

The Prop perty of o Entro opy IA AS-2.

Heat flows be etween two o reservoirss having te emperature es 1000 K and a 500 K, resspectively. If I the entrop py change of o the cold reservoir r iss 10 kJ/K, th hen what is the e entropy ch hange for th he hot reser rvoir? [IAS-2004] (a) –10 kJ/K (b) –5 kJ/K (c)) 5 kJ/K (d) 10 kJ/K

Te empera ature-Entropy Plot IA AS-3.

An ideal cyclle is shown n in the giv ven pressure e-volume diiagram:

[IAS-1997]

104 4

Enttropy S K Mondal’s

Cha apter 4

The e same cyclle on temperature-entr ropy diagram will be re epresented as:

IA AS-4.

The e therma al efficien ncy of t the hyp pothetical heat h engine e cycle show wn in the t given figure is: (a) 0.5 (b) 0.45 0 (c) 0.35 (d) 0.25

[IAS-2000] AS-5. IA

Wh hich one of the followiing pairs be est expresses a relatio onship similar to that exp pressed in the pair r “pressure e-volume” for a the ermodynam mic system und dergoing a process? [IAS-1995] (a) Enthalpy-en ntropy (b) Pressu ure-enthalpy P mperature (d) Tempeerature-entroopy (c) Pressure-tem

IA AS-6.

An ideal gas contained c in n a rigid ta ank is cooled c such h that T2 < and P2 0

(d)

For r real therm modynamic cycle: (a)

∫

dQ > 0 butt < ∞ T

(b)

∫

Q

∫δ T

≤0

[IAS-2003]

dQ 0 T dQ (c) ∫ > 0 an nd ΔS = 0 T (a)

∫

dQ = 0 and ΔS = 0 T dQ (d) ∫ < 0 and ΔS < 0 T (b)

∫

IA AS-10(ii). A cyclic c heat engine rece eives 600 kJ of heat fr rom a 1000 K source and a rejects 450 0 kJ to a 300 3 K sink. The quanttity resspectively (a) 2.1 kJ/K and d 70% nd 70% (c) + 0.9 kJ/K an

∫

dQ a and efficien ncy of the engine e are T (b)) –0.9 kJ/K and a 25% (d)) –2.1 kJ/K and a 25%

[IAS-2001]

A Applicattions off Entrop py Prin nciple IA AS-11.

Wh hich one of the t followin ng statemen nts is not co orrect? [IAS-2003] (a) Change in entropy e durin ng a reversiblle adiabatic process p is zerro (b) Entropy increases with the t addition of heat ansion processs (c) Throttling is a constant entropy expa hen a gas is heated under u consta ant pressure given by (d) Change in entropy wh

s2 − s1 = mC C p log e IA AS-12.

T2 T1

Asssertion (A): Entropy ch hange for a reversible adiabatic a pr rocess is zero. 106 6

Entropy S K Mondal’s

Chapter 4

Reason (R): There is no heat transfer in an adiabatic process. [IAS 1994] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Entropy Generation in a Closed System IAS-13.

1600 kJ of energy is transferred from a heat reservoir at 800 K to another heat reservoir at 400 K. The amount of entropy generated during the process would be: [IAS-2000] (a) 6 kJ/k (b) 4 kJ/k (c) 2kJ/k (d) Zero

IAS-14.

An electric motor of 5 kW is subjected to a braking test for 1 hour. The heat generated by the frictional forces in the process is transferred to the surroundings at 20°C. The resulting entropy change will be: [IAS-1998] (a) 22.1 kJ/K (b) 30.2 kJ/K (c) 61.4 kJ/K (d) 82.1 kJ/K

Entropy and Direction: The Second Law a Directional law of Nature IAS-15.

M1 kg of water at T1 is isobarically and adiabatically mixed with M2 kg of water at T2 (T1 > T2). The entropy change of the universe is: [IAS-2004] (a) Necessarily positive (b) Necessarily negative (c) Always zero (d) Negative or positive but not zero

IAS-16.

In which one of the following processes is there an increase in entropy with no degradation of energy? [IAS-1996] (a) Polytropic expansion (b) Isothermal expansion (c) Isochoric heat addition (d) Isobaric heat addition

107

Entropy S K Mondal’s

Chapter 4

Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. Ans. –11858 J/K, 12787 J/K. GATE-1A Ans. (b) ⎛υ ⎞ ⎛υ ⎞ ⎛2⎞ GATE-2. Ans. (b) w1− 2 = mRT In ⎜ 2 ⎟ = pυ In ⎜ 2 ⎟ = 200 × 1× In ⎜ ⎟ kJ = 138.6 kJ ⎝ 1⎠ ⎝ υ1 ⎠ ⎝ υ1 ⎠ GATE-3. Ans. (c) It is reversible process so ( ΔS )universe = 0 GATE-4. Ans. (d) GATE-5. Ans. (d) It is a case of spontaneous process i.e. irrepressibility involved that so why entropy change of the universe is positive. GATE-6. Ans. (c)

TQ = 300 K,

PQ = 50 kPa

TP = 350 K,

PP = 150 kPa T P SQ − SP = c P ln Q − R ln Q TP PP ⎛ 300 ⎞ ⎛ 50 ⎞ SQ − SP = 1.005 ln ⎜ ⎟ − 0.287 ln ⎜ ⎟ ⎝ 350 ⎠ ⎝ 150 ⎠ SQ − SP = 0.160 kJ/kg-K GATE-7. Ans. (b) If air has to flow from station P to station Q adiabatically means no entropy change in surroundings, then

P

Q

SQ − SP ≥ 0

⇒ c P ln ⇒

⇒ ⇒

TQ TP

− R ln

PQ PP

≥0

⎛ PQ ⎞ ⎛ 300 ⎞ 1.005 ln ⎜ ⎟ − 0.287 ln ⎜ ⎟≥0 ⎝ 350 ⎠ ⎝ 150 ⎠ ⎛ P ⎞ −0.287 ln ⎜ Q ⎟ ≥ 0.15492 ⎝ 150 ⎠ PQ ≤ 87.43

So, Ans. is PQ = 87 kPa

108

⇒

⇒

⎛ P ⎞ −0.15492 − 0.287 ln ⎜ Q ⎟ ≥ 0 ⎝ 150 ⎠ ⎛ P ⎞ ln ⎜ Q ⎟ ≤ − 0.53979 ⎝ 150 ⎠

Enttropy S K Mondal’s

Cha apter 4

Previious 20-Yearrs IES Answ wers ⎛ dQ ⎞ ⎟ ⎝ T ⎠R Rev

IES-1. Ans. (a)) Remember ds = ⎜

IES-2. Ans. (b b) The chang ge of entropy y may be regarded as a measure m of th he rate of av vailability of hea at for transformation intoo work. IES-2a Ans. (b b)

S gen = −

1000 492 + = −0.02233kkW / K (285 ( + 273) (5 + 273)

So cycle is im mpossible Cy ycle See in both the case Carrnot Cycle an nd Reversiblee cycle entrop py change of the t Universe wiill be zero. Irreversible cycle entrop py change willl be positive.

IES-3. Ans. (c)) IES-4. Ans. (b) Q1 Q1 Q1 − Q2 W = = = T1 T1 T1 − T2 T1 − T2 Q1 = 600kJ

Thee entropy ch hange durin ng heat add dition Q1 600 = = = 1 kJ k /K T1 600 IES-4a Ans. (b) IES-5. Ans. (c)) IES-6. Ans. (b) IES-7. Ans. (b) IES-8. Ans. (c)) IES-9. Ans. (d) AB constan nt pressure heat h addition. IES-10. Ans. (c) XA consstant pressu ure heat rejeection. XB = const. temp p. heat rejecction. XC = isen ntropic heat rejection. r IES-11. Ans. (b b)

IES-12. Ans. (b b) IES-13. Ans. (b b) b) IES-14. Ans. (b IES-15. Ans. (c c) 109 9

Enttropy S K Mondal’s IES-16. Ans. (a a) IES-17.

∫

Cha apter 4

dQ = 0 does not neccessarily meaans reversiblle process. If dQ = 0 . T

Ans. (c) In irreversible i heat rejeection process enttropy decreases. In an irreverrsible y of the univ verse process entropy ways increasses i.e. sum m of alw systtem + su urroundings will incrreases.

Con nsider the prrocess 3–4 iff it is irreeversible prrocess then also entropy will deccrease. IES-18. Ans. (c) ( In reversiible adiabaticc expansion, entropy chan nge is zero and no changee in entropy of surroundings s s. b) Entropy in ncrease in prrocess = 100 (0.4 – 0.3) = 10 kJ/kg IES-19. Ans. (b Enttropy change e of surround dings = 5 kJ/K K Thu us net entrop py increases and a the proceess is irreverrsible. IES-20. Ans. (c) ( In reversiible process entropy e chan nge is zero an nd in four figu ures it is rep presented by stra aight vertical line. Howev ver, in irreveersible processs, entropy in ncreases in all a processes (exp pansion as well w as compression). IES-21. Ans. (b b)

( T1 − T2 ) ΔS = W or ΔS =

300 = 1 kJ / k 600 − 300

IES-22. Ans. (c c)

⎛T ⎞ dQ 1 α + 2 β T IES-23. Ans. (d d) ∫ dT =∫ T = α ln ⎜ 2 ⎟ + 2 β (T2 − T1 ) T T ⎝ T1 ⎠ T1 T1 T1

T

IES-24. Ans. (c c)

IES-25. Ans. (d d)

110 0

Entropy S K Mondal’s

Chapter 4

Q 6 = 2 800 100 Q 8 = 4 600 100 Q + Q4 η = 1− 2 Q1 + Q3

3 4 + 4 3 = 0.85 = 1− 6+8 IES-26. Ans. (d) Refer to given figure, as given Engine work developed = 100 kW = 100 × 1000 × 60 = 6 × 106 J/min. Qs = total heat supplied Thus,

= 6 × 106 +3.2 × 106 = 9.2 × 106 J/min. Let reservoir at 1200 K supply Qs1 J/min. Therefore reservoir at 800 o K will supply. Qs2 = 9.2 × 106 – Qs1 Also, by data the engine is a reversible heat engine completing 600 cycles/min. and therefore entropy change after every complete cycle is zero. Thus, or

Qs1 Qs2 QR + − =0 1200 800 400 Qs1 9.2 × 106 − Q 6 × 3.2 × 106 + − =0 1200 800 400 2Qs1 + 3(9.2 × 106 − Qs1 ) − 6 × 3.2 × 106 =0 2400

or

Qs1 = 3 × 9.3 × 106 − 6 × 3.2 × 106 = 8.4 × 106 J/min

Qs2 = 9.2 × 106 − 8.4 × 106

= 0.8 × 106 J/min 8.4 Hence ratio = =10.5 0.8 IES-27. Ans. (b) It is isobaric compression.

111

Enttropy S K Mondal’s

Cha apter 4

IES-28. Ans. (d d)

IES-29

Ans. (d)

Previious 20 0-Yearrs IAS Answ wers IA AS-1. Ans. (b) IA AS-2. Ans. (b)

+Q = 10 0 500 or Q = 5000 kJ −Q −5000 S1 = = = −5kJ / k 1000 1000 1 t thesystem m is +ive ⎤ ⎡∴Heat added to ⎢ Heat rejecteed from the syystem is -ive ⎥ ⎦ ⎣ S2 =

IA AS-3. Ans. (d) 1 × ( 5 − 1) × ( 800 − 40 00 ) Work done d area1 − 2 − 3 = = 2 = 0.25 IA AS-4. Ans. (d) η = Heat ad dded areaunder curve 2 − 3 ( 5 − 1) × 800

IA AS-5. Ans. (d) That so wh hy we are usin ng p–v or T– –s diagram. IA AS-6. Ans. (a)) IA AS-7. Ans. (b b) IA AS-8. Ans. (d) IA AS-9. Ans. (b)

112 2

Entropy S K Mondal’s

Chapter 4

dQ = 0 does not necessarily means reversible process. If dQ = 0 . T Q − Q2 Q 450 = 1− 2 = 1− = 0.25 = 25% IAS-10(ii). Ans. (b) η = 1 600 Q1 Q1 IAS-10(i). Ans. (a)

∫

IAS-11. Ans. (c) Throttling is a constant enthalpy expansion process. IAS-12. Ans. (b) dQ dQ 1600 1600 − = − = 2kJ / K IAS-13. Ans. (c) Entropy generated = dsat 400K − dsat 800K = 400 800 400 800 ΔQ 5 × 3600 = kJ / K = 61.4kJ / K IAS-14. Ans. (c) ΔS = T 293 IAS-15. Ans. (a) IAS-16. Ans. (b)

113

Availability, Irreversibility S K Mondal’s

5.

Chapter 5

Availability, Irreversibility

Theory at a Glance (For GATE, IES & PSUs) That part of the low grade energy which is available for conversion is referred to as available energy, while the part which, according to the second law, must be rejected, is known as unavailable energy.

Availability The availability of a given system is defined as the maximum useful work that can be obtained in a process in which the system comes to equilibrium with the surroundings or attains a dead state. Clearly, the availability of a system depends on the condition of the system as well as those of the surroundings.

Availability: • Yields the maximum work producing potential or the minimum work requirement of a process • Allows evaluation and quantitative comparison of options in a sustainability context Availability = Maximum possible work-Irreversibility

W useful = W rev – I Irreversibility The actual work done by a system is always less than the idealized reversible work, and the difference between the two is called the irreversibility of the process.

I = Wmax – W This is also sometimes referred to as 'degradation' or 'dissipation'.

Availability and Irreversibility 1.

Available Energy (A. E.) T ⎛ T0 ⎞ T ⎞ ⎛ = Wmax = Q1 ⎜1 − ⎟ = mcp ∫ ⎜1 − 0 ⎟dT T1 ⎠ T ⎠ T0 ⎝ ⎝ = (T1 − T0 ) ΔS

Where To is surroundings temperature. 112

Availability, Irreversibility S K Mondal’s

= u − u − T (S − S 1

2

0

2

)

2

)

1

change of volume is present there.

= h − h − T (S − S 1

2

0

1

Chapter 5 (For closed system), it is NOT ( φ1 – φ2) because

(For steady flow system), it is (A1-A2) as in steady state

no change in volume in C.V. (i.e. change in availability in steady flow) 2.

Decrease in Available Energy

= T0 [ ΔS ′ − ΔS ] 3.

Take ΔS′ and ΔS both positive quantity.

Availability function

2

C A = h − T0S + + gz 2 φ = u − T0S + P0V

For open system

For closed system

Availability = maximum useful work. For steady flow

C12 Availability = A1 − A0 = ( h1 − h0 ) − T0 ( S1 − S0 ) + + gz 2 For closed system

Availability = φ1 − φ0 = u1 − u0 − T0 ( S1 − S0 ) + P0 ( V1 − V0 )

4.

Unavailable Energy (U.E.)

= T0 ( S1 − S2 )

5.

Increase in unavailable Energy = Loss in availability

6.

Irreversibility

= T0 ( ΔS )Univ

113

Availability, Irreversibility S K Mondal’s

Chapter 5

I = Wmax − Wactual = T0 ( ΔS )univ

2

7.

Irreversibility rate = I = rate of energy degradation

S gen = ∫ mds 1

= rate of energy loss (Wlost) =

8.

T0 ×

Sgen for all processes

Wactual ⇒ dQ = du + dWact → this for closed system dWact C12 C22 dQ + gz1 + = h2 + + gz2 + → this for steady flow h1 + 2 2 dm dm

9.

Helmholtz function

F = U – TS 10.

Gibb’s function

G = H – TS 11.

Entropy generation number (Ns)

Ns = 12.

s mC gen

P

To calculate dS

114

Availability, Irreversibility S K Mondal’s

Chapter 5

⎡ P V ⎤ − S1 = m ⎢CV ln 2 + CP ln 2 ⎥ P1 V1 ⎦ ⎣ For closed system

( i ) Use S

2

Tds = du + PdV dT P + dV ds = m CV T T dT dV = m CV + mR T V 2 2 2 dT dV ∫1 ds = m CV ∫1 T + mR ∫1 V

or

For steady flow system Tds = dh − VdP dT V − dP T T 2 2 2 dT dP = − ds m C mR P ∫ ∫1 ∫ T 1 1 P ds = m CP

or

PV = mRT V mR = T P

But note that and

13.

In Pipe flow Entropy generation rate

S =S − gen

Tds = du + PdV ⎫ ⎬ Both valid for closed system only. Tds = dh − VdP ⎭

sys

Q T

1

0

= m (S − S ) − 2

1

mC (T − T ) 2

P

in Kg/s

1

T

0

Due to lack of insulation it may be T1 > T2 for hot fluid or T1 < T2 for cold fluid

∴ rate of irreversibility ( I ) = T0 Sgen 14.

Flow with friction

Decrease in availability = m RT0′

2

ΔP P1

115

P,T1

P,T2

Availability, Irreversibility S K Mondal’s

Chapter 5

15. Second Law efficiency Min n . exergy intake to perform the given tast (X min ) ηII = = ηI / ηcarnot Actual exergy intake to perform the given tast (X) X min = W if work is involved T ⎞ ⎛ = Q ⎜1 − 0 ⎟ if Heat is involved. T ⎠ ⎝ A common measure on energy use efficiency is the first law efficiency, η1 . The first law efficiency is defined as the ratio of the output energy of a device to the input energy of the device. The first law is concerned only with the quantities of energy, and disregards the forms in which the energy exists. It does not also discriminate between the energies available at different temperatures. It is the second law of thermodynamics which provides a means of assigning a quality index to energy. The concept of available energy or energy provides a useful measure of energy quality. With this concept it is possible to analyze means of minimizing the consumption of available energy to perform a given process, thereby ensuring the most efficient possible conversion of energy for the required task. The second law efficiency, η11 , of a process is defined as the ratio of the minimum available energy (or energy) which must be consumed to do a task divided by the actual amount of available energy (or energy) consumed in performing the task.

minimum exergy intake to perform the given task actual exergy intake to perform the same task A ηII = min A

ηII =

or

where A is the availability or energy. A power plant converts a fraction of available energy A or Wmax to useful work W. For the desired output of W, Amin = W and A = Wmax, Here,

116

Availab A bility, Irreversibiility S K Mondal’s

Cha apter 5

I = Wmax − W and ηII = Now

ηI =

W Wmax

W Wmax W = ⋅ m QI Wmax Q1

= ηII ⋅ ηCarnot ∴

ηII =

ηI

ηCaarnot

T0 ), can n also be obttained direectly as follows T ηI W = ηII = T ⎞ ηCarnot ⎛ Q1 ⎜1 − 0 ⎟ T ⎠ ⎝

Since Wmax = Q1 (ll -

T ⎞ ⎛ If w work is invollved, Amin = W (desired) an nd if heat is involved, i Amin = Q ⎜1 − 0 ⎟ . T ⎠ ⎝

Th he general deefinition of second s law effficiency of a process can n be obtained in terms of o change in ava ailability durring the process: Seccond Law Eff fficiency = η III Law =

Producction of availability . Destru uction of avaiilability

A Available e Energ gy Refe erred to o a Cyc cle • Decrease e in Availlable Ener rgy when Heat is Transferred T d through a Finite Tempera ature Defere ence :Wh henever heat is transferrred through h finite temp perature defference, therre is a decrease in the ava ailability of energy e so tra ansferred. Let us considerr a reversible e heat enginee operating beetween T1 an nd T0 (Fig. sh hown in below w) Then

and

Q1 = T1 ΔS, Q2 = T0 ΔS, W = A.E. = (T1 – T0) Δs

h Q1 is trransferred th hrough a finiite temperatture difference from the Let us now assume that heat urce at T1 to o the enginee absorbing, heat h at T’1, lower l turn T1 (Fig.shown n in below). resservoir or sou 117 7

Availab A bility, Irreversibiility S K Mondal’s

Cha apter 5

he availability y of Q1 as recceived by thee engine at T’ T 1 can be found by allowin ng the engin ne to operate Th rev versible in a cycle between T’1 and T0, receiving Q1 and rejectin ng Q’2.

Noow

Q1 = T1 Δs = T1' Δs '

Sin nce

T1 > T1' , ∴ Δs ' > Δs

Q2 = T0 Δs Q '2 = T0 Δs ' Sin nce

Δs ' > Δ s ∴

∴

W' = Q1 - Q'2 = T'1 Δs '− T0 Δs '

an nd      

Q '2 > Q2

W = Q1  − Q2  = T1 Δs − T0  Δs

∴ ecause Q’2  > Q2   W’ < W, be Av vailable energ gy lost due to o irreversiblee heat transffer through fiinite tempera ature deferen nce between thee source and the working fluid during g the heat add dition processs is given by W - W’ = Q’2 – Q2 s = T0(Δs′ - Δs) n A.E. = T0(Δs′ - Δs) or decrease in

he decrease in n available energy e or exeergy is thus the t product of o the lowestt feasible tem mperature of Th hea at rejection and the add ditional entroopy change in i the system m while receeiving heat irreversibly, i com mpared to the t case of reversible heat h transfeer from the same sourcce. The greater is the tem mperature diifference (T1 - T’1), the greater g is the heat rejecttion Q’2 and the greater will be the un navailable pa art of the ene ergy supplied d or energy (Fig. ( above). Energy is sa aid to be deg graded each tim me, it flows th hrough a finiite temperatu ure differencee.

PR ROBLEMS & SOLUT TIONS Ex xample 1 Wh hat is the ma aximum usefful work whiich can be ob btained when n 100 kJ aree abstracted from a heat resservoir at 675 K in an environmen ntal at 288 K? What iss the loss of o useful work if, (a) a tem mperature drrop of 50°C is introduced d between th he heat sourrce and the heat h engine, on the one hand and the heat h engine and a the heat sink, on the other. (b) th he source tem mperature drops by 50°C 118 8

Availab A bility, Irreversibiility S K Mondal’s

Cha apter 5

r by 50°C C during the heat transferr process acccording to the linear law and the sink teemperature rises dQ Q = ± constan nt ? dT T So olution: Av vailability= ⎛ To ⎞ 288 8⎞ ⎛ Wm ×100 = 57.333kJ ⎟ Q1 = ⎜1 − max = ⎜1 − T1 ⎠ 5 ⎟⎠ 675 ⎝ ⎝

ΔS´ =

Q1 100 = = 0.16kJ / K T1 625

W ´ = ΔS´ (T ´1 −T ´o ) = 0.16 ( 625 − 338 ) = 45.92 2kJ n availability y = 57.33-45.92 = 11.41 kJ J. Loss in

Av verage temperature of sou urce

119 9

Availability, Irreversibility S K Mondal’s T1a =

Chapter 5

T1 + T "1 = 650 K 2

Average temperature of sink (T0 a ) = ΔS" =

T0 + T "1 = 313 K 2

Q1 100 = = 0.1538 kJ / K T1a 650

W " = ΔS " ( 650 − 313 ) = 51.83 kJ / K Loss in availability = 57.33 − 51.83 = 5.5 kJ . Example 2. A lead storage battery used in an automobile is able to deliver 5.2 MJ of electrical energy. This energy is available for starting the car. Let compressed air be considered for doing an equivalent amount of work in starting the car. The compressed air is to be stored at 7 MPa, 25°C, what is the volume of the tank that would be required to let the compressed air having an availability of 5.2 MJ? For air pv = 0.287 T, where T is in K, p in kPa, and v in m3/kg. Solution: Availability is given by a = ( u − uo ) − To ( s − so ) + po (v − vo ) 0.287 × 298 = 0.012218m3 / kg 7000 0.287 × 298 = 0.85526m3 / kg vo = 100 7000 7000 ⎞ ⎛ − 0.287ln + 100 ( 0.012218 − 0.85526 ) a = 0 − 298 ⎜ 0.287ln 7000 100 ⎟⎠ ⎝ v=

= 279 kJ / kg 5200 =18.635kg 279 V = m.v =18.635 × 0.012218 = 0.2277m3 For availability of 5.2 MJ, m =

Example 3. Air enters a compressor in steady flow at 140 kPa, 17°C and 70 m/s and leaves it at 350 kPa, 127°C and 110 m/s. The environment is at 100 kPa, 7°C. Calculate per kg of air (a) the actual amount of work required, (b) the minimum work required, and (c) the irreversibility of the process. Solution: This is the case of a steady flow system in which the maximum useful work is given by C 2 − C22 Wmax useful = φ1 − φ2 + 1 2 C 2 − C22 = ( H1 − To S1 ) − ( H 2 − To S2 ) + 1 2 As air is to be considered as a perfect gas, h = CpT Hence C 2 − C22 Wmax useful = Cp (T1 − T2 ) − To ( S1 − S2 ) + 1 2 In the above, expression (S1 - S2) is the entropy change. For a perfect gas, when pressure and temperature very, the entropy change is given by the relation

120

Availability, Irreversibility S K Mondal’s S1 − S2 = C p ln

Chapter 5

T1 P − R ln 1 T2 P2

Hence ⎛ T P ⎞ C 2 − C22 Wmax useful = C p (T1 − T2 ) − To ⎜ C p ln 1 − R ln 1 ⎟ + 1 T2 P2 ⎠ 2 ⎝ Various parameters given are: T1 = 290 K; T2 = 400 K; Cp = 1.005 kJ/kg K P1 = 140 kPa; P2 = 350 kPa C1 = 70m / s; C2 = 110 m/s; To = 280 K (a) Actual work done in case of a steady flow process is given by C 2 − 1102 Wactual = ( H1 − H 2 ) + 1 2 702 − 1102 = 1.005 ( 290 − 400 ) + = − 114.1kJ 2 × 103 The negative sign obviously indicates that work is done on the air in compression. 290 140 ⎤ 702 − 1102 ⎡ (b) Minimum work = 1.005(290 - 400) - 280 × ⎢1.005 ln − 0.287 ln + 400 350 ⎥⎦ 2 × 103 ⎣ = -97.3 kJ (c) Irreversibility= 114.1 - 97.3 = 16.8 kJ. Example 4. By how much is the available energy of 5 kg of air increased by heating reversibly at a constant pressure of 1.5 atm. from 27°C to 227°C with the lowest available temperature of 20°C? (Cp for air = 1005 J/kg°C). Solution: T Change in entropy = mC p ln 2 T1 500 = 2.567kJ / K 300 Availability = T0 ΔS = 293 × 2.567 = 752kJ. = 5 ×1.005 × ln

121

Availability, Irreversibility S K Mondal’s

Chapter 5

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions GATE-1.

A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The heat released during this process is to be used as a source of energy. The ambient temperature is 303 K and specific heat of steel is 0.5 kJ/kg K. The available energy of this billet is: [GATE-2004] (a) 490.44 MJ (b) 30.95 MJ (c) 10.35 MJ (d) 0.10 MJ

Availability GATE-2.

Availability of a system at any given state is: [GATE-2000] (a) A property of the system (b) The maximum work obtainable as the system goes to dead state (c) The total energy of the system (d) The maximum useful work obtainable as the system goes to dead state

GATE-3.

A heat reservoir at 900 K is brought into contact with the ambient at 300 K for a short time. During this period 9000 kJ of heat is lost by the heat reservoir. The total loss in availability due to this process is: [GATE-1995] (a) 18000 kJ (b) 9000 kJ (c) 6000 kJ (d) None of the above

Irreversibility GATE-4.

Consider the following two processes: a. A heat source at 1200 K loses 2500 kJ of heat to sink at 800 K b. A heat source at 800 K loses 2000 kJ of heat to sink at 500 K Which of the following statements is TRUE? (a) Process I is more irreversible than Process II (b) Process II is more irreversible than Process I (c) Irreversibility associated in both the processes is equal (d) Both the processes are reversible

[GATE-2010]

Previous 20-Years IES Questions Available Energy IES-1.

What will be the loss of available energy associated with the transfer of 1000 kJ of heat from constant temperature system at 600 K to another at 400 K when the environment temperature is 300 K? [IES-2004] (a) 150 kJ (b) 250 kJ (c) 500 kJ (d) 700 kJ

IES-2.

What is the loss of available energy associated with the transfer of 1000 kJ of heat from a constant temperature system at 600 K to another at 400 K when the environmental temperature is 300 K? [IES-2008]

122

Availability, Irreversibility S K Mondal’s

Chapter 5

(a) 150 kJ

(b) 250 kJ

(c) 166·67 kJ

(d) 180kJ

Available Energy Referred to a Cycle IES-3.

A heat source H1 can supply 6000kJ/min. at 300°C and another heat source H2 can supply 60000 kJ/min. at 100°C. Which one of the following statements is correct if the surroundings are at 27°C? [IES-2006] (a) Both the heat sources have the same efficiency (b) The first heat source has lower efficiency (c) The second heat source has lower efficiency (d) The first heat source produces higher power

Availability IES-4.

Assertion (A): The change in availability of a system is equal to the change in the Gibbs function of the system at constant temperature and pressure. Reason (R): The Gibbs function is useful when evaluating the availability of systems in which chemical reactions occur. [IES-2006] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-5.

For a steady flow process from state 1 to 2, enthalpy changes from h1 = 400 kJ/kg to h2 = 100 kJ/kg and entropy changes from s1 = 1.1 kJ/kg-K to s2 = 0.7 kJ/kg-K. Surrounding environmental temperature is 300 K. Neglect changes in kinetic and potential energy. The change in availability of the system is: [IES-2003] (a) 420 kJ/kg (b) 300 kJ/kg (c) 180 kJ/kg (d) 90 kJ/kg

IES-6.

Availability function for a closed system is expressed as: (a) φ = u + po v − To S (b) φ = du + po dv − To ds

(c) IES-7.

φ = du + po dv + To ds

(d)

φ = u + po v + To S

Consider the following statements: 1. Availability is the maximum theoretical work obtainable. 2. Clapeyron's equation for dry saturated steam

(V

g

)

− Vf =

[IES-2002]

[IES-2001] is

given

by

dTs ⎡ hg − hf ⎤ ⎢ ⎥ dQ ⎣ Ts ⎦

3. A gas can have any temperature at a given pressure unlike a vapour which has a fixed temperature at a given pressure.

⎡ ∂s ⎤ ⎥ ⎣ ∂p ⎦ h

4. Joule Thomson coefficient is expressed as μ = ⎢ Of these statements (a) 1, 2 and 3 are correct (c) 2 and 3 are correct

(b) 1, 3 and 4 are correct (d) 1, 2 and 4 are correct 123

Availability, Irreversibility S K Mondal’s

Chapter 5

IES-8.

10kg of water is heated from 300 K to 350 K in an insulated tank due to churning action by a stirrer. The ambient temperature is 300 K. In this context, match List-I and List-II and select the correct answer using the codes given below the Lists: [IES-2000] List-I List-II A. Enthalpy change 1. 12.2 kJ/kg B. Entropy change/kg 2. 1968 kJ C. Availability/kg 3. 2090 kJ D. Loss of availability 4. 656 J/kg-k Codes: A B C D A B C D (a) 3 1 4 2 (b) 2 4 1 3 (c) 3 4 1 2 (d) 2 1 4 3

IES-9.

Neglecting changes in kinetic energy and potential energy, for unit mass the availability in a non-flow process becomes a = ɸ - ɸo, where ɸ is the availability function of the [IES-1998] (a) Open system (b) Closed system (c) Isolated system (d) Steady flow process

IES-10.

Consider the following statements: 1. Availability is generally conserved 2. Availability can either be negative or positive 3. Availability is the maximum theoretical work obtainable 4. Availability can be destroyed in irreversibility Of these correct statements are: (a) 3 and 4 (b) 1 and 2 (c) 1 and 3

[IES-1996]

(d) 2 and 4

Irreversibility IES-11.

The irreversibility is defined as the difference of the maximum useful work and actual work: I = Wmax,useful- Wactual. How can this be alternatively expressed?

(a) I = To ( ΔS system + ΔS surrounding ) (c)

I = To ( ΔSsystem + ΔSsurrounding )

(b) I = To ( ΔS system − ΔS surrounding ) (d)

[IES-2005]

I = To ( ΔSsystem − ΔSsurrounding )

IES-12.

Assertion (A): All constant entropy processes are adiabatic, but all adiabatic processes are not isentropic. [IES-2006] Reason (R): An adiabatic process which resists the exchange of energy to the surroundings may have irreversibility due to friction and heat conduction. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-13.

Which of the following statement is incorrect? [IES-1992] (a) The greater the pressure difference in throttling the lesser the irreversibility (b) The primary source of internal irreversibility in power is fluid friction in rotary machines. (c) The greater the irreversibility, the greater the increase in adiabatic process 124

Availability, Irreversibility S K Mondal’s

Chapter 5

(d) The entropy of the universe is continually on the increase.

Previous 20-Years IAS Questions Available Energy IAS-1.

IAS-2.

IAS-3.

What will be the loss of available energy associated with the transfer of 1000 kJ of heat from constant temperature system at 600 K to another at 400 K when the environment temperature is 300 K? [IAS-1995] (a) 150 kJ (b) 250 kJ (c) 500 kJ (d) 700 kJ An inventor claims that heat engine has the following specifications: [IAS-2002] Power developed = 50 kW; Fuel burned per hour = 3 kg Heating value of fuel =75,000 kJ per kg; Temperature limits = 627°C and 27°C Cost of fuel = Rs. 30/kg; Value of power = Rs. 5/kWh (a) Possible (b) Not possible (c) Economical (d) Uneconomical For a reversible power cycle, the operating temperature limits are 800 K and 300 K. It takes 400 kJ of heat. The unavailable work will be: [IAS-1997] (a) 250 kJ (b) 150 kJ (c) 120 kJ (d) 100 kJ

Quality of Energy IAS-4.

Increase in entropy of a system represents (a) Increase in availability of energy (b) Increase in temperature (c) Decrease in pressure (d) Degradation of energy

[IAS-1994]

Availability IAS-5.

If u, T, v, s, hand p refer to internal energy, temperature, volume, entropy, enthalpy and pressure respectively; and subscript 0 refers to environmental conditions, availability function for a closed system is given by: [IAS-2003] (a) u + Po v – To s (b) u – Po v+ To s (c) h + Po v – Tos (d) h – Po v + To s

IAS-6.

Match List-I with List-II and below the lists: List-I A. Irreversibility B. Joule Thomson experiment C. Joule's experiment D. Reversible engines Codes: A B C (a) 1 2 3 (c) 4 3 2

select the correct answer using the codes given List-II Mechanical equivalent Thermodynamic temperature scale Throttling process Loss of availability A B C D (b) 1 2 4 3 (d) 4 3 1 2 1. 2. 3. 4.

D 4 1

125

Availability, Irreversibility S K Mondal’s

Chapter 5

Irreversibility IAS-7.

The loss due to irreversibility in the expansion valve of a refrigeration cycle shown in the given figure is represented by the area under the line. (a) GB (b) AG (c) AH (d) BH [IAS-1999]

IAS-8.

Assertion (A): When a gas is forced steadily through an insulated pipe containing a porous plug, the enthalpy of gas is the same on both sides of the plug. [IAS-1997] Reason (R): The gas undergoes an isentropic expansion through the porous plug. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Second Law efficiency IAS-9.

Assertion (A): The first-law efficiency evaluates the energy quantity utilization, whereas the second-law efficiency evaluates the energy quality utilization. Reason (R): The second-law efficiency for a process is defined as the ratio of change of available energy of the source to the change of available energy of the system. [IAS-1998] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

126

Availability, Irreversibility S K Mondal’s

Chapter 5

Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. Ans. (a) A.E =

T2

⎛ To ∫T mcp ⎜⎝1 − T 1

T2 ⎡ ⎛ T ⎞⎤ ⎞ = dT mc p ⎢( T2 − T1 ) − To ln ⎜ 2 ⎟ ⎥ ⎟ ∫ ⎠ ⎢⎣ ⎝ T1 ⎠ ⎥⎦ T1

⎡ ⎛ 1250 ⎞ ⎤ = 2000 × 0.5 ⎢(1250 − 450 ) − 303ln ⎜ ⎟ ⎥ = 490MJ ⎝ 450 ⎠ ⎦ ⎣ GATE-2. Ans. (d) Maximum useful work, i.e. total work minus pdv work. Not maximum work. GATE-3. Ans. (d) The availability of a thermal reservoir is equivalent to the work output of a Carnot heat engine operating between the reservoir and the environment. Here as there is no change in the temperatures of source (reservoir) or the sink (atmosphere), the initial and final availabilities are same. Hence, there is no loss in availability. GATE-4. Ans. (b)

Previous 20-Years IES Answers ⎧1000 1000 ⎫ IES-1. Ans. (b) Loss of available energy = To × ( ΔS )univ. = 300 ⎨ − ⎬ kJ = 250kJ 600 ⎭ ⎩ 400 IES-2. Ans. (b) Loss of Availability = T0 ΔS T0 = 300K

ΔS =

Q Q − T1 T2

⎛ 1000 1000 ⎞ ∴ Loss of Availability = 300 ⎜ − = 250 kJ 600 ⎟⎠ ⎝ 400

IES-3. Ans. (c)

η = 1−

Tource Tsurroundings

∴η1 >η2

IES-4. Ans. (b) IES-5. Ans. (c) U.E. = To (s1 − s 2 ) = 300 × (1.1 − 0.7) = 120 kJ/kg

Change in availability = (h1 − h 2 ) − (U.E.) = (400 − 100) − 120 = 180 kJ/kg IES-6. Ans. (a) IES-7. Ans. (c) The availability of a given system is defined as the maximum useful work that can be obtained in a process in which the system comes to equilibrium with the surroundings or attains a dead state. Clearly, the availability of a system depends on the condition of the system as well as those of the surroundings.

127

Availability, Irreversibility S K Mondal’s

Chapter 5

IES-8. Ans. (c) IES-9. Ans. (a) IES-10. Ans. (a) Availability is the maximum theoretical work obtainable and it can be destroyed in irreversibility. IES-11. Ans. (a) I = To × ( ΔS )universe = To × ⎡⎣ ΔSsystem + ΔSsurrounding ⎤⎦ IES-12. Ans. (d) A is false, For a process due to irreversibility entropy will increase and actual process may be 1–2' but due to heat loss to the surroundings, may 2' coincide with 2 but the process not adiabatic. So, all isentropic process is not adiabatic. IES-13. Ans. (a)

Previous 20-Years IAS Answers ⎧1000 1000 ⎫ IAS-1. Ans. (b) Loss of available energy = To × ( ΔS )univ. = 300 ⎨ − ⎬ kJ = 250kJ 600 ⎭ ⎩ 400

IAS-2. Ans. (b) Maximum possible efficiency (ηmax) = 1 −

T2 300 2 = 1− = 900 3 T1

Maximum possible Power output with this machine (Wmax) = Q × η max =

3 × 75000 2 × kW 41. 67 KW 3600 3

So above demand is impossible.

⎛ T ⎞ ⎛ 300 ⎞ IAS-3. Ans. (b) Available part of the heat (WE) = Q ⎜ 1 − 2 ⎟ = 400 ⎜ 1 − ⎟ = 250 kJ T ⎝ 800 ⎠ 1 ⎠ ⎝ Unavailable work (Wu) = 400 – 250 = 150 kJ ⎡ ⎛ 1250 ⎞ ⎤ = 2000 × 0.5 ⎢(1250 − 450 ) − 303ln ⎜ ⎟ ⎥ = 490MJ s ⎝ 450 ⎠ ⎦ ⎣ IAS-4. Ans. (d) IAS-5. Ans. (a) IAS-6. Ans. (d) 128

Availability, Irreversibility S K Mondal’s

Chapter 5

IAS-7. Ans. (d) Entropy will increase in the process AH is BH. Therefore Irreversibility (I) = To × ΔS i.e. area under the line BH. IAS-8. Ans. (c) Expansion through the porous plug is adiabatic as no heat added or rejected to the system. It is not reversible, due to large irreversibility entropy increases so it is not an isentropic process. mimimum energy int ake to perform the given task IAS-9. Ans. (c) ηII = actual energy int ake to perform the same task

129

Thermodynamic Relations S K Mondal’s

6.

Chapter 6

Thermodynamic Relations

Theory at a Glance (For GATE, IES & PSUs) Adiabatic index (γ) =1+

2 N

where N is degrees of freedom of molecules N=3 for monatomic gas N=5 for diatomic gas N=6 for try atomic gas.

Some Mathematical Theorem Theorem 1. If a relation exists among the variables x, y and z, then z may be expressed as a function of x and y, or ⎛ ∂z ⎞ ⎛ ∂z ⎞ dz = ⎜ ⎟ dx + ⎜ ⎟ dy ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x

then dz = M dx + N dy. where z, M and N are functions of x and y. Differentiating M partially with respect to y, and N with respect to x. ∂2z ⎛ ∂M ⎞ ⎜ ∂y ⎟ = ∂x.∂y ⎝ ⎠x ∂2 z ⎛ ∂N ⎞ ⎜ ∂x ⎟ = ∂y.∂x ⎝ ⎠y ⎛ ∂M ⎞ ⎛ ∂N ⎞ ⎜ ∂y ⎟ = ⎜ ∂x ⎟ ⎠y ⎝ ⎠x ⎝ This is the condition of exact (or perfect) differential.

Theorem 2. If a quantity f is a function of x, y and z, and a relation exists among x, y and z, then f is a function of any two of x, y and z. Similarly any one of x, y and z may be regarded to be a function of f and any one of x, y and z. Thus, if x = x (f, y) ⎛ ∂x ⎞ ⎛ ∂x ⎞ dx = ⎜ ⎟ df + ⎜ ⎟ dy ∂ f ⎝ ⎠y ⎝ ∂y ⎠ f

Similarly, if

y = y (f, z) ⎛ ∂y ⎞ ⎛ ∂y ⎞ dy = ⎜ ⎟ df + ⎜ ⎟ dz ⎝ ∂z ⎠ f ⎝ ∂f ⎠ z Substituting the expression of dy in the preceding equation

130

Thermodynamic Relations S K Mondal’s ⎛ ∂x ⎞ ⎛ ∂x ⎞ dx = ⎜ ⎟ df + ⎜ ⎟ ⎝ ∂f ⎠ y ⎝ ∂y ⎠ f

Chapter 6 ⎡ ⎛ ∂y ⎞ ⎤ ⎛ ∂y ⎞ ⎢ ⎜ ⎟ df + ⎜ ⎟ dz ⎥ ⎝ ∂z ⎠ f ⎥⎦ ⎢⎣ ⎝ ∂f ⎠ z

⎡ ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎤ ⎛ ∂x ⎞ ⎛ ∂y ⎞ = ⎢ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥ df + ⎜ ⎟ ⎜ ⎟ dz ⎝ ∂y ⎠ f ⎝ ∂z ⎠ f ⎢⎣ ⎝ ∂f ⎠ y ⎝ ∂y ⎠ f ⎝ ∂f ⎠ z ⎥⎦ Again ⎛ ∂x ⎞ ⎛ ∂x ⎞ dx = ⎜ ⎟ df + ⎜ ⎟ dz ⎝ ∂z ⎠ f ⎝ ∂f ⎠ z ⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂x ⎞ ⎜ ∂z ⎟ = ⎜ ∂y ⎟ ⎜ ∂z ⎟ ⎝ ⎠f ⎝ ⎠f ⎝ ⎠f ⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂z ⎞ ⎜ ∂y ⎟ ⎜ ∂z ⎟ ⎜ ∂x ⎟ = 1 ⎝ ⎠f ⎝ ⎠f ⎝ ⎠f

Theorem 3. Among the variables x, y, and z any one variable may be considered as a function of the other two. Thus x = x(y, z) ⎛ ∂x ⎞ ⎛ ∂x ⎞ dx = ⎜ ⎟ dy + ⎜ ⎟ dz ⎝ ∂z ⎠ y ⎝ ∂y ⎠ z Similarly, ⎛ ∂z ⎞ ⎛ ∂z ⎞ dz = ⎜ ⎟ dx + ⎜ ⎟ dy ∂ x ⎝ ⎠y ⎝ ∂y ⎠ x ⎤ ⎛ ∂x ⎞ ⎛ ∂z ⎞ ⎛ ∂x ⎞ ⎡ ⎛ ∂z ⎞ dx = ⎜ ⎟ dy + ⎜ ⎟ ⎢ ⎜ ⎟ dx + ⎜ ⎟ dy ⎥ ⎝ ∂z ⎠ y ⎢⎣ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ z ⎝ ∂y ⎠ x ⎥⎦ ⎡ ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂z ⎞ ⎤ ⎛ ∂x ⎞ ⎛ ∂z ⎞ = ⎢ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥ dy + ⎜ ⎟ ⎜ ⎟ dx ∂ ∂ ∂ y z y ⎝ ∂z ⎠ y ⎝ ∂x ⎠ y ⎣⎢ ⎝ ⎠ z ⎝ ⎠ y ⎝ ⎠ x ⎦⎥ ⎡ ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂z ⎞ ⎤ = ⎢ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥ dy + dx ⎣⎢ ⎝ ∂y ⎠ z ⎝ ∂z ⎠ y ⎝ ∂y ⎠ x ⎦⎥ ⎛ ∂x ⎞ ⎛ ∂z ⎞ ⎛ ∂x ⎞ ∴⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ = 0 ⎝ ∂y ⎠ z ⎝ ∂y ⎠ x ⎝ ∂z ⎠ y ⎛ ∂x ⎞ ⎛ ∂z ⎞ ⎛ ∂y ⎞ ⎜ ∂y ⎟ ⎜ ∂x ⎟ ⎜ ∂z ⎟ = −1 ⎝ ⎠z ⎝ ⎠ y ⎝ ⎠x Among the thermodynamic variables p, V and T. The following relation holds good ⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎜ ∂V ⎟ ⎜ ∂T ⎟ ⎜ ∂p ⎟ = −1 ⎝ ⎠T ⎝ ⎠p ⎝ ⎠v

Maxwell’s Equations:131

Thermodynamic Relations S K Mondal’s

Chapter 6

A pure substance existing in a single phase has only two independent variables. Of the eight quantities p, V, T, S, U, H, F (Helmholtz function), and G (Gibbs function) any one may be expressed as a function of any two others. For a pure substance undergoing an infinitesimal reversible process (a) dU = TdS - pdV (b) dH = dU + pdV + VdP = TdS + Vdp (c) dF = dU - TdS - SdT = - pdT - SdT (d) dG = dH - TdS - SdT = Vdp - SdT Since U, H, F and G are thermodynamic properties and exact differentials of the type dz = M dx + N dy, then ⎛ ∂M ⎞ ⎛ ∂N ⎞ ⎜ ⎟ =⎜ ⎟ y ∂ ⎝ ⎠ x ⎝ ∂x ⎠ y Applying this to the four equations ⎛ ∂T ⎞ ⎛ ∂p ⎞ ⎜ ∂V ⎟ = − ⎜ ∂S ⎟ ⎝ ⎠s ⎝ ⎠v ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎜ ∂P ⎟ = ⎜ ∂S ⎟ ⎝ ⎠s ⎝ ⎠p ⎛ ∂p ⎞ ⎛ ∂S ⎞ ⎜ ∂T ⎟ = ⎜ ∂V ⎟ ⎝ ⎠V ⎝ ⎠T ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ∂T ⎟ = − ⎜ ∂p ⎟ ⎝ ⎠P ⎝ ⎠T These four equations are known as Maxwell’s equations.

(i) Derive:

dS = Cv

dT ⎛ ∂p ⎞ + dV T ⎜⎝ ∂T ⎟⎠ v

[IAS - 1986]

Let entropy S be imagined as a function of T and V.

132

Thermodynamic Relations S K Mondal’s

Chapter 6

Then                S = S ( T, V ) ⎛ ∂S ⎞ ⎛ ∂S ⎞             or        dS = ⎜ dT + ⎜ ⎟ ⎟ dV ⎝ ∂T ⎠ V ⎝ ∂V ⎠ T multiplying both side by T

Since and ∴

⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ dT + T ⎜ ⎟ ⎟ dV ⎝ ∂T ⎠ V ⎝ ∂V ⎠ T ⎛ ∂S ⎞ T⎜ ⎟ = CV , heat capacity at constant volume ⎝ ∂T ⎠ V ⎛ ∂p ⎞ ⎛ ∂S ⎞ ⎜ ∂V ⎟ = ⎜ ∂T ⎟ by Maxwell 's equation ⎠V ⎝ ⎠T ⎝ ⎛ ∂p ⎞ TdS = CV dT + T ⎜ ⎟ dV ⎝ ∂T ⎠ V

dividing both side by T dS = C V

dT ⎛ ∂p ⎞ + dV proved T ⎜⎝ ∂T ⎟⎠ V

(ii) Derive:

⎛ ∂V ⎞ TdS = CpdT − T ⎜ ⎟ dp T ∂ ⎝ ⎠p

[IES-1998]

Let entropy S be imagined as a function of T and p.

Then                S = S ( T, p) ⎛ ∂S ⎞ ⎛ ∂S ⎞             or        dS = ⎜ dT + ⎜ ⎟ dp ⎟ ⎝ ∂T ⎠ p ⎝ ∂p ⎠ T multiplying both side by T ⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ dT + T ⎜ ⎟ dp ⎟ ⎝ ∂T ⎠ p ⎝ ∂p ⎠ T Since

⎛ ∂S ⎞ T⎜ ⎟ = Cp , heat capacity at constant pressure ⎝ ∂T ⎠ p

and

⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ∂p ⎟ = − ⎜ ∂T ⎟ by Maxwell 's equation ⎝ ⎠p ⎝ ⎠T

∴

⎛ ∂V ⎞ TdS = CpdT − T ⎜ ⎟ dp ⎝ ∂T ⎠ p

proved.

(iii) Derive: 133

Thermodynamic Relations S K Mondal’s

Chapter 6

k Cp dp CV β TdS = CV dT + T dV = CpdT − TVβ dp = dV + k β βV [IES-2001] We know that volume expansivity (β) = and

1 ⎛ ∂V ⎞ V ⎜⎝ ∂T ⎟⎠ p

isothermal compressibility (k) = −

1 ⎛ ∂V ⎞ V ⎜⎝ ∂p ⎟⎠ T

∴ From first TdS equation ⎛ ∂p ⎞ TdS = C V dT + T ⎜ ⎟ dV ⎝ ∂T ⎠ V ⎛ ∂V ⎞ ⎜ ∂T ⎟ β ⎝ ⎠p ⎛ ∂V ⎞ ⎛ ∂p ⎞ =− = −⎜ ⎟ ⋅⎜ ⎟ k ⎛ ∂V ⎞ ⎝ ∂T ⎠ p ⎝ ∂V ⎠ T ⎜ ∂p ⎟ ⎝ ⎠T

As

⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂p ⎞ ⎜ ∂T ⎟ ⋅ ⎜ ∂p ⎟ ⋅ ⎜ ∂V ⎟ = − 1 ⎝ ⎠p ⎝ ⎠T ⎠V ⎝

∴

⎛ ∂V ⎞ ⎛ ∂p ⎞ ⎛ ∂p ⎞ −⎜ ⋅⎜ =⎜ ⎟ ⎟ ⎟ ⎝ ∂T ⎠ p ⎝ ∂V ⎠ T ⎝ ∂T ⎠ V

or

β ⎛ ∂p ⎞ = k ⎜⎝ ∂T ⎟⎠ V

∴

TdS = CV dT + T ⋅

β ⋅ dV k

proved

From second TdS relation

⎛ ∂V ⎞ TdS = CpdT − T ⎜ ⎟ dp ⎝ ∂T ⎠ p as ∴ ∴

β=

1 ⎛ ∂V ⎞ V ⎜⎝ ∂T ⎟⎠ p

⎛ ∂V ⎞ ⎜ ∂T ⎟ = Vβ ⎝ ⎠p TdS = CpdT − TVβ dp

proved

Let S is a function of p, V ∴ S = S(p, V) ⎛ ∂S ⎞ ⎛ ∂S ⎞ ∴ dS = ⎜ ⎟ dp + ⎜ dV p ∂ ∂V ⎠⎟ p ⎝ ⎝ ⎠V

134

Thermodynamic Relations S K Mondal’s

Chapter 6

Multiply both side by T ⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ ⎟ dp + T ⎜ ⎟ dV ∂ p ⎝ ∂V ⎠ p ⎝ ⎠V or

⎛ ∂S ∂T ⎞ ⎛ ∂S ∂T ⎞ TdS = T ⎜ dp + T ⎜ ⋅ ⋅ ⎟ dV ⎟ ⎝ ∂T ∂V ⎠ p ⎝ ∂T ∂p ⎠ V

or

⎛ ∂S ⎞ ⎛ ∂T ⎞ ⎛ ∂S ⎞ ⎛ ∂T ⎞ TdS = T ⎜ dp + T ⎜ ⋅⎜ ⎟ ⎟ ⋅⎜ ⎟ dV ⎟ ⎝ ∂T ⎠ V ⎝ ∂p ⎠ V ⎝ ∂T ⎠ p ⎝ ∂V ⎠ p ⎛ ∂S ⎞ Cp = T ⎜ ⎟ ⎝ ∂T ⎠ p

∴

β ⎛ ∂p ⎞ = k ⎜⎝ ∂T ⎟⎠ V

∴

TdS = Cp

∴

β=

∴

⎛ ∂S ⎞ CV = T ⎜ ⎟ ⎝ ∂T ⎠ V

⎛ ∂T ⎞ ⎛ ∂T ⎞ TdS = Cp ⎜ ⎟ dp + C V ⎜ ∂V ⎟ dV p ∂ ⎝ ⎠p ⎝ ⎠V

From first

∴

and

k ⎛ ∂T ⎞ = β ⎜⎝ ∂p ⎟⎠ V

or

k ⎛ ∂T ⎞ dp + CV ⎜ ⎟ dV β ⎝ ∂V ⎠ p

1 ⎛ ∂V ⎞ V ⎜⎝ ∂T ⎟⎠ p

1 ⎛ ∂T ⎞ ⎜ ∂V ⎟ = βV ⎝ ⎠p C k dp C V TdS = p dV + β βV

proved.

(iv) Prove that 2

⎛ ∂V ⎞ ⎛ ∂p ⎞ Cp − Cv = − T ⎜ ⋅⎜ ⎟ ⎟ ⎝ ∂T ⎠ p ⎝ ∂V ⎠ T We know that

135

[IAS-1998]

Thermodynamic Relations S K Mondal’s

Chapter 6

⎛ ∂V ⎞ ⎛ ∂p ⎞ TdS = CpdT − T ⎜ dp = CV dT + T ⎜ ⎟ ⎟ dV ⎝ ∂T ⎠ p ⎝ ∂T ⎠ V or

or

(C

p

− Cv ) dT =

⎛ ∂V ⎞ ⎛ ∂p ⎞ T⎜ dp + T ⎜ ⎟ ⎟ dV ⎝ ∂T ⎠ p ⎝ ∂T ⎠ V

⎛ ∂V ⎞ ∂p ⎞ T⎜ dp T ⎛⎜ dV ⎟ ∂T ⎟⎠ V ⎝ ∂T ⎠ p ⎝ + dT = Cp − C V Cp − CV

− − − (i )

sin ce T is a function of p, V T = T ( p, V ) or

⎛ ∂T ⎞ ⎛ ∂T ⎞ dT = ⎜ dp + ⎜ ⎟ dV ⎟ ⎝ ∂V ⎠ p ⎝ ∂p ⎠ V

− − − ( ii )

comparing ( i ) & ( ii ) we get ⎛ ∂V ⎞ T⎜ ⎟ ⎝ ∂T ⎠ p ⎛ ∂T ⎞ =⎜ ⎟ Cp − CV ⎝ ∂p ⎠ V both these give

⎛ ∂p ⎞ T⎜ ∂T ⎠⎟ V ⎛ ∂T ⎞ ⎝ =⎜ and ⎟ Cp − CV ⎝ ∂V ⎠ p

⎛ ∂V ⎞ ⎛ ∂p ⎞ Cp − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂T ⎠ V Here

⎛ ∂p ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎛ ∂p ⎞ ⎜ ∂T ⎟ ⋅ ⎜ ∂V ⎟ ⋅ ⎜ ∂p ⎟ = − 1 or ⎜ ∂T ⎟ = − ⎜ ∂T ⎟ ⋅ ⎜ ∂V ⎟ ⎝ ⎠V ⎝ ⎠p ⎝ ⎝ ⎠V ⎝ ⎠p ⎝ ⎠T ⎠T

∵

⎛ ∂V ⎞ ⎛ ∂p ⎞ Cp − CV = − T ⎜ ⎟ ⋅⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂V ⎠ T

2

proved.

...............Equation(A)

This is a very important equation in thermodynamics. It indicates the following important facts. 2

⎛ ∂V ⎞ ⎛ ∂p ⎞ (a) Since ⎜ ⎟ is always positive, and ⎜ ∂V ⎟ for any substance is negative. (Cp – Cv) is always T ∂ ⎝ ⎠p ⎝ ⎠T positive. Therefore, Cp is always greater than Cv. (b) As T → 0 K ,C p → Cv or at absolute zero, Cp = Cv. ⎛ ∂V ⎞ (c) When ⎜ ⎟ = 0 (e.g for water at 4ºC, when density is maximum. Or specific volume minimum). ⎝ ∂T ⎠ p C p = C v.

(d) For an ideal gas, pV = mRT

136

Thermodynamic Relations S K Mondal’s

Chapter 6

mR V ⎛ ∂V ⎞ ⎜ ∂T ⎟ = P = T ⎝ ⎠p and

∴

mRT ⎛ ∂p ⎞ ⎜ ∂V ⎟ = − V2 ⎝ ⎠T C p − Cv = mR

or c p − cv = R

Equation (A) may also be expressed in terms of volume expansively (β) defined as 1 ⎛ ∂V ⎞ β = ⎜ V ⎝ ∂T ⎟⎠ p and isothermal compressibility (kT) defined as 1 ⎛ ∂V ⎞ kT = − ⎜ ⎟ V ⎝ ∂p ⎠T

⎡ 1 ⎛ ∂V ⎞ ⎤ TV ⎢ ⎜ ⎟ ⎥ ⎢⎣ V ⎝ ∂T ⎠ p ⎥⎦ C p − Cv = 1 ⎛ ∂V ⎞ − ⎜ V ⎝ ∂p ⎟⎠T C p − Cv =

2

TV β 2 kT

Ratio of heat capacities

At constant S, the two TdS equations become ⎛ ∂V ⎞ C p dTs = T ⎜ ⎟ dps ⎝ ∂T ⎠ p ⎛ ∂p ⎞ Cv dTs = −T ⎜ ⎟ dVs ⎝ ∂T ⎠v ⎛ ∂p ⎜ Cp ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂p ⎞ ⎝ ∂V = −⎜ = ⎜ ⎟ ⎜ ⎟ ⎟ Cv ⎝ ∂T ⎠ p ⎝ ∂p ⎠v ⎝ ∂V ⎠s ⎛ ∂p ⎜ ∂V ⎝

⎞ ⎟ ⎠S =γ ⎞ ⎟ ⎠T

since γ > 1, ⎛ ∂p ⎞ ⎛ ∂p ⎞ ⎜ ∂V ⎟ > ⎜ ∂V ⎟ ⎝ ⎠s ⎝ ⎠T Therefore, the slope of an isentropic is greater than that of an isotherm on p-v diagram (figure below). For reversible and adiabatic compression, the work done is 2s

Ws = h2 s − h1 = ∫ vdp 1

= Area 1- 2 s - 3 - 4 - 1

137

Therm T modyna amic Relatio R ons S K Mondal’s

Cha apter 6

(Fig. Co ompression Work in Diffferent Rev versible Pro ocess) a isotherm mal compresssion, the worrk done would d be For reversible and 2T

WT = h2T − h1 =

dp ∫ vdp 1

= Area A 1-2T - 3 - 4 - 1 WT < Ws For polytropic compression n with 1 < n < γ. the work done wiill be betweeen these twoo values. So isoothermal com mpression min nimum work. Th he adiabatic compressibili c ity (ks) is defiined as 1 ⎛ ∂V ⎞ or ks = − ⎜ ⎟ V ⎝ ∂p ⎠s 1 ⎛ ∂V ⎞ Cp V ⎝⎜ ∂p ⎟⎠T = =γ Cv 1 ⎛ ∂V ⎞ − ⎜ V ⎝ ∂p ⎟⎠s −

γ =

k ks

En nergy Equ uation For a system undergoing u an a infinitesimal reversib ble process between b two equilibrium m states, the ange of interrnal energy iss cha dU = TdS - pdV p ubstituting th he first TdS equation e Su ⎛ ∂p ⎞  dU = Cv dT + T ⎜ ⎟ dV − pdV ⎝ ∂T ⎠V ⎤ ⎡ ⎛ ∂p ⎞ − p ⎥ dV = Cv dT + ⎢T ⎜ ⎟ ⎢⎣ ⎝ ∂T ⎠V ⎦⎥ if U = (T ,V ) ⎛ ∂U ⎞ ⎛ ∂U ⎞ dT + ⎜ dU = ⎜ ⎟ dV ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎛ ∂p ⎞ ⎛ ∂U ⎞ ⎜ ∂V ⎟ = T ⎜ ∂T ⎟ − p ⎝ ⎠V ⎝ ⎠T Th his is known as a energy equ uation. Two application a o the equatioon are given belowof b 138 8

Thermodynamic Relations S K Mondal’s (a) For an ideal gas, p =

Chapter 6 nRT V

nR p ⎛ ∂p ⎞ ∴⎜ ⎟ = V =T ∂ T ⎝ ⎠V p ⎛ ∂U ⎞ ∴⎜ = T. − p = 0 ⎟ T ⎝ ∂V ⎠T U does not change when V changes at T = C. ⎛ ∂U ⎞ ⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =1 ⎝ ∂p ⎠T ⎝ ∂V ⎠T ⎝ ∂U ⎠T ⎛ ∂U ⎞ ⎛ ∂p ⎞ ⎛ ∂U ⎞ =⎜ ⎜ ⎟ ⎜ ⎟ ⎟ =0 ⎝ ∂p ⎠T ⎝ ∂V ⎠T ⎝ ∂V ⎠T ⎛ ∂U ⎞ ⎛ ∂p ⎞ since ⎜ ⎟ ≠ 0, ⎜ ∂p ⎟ = 0 V ∂ ⎝ ⎠T ⎝ ⎠T U does not change either when p changes at T = C. So the internal energy of an ideal gas is a function of temperature only. Another important point to note is that for an ideal gas ⎛ ∂p ⎞ pV = nRT and T ⎜ ⎟ −p=0 ⎝ ∂T ⎠v Therefore dU = Cv dT holds good for an ideal gas in any process (even when the volume changes). But for any other substance dU = Cv dT is true only when the volume is constant and dV = 0 Similarly dH = TdS + Vdp and

⎛ ∂V ⎞ TdS = Cp  dT − T ⎜ ⎟ dp ⎝ ∂T ⎠ p ⎡ ⎛ ∂V ⎞ ⎤ ∴ dH = C p dT + ⎢V − T ⎜ ⎟ ⎥ dp ⎝ ∂T ⎠ p ⎥⎦ ⎣⎢ ⎛ ∂H ⎞ ⎛ ∂V ⎞ ∴⎜ ⎟ = V −T ⎜ ⎟ p ∂ ⎝ ∂T ⎠ p ⎝ ⎠T

As shown for internal energy, it can be similarly proved from Eq. shown in above that the enthalpy of an ideal gas is not a function of either volume or pressure. ⎡ ⎛ ∂H ⎞ ⎤ ⎛ ∂H ⎞ = 0⎥ ⎢i.e ⎜ ⎟ = 0 and ⎜ ⎟ ⎝ ∂V ⎠T ⎣ ⎝ ∂p ⎠T ⎦ but a function of temperature alone. Since for an ideal gas, pV =  nRT ⎛ ∂V ⎞ and V −T ⎜ ⎟ =0 ⎝ ∂T ⎠ p the relation dH = Cp dT is true for any process (even when the pressure changes.) 139

Thermodynamic Relations S K Mondal’s

Chapter 6

However, for any other substance the relation dH = Cp dT holds good only when the pressure remains constant or dp = 0. (b) Thermal radiation in equilibrium with the enclosing walls processes an energy that depends only on the volume and temperature. The energy density (u), defined as the ratio of energy to volume, is a function of temperature only, or U u = = f (T )only. V The electromagnetic theory of radiation states that radiation is equivalent to a photon gas and it exerts a pressure, and that the pressure exerted by the black body radiation in an enclosure is given by u p= 3 Black body radiation is thus specified by the pressure, volume and temperature of the radiation. since. u U = uV and p = 3 1 du ⎛ ∂U ⎞ ⎛ ∂p ⎞ ⎜ ∂V ⎟ = u and ⎜ ∂T ⎟ = 3 dT ⎝ ⎠T ⎝ ⎠V By substituting in the energy Eq. T du u − u= 3 dT 3 du dT ∴ =4 u T or ln u = ln T4 + lnb or u = bT4 where b is a constant. This is known as the Stefan - Boltzmann Law. Since U = uV = VbT 4

and

⎛ ∂U ⎞ 3 ⎜ ∂T ⎟ = Cv = 4VbT ⎝ ⎠V 1 du 4 ⎛ ∂p ⎞ 3 ⎜ ∂T ⎟ = 3 dT = 3 bT ⎝ ⎠V

From the first TdS equation ⎛ ∂p ⎞ TdS = Cv dT + T ⎜ ⎟ dV ⎝ ∂T ⎠v 4 = 4VbT 3 dT + bT 4 .dV 3 For a reversible isothermal change of volume, the heat to be supplied reversibly to keep temperature constant. 4 Q = bT 4 ΔV 3 For a reversible adiabatic change of volume

140

Thermodynamic Relations S K Mondal’s

Chapter 6

4 bT 4 dV = −4VbT 3 dT 3 dV dT or = −3 V T or VT3 = const If the temperature is one-half the original temperature. The volume of black body radiation is to be increased adiabatically eight times its original volume so that the radiation remains in equilibrium with matter at that temperature.

(v) Prove that

β ⎛ ∂p ⎞ = k ⎜⎝ ∂T ⎟⎠ V

and

⎧ ⎛ ∂U ⎞ ⎫ ⎛ ∂V ⎞ Cp − CV = ⎨ p + ⎜ ⎟ ⎬⎜ ⎟ ⎝ ∂V ⎠ T ⎭ ⎝ ∂T ⎠ p ⎩

Hence show that Cp − Cv =

Here β =

1 ⎛ ∂V ⎞ V ⎜⎝ ∂p ⎟⎠ T

⎛ ∂V ⎞ ⎜ ∂T ⎟ β ⎝ ⎠p ⎛ ∂V ⎞ ⎛ ∂p ⎞ =− = −⎜ ⎟ ⋅⎜ ⎟ k ⎛ ∂V ⎞ ⎝ ∂T ⎠ p ⎝ ∂V ⎠ T ⎜ ∂p ⎟ ⎝ ⎠T ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂p ⎞ ⎜ ∂T ⎟ ⋅ ⎜ ∂p ⎟ ⋅ ⎜ ∂V ⎟ = − 1 ⎝ ⎠p ⎝ ⎠T ⎠V ⎝

we know that or ∴ ⇒

[IES - 2003]

1 ⎛ ∂V ⎞ V ⎜⎝ ∂T ⎟⎠ p

k=−

∴

β2 TV k

⎛ ∂V ⎞ ⎛ ∂p ⎞ ⎛ ∂p ⎞ −⎜ ⋅⎜ =⎜ ⎟ ⎟ ⎟ ⎝ ∂T ⎠ p ⎝ ∂V ⎠ T ⎝ ∂T ⎠ V

β ⎛ ∂p ⎞ = k ⎜⎝ ∂T ⎟⎠ V

proved.

From Tds relations

141

Thermodynamic Relations S K Mondal’s

Chapter 6

⎛ ∂V ⎞ ⎛ ∂p ⎞ TdS = CpdT − T ⎜ ⎟ dP = C V dT + T ⎜ ∂T ⎟ dV ∂ T ⎝ ⎠p ⎝ ⎠V ∴

or

(C

p

⎛ ∂V ⎞ ⎛ ∂p ⎞ − C v ) dT = T ⎜ ⎟ dP + T ⎜ ∂T ⎟ dV ⎝ ∂T ⎠ p ⎝ ⎠v

⎛ ∂V ⎞ ⎛ ∂p ⎞ T⎜ T⎜ ⎟ ∂T ⎟⎠ V ⎝ ∂T ⎠ p dT = dP + ⎝ dV − − − ( i ) Cp − C V Cp − CV

S in ce T is a function of ( p, V ) T = T ( p, V ) ∴

⎛ ∂T ⎞ ⎛ ∂T ⎞ dT = ⎜ ⎟ dp + ⎜ ∂V ⎟ dV p ∂ ⎝ ⎠p ⎝ ⎠V

− − − ( ii )

Compairing ( i ) & ( ii ) we get ⎛ ∂V ⎞ T⎜ ⎟ ⎝ ∂T ⎠ p ⎛ ∂T ⎞ =⎜ ⎟ Cp − C V ⎝ ∂p ⎠ V

as

⎛ ∂V ⎞ ⎛ ∂p ⎞ Cp − CV = T ⎜ ⎟ ⋅⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂T ⎠ V dU = dQ − pdV

∴

dU = TdS − pdV

∴

or or

and

⎛ ∂p ⎞ T⎜ ⎟ ⎝ ∂T ⎠ V = ⎛ ∂T ⎞ ⎜ ⎟ Cp − CV ⎝ ∂V ⎠ p

⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎜ ∂V ⎟ = T ⎜ ∂V ⎟ − p ⎝ ⎠T ⎝ ⎠T ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎜ ∂V ⎟ + p = T ⎜ ∂V ⎟ ⎝ ⎠T ⎝ ⎠T

From Maxwell 's Third relations ⎛ ∂p ⎞ ⎛ ∂S ⎞ ⎜ ∂T ⎟ = ⎜ ∂V ⎟ ⎝ ⎠V ⎝ ⎠T ∴

⎧ ⎛ ∂V ⎞ ⎛ ∂p ⎞ ⎛ ∂U ⎞ ⎫ ⎛ ∂V ⎞ Cp − CV = T ⎜ ⋅⎜ = ⎨p + ⎜ ⎟ ⎟ ⎟ ⎬⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂T ⎠ V ⎩ ⎝ ∂V ⎠ T ⎭ ⎝ ∂T ⎠ p

(vi) Prove that Joule – Thomson co-efficient

T2 ⎡ ∂ ⎛ V ⎞ ⎤ ⎛ ∂T ⎞ μ=⎜ ⎟ = C ⎢ ∂T ⎜ T ⎟ ⎥ ∂ p ⎝ ⎠⎦p ⎝ ⎠h p ⎣

[IES-2002]

The numerical value of the slope of an isenthalpic on a T – p diagram at any point is called the Joule – Kelvin coefficient.

142

Thermodynamic Relations S K Mondal’s ∴ Here

Chapter 6

⎛ ∂T ⎞ μ=⎜ ⎟ ⎝ ∂p ⎠h dH = TdS + Vdp

as

⎛ ∂V ⎞ TdS = CpdT − T ⎜ ⎟ dp ⎝ ∂T ⎠ p

∴

⎡ ⎛ ∂V ⎞ ⎤ ⎛ ∂V ⎞ dH = CpdT − T ⎜ dp + Vdp = CpdT − ⎢ T ⎜ − V ⎥ dp ⎟ ⎟ ⎝ ∂T ⎠ p ⎣⎢ ⎝ ∂T ⎠ p ⎦⎥

if

H = cos t.

so

⎡ ⎛ ∂V ⎞ ⎤ Cp ( dT )h − ⎢ T ⎜ − V ⎥ ( dp)h = 0 ⎟ ⎣⎢ ⎝ ∂T ⎠ p ⎦⎥

∴

⎤ T 2 ⎡ 1 ⎛ ∂V ⎞ V ⎤ T2 ⎡ ∂ ⎛ V ⎞ ⎤ 1 ⎡ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⋅ − = T V = − = ⎢ ⎜ ⎢ ⎜ ⎥ ⎟ ⎟ ⎜ ∂p ⎟ ⎢ ⎜ ⎟⎥ 2⎥ ⎝ ⎠h Cp ⎢⎣ ⎝ ∂T ⎠ p ⎥⎦ Cp ⎢⎣ T ⎝ ∂T ⎠ p T ⎥⎦ Cp ⎣ ∂T ⎝ T ⎠ ⎦ p

∴

dH = 0

(vii) Derive clausius – clapeyron equation hfg ⎛ dp ⎞ = ⎜ dT ⎟ ⎝ ⎠ T ( vg − v f ) ⎛ ∂p ⎞ ⎛ ∂S ⎞ ⎜ ∂T ⎟ = ⎜ ∂V ⎟ ⎝ ⎠V ⎝ ⎠T

h dp = fg2 dT p RT

and

[IES-2000]

Maxwells equation

When saturated liquid convert to saturated vapour at constant temperature. During the evaporation, the pr. & T is independent of volume.

sg − s f ⎛ dp ⎞ ⎜ dT ⎟ = v − v ⎝ ⎠sat g f

∴

sg − sf = sfg = or

h fg T

h fg ⎛ dp ⎞ = ⎜ dT ⎟ ⎝ ⎠sat T ( v g − v f )

→ It is useful to estimate properties like h from other measurable properties. → At a change of phage we may find h fg i.e. latent heat.

143

Thermodynamic Relations S K Mondal’s

Chapter 6

At very low pressure  v g ≈ v f g as v f very small pv g = RT ∴

vg =

or

RT p

h fg h fg h ⋅p dp = = = fg 2 RT dT T ⋅ v g T ⋅ RT p

or

dp h fg dT = ⋅ p R T2

or

⎛ p ⎞ h fg ⎛ 1 1 ⎞ ln ⎜ 2 ⎟ = ⎜ − ⎟ R ⎝ T1 T2 ⎠ ⎝ p1 ⎠

→ Knowing vapour pressure p1 at temperature T1, we may find out p2 at temperature T2.

Joule-Kelvin Effect or Joule-Thomson coefficient The value of the specific heat cp can be determined from p–v–T data and the Joule–Thomson coefficient. The Joule–Thomson coefficient μJ is defined as

⎛ ∂T ⎞

μJ = ⎜ ⎟ ⎝ ∂p ⎠h Like other partial differential coefficients introduced in this section, the Joule–Thomson coefficient is defined in terms of thermodynamic properties only and thus is itself a property. The units of μJ are those of temperature divided by pressure. A relationship between the specific heat cp and the Joule–Thomson coefficient μJ can be established to write

⎛ ∂T ⎞ ⎛ ∂p ⎞ ⎛ ∂h ⎞ ⎜ ∂p ⎟ ⎜ ∂h ⎟ ⎜ ∂T ⎟ = − 1 ⎠p ⎝ ⎠h ⎝ ⎠T ⎝ The first factor in this expression is the Joule–Thomson coefficient and the third is cp. Thus 144

Therm T modyna amic Relatio R ons S K Mondal’s

Cha apter 6 cp =

Wiith ( ∂h / ∂p )T = 1 / ( ∂p / ∂h )T

−1 μJ ( ∂p / ∂h )T

t this can be written as

cp = −

1 ⎛ ∂h ⎞ μJ ⎜⎝ ∂p ⎟⎠T

Th he partial derrivative ( ∂h / ∂p )T , called the constan nt-temperatture coefficiient, can be eliminated. Th he following expression e re esults:

cp =

⎤ 1 ⎡ ⎛ ∂v ⎞ − v⎥ ⎢T ⎜ ⎟ μJ ⎣ ⎝ ∂T ⎠ p ⎦

allows the valu ue of cp at a state to bee determined d using p–v–T T data and the value off the Joule– Th homson coeffiicient at tha at state. Let us consider next how th he Joule–Thoomson coefficcient can be fou und experimeentally. Th he numerical value of the slope of an isenthalpic i on a T-p diagram at any point p is called the JouleKelvin coefficieent and is denoted d by μJ . Thus the locus of alll points at which w μJ is zero is the version curvee. The regio on inside thee inversion curve c where μJ is positive is called the cooling inv reg gion and the region outsid de where μJ is negative is i called the heating h regioon. So,

⎛ ∂T ⎞

μJ = ⎜ ⎟ ⎝ ∂p ⎠h

G Gibbs Phase Rule R Gibbs Ph hase Rule determines wha at is expected d to define th he state of a system s

F=C–P+2 145 5

Thermodynamic Relations S K Mondal’s

Chapter 6

F = Number of degrees of freedom (i.e.., no. of properties required) C = Number of components P = Number of phases e.g., Nitrogen gas C = 1; P = 1. Therefore, F = 2 • • • • •

To determine the state of the nitrogen gas in a cylinder two properties are adequate. A closed vessel containing water and steam in equilibrium: P = 2, C = 1 Therefore, F = 1. If any one property is specified it is sufficient. A vessel containing water, ice and steam in equilibrium P = 3, C = 1 therefore F = 0. The triple point is uniquely defined.

146

Thermodynamic Relations S K Mondal’s

Chapter 6

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions Maxwell's Equations GATE-1.

Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy? [GATE-2007] (a) δQ = dU + δW (b) TdS = dU + pdV (c) TdS = dU + δW (d) δQ = dU + pdV

GATE-2.

Considering the relationship TdS = dU + pdV between the entropy (S), internal energy (U), pressure (p), temperature (T) and volume (V), which of the following statements is correct? [GATE-2003] (a) It is applicable only for a reversible process (b) For an irreversible process, TdS > dU + pdV (c) It is valid only for an ideal gas (d) It is equivalent to 1 law, for a reversible process

Difference in Heat Capacities and Ratio of Heat Capacities GATE-3.

The specific heats of an ideal gas depend on its [GATE-1996] (a) Temperature (b) Pressure (c) Volume (d) Molecular weight and structure

GATE-4.

The specific heats of an ideal gas depends on its [GATE-1996] (a) Temperature (b) Pressure (c) Volume (d) Molecular weight and structure

GATE-5.

For an ideal gas the expression ⎢T ⎜

⎡ ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎤ ⎟ −T ⎜ ⎟ ⎥ is always equal to: ⎝ ∂T ⎠v ⎥⎦ ⎢⎣ ⎝ ∂T ⎠ p [GATE-1997]

(a ) zero GATE-6.

(b)

cp

(c ) R

cv

(d ) RT

A 2 kW, 40 litre water heater is switched on for 20 minutes. The heat capacity Cp for water is 4.2 kJ/kg K. Assuming all the electrical energy has gone into heating the water, increase of the water temperature in degree centigrade is: [GATE-2003] (a) 2.7 (b) 4.0 (c) 14.3 (d) 25.25

Joule-Kelvin Effect or Joule-Thomson coefficient GATE-7.

Which combination of the following statements is correct?

147

[GATE-2007]

Thermodynamic Relations S K Mondal’s

Chapter 6

P: A gas cools upon expansion only when its Joule-Thomson coefficient is positive in the temperature range of expansion. Q: For a system undergoing a process, its entropy remains constant only when the process is reversible. R: The work done by a closed system in an adiabatic process is a point function. S: A liquid expands upon freezing when the slop of its fusion curve on Pressure Temperature diagram is negative. (a) R and S (b) P and Q (c) Q, R and S (d) P, Q and R GATE-8.

A positive value to Joule-Thomson coefficient of a fluid means (a) Temperature drops during throttling (b) Temperature remains constant during throttling (c) Temperature rises during throttling (d) None of these

GATE-9.

A gas having a negative Joule-Thompson coefficient (µ < 0), when throttled, will: [GATE-2001] (a) Become cooler (b) Become warmer (c) Remain at the same temperature (d) Either be cooler or warmer depending on the type of gas

GATE-10. Match 4 correct pairs between List-I and List-II for the questions For a perfect gas: List-I List-II (a) Isobaric thermal expansion coefficient 1. 0 (b) Isothermal compressibility 2. ∞ (c) Isentropic compressibility 3. 1/v (d) Joule – Thomson coefficient 4. 1/T 5. 1/p 6. 1/ γ p

[GATE-2002]

[GATE-1994]

Previous 20-Years IES Questions Some Mathematical Theorems IES-1.

Given: [IES-1993] p = pressure, T = Temperature, v = specific volume Which one of the following can be considered as property of a system?

(a ) ∫ pdv

(b) ∫ vdp

⎛ dT p.dv ⎞ (c ) ∫ ⎜ + ⎟ v ⎠ ⎝ T

⎛ dT v.dp ⎞ (d ) ∫ ⎜ − ⎟ T ⎠ ⎝ T

Maxwell's Equations IES-2.

Which thermodynamic property is evaluated with the help of Maxwell equations from the data of other measurable properties of a system? [IES 2007] (a) Enthalpy (b) Entropy (c) Latent heat (d) Specific heat 148

Thermodynamic Relations S K Mondal’s IES-3.

Chapter 6

Consider the following statements pertaining to the Clapeyron equation: 1. It is useful to estimate properties like enthalpy from other measurable properties. [IES-2006] 2. At a change of phase, it can be used to find the latent heat at a given pressure.

⎛ ∂p ⎞ ⎛ ∂s ⎞ ⎟ =⎜ ⎟ ⎝ ∂v ⎠T ⎝ ∂T ⎠V

3. It is derived from the relationship ⎜

Which of the statements given above are correct? (a) 1,2 and 3 (b) Only 1 and 2 (c) Only 1 and 3 IES-3a

(d) Only 2 and 3

Clapeyron’s equation is used for finding out the (a) Dryness fraction of steam only (b) Entropy of superheater vapour only (c) Specific volume at any temperature and pressure (d) Total heat of superheated steam only

[IES-2011]

TdS Equations IES-4.

T ds equation can be expressed as:

T β dv k Tk dv (c) Tds = Cv dT +

[IES-2002]

Tdv k Tβ (d) Tds = Cv dT + dp k

(a) Tds = Cv dT +

(b) Tds = Cv dT +

β

IES-5.

Which one of the following statements applicable to a perfect gas will also be true for an irreversible process? (Symbols have the usual meanings). [IES-1996] (a) dQ = du + pdV (b) dQ = Tds (c) Tds = du + pdV (d) None of the above

IES-6.

Consider the following thermodynamic relations:

1.Tds = du + pdv

2.Tds = du − pdv

3.Tds = dh + vdp

4.Tds = dh − vdp

[IES-2000]

Which of these thermodynamic relations are correct? (a) 1 and 3 (b) 1 and 4 (c) 2 and 3

(d) 2 and 4

Difference in Heat Capacities and Ratio of Heat Capacities IES-7.

Match List-l (Terms) with List-II (Relations) and select the correct answer using the codes given below the Lists: [IES-2003] List-I (Terms) List-II (Relations) A. Specific heat at constant volume, Cv 1 ⎛ ∂v ⎞ 1. ⎜ ⎟ B. Isothermal compressibility kT

2.

149

v ⎝ ∂T ⎠ p ⎛ ∂p ⎞ ⎛ ∂v ⎞ −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠v ⎝ ∂T ⎠ p

Thermodynamic Relations S K Mondal’s

Chapter 6

C. Volume expansivity β

3.

D. Difference between specific heats at constant pressure and at constant Cp – Cv Codes: (a) (c)

A 3 3

B 4 4

C 2 1

D 1 2

(b) (d)

A 4 4

4.

⎛ ∂s ⎞ T⎜ ⎟ ⎝ ∂T ⎠v 1 ⎛ ∂v ⎞ − ⎜ ⎟ v ⎝ ∂p ⎠T B 1 1

C 3 2

D 2 3

IES-8.

Assertion (A): Specific heat at constant pressure for an ideal gas is always greater than the specific heat at constant volume. [IES-2002] Reason (R): Heat added at constant volume is not utilized for doing any external work. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-9.

An insulated box containing 0.5 kg of a gas having Cv = 0.98 kJ/kgK falls from a balloon 4 km above the earth’s surface. What will be the temperature rise of the gas when the box hits the ground? [IES-2004] (a) 0 K (b) 20 K (c) 40 K (d) 60 K

IES-10.

As compared to air standard cycle, in actual working, the effect of variations in specific heats is to: [IES-1994] (a) Increase maximum pressure and maximum temperature (b) Reduce maximum pressure and maximum temperature (c) Increase maximum pressure and decrease maximum temperature (d) Decrease maximum pressure and increase maximum temperature

IES-11.

The number of degrees of freedom for a diatomic molecule (a) 2 (b) 3 (c) 4

IES-12.

The ratio

(a) n + I

Cp Cv

for a gas with n degrees of freedom is equal to:

(b) n – I

(c)

2 −1 n

[IES-1992] (d) 5 [IES-1992]

(d) 1 +

2 n

IES-13.

Molal specific heats of an ideal gas depend on [IES-2010] (a) Its pressure (b) Its temperature (c) Both its pressure and temperature (d) The number of atoms in a molecule

IES-14.

Assertion (A): Ratio of specific heats

Cp Cv

decreases with increase

in

temperature. [IES-1996] Reason (R): With increase in temperature, C p decreases at a higher rate than C v. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A

150

Thermodynamic Relations S K Mondal’s

Chapter 6

(c) A is true but R is false (d) A is false but R is true IES-15.

It can be shown that for a simple compressible substance, the relationship

⎛ ∂V ⎞ ⎛ ∂P ⎞ Cp – Cv = – T ⎜ ⎟ ⎜ ⎟ exists. ⎝ ∂T ⎠ p ⎝ ∂v ⎠T 2

[IES-1998]

Where C p and C v are specific heats at constant pressure and constant volume respectively. T is the temperature V is volume and P is pressure. Which one of the following statements is NOT true? (a) C p is always greater than C v. (b) The right side of the equation reduces to R for ideal gas.

⎛ ∂V ⎞ ⎛ ∂P ⎞ ⎟ must be positive, T ⎜ ⎟ can be either positive or negative, and ⎜ ⎝ ∂v ⎠T ⎝ ∂T ⎠ p ⎛ ∂P ⎞ must have a sign that is opposite to that of ⎜ ⎟ ⎝ ∂v ⎠T 2

(c) Since

(d) Is very nearly equal to for liquid water.

Joule-Kelvin Effect or Joule-Thomson coefficient IES-16.

Joule-Thomson coefficient is defined as:

⎛ ∂T ⎞ (a) ⎜ ⎟ ⎝ ∂p ⎠ h

⎛ ∂h ⎞ (b) ⎜ ⎟ ⎝ ∂p ⎠T

[IES-1995]

⎛ ∂h ⎞ (c) ⎜ ⎟ ⎝ ∂T ⎠ p

(d)

⎛ ∂p ⎞ ⎜ ⎟ ⎝ ∂T ⎠ h

IES-17.

The throttling of certain gasses may be used for getting the refrigeration effect. What is the value of Joule – Thomson coefficient (µ) for such a throttling process? [IES-2007] (a) µ = 0 (b) µ = 1 (c) µ < 1 (d) µ > 1

IES-18.

Which one of the following is correct? [IES 2007] When a real gas undergoes Joule-Thomson expansion, the temperature (a) May remain constant (b) Always increases (c) May increase or decrease (d) Always decreases

IES-19.

Assertion (A): Throttling process for real gases at initial temperature higher than maximum inversion temperature is accompanied by decrease in temperature of the gas. [IES-2003] Reason (R): Joule-Kelvin coefficient μj is given ( ∂T / ∂p ) h and should have a positive value for decrease in temperature during throttling process. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-20.

Match List-I (Name of entity) with List-II (Definition) and select the correct answer using the codes given below the lists: [IES-2001] List-I (Name of entity) List-II (Definition) 151

Thermodynamic Relations S K Mondal’s

Chapter 6 1 ⎛ ∂v ⎞ − ⎜ ⎟ v ⎝ ∂p ⎠T ⎛ ∂h ⎞ 2. ⎜ ⎟ ⎝ ∂T ⎠ p

A. Compressibility factor

1.

B. Joule – Thomson coefficient C. Constant pressure specific heat

3.

⎛ ∂T ⎞ ⎜ ⎟ ⎝ ∂p ⎠ h

D. Isothermal compressibility

4.

⎛ pv ⎞ ⎜ ⎟ ⎝ RT ⎠

Codes: (a) (c)

A 2 2

B 1 3

C 4 4

D 3 1

A 4 4

(b) (d)

B 3 1

C 2 2

D 1 3

IES-21.

Joule – Thomson coefficient is the ratio of [IES-1999] (a) Pressure change to temperature change occurring when a gas undergoes the process of adiabatic throttling (b) Temperature change to pressure change occurring when a gas undergoes the process of adiabatic throttling (c) Temperature change to pressure change occurring when a gas undergoes the process of adiabatic compression (d) Pressure change to temperature change occurring when a gas undergoes the process of adiabatic compression

IES-22.

The Joule – Thomson coefficient is the

[IES-1996]

⎛ ∂T ⎞ ⎟ of pressure-temperature curve of real gases ⎝ ∂p ⎠ h

(a) ⎜

⎛ ∂T ⎞ ⎟ of temperature-entropy curve of real gases ⎝ ∂p ⎠ v

(b) ⎜

⎛ ∂h ⎞ ⎟ of enthalpy-entropy curve of real gases ⎝ ∂s ⎠T ⎛ ∂V ⎞ (d) ⎜ ⎟ of pressure-volume curve of real gases ⎝ ∂T ⎠ p (c) ⎜

IES-23.

Match the following: List-I A. Work

List-II 1. Point function

B. Heat

2.

C. Internal energy

3. ⎜

D. Joule Thomson Coefficient

4.

Code:

A

B

[IES-1992]

∫ Tds ⎛ ∂u ⎞ ⎟ ⎝ ∂T ⎠ h

C

D

∫ pdv

A

152

B

C

D

Thermodynamic Relations S K Mondal’s (a) (c)

Chapter 6 4 4

2 1

1 2

3 3

(b) (d)

1 2

2 1

4 4

3 3

Clausius-Clapeyron Equation IES-24.

Consider the following statements in respect of the Clausius – Clapeyron equation: [IES-2007] 1. It points to one possible way of measuring thermodynamic temperature. 2. It permits latent heat of vaporization to be estimated from measurements of specific volumes of saturated liquid, saturated vapour and the saturation temperatures at two nearby pressures. 3. It does not apply to changes from solid to the liquid phase and from solid to the Vapour phase. Which of the statements given above are correct? (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only

IES-25.

The equation relating the following measurable properties: [IES-2005] (i) The slope of saturation pressure temperature line (ii) The latent heat, and (iii) The change in volume during phase transformation; is known as: (a) Maxwell relation (b) Joules equation (c) Clapeyron equation (d) None of the above

IES-26.

The variation of saturation pressure with saturation temperature for a liquid is 0.1 bar/K at 400 K. The specific volume of saturated liquid and dry saturated vapour at 400 K are 0.251 and 0.001 m3/kg What will be the value of latent heat of vaporization using Clausius Clapeyron equation? [IES-2004] (a) 16000 kJ/kg (b) 1600 kJ/kg (c) 1000 kJ/kg (d) 160 kJ/kg If h, p, T and v refer to enthalpy, pressure, temperature and specific volume respectively and subscripts g and f refer to saturation conditions of vapour and liquid respectively then Clausius-Clapeyron equation applied to change of phase from liquid to vapour states is: [IES-1996, 2006]

IES-27.

(a)

dp (hg − h f ) = dt (vg − v f )

(b)

dp (hg − h f ) = dt T (vg − v f )

(c)

dp (hg − h f ) = dt T

(d)

dp (vg − v f )T = dt (hg − h f )

IES-28.

Which one of the following functions represents the Clapeyron equation pertaining to the change of phase of a pure substance? [IES-2002] (a) f (T, p, hfg) (b) f (T, p, hfg, vfg) (c) f (T, p, hfg, sfg) (d) f (T, p, hfg, sfg, vfg)

IES-29.

The Clapeyron equation with usual notations is given by:

⎛ dT ⎞ (a) ⎜ ⎟ = ⎝ dP ⎠sat Tv fg h fg

IES-30.

⎛ dP ⎞ (b) ⎜ ⎟ = ⎝ dT ⎠sat Tv fg h fg

⎛ dT ⎞ (c) ⎜ ⎟ = ⎝ dP ⎠ sat v fg

Th fg

[IES-2000]

Th fg ⎛ dP ⎞ (d ) ⎜ ⎟ = ⎝ dT ⎠ sat v fg

Clausius-Clapeyron equation gives the 'slope' of a curve in [IES-1999] (a) p–v diagram (b) p–h diagram (c) p–T diagram (d) T–S diagram

153

Thermodynamic Relations S K Mondal’s IES-31.

Chapter 6

The thermodynamic parameters are: [IES-1997] I. Temperature II. Specific Volume III. Pressure IV. Enthalpy V. Entropy The Clapeyron Equation of state provides relationship between: (a) I and II (b) II, III and V (c) III, IV and V (d) I, II, III and IV

Gibbs Phase Rule IES-32.

Number of components (C), phase (P) and degrees of freedom (F) are related by Gibbs-phase rule as: [IES-2001] (a) C – P – F = 2 (b) F – C – P = 2 (c) C + F – P = 2 (d) P + F – C = 2

IES-33.

As per Gibb's phase rule, if number of components is equal to 2 then the number of phases will be: [IES-2002] (a) ≤ 2 (b) ≤ 3 (c) ≤ 4 (d) ≤ 5

IES-34.

Gibb's phase rule is given by: [IES-1999] (F = number of degrees of freedom; C = number of components; P = number of phases) (a) F = C + P (b) F = C + P – 2 (c) F = C – P – 2 (d) F = C – P + 2

IES-35.

Gibb's free energy 'c' is defined as: (a) G = H – TS (b) G = U – TS

IES-36.

Which one of the following relationships defines the Helmholtz function F? [IES-2007] (a) F = H + TS (b) F = H – TS (c) F = U – TS (d) F = U +TS

IES-37.

Assertion (A): For a mixture of solid, liquid and vapour phases of a pure substance in equilibrium, the number of independent intrinsic properties needed is equal to one. [IES-2005] Reason(R): The Three phases can coexist only at one particular pressure. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-38.

Consider the following statements: [IES-2000] 1. Azeotropes are the mixtures of refrigerants and behave like pure substances. 2. Isomers refrigerants are compounds with the same chemical formula but have different molecular structures. 3. The formula n + p + q = 2m is used for unsaturated chlorofluorocarbon compounds (m, n, p and q are the numbers atoms of carbon, hydrogen, fluorine and chlorine respectively). Which of these statements are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 2 (d) 1, 2 and 3

154

(c) G = U + pV

[IES-1999] (d) G = H + TS

Thermodynamic Relations S K Mondal’s

Chapter 6

Previous 20-Years IAS Questions Maxwell's Equations IAS-1.

According to the Maxwell relation, which of the following is/are correct?

⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂P ⎠T ⎛ ∂P ⎞ ⎛ ∂s ⎞ (c) ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠v ⎝ ∂v ⎠T

⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎟ =⎜ ⎟ ⎝ ∂v ⎠T ⎝ ∂T ⎠v

(a)

(b) ⎜

[IAS-2007]

(d) All of the above

TdS Equations IAS-2.

Which one of the following expressions for T ds is true for a simple compressible substance? (Notations have the usual meaning) [IAS-1998] (a) dh – vdp (b) dh + vdp (c) dh – pdv (d) dh + pdv

Difference in Heat Capacities and Ratio of Heat Capacities IAS-3.

The specific heat Cp is given by:

⎛ ∂v ⎞ (a) T ⎜ ⎟ ⎝ ∂T ⎠ p IAS-4.

[IAS-2000]

⎛ ∂T ⎞ (b) T ⎜ ⎟ ⎝ ∂s ⎠ p

⎛ ∂s ⎞ (c) T ⎜ ⎟ ⎝ ∂T ⎠ p

(d)

⎛ ∂T ⎞ T⎜ ⎟ ⎝ ∂v ⎠ p

⎡ ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎤ −T⎜ ⎟ ⎟ ⎥ is always equal to: ⎝ ∂T ⎠ v ⎥⎦ ⎢⎣ ⎝ ∂T ⎠ p

For an ideal gas the expression ⎢T ⎜

(a ) zero

(b)

cp

(c ) R

cv

(d ) RT

[IAS-2003]

IAS-5.

Assertion (A): Specific heat at constant pressure for an ideal gas is always greater than the specific heat at constant volume. [IAS-2000] Reason (R): Heat added at constant volume is not utilized for doing any external work. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IAS-6.

Match List-I with List-II and select the correct answers using the codes given below the lists. [IAS-2002] List-I List-II A. Joule Thomson co-efficient

1.

B. Cp for monatomic gas

5 R 2

2. Cv 155

Thermodynamic Relations S K Mondal’s

Chapter 6

C. Cp – Cv for diatomic gas

3. R

⎛ ∂U ⎞ ⎟ ⎝ ∂T ⎠v

⎛ ∂T ⎞ 4. ⎜ ⎟ ⎝ ∂P ⎠ h

D. ⎜

Codes: (a) (c)

IAS-7.

A 3

B 2

C 4

D 1

(b)

A 4

B 1

C 3

D 2

3

1

4

2

(d)

4

2

3

1

Ratio of specific heats for an ideal gas is given by (symbols have the usual meanings) [IAS-1999] (a)

1 1−

R Cp

(b)

1 C 1− p R

(c)

1 C 1+ p R

(d)

1 1+

R Cp

Joule-Kelvin Effect or Joule-Thomson coefficient IAS-8.

Which one of the following properties remains unchanged for a real gas during Joule-Thomson process? [IAS-2000] (a) Temperature (b) Enthalpy (c) Entropy (d) Pressure

Clausius-Clapeyron Equation IAS-9.

If h, p, T and v refer to enthalpy, pressure, temperature and specific volume respectively and subscripts g and f refer to saturation conditions of vapour and liquid respectively then Clausius-Clapeyron equation applied to change of phase from liquid to vapour states is: [IAS-2003] (a)

dp (hg − h f ) = dt (vg − v f )

(b)

dp (hg − h f ) = dt T (vg − v f )

(c)

dp (hg − h f ) = dt T

(d)

dp (vg − v f )T = dt (hg − h f )

IAS-10.

Which one of the following is the correct statement? [IAS-2007] Clapeyron equation is used for: (a) Finding specific volume of vapour (b) Finding specific volume of liquid (c) Finding latent heat of vaporization (c) Finding sensible heat

IAS-11.

Assertion (A): Water will freeze at a higher temperature if the pressure is increased. [IAS-2003] Reason (R): Water expands on freezing which by Clapeyron's equation gives negative slope for the melting curve. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true 156

Thermodynamic Relations S K Mondal’s IAS-12.

Chapter 6

Match List-I with List-II and select the correct answer using the codes given below the lists [IAS-1994] List-I List-II A. Mechanical work 1. Clausius-Clapeyron equation B.

∫

dQ ≤0 T

C. Zeroth Law D. H–TS Codes: A (a) 1 (c) –

2. Gibb's equation

B 3 2

C 2 3

3. High grade energy 4. Concept of temperature A B C (b) 3 – 2 (d) 3 – 4

D 4 1

D 4 2

Gibbs Phase Rule IAS-13.

Which one of the following relationships defines Gibb's free energy G? [IAS-2007] (a) G = H + TS (b) G = H – TS (c) G = U + TS (d) G = U – TS

IAS-14.

The Gibbs free-energy function is a property comprising [IAS-1998] (a) Pressure, volume and temperature (b) Ethalpy, temperature and entropy (c) Temperature, pressure and enthalpy (d) Volume, ethalpy and entropy

157

Thermodynamic Relations S K Mondal’s

Chapter 6

Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. Ans. (d) GATE-2. Ans. (d) GATE-3. Ans. (a) GATE-4. Ans. (d) ⎛ ∂S ⎞ ⎛ T∂S ⎞ ⎛ dQ ⎞ GATE-5. Ans. (c) T ⎜ =⎜ =⎜ ⎟ ⎟ ⎟ = CP ⎝ ∂T ⎠P ⎝ ∂T ⎠P ⎝ ∂T ⎠P ⎛ ∂S ⎞ ⎛ T∂S ⎞ ⎛ dQ ⎞ T⎜ ⎟ = ⎜ ∂T ⎟ = ⎜ ∂T ⎟ = C V T ∂ ⎝ ⎠V ⎝ ⎠V ⎝ ⎠V ⎛ ∂S ⎞ ⎛ ∂S ⎞ ∴T⎜ − T⎜ ⎟ ⎟ = CP − CV = R ⎝ ∂T ⎠P ⎝ ∂T ⎠ V GATE-6. Ans. (c) Heat absorbed by water = Heat supplied by heater. m w c pw ( ΔT ) w = P × t or 40 × 4.2 × ( ΔT ) w = 2 × 20 × 60 or ( ΔT ) w = 14.3o C

GATE-7. Ans. (b) ⎛ ∂T ⎞ GATE-8. Ans. (a) μ = ⎜ ⎟ i,e. μ > o, ∂P is ( −ive ) so ∂T must be − ive. ⎝ ∂P ⎠h ⎛ ∂T ⎞ ⎛ ∂T ⎞ GATE-9. Ans. (b) Joule-Thomson co-efficient ⎜ ⎟ . Here, ∂p, − ive and ⎜ ∂P ⎟ , -ive so ∂T must be P ∂ ⎝ ⎠h ⎝ ⎠h +ive so gas will be warmer. GATE-10. Ans. (a) – 4, (b) –5, (c) –6, (d) – 1

Previous 20-Years IES Answers IES-1. Ans. (d) P is a function of v and both are connected by a line path on p and v coordinates. Thus

∫ pdv and ∫ vdp are not exact differentials and thus not properties.

If X and Y are two properties of a system, then dx and dy are exact differentials. If the

⎡ ∂M ⎤ ⎡ ∂N ⎤ =⎢ ⎥ ⎥ ⎣ ∂y ⎦ x ⎣ ∂x ⎦ y

differential is of the form Mdx + Ndy, then the test for exactness is ⎢ Now applying above test for

2 R ⎛ dT p.dv ⎞ ⎡ ∂ (1/ T ) ⎤ ⎡ ∂ ( p / v) ⎤ ⎡ ∂ ( RT / v ) ⎤ + , = = ∫ ⎜⎝ T v ⎟⎠ ⎢⎣ ∂v ⎥⎦T ⎢⎣ ∂T ⎥⎦ v ⎢⎣ ∂T ⎥⎦ or 0 = v 2 v

This differential is not exact and hence is not a point function and hence

⎛ dT

∫ ⎜⎝ T

+

p.dv ⎞ ⎟ is not a point function and hence not a property. v ⎠ 158

Thermodynamic Relations S K Mondal’s

Chapter 6

v.dp ⎞ ⎡ ∂ (1/ T ) ⎤ ⎡ ∂ ( −v / T ) ⎤ ⎡ ∂ (− R / P ) ⎤ =⎢ =⎢ ⎟⎢ ⎥ ⎥ ⎥⎦ or 0 = 0 T ⎠ ⎣ ∂p ⎦T ⎣ ∂T ⎦ P ⎣ ∂T P ⎛ dT v.dp ⎞ − Thus ∫ ⎜ ⎟ is exact and may be written as ds, where s is a point function and T ⎠ ⎝ T And for

⎛ dT

∫ ⎜⎝ T

−

hence a property IES-2. Ans. (a) From Maxwell relation Clapeyron equation comes.

IES-3. Ans. (b) 3 is false. It is derived from the Maxwell’s 3rd relationship

⎛ ∂p ⎞ ⎛ ∂s ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠v ⎝ ∂v ⎠T

IES-3a Ans. (c) IES-4. Ans. (a) IES-5. Ans. (c) The relations in (a) and (b) are applicable for a reversible processes and (c) Tds = du + pdV is a relation among properties which are independent of the path. IES-6. Ans. (b) IES-7. Ans. (c) IES-8. Ans. (a) IES-9. Ans. (c) Potential energy will converted to heat energy. gh 980 × 4000 = = 40K mgh = mc v ΔT or ΔT = cv 980 IES-10. Ans. (b) IES-11. Ans. (d) A diatomic gas (such as that of oxygen) has six degrees of freedom in all-three corresponding to translator motion, two corresponding to rotatory motion and one corresponding to vibratory motion. Experiments have shown that at ordinary temperatures, the vibratory motion does not occur. Hence, at 27°C, an oxygen molecule has just five degrees of freedom. IES-12. Ans. (d)

IES-13. Ans. (b) For ideal gas CP and CV are constant but mole is depends on the number of atoms in a molecule. IES-14. Ans. (c). A is correct but R is false. We know that C p = a+KT+K1T2+K2T3 C v = b+ KT+K1T2+K2T3 See Cp and C v both increase with temperature and by same amount. As Cp > C v then C percentage increase of Cp is less than C v. So p decreases with temperature. Cv IES-15. Ans. (c) Sign of T must be positive

⎛ ∂P ⎞ ⎜ ⎟ is always negative. ⎝ ∂v ⎠T

IES-16. Ans. (a) IES-17. Ans. (d) Actually Joule – Thomson coefficient will be positive. IES-18. Ans. (c) For ideal gas µ = 0 and for real gas µ may be positive (N2, O2, CO2 etc.) or negative (H2). 159

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IES-19. Ans. (d) When a real gas whiich is initia ally at a tem mperature lower than the maximum inversion tem mperature is throttled, tem its mperature decreases.

IES-20. Ans. (b b) IES-21. Ans. (b) ( Joule Th homson coeffiicient is the ratio of temperature cha ange to presssure change wheen a gas und dergoes adiab batic throttlin ng. IES-22. Ans. (a) The slope of the issenthalpic cu urve at any point is kn nown as Jou ule-Thomson coeffficient and is i expressed as, a

⎛ ∂T ⎞ ⎟ ⎝ ∂p ⎠ h

μ =⎜

IES-23. Ans. (a a) IES-24. Ans. (b b) IES-25. Ans. (c c) hfg ⎛ dP ⎞ IES-26. Ans. (c = c) ⎜ ⎟ ⎝ dT ⎠sat T ( Vg − Vf )

⎛ dP ⎞ 5 or hfg = T ( Vg − Vf ) × ⎜ ⎟ = 400 × ( 0251 − 0.001) × 0.1× 10 J / kg = 1000kJ / kg ⎝ dT ⎠ sat IES-27. Ans. (b b) b) IES-28. Ans. (b b) IES-29. Ans. (b c) IES-30. Ans. (c n state proviides relation nship betweeen temperatu ure, specific IES-31. Ans. (d) Clapeyrron equation volu ume, pressurre and enthallpy. d) IES-32. Ans. (d c) IES-33. Ans. (c d) F = C – P + 2 IES-34. Ans. (d ( Gibb's fre ee energy 'G' is defined ass G = H – TS.. IES-35. Ans. (a) c) IES-36. Ans. (c d) F = C – P + 2 IES-37. Ans. (d C = 1, P = 3 or F = 1 – 3 + 2 = 0 IES-38. Ans. (d) Isomers:: Compounds with the same s chemiccal formula but differen nt molecular stru ucture.

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Previous 20-Years IAS Answers IAS-1. Ans. (c)

⎛ ∂P ⎞ ⎛ ∂S ⎞ ⎜ ⎟ =⎜ ⎟ to memorize Maxwell’s relation remember T V P S, –ive and S S V ⎝ ∂T ⎠V ⎝ ∂V ⎠T

P see highlights. IAS-2. Ans. (a) dQ = dh – Vdp or Tds = dh – Vdp dQp ⎛ ∂s ⎞ IAS-3. Ans. (c) Cp = = T⎜ [∵ dQ = TdS] ⎟ ∂T ⎝ ∂T ⎠p IAS-4. Ans. (c) ⎛ ∂S ⎞ ⎛ T∂S ⎞ ⎛ dQ ⎞ T⎜ ⎟ =⎜ ⎟ =⎜ ⎟ = CP ⎝ ∂T ⎠P ⎝ ∂T ⎠P ⎝ ∂T ⎠P ⎛ ∂S ⎞ ⎛ T∂S ⎞ ⎛ dQ ⎞ T⎜ =⎜ =⎜ ⎟ ⎟ ⎟ = CV ⎝ ∂T ⎠ V ⎝ ∂T ⎠ V ⎝ ∂T ⎠ V

⎛ ∂S ⎞ ⎛ ∂S ⎞ ∴T⎜ ⎟ − T ⎜ ∂T ⎟ = CP − CV = R ⎝ ∂T ⎠P ⎝ ⎠V IAS-5. Ans. (a) Both A and R correct and R is the correct explanation of A IAS-6. Ans. (b) Cp – Cv for all ideal gas is R, So C-3, (a) & (c) out. A automatically match 4, and Cp =

5 5 γ R for monatomic gas γ = . So, γ = R . 3 2 γ −1

IAS-7. Ans. (a) Cp − Cv = R and γ =

Cp Cv

=

Cp Cp − R

=

1 R 1− Cp

IAS-8. Ans. (b) IAS-9. Ans. (b) IAS-10. Ans. (c) IAS-11. Ans. (a) IAS-12. Ans. (d) IAS-13. Ans. (b) IAS-14. Ans. (b)

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Pure e Subs stanc ces

Theo ory at a Glance (Fo or GAT TE, IES S & PS SUs) A pure substan nce is define ed as one th hat is homog geneous and invariable in i chemical composition thrroughout its mass. The relative r prop portions of th he chemical elements e con nstituting the substance aree also consta ant. Atmosph heric air, steeam – water mixture and d combustion n products of o a fuel are reg garded as pu ure substance es. But the mixture m of airr and liquid air a is not a pure substancce, since the rellative proportions of oxyg gen and nitrogen differ in the gas and liquid phases in equilibriium. Miixtures are not pure substtances. (e.g., Humid air) Ex xception!! Air A is treated d as a pure substance though t it is a mixture of o gases. In a majority of o cases a min nimum of tw wo properties are required d to define th he state of a system. s The besst choice is an extensive property p and d an intensivee property. Th he state of a pure substa ance of given n mass can be b fixed by sp pecifying twoo independen nt intensive prooperties, provided the sy ystem is in equilibrium. e T This is know wn as the 'tw wo-property y rule'. The sta ate can thuss be represented as a point p on therrmodynamic property diiagrams. On nce any two prooperties of a pure substa ance are knoown, other properties p can n be determ mined from th he available theermodynamicc relations.

p--v Diag gram for a Pure Subs stance

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Trriple po oint Triiple point is a point wherre all the threee phases exiist in equilibrrium. u = 0 for water at triple poin nt. (You can assign any number n you liike instead of o 0). Sin nce, p and v for f water at triple t point are a known yoou can calcula ate h for water at triple point p (it will nott be zero). The entropy of sa aturated waterr is also chosen n to be zero at the t triple point.

Gibbs Pha ase Rule e at Triplle Point If C = 1, φ = 3, then t f = 0. Th he state is th hus unique foor a substancce; and referss to the triplee point wh here all the th hree phases exist e in equillibrium. •

ln p = A + B/T valid d only near th he triple poin nt.(Called An ntoine’s equ uation)

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p--T Diag gram fo or a Purre Subs stance

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p--v-T Su urface

TT-s Diagram for a Pure e Substance

C Critical Point P A state at whiich a phase change begin ns or ends is called a sa aturation state. s The doome shaped reg gion composeed of the two-phase liquid d–vapor statees is called th he vapor do ome. The linees bordering thee vapor domee are called saturated s liq quid and satu urated vapor lines. At thee top of the dome, d where thee saturated liquid l and sa aturated vapoor lines meett, is the critical point. The critical temperature Tc of a pure su ubstance is th he maximum temperaturee at which liq quid and vap por phases ca an coexist in equ uilibrium. Th he pressure at a the critica al point is callled the critical pressure, pc. The specific volume at this state is the critical specific sp volum me.

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Th he isotherm passing p through the crittical point is called thee critical isotherm, and the n as the corrresponding temperaturre is known criitical tempeerature (tc). The presssure and vollume at the critical poin nt are know wn as the criitical pressurre (Pc) and th he critical voolume (vc) resspectively. Foor water

Pc = 221.2 bar tc = 374 4.15°C vc = 0.0 00317 m3/kg

Fig. F Criticall Point on P–v, P T–S, h–ss diagram

Ch haracter ristics off the crittical poiint: 1. 2. 3. 4.

It is the highest h temp perature at which w the liqu uid and vapou ur phases can n coexist. At the criitical point hfg, ufg and vfg are zero. Liquid va apour menisccus will disap ppear. Specific heat h at consta ant pressure is infinite. • A ma ajority of eng gineering app plications (e.g g., steam bassed power geeneration; Reefrigeration, gas liiquefaction) involve i therm modynamic processes closse to saturatiion. • For saturated pha ase often it en nthalpy is an n important property. p alpy-pressurre charts are used for refrrigeration cyccle analysis. • Entha • Entha alpy-entropy y charts for water w are used for steam cycle c analysiss. • Note:: Unlike presssure, volume and tempeerature which h have speciffied numbers associated with it, in the casse of interna al energy, en nthalpy (and entropy) onlly changes arre required. Conseequently, a base b (or datum m) is defined d - as you hav ve seen in thee case of water. • For example e for NIST N steam tables u = 0 for water at a triple poin nt. (You can n assign any numb ber you like instead i of 0). [Don’t be su urprised if tw wo different sets s of steam m tables give differrent values fo or internal en nergy and en nthalpy].

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Sincee, p and v forr water at trriple point arre known you u can calcula ate h for watter at triple point (it will not be b zero).

h--s Diagrram or Mollier M Diagram m for a Pure Su ubstanc ce Froom the first and a second la aws of therm modynamics, the t following g property rellation was ob btained.

Tdss = dh - vdp

or

⎛ ∂h ⎞ ⎜ ∂s ⎟ = T ⎝ ⎠p

Th his equation forms the basis b of the h-s diagram m of a pure substance, also a called the t Mollier dia agram. Thee slope of an a isobar on n the h-s cooordinates iss equal to the t absolutee saturation tem mperature (tsat re. If the tem mperature rem mains consta ant the slope will remain s + 273) at that pressur con nstant. If thee temperature increases, the t slope of the t isobar willl increase.

Q Quality or o Dryn ness Fraction • • • • • • • • •

The zon ne between th he saturated d liquid and the saturated d vapour region is called the t two phase region r - wherre the liquid and a vapour can c co-exist in n equilibrium m. Drynesss fraction: It is the mass fraction of va apour in the mixture. Normallly designate ed by ‘x’. On the saturated liquid l line x = 0 On the saturated vapour v line x = 1 x can have a value only o between n 0 and 1 Data ta ables will listt properties at a the two ends of saturattion. To calcu ulate propertties in the tw wo-phase regiion: p, T willl be the sam me as for saturated liquid or saturated vapour.

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Chapter 7

v = x v + (1 − x ) v g

f

h = x h + (1 − x ) h

f

u = x u + (1 − x ) u

f

g

g

AND

v = vf + xvfg u = uf + xufg h = hf + xhfg s = sf + xsfg Saturation States When a liquid and its vapour are in equilibrium at a certain pressure and temperature, only the pressure or the temperature is sufficient to identify the saturation state. If the pressure is given, the temperature of the mixture gets fixed, which is known as the saturation temperature, or if the temperature is given, the saturation pressure gets fixed. Saturated liquid or the saturated vapour has only one independent variable, i.e, only one property is required to be known to fix up the state. a Steam Table(saturated state) give the properties of sutured liquid and saturated vapour. In steam table the independent variable is temperature. At a particular temperature, the values of saturation pressure p, and states, Vg ,h g and S g refer to the saturated vapour state; and v fg ,h fg , and s fg refer to the changes in the property values during evaporation (or condensation) at the temperature.  where v fg = v g − v f and s fg = s g − s f . If steam Table the independent variable is pressure. At a particular pressure, the values of saturation temperature t, and  v f , v g , h f , h fg , s f , and s g are given. whether the pressure or the temperature is given, either steam Table can be conveniently used for computing the properties of saturation states. 169

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If data a are required for interrmediate tem mperature orr pressures, linear interrpolation is norrmally accurrate. The reasson for the tw wo tables is to t reduce the amount of in nterpolation required.

Liq quid vapou ur Mixtures:: Let us considerr a mixture of o saturated liquid waterr and water vapour in eq quilibrium att pressure p and temperatu ure t. The com mposition of the mixture by mass willl be within the t vapour dome d figure. Th he properties of the mixtu ure are as giv ven in Article. i.e

Fig. wh here v , v , h , s and d s fg are the saturation properties at the t given preessure and teemperature. f

fg

fg

f

If p or t and th he quality of the t mixture are given, th he properties of the mixtu ure (v, u, h an nd s) can be eva aluated from m the above equations. e Soometimes instead of qua ality, one of the t above eq quations say speecific volumee v and presssure or temp perature are given. In tha at case, the quality of th he mixture x has to be calcu ulated from the given v and p or t and then x being know wn, other prooperties are aluated. eva If p or t and th he quality of the t mixture are given, th he properties of the mixtu ure (v, u, h an nd s) can be eva aluated from m the above equations. Soometimes, insstead of quallity, one of th he above prop perties, say, speecific volumee v and presssure or temp perature are given in tha at case, the quality q of th he mixture x has to be calcu ulated from the given v and p or t and then x being know wn, other prooperties are aluated. eva

Su uperheated Vapour: Wh hen the temp perature of the t vapour is greater tha an the saturration tempeerature corresponding to thee given presssure, the vap pour is said to t be superheated (state 1 in figure). The differen nce between thee temperaturre of the superheated va apour and th he saturatioon temperatu ure at that pressure is callled the supeerheat or the degree of su uperheat. As shown in fig gure below, the t differencee (t1 - tsat) is thee superheat.

(Fig. - Superheat an nd Sub-coolling.) 170 0

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Chapter 7

In a superheated vapour at a given pressure, the temperature may have different values greater than the saturation temperature. Steam Table gives the values of the properties (volume, enthalpy, and entropy) of superheated vapour for each tabulated pair of values of pressure and temperature, both of which are now independent. Interpolation or extrapolation is to be used for pairs of values of pressure and temperature not given.

Compressed Liquid: When the temperature of a liquid is less than the saturation temperature at the given pressure, the liquid is called compressed liquid (state 2 in figure above). The pressure and temperature of compressed liquid may vary independently, and a table of properties like the superheated vapour table could be arranged, to give the properties at any p and t. However, the properties of liquids vary little with pressure. Hence the properties are taken from the saturation tables at the temperature of the compressed liquid. When a liquid is cooled below its saturation temperature at a certain pressure it is said to be sub-cool. The difference in saturation temperature and the actual liquid temperature is known as the degree of Sub-cooling, or simply, Sub-cooling.

Charts of Thermodynamic Properties • •

One of the important properties is the change in enthalpy of phase transition hfg also called the latent heat of vaporisation or latent heat of boiling. It is equal to hg - hf. Similarly ufg - internal energy change due to evaporation and v fg - volume change due to

• •

evaporation can be defined (but used seldom). The saturation phase depicts some very interesting properties: The following saturation properties depict a maximum:

• •

4. Tc ( pc − p)

1. T ρ f

2. T ( ρ f − ρ g)

3. T h fg

5. p ( Tc − T )

6. p ( v g − v f )

7. T ( ρ c − ρ f ρ g ) 2

8. hg

The equation relating the pressure and temperature along the saturation is called the vapour pressure curve. Saturated liquid phase can exist only between the triple point and the critical point.

Measurement of Steam Quality Throttling calorimeter: Throttling calorimeter is a device for determining the quality of a twophase liquid–vapor mixture.

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Inttermediate states are non n-equilibrium m states not describable by thermody ynamic coord dinates. The iniitial state (weet) is given by b p1 and x1 , and a the finall state by p2 and a t2 (superrheated). Now w

since

h1 = h 2 H fp1 + x 1 h fg p1 = h 2

or

x1 =

h 2 − h fp 1 h fggp1

Wiith p2 and t2 being known n, h2 can be found out froom the superrheated steam m table. The values of hf and hfg are ta aken from th he saturated d steam tablle correspond ding to presssure p1 . Therefore, the w steam x1 can be calcu ulated. quality of the wet

T The Thrott T tling g Calo orime eter Example e A supplly line carrie es a two-phasse liquid–vap por mixture of o steam at 20 2 bars. A sm mall fraction of the flow in the line is diverrted through h a throttling g calorimeter and exhau usted to the phere at 1 bar. b The teemperature of o the exha aust steam is i measured d as 120ºC. atmosp Determ mine the quality of the steeam in the su upply line.

Solution n Kn nown: Steam m is diverted from a supply line throu ugh a throttliing calorimetter and exhau usted to the atm mosphere. Fin nd: Determin ne the qualitty of the steam in the sup pply line.

Sc chematic c and Giiven Datta:

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Asssumptio ons: 1. 2.

The contrrol volume sh hown on the accompanyin a ng figure is at steady statte. The diverrted steam un ndergoes a th hrottling process.

An nalysis: For a throttling process, the energy and mass m balances reduce to give h2 = h1, Thus, with sta ate 2 fixed, the t specific enthalpy e in the t supply liine is known n, and state 1 is fixed by y the known vallues of p1 an nd h1. As shown on the t accompan nying p–v diiagram, statee 1 is in thee two-phase liquid–vapor l r region and sta ate 2 is in thee superheated vapor regioon. Thus

h2 = h1 = hf1 + x1 ( hg1 _ hf1 ) Solving for x1 x1 =

h2 − hf 1 hg1 − hf 1

Froom steam ta able at 20 ba ars, hf1 = 908 8.79 kJ/kg an nd hg1 = 279 99.5 kJ/kg. At A 1 bar and 120ºC, h2 = 276 66.6 kJ/kg frrom steam table into the above a expresssion, the qua ality in the line is x1 = 0.9 956 (95.6%). For throttling calorimeterss exhausting to the atmoosphere, the quality in th he line mustt be greater tha an about 94% % to ensure th hat the steam m leaving thee calorimeterr is superhea ated.

Throttlin ng Wh hen a fluid flows f throug gh a constriccted passagee, like a parttially opened d valve, an orifice, o or a porrous plug, th here is an ap ppreciable drrop in pressu ure, and the flow is said d to be throtttled. Figure shoows the process of throttlling by a parrtially opened d valve on a fluid f flowing g in an insula ated pipe. In thee steady-flow w energy equa ation.

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dW Wx =0 dm m

dQ d = 0, 0 dm d

e very small and a ignored. Thus, the S.F.E.E. reducces to and the changees in P.E. are

C12 C22 h1 + = h2 + 2 2

Oft ften the pipe velocities in throttling arre so low thatt the K.E. terrms are also negligible. So

h1 = h2 or the enthalpy y of the fluid before throtttling is equall to the entha alpy of the flu uid after throottling.

Coonsider a throttling proc cess (also refferred to as wire w drawing g process) 1

here is no worrk done (risin ng a weight) Th If tthere is no heeat transfer Coonservation of mass requires that Sin nce 1 and 2 are a at the sam me level

2

W=0 Q=0 C1 = C2 Z1 = Z2 174 4

Pure Substances S K Mondal’s From SFEE it follows that

Chapter 7 h1 = h2

Conclusion: Throttling is a constant enthalpy process (isenthalpic process)

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Chapter 7

Properties of Pure Substances Highlights 1.

Triple Point On P – T diagram it is a Point. On P – V diagram it is a Line. Also in T – S diagram it is a Line. On U – V diagram it is a Triangle.

2.

Triple point of water T = 273.16 K P = 0.00612 bar = 0.01ºC

3.

= 4.587 mm of Hg

Entropy (s) = 0 Internal energy (u) =0 Enthalpy = u + PV = slightly positive.

Triple point of CO2 P 5 atm and T = 216.55 K = − 56.45°C that so why sublimation occurred.

4.

Critical point For water pc = 221.2 bar ≈ 225.5 kgf/cm2 tc = 374.15ºC ≈ 647.15 K vc = 0.00317 m3/kg At critical point

hfg = 0 ;        v fg = 0 ;        sfg = 0 5.

Mollier Diagram

Basis of the h – s diagram is

6.

⎛ ∂h ⎞ ⎜ ∂s ⎟ = T ⎝ ⎠P

⎡∵ Tds = dh − vdp⎤ ⎢ ⎥ ⎢∴ ⎛ ∂h ⎞ = T ⎥ ⎢⎣ ⎜⎝ ∂s ⎟⎠ P ⎥⎦

∴ The slope of an isobar on the h–s co-ordinates is equal to the absolute saturation temperature at that pressure. And for that isobars on Mollier diagram diverges from one another. Dryness fraction

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Chapter 7

m x= m +m v

v

7.

v = (1 – x) vf + xvg u = (1 – x) uf + xug h = (1 – x) hf + xhg s = (1 – x) sf + xsg

and

L

v = vf + xvfg u = uf + xufg h = hf + xhfg s = sf + xsfg

8.

Super heated vapour: When the temperature of the vapour is greater than the saturation temperature corresponding to the given pressure.

9.

Compressed liquid: When the temperature of the liquid is less than the Saturation temperature at the given pressure, the liquid is called compressed liquid.

10.

In combined calorimeter x = x1 × x 2 x1 = from throttle calorimeter. x 2 = from separation calorimeter.

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ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions Common data for Question Q1 – Q3

In the figure shown, the system is a pure substance kept in a piston-cylinder arrangement. The system is initially a two-phase mixture containing 1 kg of liquid and 0.03 kg of vapour at a pressure of 100 kPa. Initially, the piston rests on a set of stops, as shown in the figure. A pressure of 200 kPa is required to exactly balance the weight of the piston and the outside atmospheric pressure. Heat transfer takes place into the system until its volume increases by 50%. Heat transfer to the system occurs in such a manner that the piston, when allowed to move, does so in a very slow (quasi-static I quasi-equilibrium) process. The thermal reservoir from which heat is transferred to the system has a temperature of 400°C. Average temperature of the system boundary can be taken as 17°C. The heat transfer to the system is I kJ, during which its entropy increases by 10 J/K. Atmospheric pressure. Specific volumes of liquid (vf) and vapour (vg) phases, as well as values of saturation temperatures, are given in the table below.

Pressure (kPa)

Saturation temperature, Tsat (°C)

vf(m3/kg)

vg(m3/kg)

100

100

0.001

0.1

200

200

0.0015

0.002

GATE-1.

At the end of the process, which one of the following situations will be true? (a) Superheated vapour will be left in the system [GATE-2008] (b) No vapour will be left in the system (c) A liquid + vapour mixture will be left in the system (d) The mixture will exist at a dry saturated vapour state

GATE-2.

The work done by the system during the process is: (a) 0.1 kJ (b) 0.2 kJ (c) 0.3 kJ

GATE-3.

The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to: [GATE-2008] (a) 7.5 J/K (b) 7.7 J/K (c) 8.5 J/K (d) 10 J/K 178

[GATE-2008] (d) 0.4kJ

Pu ure Su ubstan nces S K Mondal’s

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TT-s Diagram for a Pure e Substance Co ommon Data for Qu uestions GATE-4 G - GA ATE-5

A tthermodyna amic cycle with w an ideal gas as wo orking fluid d is shown below. b

[GA ATE-2007]

GA ATE-4.

The e above cyc cle is repressented on T--S plane by

GA ATE-5.

If the t specific heats of th he working fluid are co onstant and d the value of specific hea at ratio γ is 1.4, the the ermal efficie ency (%) of the cycle iss: [G GATE-2007] (a) 21 (b) 40.9 (c) 42.6 (d) 59.7

GA ATE-6. The slopes of constant c vo olume and constant c pr ressure line es in the T--s diagram are e….. and….. respectivelly. [G GATE-1994] 179 9

Pure Substances S K Mondal’s

Chapter 7

h-s Diagram or Mollier Diagram for a Pure Substance GATE-7. Constant pressure lines in the superheated region of the Mollier diagram have what type of slope? [GATE-1995] (a) A positive slope (b) A negative slope (c) Zero slope (d) May have either positive or negative slopes

Quality or Dryness Fraction GATE-8. Consider a Rankine cycle with superheat. If the maximum pressure in tile cycle is increased without changing the maximum temperature and the minimum pressure, the dryness fraction of steam after the isentropic expansion will increase. [GATE-1995]

Throttling Statement for Linked Answer Questions Q9 & Q10:

The following table of properties was printed out for saturated liquid and saturated vapour of ammonia. The titles for only the first two columns are available. All that we know is that the other columns (columns 3 to 8) contain data on specific properties, namely, internal energy (kJ/kg), enthalpy (kJ/kg) and entropy (kJ/kgK) [GATE-2005]

GATE-9.

The specific enthalpy data are in columns (a) 3 and 7 (b) 3 and 8 (c) 5 and 7

[GATE-2005] (d) 5 and 8

GATE-10. When saturated liquid at 40°C is throttled to -20°C, the quality at exit will be [GATE-2005] (a) 0.189 (b) 0.212 (c) 0.231 (d) 0.788 GATE-11. When wet steam flows through a throttle valve and remains wet at exit (a) its temperature and quality increases [GATE-1996] (b) its temperature decreases but quality increases (c) its temperature increases but quality decreases (d) its temperature and quality decreases GATE-12. When an ideal gas with constant specific heats is throttled adiabatically, with negligible changes in kinetic and potential energies [GATE-2000]

( a ) Δh = 0, ΔT = 0

( b ) Δh > 0, ΔT = 0 ( c ) Δ h > 0, ΔS > 0

( d ) Δ h = 0, Δ S > 0

Where h, T and S represent respectively, enthalpy, temperature and entropy, temperature and entropy

GATE-13. One kilomole of an ideal gas is throttled from an initial pressure of 0.5 MPa to 0.1 MPa. The initial temperature is 300 K. The entropy change of the universe is: [GATE-1995] 180

Pu ure Su ubstan nces S K Mondal’s (a) 13.38 kJ/K

Cha apter 7 (b)401.3 kJ/K K

(c)) 0.0446 kJ/K K

(d) -0.044 46 kJ/K

Previo ous 20 0-Years s IES Questi Q ions IES-1.

Asssertion (A): Water is no ot a pure su ubstance. [IES-1999] Rea ason (R): The term pure subsstance dessignates a substance which is hom mogeneous and has the e same chem mical composition in all a phases. (a) Both A and R are individ dually true and a R is the correct c explan nation of A dually true but b R is NOT T the correct explanation e o A of (b) Both A and R are individ (c) A is true bu ut R is false (d) A is false bu ut R is true

IES-2.

The e given diag gram showss an isomettric coo oling proc cess 1–2 of a pu ure sub bstance. Th he ordinate and abscisssa are e respective ely (a) Pressure an nd volume (b) Enthalpy an nd entropy (c) Temperaturre and entrop py (d) Pressure an nd enthalpy [IES-1998]

IES-3.

The e ordinate and absc cissa in the giv ven figure showing s th he saturated d liqu uid and vap pour region ns of a pur re sub bstance represent: (a) Temperaturre and pressu ure (b) Enthalpy an nd entropy (c) Pressure an nd volume (d) Pressure an nd enthalpy [IES-1997]

IES-4.

IES-5.

The e given dia agram show ws the thro ottling pro ocess of a pu ure substan nce. abscissa The e ordina ate and are resspectively (a) Pressure an nd volume (b) Enthalpy an nd entropy (c) Temperaturre and entrop py (d) Pressure an nd enthalpy

[IES-1995]

Wh hich one of the following systems can be con nsidered to be contain ning a pure sub bstance? [IES-1993]

181 1

Pu ure Su ubstan nces S K Mondal’s

Cha apter 7

IES-6.

[IES-2009] Con nsider the following: f 3. Steam 1. Air A 2. Gaseo ous combusstion produc cts Wh hich of these e are pure substances, s assuming there t is no phase p chang ge? (a) 1 and 2 only (b) 1 and d 3 only (c) 2 and 3 only (d) 1, 2 and 3

IES-7.

Asssertion (A): At a given n temperatu ure, the enthalpy of super-heate s d steam is the e same as th hat of satura ated steam. [IES-1998] Rea ason (R): The T enthallpy of vapour at low wer pressur res is depe endent on tem mperature alone. a (a) Both A and R are individ dually true and a R is the correct c explan nation of A (b) Both A and R are individ dually true but b R is NOT T the correct explanation e o A of (c) A is true bu ut R is false (d) A is false bu ut R is true

IES-8.

Con nsiders the following properties p o vapour: of [IES-2009] 1. Pressure P 2. Tempe erature 3. Dryness D fraction 4. Specifiic volume Wh hich of these two prope erties alone e are not su ufficient to specify the e condition of a vapour? (a) 1 and 2 (b) 1 and 3 (c)) 2 and 3 (d) 3 and 4

IES-9.

Wh hich one of the t followin ng is correct? [IES-2008] The e specific vo olume of wa ater when heated h from m 0°C (a) First increasses and then decreases (b) First decreases d and d then increa ases (c) Increases I ste eadily (d) Decrea ases steadily

p– –v Diag gram fo or a Purre Subs stance IES-10.

Wh hich p–v diiagram for steam illu ustrates co orrectly the e isothermal process [IES 1995, 2007] und dergone by wet steam till it becom mes superhe eated?

182 2

Pu ure Su ubstan nces S K Mondal’s

Cha apter 7

p– –T Diag gram fo or a Purre Subs stance IES-11.

Con nsider the e phase diagram off a cer rtain substtance as shown in the t giv ven figure. Match List-I (Proce ess) witth List-II (Curves/line es) and select the e correct answer a usin ng the cod des giv ven below th he lists: List--II (Cur rves/lines) 1. EF F 2. EG G 3. ED D

Lisst-I (Pr rocess) A. Vaporization n B. Fusion n C. Sublimation Cod des: (a) (c)

A 1 3

B 3 2

C 2 1

(b) (d)

183 3

A 1 3

B 2 1

C 3 2

[ [IES-2001]

Pure Substances S K Mondal’s

Chapter 7

p-v-T Surface IES-12.

The p-v-T surface of a pure substance is shown in the given p figure. The two-phase regions are labelled as: (a) R, T and X (b) S, U and W (c) S, W and V (d) R, T and V

[IES-1999]

T-s Diagram for a Pure Substance IES-13.

The conversion of water from 40°C to steam at 200°C pressure of 1 bar is best [IES-1994] represented as

184

Pu ure Su ubstan nces S K Mondal’s IES-14.

Cha apter 7

The e followin ng figure shows th he T-s dia agram for steam. s With h respect to t this figu ure, match List I with List II and select the e correct an nswer using g the codess given bellow the Listts: List-II Lisst-I A. Curve C I 1. Satu urated liquid line B. Curve C II 2. Satu urated vapourr line C. Curve C III 3. Consstant pressurre line D. Curve C IV 4. Consstant volume line Cod des: A B C D (a) 2 1 4 3 (b) (c) 1 2 3 4 (d)

[IES-1994] A 2 1

B 1 2

C 3 4

D 4 3

C Critical Point P IES-15.

Wh hich one of the t followin ng is correct? At critical point the enthalpy of vapo orization iss (a) dependent on n temperaturre only (b)) maximum m (d)) zero (c) minimum

[ [IES-2008]

IES-16.

Con nsider the following f sttatements about a critica al point of water: w 1. The T latent heat h is zero.. 2. The T liquid is denser tha an its vapou ur. 3. Steam S gener rators can operate o abo ove this poin nt. Of these statem ments (a) 1, 2 and 3 arre correct (b) 1 and a 2 are corrrect c (d) 1 and a 3 are corrrect (c) 2 and 3 are correct

[ [IES-1993]

IES-17.

Wh hich one of the t followin ng statemen nts is correc ct when satturation pre essure of a vap pour increa ases? [IES 2007] (a) Saturation temperature decreases d (b) Enthalpy of evaporation decreases E of evaporation e i increases (c) Enthalpy (d) Specific volu ume change of o phase increeases

IES-18.

Ma atch List I with w List III and selec ct the corre ect answer using the code c given bellow the Listts: [IES-2005] List-I List-III A. Critical poin nt 1. All thee three pha ases - solid, liquid and vapourr co-exists in equilibrium B. Sublimation n 2. Phase change c form solid to liquiid C. Triple pointt 3. Properties of saturated liquid and saturatted vapour are identical D. Melting 4. Heatin ng process where w solid gets g directly transfoormed to gaseeous phase des: A B C D A B C D Cod (a) 2 1 4 3 (b) 3 4 1 2 4 1 3 (d) 3 1 (c) 2 4 2 185 5

Pure Substances S K Mondal’s

Chapter 7

IES-19.

With increase of pressure, the latent heat of steam [IES-2002] (a) Remains same (b) Increases (c) Decreases (d) Behaves unpredictably

IES-20.

List-I gives some processes of steam whereas List-II gives the effects due to the processes. Match List I with List II, and select the correct answer using the codes given below the lists: [IES-1995] List-I List-II A. As saturation pressure increases 1. Entropy increases. B. As saturation temperature increases 2. Specific volume increases. C. As saturation pressure decreases 3. Enthalpy of evaporation decreases. D. As dryness fraction increases 4. Saturation temperature increases. Code: A B C D A B C D (a) 1 3 2 4 (b) 4 3 2 1 (c) 4 3 1 2 (d) 2 4 3 1

h-s Diagram or Mollier Diagram for a Pure Substance IES-21.

Which one of the following represents the condensation of a mixture of saturated liquid and saturated vapour on the enthalpy-entropy diagram? [IES-2004] (a) A horizontal line (b) An inclined line of constant slope (c) A vertical line (d) A curved line

Measurement of Steam Quality IES-22.

Saturated liquid at a high pressure P1 having enthalpy of saturated liquid 1000 kJ/kg is throttled to a lower pressure P2. At pressure p2 enthalpy of saturated liquid and that of the saturated vapour are 800 and 2800 kJ/kg respectively. The dryness fraction of vapour after throttling process is: [IES-2003] (a) 0.1 (b) 0.5 (c) 18/28 (d) 0.8

IES-23.

Consider the following statements regarding the throttling process of wet steam: [IES-2002] 1. The steam pressure and temperature decrease but enthalpy remains constant. 2. The steam pressure decreases, the temperature increases but enthalpy remains constant. 3. The entropy, specific volume, and dryness fraction increase. 4. The entropy increases but the volume and dryness fraction decrease. Which of the above statements are correct? (a) 1 and 4 (b) 2 and 3 (c) 1 and 3 (d) 2 and 4

IES-24.

Match List-I (Apparatus) with List-II (Thermodynamic process) and select the correct answer using the code given below the Lists: [IES-2006] List-I List-II 1. Adiabatic process A. Separating calorimeter B. Throttling calorimeter 2. Isobaric process C. Sling psychrometer 3. Isochoric process 186

Pu ure Su ubstan nces S K Mondal’s D. Gas thermometer Cod des: A B (a) 1 3 4 (c) 1 IES-25.

Cha apter 7 C 2 2

4. Isentha alpic process A B C (b) 2 4 1 (d) 2 3 1

D 4 3

Sellect the corr rect answer r using the codes given n below the Lists: List-I List-III A. Bomb caloriimeter 1. Pressure B. Exhaust gass calorimeterr 2. Enthallpy C. Junker gas calorimeter 3. Volumee D. Throttling calorimeter c 4. Specific heats Cod de: A B C D A B C (a) 3 4 1 2 (b) 2 4 1 (c) 3 2 1 4 2 (d) 4 3

D 3 4 [IES-1998]

D 3 1

Throttlin ng IES-26.

In a throttlin ng processs, which on ne of the following parameter rs remains con nstant? [IES-2009] (a) Temperature e (b) Pressure (c)) Enthalpy (d) Entropy E

IES-27.

Con nsider the following f sttatements: [IES-2000] Wh hen dry satturated ste eam is thro ottled from m a higher pressure to t a lower pre essure, the 1. Pressure decreases d an nd the volum me increase es 2. Temperatu ure decreases and the steam s becom mes superh heated 3. Temperatu ure and the dryness fra action incre ease 4. Entropy in ncreases witthout any change in en nthalpy Wh hich of these e statementts are correct? (a) 1and 4 (b) 1, 2 and a 4 (c) 1 and 3 (d) 2 and 4

IES-28.

The e process 1--2 for steam m shown in the t giv ven figure is s (a) Isobaric (b) Isentrop pic (c) Issenthalpic (d) Isotherm mal

[IIES-2000] IES-29.

A fluid fl flowing g along a piipe line und dergoes a th hrottling process from 10 bar to 1 Bar r in passing through a partially open valve e. Before th hrottling, th he specific vollume of the fluid is 0.5 m3 /kg and after throtttling is 2.0 m3 /kg. Whatt is the Change in spe ecific intern nal energy during d the throttling t process? [IES 2007] (a) Zero (b) 100 0 kJ/kg (c) 200 kJ//kg (d) 300 kJ/k kg

187 7

Pure Substances S K Mondal’s IES-30.

Chapter 7

The throttling process undergone by a gas across an orifice is shown by its states in the following figure:

[IES-1996] IES-31.

In the figure shown, throttling process is represented by (a) a e (b) a d (c) a c (d) a b

[IES-1992] IES-32.

Match List-l with List-Il and select the correct answer using the code given below the lists: [IES-2009] List-lI List-l 1. Energy is always constant A. Isolated system B. Nozzle 2. Increase in velocity at the expense of its C. Throttling device pressure drop D. Centrifugal compressor 3. Appreciable drop in pressure without any change in energy 4. Enthalpy of the fluid increases by the amount of work input Codes: (a) (c)

A 4 4

B 3 2

C 2 3

D 1 1

(b) (d)

188

A 1 1

B 3 2

C 2 3

D 4 4

Pu ure Su ubstan nces S K Mondal’s

Cha apter 7

Previo ous 20 0-Years s IAS Quest Q ions IA AS-1.

Wh hich one of the following systems can be con nsidered to be contain ning a pure sub bstance? [IAS 1998]

IA AS-2.

Asssertion (A): On the enthalpy-en e ntropy diag gram of a pure subsstance the con nstant dryn ness fraction n lines startt from the critical c poin nt. [IAS-2001] Rea ason (R): Alll the three phases co-e exist at the critical point. (a) Both A and R are individ dually true and a R is the correct c explan nation of A dually true but b R is NOT T the correct explanation e o A of (b) Both A and R are individ (c) A is true bu ut R is false (d) A is false bu ut R is true

IA AS-3.

Asssertion (A): Air, a mixtu ure of O2 an nd N2, is a pure substan nce. [IAS-2000] Rea ason(R): Air is homo ogeneous in composition and uniform u in n chemical agg gregation. (a) Both A and R are individ dually true and a R is the correct c explan nation of A dually true but b R is NOT T the correct explanation e o A of (b) Both A and R are individ (c) A is true bu ut R is false (d) A is false bu ut R is true

IA AS-4.

If a pure substance conta ained in a rigid r vessel passes thro ough the critical state on heating, its s initial statte should be e: [IAS-1998] (a) Subcooled water (b) Saturated d water (c)) Wet steam (d) Saturateed steam

IA AS-5.

Asssertion (A): Air is a pu ure substan nce but a mixture m of air a and liqu uid air in a cylinder is nott a pure sub bstance. [IAS-1996] Rea ason (R): Aiir is homog geneous in composition c n but a mix xture of air and liquid air is heteroge eneous. (a) Both A and R are individ dually true and a R is the correct c explan nation of A (b) Both A and R are individ dually true but b R is NOT T the correct explanation e o A of (c) A is true bu ut R is false (d) A is false bu ut R is true

189 9

Pu ure Su ubstan nces S K Mondal’s IA AS-6.

Cha apter 7

Asssertion (A): Temperatu ure and pressure are sufficient to o fix the statte of a two pha ase system. [IAS-1995] Rea ason(R): Tw wo indepen ndent and intensive properties are requiired to be kno own to defin ne the state e of a pure substance. s (a) Both A and R are individ dually true and a R is the correct c explan nation of A (b) Both A and R are individ dually true but b R is NOT T the correct explanation e o A of (c) A is true bu ut R is false (d) A is false bu ut R is true

p– –v Diag gram fo or a Purre Subs stance IA AS-7.

Tw wo-phase regions r in n the given g pre essure-volum me diagra am of a pure sub bstance are represente ed by (a) A, A E and F (b) B, C and D (c) B, B D and F (d) A, C and E [IAS-1999]

IA AS-8.

A cyclic c proce ess ABC is shown s on a V–T dia agram in fig gure. The e same proc cess on a P––V diagram m will be represent as: a

[ [IAS-1996]

190 0

Pure Substances S K Mondal’s IAS-9.

Chapter 7

The network done for the closed shown in the given pressure-volume diagram, is (a) 600kN-m (b) 700kN-m (c) 900kN-m (d) 1000kN-m

[IAS-1995]

Triple point IAS-10.

Triple point temperature of water is: (a) 273 K (b) 273.14 K (c) 273.15K

[IAS-2000] (d) 273.16 K

p–T Diagram for a Pure Substance IAS-11.

In the following P-T diagram of water showing phase equilibrium lines, the sublimation line is: (a) p (b) q (c) r (d) S

[IAS-1998]

T-s Diagram for a Pure Substance IAS-12.

Entropy of a saturated liquid at 227°C is 2.6 kJ/kgK. Its latent heat of vaporization is 1800 kJ/kg; then the entropy of saturated vapour at 227°C would be: [IAS-2001] (a) 2.88 kJ/kg K (b) 6.2 kJ/kg K (c) 7.93 kJ/kg K (d) 10.53 kJ/kg K

191

Pure Substances S K Mondal’s IAS-13.

Chapter 7

Two heat engine cycles (l - 2 3 - 1 and l' - 2' - 3' - l’) are shown on T-s co-ordinates in

[IAS-1999]

IAS-14.

The mean effective pressure of the thermodynamic cycle shown in the given pressurevolume diagram is: (a) 3.0 bar (b) 3.5 bar (c) 4.0 bar (d) 4.5 bar

[IAS-1999]

192

Pure Substances S K Mondal’s IAS-15.

Chapter 7

The given figure shows a thermodynamic cycle on T-s diagram. All the processes are straight times. The efficiency of the cycle is given by (a) (0.5 Th – Te)/ Th (b) 0.5 (Th – Te)/ Th (c) (Th – Te)/ 0.5 Th (d) (Th – 0.5 Te)/ Th

[IAS-1996]

h-s Diagram or Mollier Diagram for a Pure Substance IAS-16.

Constant pressure lines in the superheated region of the Mollier diagram have what type of slope? [IAS-2007] (a) A positive slope (b) A negative slope (c) Zero slope (d) May have either positive or negative slopes

IAS-17.

Assertion (A): In Mollier chart for steam, the constant pressure lines are straight lines in wet region. Reason (R): The slope of constant pressure lines in wet region is equal to T. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false [IAS-1995] (d) A is false but R is true

Quality or Dryness Fraction IAS-18.

Dryness fraction of steam means the mass ratio of [IAS-2001] (a) Wet steam, to dry steam (b) Dry steam to water particles in steam (c) Water particles to total steam (d) Dry steam to total steam

Throttling IAS-19.

Assertion (A): Throttle governing is thermodynamically more efficient than nozzle control governing for steam turbines. [IAS-2000] Reason (R): Throttling process conserves the total enthalpy. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

193

Pu ure Su ubstan nces S K Mondal’s

Cha apter 7

Answ wers with Expla anatio on (Ob bjectiv ve) Previo ous 20-Years s GATE E Answ wers GA ATE-1. Ans. (a) Initial Volume V (V1) = 0.001 + 0.03 3 × 0.1 m3 = 0.004 0 m3 Lett dryness fracction = x Theerefore 0.004 4 × 1.5 = (1 – x) × 0.0015 × 1.03 + x × 0.002 × 1.0 03 Tha at gives an absurd value of x = 8.65 (IIt must be leess than equa al to unity). So S vapour is sup perheated. GA ATE-2. Ans. (d) Work do one = first con nstant volum me heating +

∫ pdv

= 0 + P (V2-V1) = 200 × (0.006-0.004) = 0.4 kJ

GA ATE-3. Ans. (c) ( ΔS = (Δ ΔS ) syatem + (ΔS) surroundings = 10 –

10000 = 8.51 J/K (273 + 400)

GA ATE-4. Ans. (c) GA ATE-5. Ans. (b) GA ATE-6. Ans.. Higher, Low wer GA ATE-7. Ans. (a) Mollier diagram d is a h-s plot. Tdss= dh - υ dp

⎛ ∂h ⎞ = T = slope ⎜ orr ⎝ ∂s ⎟⎠ P

T iss always + ive so slope allways +ive. Not N only this if T ↑ then sllope ↑ GA ATE-8. Ans. False

GA ATE-9. Ans. (d) ATE-10. Anss. (b) GA h 40 = h−20 = (1 − x ) hf − 20 + xhg

or 371.43 = (1 − x ) 89.05 + x × 1418.0 or x = 0.212 3 194 4

Pure Substances S K Mondal’s

Chapter 7

GATE-11. Ans. (b) GATE-12. Ans. (d) Δh = o

Δs > 0 ΔT < 0

GATE-13. Ans. (a) S2 − S1 = C pav ln

T2 P − Ru ln 2 T1 P1

Change in entropy of the universe = − Ru ln

P2 P1

0.1 kJ = 13.38 0.5 K For an ideal gas change in enthalpy is a function of temperature alone and change in enthalpy of a throttling process is zero. = −8.314 ln

Previous 20-Years IES Answers IES-1. Ans. (d) Water for all practical purpose can be considered as pure substance because it is homogeneous and has same chemical composition under all phases. IES-2. Ans. (d) IES-3. Ans. (d) The ordinate and abscissa in given figure are pressure and enthalpy. Such diagram is common in vapour compression refrigeration systems. IES-4. Ans. (d) The throttling process given in figure is on pressure-enthalpy diagram. IES-5. Ans. (d) IES-6. Ans. (b) A pure substance is one whose chemical composition does not change during thermodynamic processes. • Pure Substance is one with uniform and invariant chemical composition. • Eg: Elements and chemical compounds are pure substances. (water, stainless steel) • Mixtures are not pure substances. (eg: Humid air) • Exception!! Air is treated as a pure substance though it is a mixture of gases. Gaseous combustion products are a mixture of gases and not a pure substance. IES-7. Ans. (d) IES-8. Ans. (a) IES-9. Ans. (b) The largest density of water near atmospheric pressure is at 4°c. IES-10. Ans. (c) Up to saturation point pressure must be constant. After saturation its slope will be dp p =− –ive, as pv = RT or pv = const. or vdp + pdv = 0 or dv v 195

Pure Substances S K Mondal’s

Chapter 7

IES-11. Ans. (c) IES-12. Ans. (c) IES-13. Ans. (a) IES-14. Ans. (c) IES-15.Ans.(d) Characteristics of the critical point 1. It is the highest temperature at which the liquid and vapour phases can coexist. 2. At the critical point hfg, ufg and vfg are zero. 3. Liquid vapour meniscus will disappear. 4. Specific heat at constant pressure is infinite. IES-16. Ans. (d) At critical point, the latent heat in zero and steam generators can operate above this point as in the case of once through boilers. The density of liquid and its vapour is however same and thus statement 2 is wrong. IES-17. Ans. (b) IES-18. Ans. (b) IES-19. Ans. (c) IES-20. Ans. (c) IES-21. Ans. (b) ⎛ ∂h ⎞ Tds = dh – Vdp or ⎜ ∂s ⎟ = T ⎝ ⎠P The slope of the isobar on the h–s diagram is equal to the absolute temp, for condensation T is cost so slope is const, but not zero so it is inclined line.

IES-22. Ans. (a) For throttling process (1 − 2), h1 = h2

h1 = hf = 1000 kJ/kg at pressure P1 h2 = hf + x (hg − hf ) at pressure P2 ∴ 1000 = 800 + x (2800 − 800) or x = 0.1 IES-23. Ans. (c) IES-24. Ans. (c) IES-25. Ans. (a) IES-26. Ans. (c) Consider a throttling process (also referred to as wire drawing process)

There is no work done (rising a weight) If there is no heat transfer Conservation of mass requires that Since 1 and 2 are at the same level 196

W=0 Q=0

C1 = C 2 Z1 = Z 2

Pure Substances S K Mondal’s

Chapter 7

From SFEE it follows that h1 = h2 Conclusion: Throttling is a constant enthalpy process (isenthalpic process) IES-27. Ans. (b) Temperature decreases and the steam becomes superheated.

IES-28. Ans. (d) IES-29. Ans. (d) Throttling is a isenthalpic process h1 = h2 or u1 + p1v1 = u2 + p2v2 or u2 – u1 = p1v1 – p2v2 = 1000 × 0.5 – 100 × 2 = 300 kJ/kg IES-30. Ans. (d) The throttling process takes places with enthalpy remaining constant. This process on T–S diagram is represented by a line starting diagonally from top to bottom. IES-31. Ans. (b) IES-32. Ans. (d)

Previous 20-Years IAS Answers IAS-1. Ans. (d) IAS-2. Ans. (c) Only two phase liquid-vapour is co-exists at the critical point, but at triple point-all three phase are co-exists. IAS-3. Ans. (a) A pure substance is a substance of constant chemical composition throughout its mass. IAS-4. Ans. (c)

IAS-5. Ans. (a) IAS-6. Ans. (d) A is false but R is true. IAS-7. Ans. (c) IAS-8. Ans. (d) IAS-9. Ans. (d) Network done is area of closed loop ABCD = Area of trapezium AB32 + Area BC63 – Area CD56 – Area AD52 ⎛4+6⎞ ⎛6+4⎞ ⎛1 + 4 ⎞ ⎛1 + 4 ⎞ =⎜ ⎟ × (3 − 2) + ⎜ 2 ⎟ × ( 6 − 3) − ⎜ 2 ⎟ × ( 6 − 5) + ⎜ 2 ⎟ × (5 − 2) 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 5 × 1 + 5 × 3 − 2.5 × 1 − 2.5 × 3 = 10 bar m3 N × m3 = 106 Nm = 1000 kNm m2 IAS-10. Ans. (d) Remember: Triple point temperature of water = 273.16 K = 0.01°C = 10 × 105

197

Pure Substances S K Mondal’s

Chapter 7

IAS-11. Ans. (a) IAS-12. Ans.

Sg = S f +

h fg Tsat

= 2.6 +

1800 = 6.2 kJ / kgK 500

IAS-13. Ans. (d) IAS-14. Ans. (a) Work (W) = ( 0.03 − 0.01) × ( 400 − 200 ) + W = pm × ΔV or pm =

1 × ( 600 − 400 ) × ( 0.03 − 0.01) = 6kJ 2

W 6 = kPa = 3bar ΔV ( 0.03 − 0.01)

1 × ( Th − Tc ) × ( S2 − S1 ) 2 Heat added = Area under 1 − 2 = Th ( S2 − S1 )

IAS-15. Ans. (b) Work output = Area 123 =

1 ( Th − Tc )( S2 − S1 ) ∴η = 2 = 0.5 ( Th − Tc ) / Th Th ( S2 − S1 ) IAS-16. Ans. (a) Mollier diagram is a h-s plot.

Tds= dh – υ dp

or

⎛ ∂h ⎞ ⎜ ⎟ = T = slope ⎝ ∂s ⎠ P

T is always + ive so slope always +ive. Not only this if T ↑ then slope ↑ IAS-17. Ans. (a) Both A and R are true and R is the correct explanation of A IAS-18. Ans. (d) IAS-19. Ans. (d) If throttle governing is done at low loads, the turbine efficiency is considerably reduced. The nozzle control may then be a better method of governing.

198

Prop pertie es of Gasses G s & Ga as Mix. S K Mondal’s

8.

Cha apter 8

Prop pertie es of Gass G es an nd Ga as Mixtu ure

Theo ory at a Glance (Fo or GAT TE, IES S & PS SUs) 1.

The fun nctional relattionship amoong the indep pendent prop perties, presssure P, mola ar or specific volumee v, and temp perature T, iss known as ‘E Equation of sttate’ i.e. PV = RT for gasses.

2.

A hypothetical gas which obeys the law PV = RT at all temperatures t s and pressures is called an ‘idea al gas’ • as’ has no forcces of interm molecular attrraction. An ‘ideal ga • The specific heat capacitties are consttant.

3.

‘Real gas’ g does nott conform to equation of state with coomplete accu uracy. As PÆ0 or T Æ∞ , the real gas approacches the idea al gas behaviour.

4.

Joule’ss law state es that the specific intternal energ gy of a gass depends only o on the temperature of the gas and is in ndependent off both pressu ure and volum me.

PR ROCESS Constant Vo olume Proce ess: – The prrocess is reprresented on a pv diagram as shown in Figure. •C

F Fig. Here or Therefore or work trransfer

v1 = v2 dv = 0 pdv = 0 W=0 200 0

Prop pertie es of Gasses G s & Ga as Mix. S K Mondal’s

Cha apter 8

Therefore d Q = dE = du = Cv (T2 – T1) or Q1-2 = Cv (T2 – T1) (i.e.,) duriing constant volume proccess, the heatt transfer is the change in internal en nergy. From gass equation pv p1v1 = 2 2 T2 T1 We get, p1 T2 = p2 T1 • Constant Pressure P Pr rocess: Thiss process is represented r o a pv diagram as shown on n in Figure. In this process.. p1 = p2 Froom gas equattion we get p1v1 pv = 2 2 T1 T2

Fig.-v1 T2 = v2 T1 d W = pdv = p(v2 – v1) d Q = dE + d W = du + d W = d(u + pv) = d (h)) = CP (T T2 – T1) (i.e.,) heatt transfer is the t change in n enthalpy in n the case of a constant prressure proceess. (or)

  

Constant tem mperature (or) Isother rmal processs:•C The process is represented in figurre(below). Here T1 = T2 d W = pdv 2

dv ; V 1

W = ∫ p dv v = C∫

since pv = C

201 1

Prop pertie es of Gasses G s & Ga as Mix. S K Mondal’s

Cha apter 8

F Fig. ⎛v ⎞ ∴ W = p1v1 l n ⎜ 2 ⎟ = p1v1  n l (r ) ⎝ v1 ⎠ v r= 2 where v1 Since du =

T1  = T2 and dT T = 0.

(u2 – u1 ) =

Cv  ( T2  – T1 ) = 0.

Therefore, by the Firstt Law of Thermodynamiccs Q = W = p1v1 ln (r)

Adiabatic pr rocess: •A Th his is also ca alled isentro opic process as entropy y in this proocess will rem main constan nt and heat tra ansfer in thiss process is allso zero. The process is reepresented in n pv diagram m shown in Fiigure below.

F Fig.The equattion for this process p becom mes, γ γ p1v1 = p2v2 where γ = ratio of speccific heats C (i.e.,) = P Cv For this process Q = 0 Hence 0 = d E + d W = du + d W or d W = – du = – mCv dT d W = pdv or pdv = – du = – mCv dT d or pdv + mCvdT = 0

( eqquation.....1)

But pv = mRT differentia ating; pdv + vdp v = mR dT T From the above equatiion, pdv + vdp dT = mR R 202 2

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

In equation-1, value of dT is substituted. ⎛ pdv + vdp ⎞ pdv + mC v ⎜ ⎟= 0 mR ⎝ ⎠ R.pdv + Cv (pdv + vdp) = 0 But R = (CP – Cv) Therefore (CP – Cv) pdv + Cv (pdv + vdp) = 0 or Cv vdp + CP pdv = 0 Dividing by (Cv p v) we get dp CP dv . + =0 p Cv V

But          

CP =γ Cv

dp dv +γ = 0 p V Integration and rearranging we get, ln p + γ ln v = C Or pvγ = C p1v1γ = p2v2γ (i.e.) From the gas equation pv = mRT, we can obtain the relationships between pressure and temperature and volume and temperature as given below. Therefore

γ −1

γ −1

⎛v ⎞ ⎛p ⎞γ T1 = ⎜ 2⎟ = ⎜ 1⎟ T2 ⎝ v1 ⎠ ⎝ p2 ⎠ Work transfer d W = pdv 2

W = ∫ pdv 1

pvγ = C p=

C vγ

Therefore 2

dv γ 1V

W = C∫

But

⎡ v −γ +1 − v1−γ +1 ⎤ = C⎢ 2 ⎥ -γ +1 ⎣ ⎦ γ γ C = p1v1 = p2v2

⎛ p v − p1v1 ⎞ W =⎜ 2 2 ⎟ ⎝ −γ + 1 ⎠ ⎛ p v − p2v2 ⎞ W =⎜ 1 1 ⎟ γ −1 ⎠ ⎝ Also first law becomes W = – ΔE = – ΔU

203

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

General process or polytropic process: In the adiabatic process the index for V is γ. In a most general case this γ can be replaced by n. The values of n for different processes are indicated below : if n = 0 the process is constant pressure if n = 1 the process is isothermal if n = ∞ the process is constant volume if n = γ the process is adiabatic. and for any process other than the above, n becomes a general value n. All the equations of adiabatic process can be assumed for this process with γ replaced by n. n −1

⎛p ⎞ T1 ⎛ v2 ⎞ =⎜ ⎟ =⎜ 1⎟ T2 ⎝ v1 ⎠ ⎝ p2 ⎠ p v − p2v2 W = 1 1 n −1 But

n −1 n

W = is not true as Q is not zero.

By the first law of Thermodynamics, Q = pdv + ( u2 – u1 )

p1v1 − p2v2 + mCv (T2 − T1 ) n −1 But by gas equation p1v1 = mRT1 and p2v2 = mRT2 =

so,

But,

mR(T2 − T1 ) 1−n R ⎞ ⎛ = m(T2 − T1 ) ⎜ Cv + 1 − n ⎟⎠ ⎝ C − Cv ⎞ ⎛ Q = m ( T2 – T1 ) ⎜ Cv + P ⎟ 1−n ⎠ ⎝

Q = mCv ( T2 – T1 ) +

C − Cv ⎞ ⎛ = m ⎜ Cv + P ⎟ ( T2 – T1 ) 1−n ⎠ ⎝ ⎛ C − nCv ⎞ Q=⎜ P ⎟ ( T2 – T1 ) ⎝ 1−n ⎠ Substituting CP  = γ Cv ⎛γ −n⎞ Q=⎜ ⎟ Cv ( T2 – T1 ) ⎝1−n ⎠ Hence the specific heat for a polytropic process is Cn = Cv ⎛γ −n⎞ = Cn ⎜ ⎟ ⎝1−n ⎠

4.

( T2 – T1 )

For minimum work in multistage compression P2 = √P1P3 a. Equal pressure ratio i.e. P2 = P3 P1 P2

204

Prop pertie es of Gasses G s & Ga as Mix. S K Mondal’s b. c. 5.

Cha apter 8

Equal discha E arge temper rature i.e. T2 =T3 E Equal work required fo or both the stages. s

Equatiion of states s for real ga as

a )(v-b)) = RT v2

a.

V der waalss equation (p Van p+

b.

The coefficien T nt a is intrroduced to account a for the t existencce of mutual attraction 2 between the molecules. m Th he term a/v is i called the force of coh hesion. The coefficient b iss introduced to account foor the volumees of the moleecules, and iss known as co-volume. c B Beattie Brid dgeman equation p=

A RT (1 − e)( ) v + B) v2 v2

a ) v2 b B = B0 (1- ) v c e= vT 3

Where A = A0 (1-

T This equation n does not giv ve satisfactory results in the t critical pooint region.

PV RT

6.

The rattio

7.

Value of o compressib bility factor (Z Z) at critical point is 0.37 75 for Van der waals gas. For ide eal gas z = 1 Critica al Propertie es:

a =3PcVc2,

is called c the com mpressibilitty factor.

b=

8 PcVc Vc , and R= = 3 Tc 3

Where Pc, Vc and Tc are critica al point presssure, volumee and tempera ature respecttively. At Critical Po oint (i) Three rea al roots of Va ander Waal equation coincide. (ii))

⎛ ∂p ⎞ ⎜ ⎟ = 0 i.e. Slope of p-v diagraam is zero. ⎝ ∂v ⎠Tc

(iiii)

⎛ ∂2 p ⎞ nge of slope also a zero. ⎜ 2 ⎟ = 0 i.e. Chan ⎝ ∂v ⎠Tc

Fig.

(iv v)

⎛ ∂3 p ⎞ e to -9pc ⎜ 3 ⎟ < 0 i.e. neegative, and equal ⎝ ∂v ⎠Tc

8.

Boyle’ss Temperatu ure (TB) =

a b bR 205 5

Critical properties diagram

on

p–v

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

Boyle’s Law is obeyed fairly accurately up to a moderate pressure and the corresponding temperature is called the Boyle’s Temperature. 9.

Dalton’s Law a. The pressure of a mixture of gases is equal to the sum of the partial pressures of the constituents. b. The partial pressure of each constituent is that pressure which the gas would exert if it occupied alone that volume occupied by the mixture at the same temperature.

10.

Gibbs-Dalton Law a. The internal energy, enthalpy and entropy of a gaseous mixture are respectively equal to the sum of the internal energies, enthalpies, entropies of the constituents. b. Each Constituent has that internal energy, enthalpy and entropy, which it would have if it occupied alone that volume occupied by the mixture at the same temperature.

11.

Equivalent molecular weight (Me) = x1M1+x2M2+------------+ xnMn Equivalent gas constant (Re) = x1R1+x2R2+-------------+xnRn Equivalent constant volume specific heat (Cve) = x1Cv1+x2Cv2+------------+xnCnv Equivalent constant pressure specific heat (Cpe) = x1Cp1+x2Cp2+------------+xnCpn xi =

12.

mi = mass fraction of a constituent m

The value of Universal Gas constant R = 8.3143 KJ/Kg mole K.

PROBLEMS & SOLUTIONS Example 1. Ten moles of ethane were confined in a vessel of volume 4.86 litres at 300K. Predict the pressure of gaseous ethane under these conditions with the use of equation of state of Van der Waals when a = 5.44 litre2 atm mol-2; b = 64.3 millititre mol-1; R = 0.08205 atm mol-1 deg-1 Solution: Using van der Waals equation, (for total volume V = nv) nRT an2 10.00 × 0.08205 × 300 5.44 × 10.02 p= − 2 = − V − nb V 4.86 − 10.0 × 0.0643 4.862 = 58.4 − 23 = 35.4 atm. Example 2. A rigid vessel of 0.3 m3 volume contains a perfect gas at a pressure of 1 bar. In order to reduce the pressure, it is connected to an extraction pump. The volume flow rate though the pump is 0.014 m3/min. Assuming the gas temperature remains constant, calculate the time taken to reduce the pressure to 0.35 bar. Solution: Vdp m = pV / RT ; dm = RT − p. (Ve dt ) Mass of gas removed = RT Where Ve is the rate of extraction and dt is the time interval. Equating and simplifying

206

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

dp ⎛V ⎞ = − ⎜ e ⎟ dt p ⎝V ⎠ p2

t

dp ⎛ −Ve ⎞ ∫p p = ∫0 ⎜⎝ V ⎟⎠ dt 1

ln

P2 ⎛V ⎞ = − ⎜ e ⎟t P1 ⎝V ⎠

⎛V ⎞ P ∴ t = ⎜ ⎟ ln 1 ⎝ Ve ⎠ P2 ⎛ 0.3 ⎞ ⎛ 1.0 ⎞ =⎜ ⎟ × ln ⎜ 0.35 ⎟ = 22.5min. 0.014 ⎝ ⎠ ⎝ ⎠

Example 3. The value of characteristic constant for a gas is 4.1 kJ/kgK and specific heat at constant pressure is 14.28 kJ/kg K. 5 cubic meters of this gas at a pressure of 100 kPa and 20°C are compressed adiabatically to 500 kPa. The compressed gas is then expanded isothermally to original volume. Calculate: (a) The final pressure of the gas after expansion, (b) The quantity of heat added from the beginning of compression to the end of expansion. Solution: pV 100 ×103 × 5 m= = = 0.4162kg RT 4.1 × 103 × 293 C p = 14.28 kJ / kg K Cv = C p − R = 10.18

γ = 1.4028 1

1

⎛ p ⎞γ ⎛ 1 ⎞1.4028 V2 = V1 ⎜ 1 ⎟ = 5 × ⎜ ⎟ = 1.5874m3 p 5 ⎝ ⎠ ⎝ 2⎠ V 5 ×105 × 1.5874 p3 = 2 × p2 = = 158.74 kPa V1 5 γ −1

⎛p ⎞γ T2 = T1 ⎜ 2 ⎟ ⎝ p1 ⎠ ⎛V Q = p2V2 ln ⎜ 1 ⎝ V2 = 910.6 kJ .

= 465 K ⎞ 5 5 ⎟ = 5 ×10 ×1.5874 ln 1.5874 ⎠

207

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions Equation of State of a Gas GATE-1.

Nitrogen at an initial stage of 10 bar, 1 m3 and 300K is expanded isothermally ⎛ a ⎞ ⎟ v = RT , where a > 0. to a final volume of 2 m3. The P-V-T relation is ⎜ P + 2 ⎜ v ⎟⎠ ⎝ The final pressure will be: [GATE-2005] (a) Slightly less than 5 bar (b) Slightly more than 5 bar (c) Exactly 5 bar (d) Cannot be ascertained.

Adiabatic Process GATE-2.

A mono-atomic ideal l gas (γ = 1.67, molecular weight = 40) is compressed adiabatically from 0.1 MPa, 300 K to 0.2 MPa. The universal gas constant is 8.314 kJ kmol-1K-1. The work of compression of the gas (in kJ kg-1) is: [GATE-2010] (a) 29.7 (b) 19.9 (c) 13.3 (d) 0

Statement for Linked Answer Questions Q3 & Q4: A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C. When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2500 cm3. GATE-3.

The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal [GATE-2006] (a) 30.6 J, 1.94 bar (b) 21.8 J, 0.93 bar (c) 61.1 J, 1.94 bar (d) 43.7 J, 0.93 bar

GATE-4.

Gauge pressure of air to which the ball must have been originally inflated so that it would equal 1 bar gauge at the stadium is: [GATE-2006] (a) 2.23 bar (b) 1.94 bar (c) 1.07 bar (d) 1.00 bar

GATE-5.

A 100 W electric bulb was switched on in a 2.5 m × 3 m × 3 m size thermally insulated room having a temperature of 20°C. The room temperature at the end of 24 hours will be [GATE-2006] (a) 321°C (b) 341°C (c) 450°C (d) 470°C

208

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

Isothermal Process GATE-6.

A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3. It expands quasi-statically at constant temperature to a final volume [GATE-2009] of 0.030 m3. The work output (in id) during this process will be: (a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00

Properties of Mixtures of Gases GATE-7.

2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by [GATE-2008] (A) –Rln2 (B) 0 (C) Rln2 (D) Rln4

Previous 20-Years IES Questions Avogadro's Law IES-1.

Assertion (A): The mass flow rate through a compressor for various refrigerants at same temperature and pressure, is proportional to their molecular weights. [IES-2002] Reason (R): According to Avogardo’s Law all gases have same number of moles in a given volume of same pressure and temperature. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Ideal Gas IES-2.

Assertion (A): A perfect gas is one that satisfies the equation of state and whose specific heats are constant. [IES-1993] Reason (R): The enthalpy and internal energy of a perfect gas are functions of temperature only. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-3.

In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3 to 2 bar and 10m3, during the process the heat supplied is 100 kW. What is the work done during the process? [IES-2009] (a) 33.3 kW (b) 100 kW (c) 80 kW (d) 20 kW

IES-4.

Consider an ideal gas contained in vessel. If intermolecular interaction suddenly begins to act, which of the following happens? [IES-1992]

209

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

(a) The pressure increase (c) The pressure increase

(b) The pressure remains unchanged (d) The gas collapses

IES-5.

Which of the following statement is correct? [IES-1992] (a) Boilers are occasionally scrubbed by rapidly and artificially circulating water inside them to remove any thin water film may have formed on their inside (b) A sphere, a cube and a thin circular plate of the same mass are made of the same material. If all of them are heated to the same high temperature, the rate of cooling is maximum for the plate and minimum for the sphere. (c) One mole of a monoatomic ideal gas is mixed with one mole of diatomic ideal gas. The molar specific heat of the mixture a constant volume is 2R, where R is the molar gas constant. (d) The average kinetic energy of 1 kg of all ideal gases, at the same temperature, is the same.

IES-6.

Consider the following statements: A real gas obeys perfect gas law at a very 1. High temperature 2. High-pressure Which of the following statements is/are correct? (a) 1 alone (b) 1 and 3 (c) 2 alone

[IES-2000] 3. Low pressure

(d) 3 alone

Equation of State of a Gas IES-7.

The correct sequence of the decreasing order of the value of characteristic gas constants of the given gases is: [IES-1995] (a) Hydrogen, nitrogen, air, carbon dioxide (b) Carbon dioxide, hydrogen, nitrogen, air (c) Air, nitrogen, carbon dioxide, hydrogen (d) Nitrogen, air, hydrogen, carbon dioxide

IES-8.

If a real gas obeys the Clausius equation of state p(v – b) = RT then,

(a)

⎛ ∂u ⎞ ⎜ ⎟ ≠0 ⎝ ∂v ⎠T

(b)

⎛ ∂u ⎞ ⎜ ⎟ =0 ⎝ ∂v ⎠T

(c)

[IES-1992]

1 ⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎜ ⎟ = 1 (d) ⎜ ⎟ = ⎝ ∂v ⎠T p ⎝ ∂v ⎠T

IES-9.

Pressure reaches a value of absolute zero (a) At a temperature of -273K (b) Under vacuum condition (c) At the earth’s centre (d) When molecular momentum of system becomes zero

[IES-2002]

IES-9a.

Reduced pressure is (a) Always less than atmospheric pressure (c) An index of molecular position of a gas

[IES-2011]

210

(b) Always unity (d) Dimensionless

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

Van der Waals equation IES-10.

Which one of the following is the characteristic equation of a real gas? [IES-2006]

a⎞ ⎛ ⎟ ( v − b ) = RT v2 ⎠ ⎝ (c) pv = RT

a⎞ ⎛ ⎟ ( v + b ) = RT v2 ⎠ ⎝ (d) pv = nRT

(a) ⎜ p +

(b) ⎜ p −

IES-11.

Which of the following statement about Van der waal's equation i valid? (a) It is valid for all pressure and temperatures [IES-1992] (b) It represents a straight line on pv versus v plot (c) It has three roots of identical value at the critical point (d) The equation is valid for diatomic gases only.

IES-12.

The

internal

energy

of

a

gas

obeying

Van

der

Waal’s

a⎞ ⎛ ⎜ P + 2 ⎟ ( v − b ) = RT , depends on v ⎝ ⎠

IES-13.

equation [IES-2000]

(a) Temperature

(b) Temperature and pressure

(c) Temperature and specific volume

(d) Pressure and specific volume

a⎞ ⎛ Van der Waal’s equation of state is given by ⎜ P + 2 ⎟ ( v − b ) = RT . The constant v ⎝ ⎠

‘b’ in the equation in terms of specific volume at critical point Vc is equal to: [IES-2003]

(a) Vc/3

(b) 2 Vc

(c) 3 Vc

(d)

8a 27VcR

Compressibility IES-14.

Consider the following statements: [IES-2007] 1. A gas with a compressibility factor more than 1 is more compressible than a perfect gas. 2. The x and y axes of the compressibility chart are compressibility factor on y-axis and reduced pressure on x-axis. 3. The first and second derivatives of the pressure with respect to volume at critical points are zero. Which of the statements given above is/are correct? (a) 2 and 3 only (b) 1 and 3 only (c) 1 and 2 only (d) 1, 2 and 3

IES-15.

Which one of the following statements is correct? (a) Compressibility factor is unity for ideal gases (b) Compressibility factor is zero for ideal gases (c) Compressibility factor is lesser than unity for ideal gases 211

[IES-2007]

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

(d) Compressibility factor is more than unity for ideal gases IES-16.

Assertion (A): At very high densities, compressibility of a real gas is less than one. [IES-2006] Reason (R): As the temperature is considerably reduced, the molecules are brought closer together and thermonuclear attractive forces become greater at pressures around 4 MPa. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-17.

The value of compressibility factor for an ideal gas may be: [IES-2002] 1. less or more than one 2. equal to one 3. zero 4. less than zero The correct value(s) is/are given by: (a) 1 and 2 (b) 1 and 4 (c) 2 only (d) 1 only

IES-18.

Assertion (A): The value of compressibility factor, Z approaches zero of all isotherms as pressure p approaches zero. [IES-1992] Reason (R): The value of Z at the critical points is about 0.29. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Adiabatic Process IES-19.

IES-20.

IES-21.

Assertion (A): An adiabatic process is always a constant entropy process. Reason(R): In an adiabatic process there is no heat transfer. [IES-2005] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true A control mass undergoes a process from state 1 to state 2 as shown in the given figure. During this process, the heat transfer to the system is 200 KJ. IF the control mass returned adiabatically. From state 2 to state 1 by another process, then the work interaction during the return process (in kNm) would be: (a) –400 (b) –200 (c) 200 (d) 400

[IES-1998]

A gas expands from pressure P1 to pressure P2 (P2 = p1/10). If the process of expansion is isothermal, the volume at the end of expansion is 0.55 m3. If the process of expansion is adiabatic, the volume at the end of expansion will be closer to: [IES-1997]

212

Properties of Gasses & Gas Mix. S K Mondal’s (a) 0.45 m3 IES-22.

Chapter 8 (b) 0.55 m3

(c) 0.65 m3

(d) 0.75 m3

A balloon which is initially collapsed and flat is slowly filled with a gas at 100 kPa so as to form it into a sphere of 1 m radius. What is the work done by the gas in the balloon during the filling process? [IES-2008] (a) 428·9 kJ (b) 418·9 kJ (c) 420·9 kJ (d) 416·9 kJ

Isothermal Process IES-23.

IES-24.

An ideal gas undergoes an isothermal expansion from state R to state S in a turbine as shown in the diagram given below: The area of shaded region is 1000 Nm. What is the amount is turbine work done during the process? (a) 14,000 Nm (b) 12,000 Nm

(c) 11,000Nm

The work done in compressing a gas isothermally is given by γ −1 ⎡ ⎤ ⎛ ⎞ p2 γ γ ⎢ (a) p1v1 ⎜ ⎟ − 1⎥ ⎢⎝ p1 ⎠ ⎥ γ −1 ⎢⎣ ⎥⎦

⎛p ⎞ (b) mRT1 log e ⎜ 2 ⎟ ⎝ p1 ⎠

(c) mc p (T2 − T1 ) kJ

⎛ T ⎞ (d ) mRT1 ⎜ 1 − 2 ⎟ kJ ⎝ T1 ⎠

[IES-2004] (d) 10,000Nm [IES-1997]

IES-25.

The slope of log P-log V graph for a gas for isothermal change is m1 and for [IES-1992] adiabatic changes is m2. If the gas is diatomic gas, then (a) m1m2 (c) m1 + m2 = 1.0 (d) m1 = m2

IES-26.

The work done during expansion of a gas is independent of pressure if the expansion takes place [IES-1992] (a) Isothermally (b) Adiabatically (c) In both the above cases (d) In none of the above cases

IES-26a

Air is being forced by the bicycle pump into a tyre against a pressure of 4-5 bars. A slow downward movement of the piston can be approximated as (a) Isobaric process (b) Adiabatic process (c) Throttling process (d) Isothermal process [IES-2011]

IES-27.

Three moles of an ideal gas are compressed to half the initial volume at a constant temperature of 300k. The work done in the process is [IES-1992] (a) 5188 J (b) 2500 J (c) –2500 J (d) –5188 J

213

Properties of Gasses & Gas Mix. S K Mondal’s IES-28.

Chapter 8

The change in specific entropy of a system undergoing a reversible process is given by s2 − s1 = c p − cv ln ( v2 / v1 ) . This is valid for which one of the following?

(

(a) (b) (c) (d)

)

Adiabatic process undergone by an ideal gas Isothermal process undergone by an ideal gas Polytropic process undergone by a real gas Isobaric phase change from liquid to vapour

[IES-2008]

Polytropic Process IES-29.

IES-30.

Assertion (A): Though head is added during a polytropic expansion process for which γ > n> 1, the temperature of the gas decreases during the process. Reason (R): The work done by the system exceeds the heat added to the system. (a) Both A and R are individually true and R is the correct explanation of A [IES 2007] (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true ⎛ γ − n ⎞ ⎧ p1v1 − p2 v2 ⎫ In a polytropic process, the term ⎜ [IES-2005] ⎬ is equal to: ⎟⎨ ⎝ γ − 1 ⎠ ⎩ (n − 1) ⎭

(a) Heat absorbed or rejected (c) Ratio of T1/T2 IES-31.

(b) Change in internal energy (d) Work done during polytropic expansion

The heat absorbed or rejected during a polytropic process is equal to: 1/2

⎛γ −n⎞ (a) ⎜ ⎟ ⎝ γ −1 ⎠

⎛γ −n⎞ (b) ⎜ ⎟ x work done ⎝ n −1 ⎠

x work done

⎛γ −n⎞ (c) ⎜ ⎟ x work done ⎝ γ −1 ⎠

[IES-2002]

2

⎛γ −n⎞ (d ) ⎜ ⎟ x work done ⎝ γ −1 ⎠

Constant Pressure or Isobaric Process IES-32.

Change in enthalpy in a closed system is equal to the heat transferred, if the reversible process takes place at [IES-2005] (a) Temperature (b) Internal energy (c) Pressure (d) Entropy

IES-33.

Which one of the following phenomena occurs when gas in a piston-in-cylinder assembly expands reversibly at constant pressure? [IES-2003] (a) Heat is added to the gas (b) Heat is removed from the gas (c) Gas does work from its own stored energy (d) Gas undergoes adiabatic expansion

IES-34.

A saturated vapour is compressed to half its volume without changing its temperature. The result is that: [IES-1997] (a) All the vapour condenses to liquid (b) Some of the liquid evaporates and the pressure does not change (c) The pressure is double its initial value (d) Some of the vapour condenses and the pressure does not change

214

Properties of Gasses & Gas Mix. S K Mondal’s IES-35.

Chapter 8

An ideal gas at 27°C is heated at constant pressure till its volume becomes three times. [IES-2008] What would be then the temperature of gas? (a) 81° C (b) 627° C (c) 543° C (d) 327° C

Constant Volume or Isochoric Process IES-36.

In which one of the following processes, in a closed system the thermal energy transferred to a gas is completely converted to internal energy resulting in an increase in gas temperature? [IES-2008] (a) Isochoric process (b) Adiabatic process (c) Isothermal process (d) Free expansion

IES-37.

Which one of the following thermodynamic processes approximates the steaming of food in a pressure cooker? [IES-2007] (a) Isenthalpic (b) Isobaric (c) Isochoric (d) Isothermal

IES-37a

Assertion (A): The constant pressure lines are steeper than the constant volume lines for a perfect gas on the T-S plane. [IES-2010] Reason (R): The specific heat at constant pressure is more than the specific heat at constant volume for a perfect gas. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-38.

Consider the four processes A, B, C and D shown in the graph given above: Match List-I (Processes shown in the graph) with List-II (Index ‘n’ in the equation pvn = Const) and select the correct answer using the code given below the lists: [IES-2007, 2011] List-I A. A B. B C. C D. D Codes: (a) (c)

IES-38a

List-II 1. 0 2. 1 3. 1.4 4. ∞ A 4 1

B 2 3

C 3 2

D 1 4

(b) (d)

A 1 4

B 2 3

C 3 2

D 4 1

Match List I with List II and select the correct answer using the code given below the lists: List I List II (Process index ‘n’) (T-s traces) A. 0 215

Properties of Gasses & Gas Mix. S K Mondal’s Code :

(a) (b) (c) (d)

Chapter 8

B. 1.0 C. 1.4 D. ∞

A 3 2 3 2

1

B 1 1 4 4

C 4 4 1 1

2 3

D 2 3 2 3

T 4

s

IES-39.

[IES-2010] Match List-I (process) with List-II (index n in PVn = constant) and select the correct answers using the codes given below the lists. [IES-1999] List-I List-II 1. n = infinity A. Adiabatic B. Isothermal 2. n = C p Cv

C. Constant pressure D. Constant volume

3. n = 1 4. n = C p -1

Codes: (a) (c)

5. n = zero A (b) 3 (d) 2

Cv

A 2 2

B 3 3

C 5 5

D 4 1

B 2 5

C 1 3

D 5 1

IES-40.

A system at a given state undergoes change through the following expansion processes to reach the same final volume [IES-1994] 1. Isothermal 2. Isobaric 4. Polytropic(n =1.3) 3. Adiabatic ( γ = 1.4) The correct ascending order of the work output in these four processes is (a) 3, 4, 1, 2 (b) 1, 4, 3, 2 (c) 4, 1, 3, 2 (d) 4, 1, 2, 3

IES-41.

Match the curves in Diagram-I with the curves in Diagram-II and select the correct answer. [IES-1996] Diagram-I (Process on p-V plane) Diagram-II (Process on T-s plane)

Code: (a)

A 3

B 2

C 4

D 5

(b) 216

A 2

B 3

C 4

D 5

Properties of Gasses & Gas Mix. S K Mondal’s (c)

Chapter 8 2

3

4

1

(d)

1

4

2

3

IES-42.

Four processes of a thermodynamic cycle are shown above in Fig.I on the T-s plane in the sequence 1-2-3-4. The corresponding correct sequence of these processes in the p- V plane as shown above in Fig. II will be [IES-1998] (a) C–D–A–B (b) D–A–B–C (c) A–B–C–D (d) B–C–D–A IES-43.

Match List-I with List-II and select the correct answer List-I List-II

IES-44.

− ∫ vdp

A. Work done in a polytropic process

1.

B. Work done in a steady flow process

2. Zero

C. Heat transfer in a reversible adiabatic process

3.

D. Work done in an isentropic process

4.

Codes: (a) (c)

A 4 4

B 1 1

C 3 2

D 2 3

(b) (d)

A 1 1

[IES-1996]

p1V1 − p2V2 γ −1 p1V1 − p2V2 n −1

B 4 2

C 2 3

D 3 4

A perfect gas at 27°C was heated until its volume was doubled using the following three different processes separately [IES-2004] 1. Constant pressure process 2 Isothermal process 3. Isentropic process Which one of the following is the correct sequence in the order of increasing value of the final temperature of the gas reached by using the above three different processes? (a) 1 – 2 – 3 (b) 2 – 3 – 1 (c) 3 – 2 – 1 (d) 3 – 1 – 2

Previous 20-Years IAS Questions Ideal Gas IAS-1.

Variation of pressure and volume at constant temperature are correlated through [IAS-2002] (a) Charles law (b) Boyle’s law (c) Joule’s Law (d) Gay Lussac’s Law

217

Properties of Gasses & Gas Mix. S K Mondal’s IAS-2.

Chapter 8

Assertion (A): For a perfect gas, hyperbolic expansion is an isothermal expansion. [IAS-2007] Reason (R): For a perfect gas,

(a) (b) (c) (d)

Pv = constant. T

Both A and R are individually true and R is the correct explanation of A Both A and R are individually true but R is not the correct explanation of A A is true but R is false A is false but R is true

IAS-3.

Variation of pressure and volume at constant temperature are correlated through [IAS-2002] (a) Charle’s law (b) Boyle’s law (c) Joule’s law (d)Gay Lussac’s law

IAS-4.

An ideal gas with initial volume, pressure and temperature of 0.1 m3, 1bar and 27°C respectively is compressed in a cylinder by a piston such that its final volume and pressure are 0.04 m3 and 5 bars respectively, then its final temperature will be: [IAS-2001] (a) –123°C (b) 54°C (c) 327°C (d) 600°C

Equation of State of a Gas IAS-5.

The volumetric air content of a tyre at 27°C and at 2 bars is 30 litres. If one morning, the temperature dips to -3°C then the air pressure in the tyre would be: [IAS-2000] (a) 1.8 bars (b) 1.1 bars (c) 0.8 bars (d) The same as at 27°C

IAS-6.

An Ideal gas with initial volume, pressure and temperature of 0.1m3, 1 bar and 27°C respectively is compressed in a cylinder by piston such that its final volume and pressure 0.04 m3 and 5 bar respectively, then its final temperature will be: [IAS-2001] (a) –123°C (b) 54°C (c) 327°C (d)600°C

IAS-7.

Which one of the following PV-T diagrams correctly represents the properties of an ideal gas? [IAS-1995]

218

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

Van der Waals Equation IAS-8.

If a gas obeys van der Waals' equation at the critical point, then

RTc is equal pc vc

to which one of the following (a) 0 (b) 1

IAS-9.

[IAS-2004; 2007] (c) 1·5 (d) 2·67 a⎞ ⎛ In Van der Waal’s gas equation ⎜ P + 2 ⎟ ( v − b ) = RT (R = Universal gas v ⎠ ⎝ constant) the unit of ‘b’ is:

(a) Liter/mole°C IAS-10.

[IAS-1997]

(b) m3/mole

(c) kg-liter/mole

(d) Dimensionless

A higher value of Van der waal’s constant for a gas indicates that the [IAS-2003] (a) Molecules of the gas have smaller diameter (b) Gas can be easily liquefied (c) Gas has higher molecular weight (d) Gas has lower molecular weight

Critical Properties IAS-11.

The mathematical conditions at the critical point for a pure substance are represented by: [IAS-1999]

(a) (c) IAS-12.

δp δ2p δ3p < 0, 2 = 0 and 3 = 0 δv δv δv 2 δp δ p δ3p = 0, 2 = 0 and 3 < 0 δv δv δv

(b) (d)

δp δ2p δ3p = 0, 2 < 0 and 3 = 0 δv δv δv 2 δp δ p δ3p = 0, 2 = 0 and 3 = 0 δv δv δv

In the above figure, yc corresponds to the critical point of a pure substance under study. Which of the following mathematical conditions applies/apply at the critical point?

⎛ ∂P ⎞ ⎟ =0 ⎝ ∂v ⎠Tc

(a) ⎜

⎛ ∂2 P ⎞ =0 (b) ⎜ 2 ⎟ ⎝ ∂v ⎠Tc ⎛∂ P⎞ ⎜ 3 ⎟ 0 ∴P2 is slightly more than 5 bar. GATE-2. Ans. (a) w.d =

1

γ −1

( p2 v2 − p1v1 )

W .K .T PV = mRT R T1 ⎡ R = 8.314kJ kmol −1 K −1 ⎤⎦ m ⎣ 8.314 0.1× 103 × V1 = × 300 40 ∴V1 = 0.623 m3

∴ PV 1 1 =

1

⎛ P1 ⎞ γ V2 ⇒ V2 = 0.41m3 ⎜ ⎟ = V1 ⎝ P2 ⎠ (.2 × 0.41 − 0.1× 0.623) ×103 ∴W .d = = 29.7 0.67 GATE-3. Ans. (d) Heat lost = n Cv d T GATE-4. Ans. (c) GATE-5. Ans. (c) Heat produced by electric bulb in 24 hr. = 100 × 24 × 60 × 60 J = 8640kJ

Volume of air = 2.5 × 3 × 3 = 22.5 m 3 Density (ρ) = 1.24 kg/m3 ΔQ 8640 ΔQ = mCv Δt or Δt = = = 430 o C ∴t = 430 + 20 = 450 o C mCv 22.5 × 1.24 × 0.716

⎛ V2 ⎞ ⎟ ⎝ V1 ⎠

GATE-6. Ans. (a) Iso-thermal work done (W) = RT1 ln ⎜

222

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8 ⎛ V2 ⎞ = PV ⎟ 1 1 ln ⎜ ⎝ V1 ⎠ ⎛ 0.030 ⎞ = 800 × 0.015 × ln ⎜ ⎟ = 8.32kJ/Kg ⎝ 0.015 ⎠

GATE-7. Ans. (b) Remember if we mix 2 mole of oxygen with another 2 mole of other gas the

volume will be doubled for first and second constituents ΔS = nR ln

Vtotal = 2 R ln 2 ∴ Vinitial

Total Entropy change = 4Rln2 So, Entropy change per mole=Rln2. And it is due to diffusion of one gas into another.

Previous 20-Years IES Answers IES-1. Ans. (a) Both A and R correct and R is the correct explanation of A IES-2. Ans. (b) For perfect gas, both the assertion A and reason R are true. However R is not the explanation for A. A provides definition of perfect gas. R provides further relationship for enthalpy and internal energy but can't be reason for definition of perfect gas. IES-3. Ans. (b) As internal energy is a function of temperature only. In isothermal expansion process no temperature change therefore no internal energy change. A Reversible isothermal expansion process is constant internal energy process i.e. dU = 0 ∵ dQ = dU + dW

∴

dQ = dW (∵ dU = 0 )

Work done during the process = 100kW ∴ IES-4. Ans. (a) IES-5. Ans. (d) (a) True. A water film, if formed, will act as a very poor conductor of heat and will not easily let the heat of the furnace pass into the boiler. An oil film if present, is even worse than water film and the formation of such films inside the boiler must be avoided. (b) Since the mass and material are the same, the volumes must also be the same. For the same volume, the surface area of the plate is the greatest and that of the sphere is the least. The rate of loss of heat by radiation being proportional to the surface area, the plate cools the fastest and the sphere the slowest. 3 5 (c) True, for a monoatomic gas, C1 = R and for a diatomic gas, C1 = R. 2 2 1⎛3 5 ⎞ Since the mixture has two moles, the value of C1 for the mixture = ⎜ R + R ⎟ = 2 R 2⎝2 2 ⎠ 3 RT (d) False, The average kinetic energy of 1 g of an ideal gas = 2 M Where M is the molecular weight of the gas and it is different gases, as the value of M will be different. IES-6. Ans (b) In Perfect gas intermolecular attraction is zero. It will be only possible when intermolecular distance will be too high. High temperature or low pressure or both cause high intermolecular distance so choice 1 and 3. IES-7. Ans. (a) The correct sequence for decreasing order of the value of characteristic gas constants is hydrogen, nitrogen, air and carbon dioxide. IES-8. Ans. (b)

223

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

IES-9. Ans. (d) we know that P =

1 ρ C 2 If momentum is zero then C must be zero. Hence P 3

would be zero. That will occur at absolute zero temperature. But note here choice (a) has in defined temp. –273K which is imaginary temp. IES-9a. Ans. (d) IES-10. Ans. (a) IES-11. Ans. (c) IES-12. Ans. (b) Joule’s law states that for an Ideal gas internal energy is a function of temperature only. u = ƒ(T). But this is not Ideal gas it is real gas. IES-13. Ans. (a) We know that at critical point a = 3PcVc2

;

b = Vc/3 and R =

8 PcVc 3Tc

IES-14. Ans. (a) 1 is false. At very low pressure, all the gases shown have z ≈ 1 and behave nearly perfectly. At high pressure all the gases have z > 1, signifying that they are more difficult to compress than a perfect gas (for a given molar volume, the product pv is greater than RT). Repulsive forces are now dominant. At intermediate pressure, must gasses have Z < 1, including that the attractive forces are dominant and favour compression. IES-15. Ans. (a) IES-16. Ans. (d) IES-17. Ans. (c) IES-18. Ans. (d) IES-19. Ans. (d) IES-20. Ans. (b) During adiabatic process, work done = change in internal energy. Since control man (so case of closed system). Intercept of path on X-axis is the work done by the process. W = area of ∆A12 + area of □A2CB 1 х (3 - 1) х 200 + 100 х (3 - 1) W= 2 = 200 + 200 = 400 kJ

1 (300 - 100) х 2 + 100 х 2 2 = 200 + 200 = 400 kJ ∴ From 1 → 2. U1 + Q = U2 + W. U1 – U2 = W – Q = 400 – 200 = 200 kJ. From 2 → 1 Work done will be same Since adiabatic So Q = 0 U2 + Q = U1 + W W = U2 – U1 = – ( U1 – U2 ) = – 200 kJ.

W=

IES-21. Ans. (a) For isothermal process, p1v1 = p2 v2 , or p1v1 =

For adiabatic process 224

p1 × 0.55, v1 = 0.055 m3 10

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

p1v11.4 = p2 v1.4 2 , or p1 ( 0.055 )

1.4

=

p1 1.4 × v2 or v2 = 0.0551.4 10 = 0.45 m3 10

IES-22. Ans. (b) Work done by the gas during filling process = − ∫ vdp

(

)

= 4 π13 (100 ) = 418.9 kJ 3

IES-23. Ans. (c) Turbine work = area under curve R–S = ∫ P dv

= 1 bar × ( 0.2 − 0.1) m3 + 1000 Nm

= 105 × ( 0.2 − 0.1) Nm + 1000Nm = 11000Nm IES-24. Ans. (b) IES-25. Ans. (a) PV = constant, C ⇒ log P + log V = log C m1 = -1 Pv 4 = C ⇒ log P + q log V = log C m2 = -q = - 1.4 ∴ m2 > m1 IES-26. Ans. (d) IES-26a Ans. (d) IES-27. Ans. (d) Since the temperature remains constant, the process is isothemal. ⎛V ⎞ ∴ Work-done in the process, W = 2.303 nRT log ⎜ 2 ⎟ ⎝ V1 ⎠ ⎛1 ⎞ = 2.303 х 3 х 8.315 х 8.315 х 300 log ⎜ ⎟ ⎝2⎠ = – 5188 J. The negative sign indicates that work is done on the gas. IES-28. Ans. (b) Tds = du + pdv

⇒ Τds = C V dT + pdv

dT P + dv T T dT R ⇒ ds = C V + dv T V ⇒ ds = C V

Integrating the above expression

⇒ S2 − S1 = C V In

T2 V + RIn 2 T1 V1

For isothermal process undergone by ideal gas.

⇒ S2 − S1 = ( CP − C V ) In

V2 V1

IES-29. Ans. (a) IES-30. Ans. (a) IES-31. Ans. (c) IES-32. Ans. (c) dQ = du + pdυ + υpd − υ dp = d ( u + pυ ) − υ dp = dh − υ dp

225

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

if dp = 0 or p = const. these for ( dQ )p = ( dh )p

IES-33. Ans. (a)

IES-34. Ans. (d) By compressing a saturated vapour, its vapours condense and pressure remains unchanged. Remember it is not gas. V V V1 3V1 IES-35. Ans. (b) 1 = 2 ⇒ = T1 T2 ( 273 + 27 ) T2 ⇒ T2 = 300 × 3 = 900 K = 627°C IES-36. Ans. (a) Constant volume (isochoric) process: An example of this process is the heating or cooling of a gas stored in a rigid cylinder. Since the volume of the gas does not change, no external work is done, and work transferred ΔW is zero. Therefore from 1st law of thermodynamics for a constant volume process:

W2 = 0

1

2

1Q 2 = ∫ dU = U2 − U1 1

IES-37. (c) In a pressure cooker, the volume of the cooker is fixed so constant volume process but for safety some of steam goes out to maintain a maximum pressure. But it occurs after proper steaming. IES-37a Ans. (d) IES-38. Ans. (b) IES-38a Ans. (c) IES-39. Ans. (c) IES-40. Ans. (a) IES-41. Ans. (b) IES-42. Ans. (d) IES-43. Ans. (c) IES-44. Ans. (c) Perfect gas T1 = 27 + 273 = 300 k V1 = initial volume V2 = final volume, V2 = 2V1 T2 = ?

226

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

(i) Constant Pressure Process Vα T V1 V2 = T1 T2 V1 2V1 = T1 T2

⇒ T2 = 2T1 = 600  k (ii) Isothermal Process T = Constant T1 = T2 = 300  k (iii) Isentropic Process T1 V1r −1 = T2 V2r −1 r −1

1.4 −1

⎛V ⎞ ⎛ V ⎞ T2 = T1 ⎜ 1 ⎟ = T1 ⎜ 1 ⎟ ⎝ V2 ⎠ ⎝ 2V2 ⎠ T2 = 0.757 T1 = 227.35K

Previous 20-Years IAS Answers IAS-1. Ans. (b) Boyle’s law: It states that volume of a given mass of a perfect gas varies inversely as the absolute pressure when temperature is constant. IAS-2. Ans. (a) IAS-3. Ans. (b) IAS-4. Ans. (c)

PV PV 1 1 = 2 2 T1 T2

or T2 =

PV 5 × 0.04 2 2 × T1 = × (300) = 600 K = 327°C 1× 0.1 PV 1 1

IAS-5. Ans. (a) Apply equation of states ( 273 − 3 ) P1V1 P2 V2 T [∵V1 = V2 ] or P2 = P1 × 2 = 2 × = = 1.8bar T1 T2 T1 ( 273 + 27 ) IAS-6. Ans. (c ) Apply equation of states

∴T2 = (

P1V1 P2V2 PV = or T2 = 2 2 xT1 T1 T2 P1V1

5 0.04 )×( ) × (273 + 27) = 600K = 327°C 1 0 .1

IAS-7. Ans. (c) For an ideal gas PV = MRT i.e. P and T follow direct straight line relationship, which is depicted in figure (c). IAS-8. Ans. (d) a = 3 pc Vc2, b =

Vc 8 PV c c , R= 3 3 Tc

IAS-9. Ans. (b) According to dimensional homogeneity law unit of molar-volume and ‘b’ must be

same. i.e. m3/mole IAS-10. Ans. (b) IAS-11. Ans. (c) IAS-12. Ans. (d) Van der Waals equation

227

Properties of Gasses & Gas Mix. S K Mondal’s a ⎞ ⎛ ⎜ P + 2 ⎟ (υ − b ) = RT υ ⎠ ⎝

Chapter 8

RT a − 2 υ −b υ V 8 PV c c At critical point a= 3pcVc2, b= c , R = 3 3 Tc − RTc 2a ⎛ ∂P ⎞ = + 3 =0 ⎜ ⎟ 2 ⎝ ∂V ⎠T =Tc (Vc − b ) Vc or P =

⎛ ∂2 P ⎞ 2.RTc 6a = − 4 =0 ⎜ 2⎟ 3 ⎝ ∂V ⎠T =Tc (Vc − b ) Vc ⎛ ∂3 P ⎞ 6 RTc 24a =− − 5 = −9 pc i.e.-ive And ⎜ 3 ⎟ ( vc − b ) vc ⎝ ∂V ⎠T =Tc IAS-13. Ans. (a) In adiabatic mixing there is always increase in entropy so large amount of irreversibility is these.

IAS-14. Ans. (c) IAS-15. Ans. (d) For reversible isothermal expansion heat supplied is equal to work done during the

⎛ v2 ⎞ ⎟ ⎝ v1 ⎠

process and equal to Q = W = mRT1 ln ⎜

Temperature constant so no change in internal energy dQ = dU + dW; dU = 0 Therefore dQ = dW. IAS-16. Ans. (d) In reversible isothermal process temperature constant. No change in internal energy. So internal energy constant dQ = δ u + δ W as δ u = 0 , dQ = dW IAS-17. Ans. (d) IAS-18. Ans. (d) ∵

228

Properties of Gasses & Gas Mix. S K Mondal’s

Chapter 8

IAS-19. Ans. (b) Work requirement 1. Isothermal – area under 121B1A 2. Adiabatic – area under 122B2A 3. pv1.1 = c – area under 123B3A

IAS-20. Ans. (c) IAS-21. Ans. (a) IAS-22. Ans. (b)

229