Concepts of Thermodynamics Chem
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Introduction and First Law of Thermodynamics
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First law of Thermodynamics
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Introduction and First Law of Thermodynamics
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CHEMISTRY (T HERMODYNAMICS )
TERMS USED IN THERMODYNAMICS System, Surrounding and Boundary : A system is defined as a specified part of the universe or specified portion of the matter which is under experimental investigation and the rest of the universe, i.e., all other matter which can interact with the system, is surrounding. Boundary: Anything which separates system and surrounding is called boundary. (i) Adiabatic and diathermic : The terms diathermic wall and adiabatic wall are used for con ductor and non-conductor of heat boundaries respectively. (ii) Real and imaginary (iii) Rigid and non-rigid
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TYPES OF SYSTEM :There are three types of system
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A.
Isolated system : This type of system has no interaction with its surrounding. Neither matter nor energy can be exchanged with surroundings.
B.
Closed system : This type of system can exchange energy in the form of heat, work or radiations but not matter with its surrounding.
C.
Open system : This type of system can exchange matter as well as energy with surroundies.
(i). Homogeneous system : A system is said to be homogeneous when it is completely uniform throughout. A homogeneous system is made of one-phase only. Examples are : a pure single solid, liquid or gas, mixture of gases and a true solution (ii). Heterogeneous : A system is said to be heterogeneous when it is not uniform throughout, i.e., it consists two or more phases. Examples are : ice in contact with water, two or more immiscible liquids, insoluble solid in contact with a liquid, a liquid in contact with vapour, etc.
THERMODYNAMIC PROPERTIES : These are of two types : 1.Intensive properties : The properties which do not depend upon the quantity of matter present in the system or size of the system are called intensive properties. Pressure, temperature, density, specific heat, surface tension, refractive index, viscosity, melting point, boiling point, volume per mole, concentration, etc., are the example of intensive properties of the system. 2.Extensive properties : The properties whose magnitude depends upon the quantity of matter present in the system are called extensive properties.
Introduction and First Law of Thermodynamics
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Extensive properties follow addition property but intensive propreties are non additive. The product , sum and ratio of properties are also intensive properties. Extensive property when expressed per mol or per gram becomes intenisve property. A A and are intensive variables. B B
Let A and B be two extensive properties, then In t e ns ive p r o p e rt ie s M ola r ity M ola lity C on c e ntr a tio n D en sit y T e m pe r a tur e P r es su re M ole f r a c tion M ola r en tha lp y M ola r en tr opy M ola r in te rn a l e ne r gy R e fr a c tiv e in de x S p ec if ic he a t V isc os it y S u rf a c e te ns ion D ie le ctr ic c ons ta nt EM F
E xt e n siv e p r o p e r tie s He a t c a pa c ity Mas s Vo lu m e Nu m b e r o f m ole s Ent ha lp y Ent ro py Int er na l e n er g y Gib bs fr e e e n rg y
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State of System When the total mass, temperature, volume, number of moles and composition have definite values, the system is said to be in a definite state. when there is any change in any one of these properties, it is said that the system has undergone a change of state.
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State function or State Variables : Fundamental properties which determine the state of a system are referred to as state variables or state function or thermodynamic parameters. The change in the state properties depends only upon the initial and final states of the system, but is independent of the manner in which the change has been brought about. In other words, the state properties do not depend upon a path followed. Following are the state variable that are commonly used to describe the state of the thermodynamic system : 1. Pressure (P) 2. Themperature (T) 3. Volume (V) 4. Internal energy (E) 5. Enthalpy (H) 6. Entropy (S) 7. Free energy (G) 8. Number of moles (n) Path Functions : The value of these functions are dependent on the path followed in bringing about In the change. e.g.Heat and work Note :- (i) In adiabatic process, E = W, so here W is a state function.. (ii) In isochoric process E = Q , so here Q is a state function. (iii) H and E are path functions whereas H and E are state functions..
Introduction and First Law of Thermodynamics
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Thermodynamic Processes : When the thermodynamic system changes its state, the path followed in order to do so is called a process. The various types of the processes are : A. Isothermal process : The process is termed isothermal if temperature remains fixed, i.e., operation is done at constant temperature. This can be achieved by placing the system in constant temperature bath, i.e., thermostat. For an isothermal process dT = 0, i.e., heat is exchanged with the surrounding and the system is not thermally isolated.
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B. Adiabatic process : If a process is carried out under such condition that no exchange of heat takes place between the system and surrounding, the process is termed adiabatic. The system is thermally isolated,i.e., dQ = 0. This can done by keeping the system in an insulated container, i.e., thermos flask. In adiabatic process, the temperature of the system varies.
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Isothermal
The slope of adiabatic curve is time the slope of isothermal curve. C. Isobaric process : The process is known as isobaric in which the pressure remains constant throughout the change, i.e., dP = 0. D. Isochoric process : The process is termed as isochoric in which volume remains constant throughout the process i.e., dV = 0. E. Cyclic process : When a system undergoes a number of different processes and finally returns to its initial state, it is termed as cyclic process. For a cyclic process dE = 0 and dH =0.
Introduction and First Law of Thermodynamics
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Isobaric process
Isochoric process
F. Reversible process : A process which occurs infinitesimally slowly, i.e., opposing force is infinitesimally smaller than driving force and when infinitesimal increase in the opposing force can reverse the process, it is said to be reversible process. A reversible process is an ideal process and cannot be realised in practice. Both system and surroundings return to their original state in a reversible process.
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G. Irreversible process : When the process goes from initial to final state in single step in finite time and cannot be reversed, it is termed as irreversible process. An irreversible process is spontaneous in nature. All natural processes are irreversible in nature.
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WORK (W) Work is the mode of transfer of energy between system and surrpundings.Work associated with change in volume of a system against external pressure is called mechanical work Pext (V2 – V2) = Pext V M
e
c
h
a
n
i
c
a
B l
w
o
r
k
=
Work (w) is a path-dependent function. Work done on the system, w = +ve Work done by the system, w = –ve
HEAT It may be defined as the quantity of energy which flows between a system and its surroundings due to temperature difference. (i) (ii)
Heat absobed by the system, Q = +ve Heat released by the system, Q = –ve
Unit of Heat and Work : The unit of heat is calorie (cal).. SI unit of heat is joule (J). 1 Joule = 0.2390 cal 1 calorie = 4.184 J 1 kcal = 4.184 kJ 1 litre-atm = 101.3 J. = 1.013 × 109 erg = 24.206 cal
Introduction and First Law of Thermodynamics
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INTERNAL ENERGY (U) Total of all possible kinds of energy of a system is called its internal energy*. U = UKinetics + UPotential + UElectronic + Unuclear + ..... Characteristics of Internal Energy : (i) Internal energy of a system is an extensive property. (ii) Internal energy is a state function i.e. U = Ufinal – Uinitial (iii) There is no change in internal energy in a cyclic process. * For a given system , U can be represented as a function of T and V. U = f (T, V)
U . dT + dU = T v *
U dV V T
For isochoric process : dV = 0
U dT dU = T v dU = Cv . dT U C v .dT
For an ideal gas U 0 V T
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because for ideal gas When there are no interactions between the molecules, the internal energy is independent of their separation and hence independent of the volume of the sample.
U 0 V T U 0 V T
Perfect gas
U 0 V T Attractions dominate
Volume,V
Introduction and First Law of Thermodynamics
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dU = Cv . dT U C v .dT
ZEROTH LAW OF THERMODYNAMICS Two systems in thermal equilibrium with a third system are also in thermal equilibrium with each other. FIRST LAW OF THERMODYNAMICS : It is based upon the law of conservation of energy.The total energy of the universe is constant. When a system is changed from initial state to the final state, it undergoes a change in the internal energy from Uin to Uf . Thus, U can be written as : U = Uf – Uin The change in internal energy can be brought about in two ways : (i) Either by allowing the heat to flow into the system (absorption) or out of the system (evolution) (ii) By doing work on the system or the work done by the system. U = q + w
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Enthalpy (H): Heat content of a system at constant pressure is called enthalpy denoted by ‘H’ Enthalpy function can be written as H = U+ PV. Enthalpy of a substance depends on Temperature and state of substance. From first law of thermodynamics ; Q = U + PV ......(i) Change in heat Q = U + PV + VP
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Heat change at constant pressure can be give as Q = U + PV
......(ii)
At constant pressure heat can be replaced by enthalpy H = U + PV ......(iii) At constant volume, V = 0; thus equation (ii) can be written as Q = U H = Heat change or heat of reaction at constant pressure U = Heat change or heat of reaction at constant volume. (i) In case of solids and liquids participating in a reaction, H E (PV 0) (ii) Difference between H and E is significant when gases are involved in a chemistry reaction. H = E + PV H = E + ngRT Here, ng = Number of gaseous moles of products – Number of gaseous moles of reactants.
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Introduction and First Law of Thermodynamics For a given system Enthalpy can be represented as a function of T and P H = f (T, P)
H H . dT + dP dH = T P P T
For isobaric process : dP = 0
H dT dH = T P dH = CP . dT
H C P .dT
For an ideal gas
H 0 P T dH = CP . dT
H C P .dT
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HEAT CAPACITY
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Heat capacity is defined as the amount of heat required to raise the temperature of the system by one degree.
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dq dT Heat capacity is an extensive property as the amount of heat required (q) to raise the temperature by T depends on the mass of the substance.
Heat Capacity (C) =
Molar heat capacity is the heat capacity per mole of the substance and it is an intensive property. Its unit is JK-1mol-1. Specific heat capacity is the heat capacity per unit mass of the substance and it is also an intensive property. Its unit is J K-1 Kg-1 Following two types of heat capacities are used-
U and thus U C v .dT (i) Heat capacity at constant volume (CV) = T V H and thus H C .dT (ii) Heat capacity at constant pressure(CP) = P T P Cp - Cv =PV = R
Cp Cv
1
2 , where f is the degree of freedom for a given gas f
Introduction and First Law of Thermodynamics
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Cv
H i g h T e m p e r a tu r e
Cp
Cv
CP
He
3 R 2
5 R 2
5 = 1 .6 7 3
3 R 2
5 R 2
5 = 1.6 7 3
H2
5 R 2
7 R 2
7 1 .4 5
7 R 2
9 R 2
9 1 .2 9 7
CO2
5 R 2
7 R 2
7 1 .4 5
13 R 2
15 R 2
15 1 .1 5 13
SO2
3R
4R
4 1 .3 3 3
6R
7R
7 1 .1 7 6
CH4
3R
4R
4 1 .3 3 3
1 2R
13 R
13 1 .0 9 12
Heat capacity of mixtures:- let n1 and n2 moles of two non-reacting gases A and B are mixed than
Cv mix
n1 C v 1 n2 (C v ) 2 n1 n2
Change in Enthalpy with temperature
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qp = H =
For a finite change ,
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C
p
dT
T1
if Cp is independent of T in the range T1 to T2, then H = Cp(T2 - T1) If Cp depends on temperature as = a + bT + cT2 + ....... p C
B
where a , b, c .. are constants then
T2
H =
(a bT cT
2
......)dT
T1
Example:1. 10 dm3 of O2 at 101.325 kPa and 298 K is heated to 348 K. Calculate the heat absorbed, H and U for this process at (i) Costant pressure (ii) at constant volume . Given Cp = 25.72 + 0.013(T/K) - 3.86 x 10-6 Solution:-
No. of moles of gas (n) =
PV 102.325 10 0.409 RT 8.314 298
(a) at constant pressure T2
2 3 T 6 T n 25 . 72 T 0 . 013 3 . 86 10 .... n C dT p qp = H = = 2 3 T1 T1
T2
after substituting the values we get
qp = H = 603.59 J
U = H - nRT = 603.59 - 0.409 x 8.314 x 50 = 433.57 J (b) at constant volume T2
T2
T2
T2
T1
T1
T1
T1
qv = U = n C v dT n (C p R ) dT n C p dT - nR dT = 433.57 J
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Introduction and First Law of Thermodynamics
H = U + nRT = 603.59 J This question also proves that q is a path function because the value of heat is different in both cases whileU and H which are state functions have same values.
THERMODYNAMICAL ANALYSIS OF VARIOUS PROCESSES (1) Isothermal Process : In an isothermal expansion, heat is allowed to flow into or out of the system so that temperature remains constant throughout the process of expansion. E = 0 According to first law of thermodynamics, E = q + w Since, for isothermal process, E = 0, hence q = –w
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Work done in reversible isothermal expansion : w = 2.303 nRT log
V2 P - 2.303 nRT log 1 V1 P2
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Work done in irreversible isothermal expansion : The work done when volume changes from V1 toV2 is given by w = – Pext ×dV = –Pext (V1 – V2)
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Maximum work : Maximum work is done by the gas on surrounding when Pext = P gas (2)Adiabatic Expansion : In adiabatic expansion, no heat is allowed to enter or leave the system, hence, q = 0. According to first law of thermodynamics, E = q + w, we get E =w. Reversible adiabatic expansion : PV = constant R
Work done = C . T = C(T2 – T1) = ( – 1) (T2 – T1) = nR
For n moles = ( – 1) (T2 – T1) Irreversible adiabatic expansion :
RT RT w Pext (V2 V1 ) Pext 2 1 P1 P2
P2V2 P1V1 1
Introduction and First Law of Thermodynamics
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w = C (T2 – T1) = – RPext (
P1T2 P2
P2T1 ) P1
Note:- Work done in any process is the area under the P-V curve. Compression
Expansion
Adiabatic
Isothermal
P
P Isothermal
Adiabatic V
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When initial and final volume are same for both the processes then (i) For expansion Wisothermal > Wadiabatic (ii) For compression Wadiabatic > W isothermal
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Magnitude of Work done in a reversible expansion is more than irreversible expansion T1 P2 T2 P1 whereas work done in reversible compression is lesser than work done in irreversible P P 1 2 compression.It therefore follows that the temperature decrease in an irreversible expansion will be less than the corresponding temperature decrease in reversible expansion.where as temperature rise will be more in irreversible compression than the reversible one.
B
Example: 2 5 moles of an ideal gas at 293K are expanded isothermally from an initial pressure of 0.4053 MPa to a final pressure of 0.1013 MPa against a constant external pressure of 0.1013 MPa. Calculate q,W, U and H. Calculate the corresponding values if the above process is carried out reversibly. Solution For an isothermal expansion against constant pressure work done
nRT nRT 1 1 Pext nRT w = -Pext(V2 -V1) = Pext p2 p1 p1 p2 substituting the given values w = -9.136 kJ Since the process is isothermal i.e. T = constant , so U = 0, H = 0 and q = w If the process is carried out reversibly than for reversible isothermal expansion we have w = - 2.303 nRT log
P1 P2
w = -16.889 kJ Example: 3 A sample of a fluorocarbon was allowed to expand reversibly and adiabatically to twice its volume. In the expansion the temperature dropped from 298.15 K to 248.44 K. Assume the gas behaves perfectly estimate the value of CV.
Introduction and First Law of Thermodynamics According to the equation of state of adiabatic process
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Solution:
TV 1 constant 1
T1 V2 T2 V1 after substituting the values 1.2 (2) -1 log1.2 ( 1) log 2 1.262 R R Now C v 3.816R 31.7 JK -1 mol 1 1 0.262
(3)Free Expansion :- It is an irreversible process in which expansion takes place agsinst zero pressure i.e. Pext=0 in free expansion w = PextdV = 0 Real gases experience a temperature change during free expansion. For an ideal gas, the temperaturE doesn’t change, and the conditions before and after adiabatic free expansion satisfy. PiVi Pf V f
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where p is the pressure, V is the volume, and i and f refer to the initial and final states.
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Since initial and final states are same so H = U = 0 amd thus q = -w =0 For an adiabatic process H = U
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Calculation of temperature in adiabatic process:For an irreversible adiabatic expansion work done w = P(V2-V1) = nCv(T2-T1)
B
nRT1 nRT2 nCv(T2-T1) = Pext p2 p1
, from this equation final temperature can be calculated.
Example: 4 An ideal monoatomic gas (Cv = 1.5 R) initially at 298 K and 1.013 MPa pressure expands adiabatically and irreversibly until it is in equillibrium with a constant external pressure of 0.1013 Mpa. What is the final temperature of the gas ? Solution:-
nRT1 nRT2 For an adiabatic irreversible process nCv(T2-T1) = Pext p2 p1
T T T T n 1.5R(T2 T1 ) nRPext 1 2 1.5(T2 T1 ) Pext 1 2 p1 p2 p1 p2 On substituting the values we get T2 = 190.7 K
Introduction and First Law of Thermodynamics
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V2 nb For any real gas expression of q for isothermal revrsible change q nRT ln V1 nb 1 2 1 For any real gas expression of U for isothermal revrsible change U n a V2 V1 For any real gas expression of w for isothermal revrsible change V nb 1 1 n 2 a ( w ) nRT ln 2 V1 nb V 2 V1 For any expansion process we know that V1 V2
V1 nb V2 nb - V1 nb - V2 nb
V2 V1 - V1 nb V2 V1 - V2 nb Thus
(V2 nb) V2 (V1 nb) V1
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V1 (V2 nb) V2 (V1 nb)
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So we can conclude that q real > qideal i.e. for same change in volume heat absorbed by a real gas is greater than the heat absorbed by an ideal gas.
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JOULE-THOMSON EFFECT For an ideal gas there is no attraction or repulsion between the gas molecules if this gas expands adiabatically into vaccum no cooling is produced in the process which menas that there is no change in internal energy it depends only on temperature. Where as a real gas on expansion from high to low pressure gets cooled on the side of lower pressure. The greater the difference in the pressure higher will be difference in temperature.
B
Thus when a gas expands adiabatically from a region of high pressure to low pressure it gets cooled and this phenomenon is known as Joule-thomson effect. Most of the gases are found to undergo cooling on expansion except Helium & Hydrogen. If the initial pressure and volume are P1 and V1 and final values are P2 and V2 respectively than Total work done by gas (w) = P1V1 - P2V2 Since q =0 , the work done by the gas lowers its internal energy and hence temperature falls. Thus from first Law of Thermodynamics U = w = P1V1 - P2V2
E 2 E1 P1V1 P2V2 ( E 2 P2V2 ) ( E1 P1V1 ) 0 H 2 H 1 0 H 0 i.e.Joule - Thomson effect occurs at constant Enthalpy and thus it is isenthalpic .
Introduction and First Law of Thermodynamics
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Joule - Thomson Coefficient ()
dT 1 2a b dP C p RT
if >0, than gas cools on expansion and if 0 and H > 0
(B) H = 0 (D) Both E = 0 and H = 0
Q.5.
For the reaction of one mole zinc dust with one mole sulphuric acid in a bomb calorimeter, U and w correspond to (A) U < 0, w = 0 (B) U < 0, w < 0 (C) U > 0, w = 0 (D) U > 0, w > 0
Q.6.
The work done in ergs for a reversible expansion of one mole of an ideal gas from a volume of 10 litre to 20 litre at 25°C is : (A) 2.303 × 8.31 × 107 × 298 log2 (B) 2.303 × 0.0821 × 298 log 2 (C) 2.303 × 0.0821 × 298 log 0.5 (D) 2.303 × 2 × 298 log 2
Q.7.
What is the change in internal energy when a gas contracts from 377 ml to 177 ml under a constant pressure of 1520 torr, while at the same time being cooled by removing 124 J heat? [Take : (1 L atm) = 100 J ] (A) –24 J (B) – 84 J (C) – 164 J (D) –248 J
B
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(D) Viscosity
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Q.8.
The value of H – U for the following reaction at 27°C will be : 2NH3(g) N2(g) + 3H2(g) (A) 8.314 × 273 × (–2) (B) 8.314 × 300 × (–2) (C) 8.314 × 273 × (–2) (D) 8.314 × 300 × 2
Q.9.
For an endothermic reaction, where H represents the enthalpy of reaction, the minimum value for the energy of activation will be : (A) Less than H (B) Zero (C) Equal to H (D) More than H
Q.10. H and U for the reaction, S(s) + (A) H = U – 0.5 RT (C) H = U + RT
3 O (g) SO3(g) are related as : 2 2
(B) H = U – 1.5 RT (D) H = U + 1.5 RT
Q.11. The word ‘standard’ in standard molar enthalpy change implies : (A) Temperature 298 K (B) Pressure 1 atm (C) Temperature 298 K and pressure 1 atm (D) All temperatures and all pressure
Introduction and First Law of Thermodynamics
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Q.12. Which one of the following statements is false ? (A) Work is a state function (B) Temperature is a state function (C) Work appears at the boundary of the system (D) Change in the state is completely defined when the initial and final are specified. Q.13. Which one of the following equations does not correctly represent the first law of thermodynamics for the given process ? (A) Isothermal process : q=–w (B) Cyclic process : q=–w (C) Isochoric process : U = q (D) Adiabatic process : U = –w (E) Expansion of gas into vacuum : U = q Q.14. In which of the following pairs, both properties are intensive ? (A) number of moles, temperature (B) Density, volume (C) Temperature, density (D) Pressure, volume
By
PR
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R I S B U
A
Q.15. From the given graph interpret which of the following is correct ? T
(A) Cv(A) > Cv(B) (B) Cp(A) >Cp(B) (C) Cv(A) = Cv(B) (D) Cv(A) 0 for isothermal expansion. (B) Heat absorbed in isothermal reversible process is greater than isothermal irreversible process. (C) (w)irev > (w) rev for isothermal compression. (D) Heat rejected in reversible isothermal process is greater in (n–1)th stage as compare to nth stage work.
Q.3.
Select the False statement(s): (A) E = q + w for every thermodynamic system at rest in the absence of external field. (B) q = 0 for every cyclic process (C) In isothermal free expansion q = 0, w = 0, E = 0, H = 0
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H = V where V is molar volume of solid, liquid or gas. (D) m m P T
B
Q.4.
Select the correct option(s). (A) Entropy change will be same for the adiabatic reversible and irreversible process if carried out from same initial states to same final volume. (B) A closed system having all adiabatic boundaries must be isolated system. (C) Work done in a cyclic process may be zero (D) Area enclosed by a cyclic process on PV diagram is same as that of TS diagram.
Q.5
Which of the following is true for reversible adiabatic process involving an ideal gas? (A) Gas with higher has high magnitude of slope in a P (y-axis) v/s T (x-axis) curve (B) Gas with higher has high magnitude of slope in a V (y-axis) v/s T (x-axis) curve (C) Gas with higher has high magnitude of slope in a P (y-axis) v/s V (x-axis) curve (D) Gas with higher has low magnitude of slope in a P (y-axis) v/s T (x-axis) curve
Q.6.
Select the correct option. (A) Adiabatic free expansion of a real gas having temperature less than inversion temperature results in decrease in kinetic energy of gas. (B) More heat is absorbed by an ideal gas if subjected to reversible isothermal process than an irreversible isothermal process between same states. (C) Final temperature of an ideal gas will be more for adiabatic reversible expansion process than
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Q.7.
Q.8.
Introduction and First Law of Thermodynamics isothermal reversible expansion process if carried out from same initial state to same final pressure. (D) On isothermal expansion of an ideal gas upto same final volume from same initial state, final pressure will be more for irreversible process than reversible process. Which of the following is/are correct statement(s): (A) In Adiabatic expansion heat is released. (B) In adiabatic compression heat is released. (C) In adiabatic compression heat is absorbed. (D) H = nCpT is applicable to only reversible adiabatic process. 2 moles of an ideal monoatomic gas expand irreversibly and adiabatically from an intial pressure of 10 atm against a constant external pressure of 1 atm, until the temperature drops from initial value of 325 K to a final value of 275 K. Which of the foolowing is correct (A) w = +1247 J (B) U = -1247 J (C) Vfinal = 5.33 L (D) Vfinal = 17.63 L
Q.9. Isothermal reversible expansion
p,V,T
1
Adiabatic reversible expansion Adiabatic irreversible expansion
PR
Select the correct statement(s): (A) T2>T (B) p2 wirr. Graphically work is calculated by calculating the area under the P-V graph. On the basis of above information answer the following : Page 29
Q.1.
If w1, w2, w3 and w4 are work done in isothermal, adiabatic, isobaric and isochoric reversible processes respectively then the correct sequence would be (A) w1>w2>w3>w4
A B
(D)w3>w1>w2>w4
(A) H2 and O2
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Plot A and plot B should correspond to:
V o lu m e
Q.3.
(C) w3>w2>w4>w1
P-V plots for two gases during adiabatic processes are given as
p re s su re
Q.2.
(B) w3>w2>w1>w4
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(B) He and Ar
B
(C) O2 and He (D) O2 and F2
Select the incorrect statment (A) The temperature decrease in reversible process is more than irreversible process (B) Work done in reversible compression is less than the work done in irreversible process (C) Maximum work in an isothermal expansion is obtained when Pext = Pgas (D) The heat released in reversible process is less than irreversible process.
C-(4) For any given gas the change in internal energy follows the given differential equation -
Q.1. Q.2.
Q.3.
U U dU dT dV T V V T the expression for internal energy of a gas was found to be U = aV2 + bTV Where a = 1 Jdm-6 and b = 0.5JK-1dm-3. If initial volume of the gas is 1LthanFor an isothermal change the change in internal energy when volume increases by 0.5 L at 27oC (A) Zero (B) 76.5 J (C) 65 J (D) 85.5 J For an ideal gas if temperature changes by 50oC and volume changes by 0.5 L than internal energy will change by (A) 15 J (B) No change (C) 37.5 J (D) 2.5kJ o For a real gas at 27 C, going through an adiabatic change the temperature changes by 10oC and volume changes by 0.5 L, then change in internal energy will be(A) Zero (B) 35.5J (C) 85.5J (D)42.87J
Introduction and First Law of Thermodynamics
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Q.1.
Q.2.
“MATCH YOUR SKILLS” Column I gives the reaction of internal energy change dU and column II gives the gases with requisite condition. [Assume CV to be independent of T] Column I Column II (A)
dU = nCV dT
(B)
dU = nCVdT + n2 a
(C)
dU = n2 a
dV V2
dV V2
(P)
Ideal gas
(Q)
Vander Waal's gas
(R)
Vander Waal's gas under isothermal condition.
(S)
Vander Waal's gas under isochoric condition.
Column I
Column II
(A) w = 0
(P) Isothermal free expansion
(B) w =
P2V2 P1V1 1
p CV R 2 p1 T2 T1 (C) Cp
B
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E C IN
R I S
(Q) Irreversible adiabatic expansion
(R) Adiabatic free expansion (S) Reversible adiabatic expansion.
Q.3.
Given a P-V curve for different types of gases. Match the following -
Column I
Column II
(A) Curve-1
(P) CO2
(B) Curve-2
(Q) H2
(C) Curve -3
(R) He (S) SO2
Introduction and First Law of Thermodynamics
Page 31
Q.4.
from the given graph match the following
Column I
Column II
(A) -2026
(P) TB
(B) -622.06
(Q) WAB
(C) 0
(R) Wtotal
Q.5.
E C IN (S) U
(D) 121.8
R I S
One mole of a certain gas obeys the equation of state P(V - b) = RT and has a constant molar heat capacity CV. Parameter b is also a constant . Match the following on the basis of given information-
B
Column I(Process)
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Column II(H)
(B) Isobaric reversible process
(CV R) P(V2 V1 ) R (Q) Cp (T2 - T1) + b(P2 - P1)
(C) Isochoric reversible process
(R)
(D) Adiabatic irreversible process
(S) b (P2 - P1)
(A) Isothermal reversible process
(P)
CP (V b) (P2 P1 ) b (P2 P1 ) R
Introduction and First Law of Thermodynamics “ASSERTIONS AND REASONS” (A) Asseration is true; reason is true; reason is a correct explanation for assertion. (B) Asseration is true; reason is true; reason is not a correct explanation for assertion. (C) Asseration is true; reason is false (D) Asseration is false; reason is true Page 32
1.
Assertion : Reason :
2.
3.
4.
5.
6.
7.
Assertion : Reason :
The heat absorbed during the isothermal expansion of an ideal gas against vaccum is zero. The volume occupied by the molecules of an ideal gas is zero.
Assertion :
Absolute values of internal energy of substances can not be determined.
Reason :
It is impossible to determine exact values of constituent energies of the substances.
Assertion : Reason :
In an isothermal process work done by the system equals the heat released by the system. Enthalpy change is zero for an isothermal process.
Assertion :
H and E are state functions.
Reason :
Values of H and E depends on initial and final state of the system.
Assertion :
A reversible process occures very slowly.
Reason :
Driving force is infinitesimally smaller than opposing force in a reversible process.
Assertion :
U 0 for an ideal gas. V T There is no interaction between the molecules in an ideal gas.
Reason : 8.
9.
10.
The extensive property of a single pure substance depends upon the number of mole of the substance present. Any extensive property expressed per mole becomes intensive.
Assertion :
B
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R I S
Reason :
When Hydrogen gas at high pressure and room temperature expands adiabatically into a region of low pressure, there is a decrease in temperature. Hydrogen gas at room temperature is above its inversion temperature.
Asserton :
When a gas at high pressure expands against vaccum, the work done is maximum.
Reason :
Work done depends on external pressure and increase in volume.
Assertion : Reason :
For a thermodynamic process Wrev > Wirr. Temperature decrease in reversible process is more than in irreversible process.
Introduction and First Law of Thermodynamics
Page 33
“FEEL THE HEAT” Q.1
Among the following, the state function(s) is (are) (A) Internal energy (B) Irreversible expansion work (C) Reversible expansion (D) Molar enthalpy
[JEE 2009]
Q.2
Statement-1: There is a natural asymmetry between converting work to heat and converting heat to work. and Statement-2: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. [JEE 2008]
Q.3
F
o
r
t
h
e
p
r
o
c
e
s
s
H
parameters is (A) G = 0, S = +ve (C) G = +ve, S = 0 Q.4
R I S
2O (l) (1 bar, 373 K) H2O (g) (1 bar, 373 K), the correct set of thermodynamic
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E C IN
(B) G = 0, S = –ve (D) G = –ve, S = +ve
[JEE 2007]
A process A B is difficult to occur directly instead it takes place in three successive steps. S (A C ) = 50 e.u. S (C D ) = 30 e.u. S (B D ) = 20 e.u.
B
where e.u. is entropy unit. Then the entropy change for the process S (A B) is (A) +100 e.u. (B) –60 e.u. (C) –100 e.u.
(D) + 60 e.u. [JEE 2006]
Q.5
The molar heat capacity of a monoatomic gas for which the ratio of pressure and volume is one. (A) 4/2 R (B) 3/2 R (C) 5/2 R (D) zero [JEE 2006]
Q.6
One mole of monoatomic ideal gas expands adiabatically at initial temp. T against a constant external pressure of 1 atm from one litre to two litre. Find out the final temp. (R = 0.0821 litre. atm K–1 mol–1) (A) T
(C) T – Q.7
(B)
2 3 0.0821
T 5 1 ( 2) 3
(D) T +
2 3 0.0821
[JEE 2005]
Two mole of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is
Introduction and First Law of Thermodynamics
Page 34
Q.8
(A) 11.4 kJ (B) –11.4 kJ (C) 0 kJ (D) 4.8 kJ [JEE 2004] –1 The enthalpy of vapourization of a liquid is 30 kJ mol and entropy of vapourization is 75 J mol –1 K. The boiling point of the liquid at 1 atm is (A) 250 K (B) 400 K (C) 450 K (D) 600 K [JEE 2004]
Q.9
One mol of non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) to (4.0 atm, 5.0 L, 245 K) with a change in internal energy (U) = 30.0 L-atm. The change in enthalpy (H) of the process in L-atm. (A) 40.0 (B) 42.3 (C) 44.0 (D) not defined, because pressure is not constant [JEE 2002]
Q.10
Which of the following statement is false? (A) Work is a state function (B) Temperature is a state function (C) Change of state is completely defined when initial and final states are specified. (D) Work appears at the boundary of the system.
SUBJECTIVE
Q.11
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R I S
[JEE 2001]
Onemole of an ideal gas is taken from a to b along two paths denoted bythe solid and the dashed l lines as shown in the graph below. If thework done along the solid line path isws and that along the dotted line path is Wd, then the integer closet to the ratio Wd /Ws is[JEE 2010]
B
Q.12. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 k to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJK-1, the numerical value for the enthalpy of combustion of the gas in kJmol-1 is ?
Introduction and First Law of Thermodynamics
Page 35
Q.13
For the reaction, 2CO(g) + O2(g) 2CO2(g); H = – 560 kJ mol–1 In one litre vessel at 500 K the initial pressure is 70 atm and after the reaction it becomes 40 atm at constant volume of one litre. Calculate change is internal energy. All the above gases show significant deviation from ideal behaviour. (1 L atm = 0.1 kJ) [Ans = -557 kJ/mol] [JEE 2006]
Q.14
One mole of a liquid (1 bar, 100 ml) is taken in an adiabatic container and the pressure increases steeply to 100 bar. Then at a constant pressure of 100 bar, volume decreases by 1 ml. Find U and H [JEE 2004]
Q.15 (a) (b) (c)
Two moles of a perfect gas undergoes the following processes : a reversible isobaric expansion from (1.0 atm, 20.0 L) to (1.0 atm, 40.0 L); a reversible isochoric change of state from (1.0 atm, 40.0 L) to (0.5 atm, 40.0 L); a reversible isothermal compression from (0.5 atm, 40.0 L) to (1.0 atm, 20.0 L); (i) (ii) (iii)
Q.16
[JEE 2002]
Sketch with labels each of the processes on the same P-V diagram. Calculate the total work (w) and the total heat change (q) involved in the above processes. What will be the values of U, H and S for the overall process ?
E C IN
R I S
A sample of argon gas at 1 atm pressure and 27°C expands reversibly and adiabatically from 1.25 dm3 to 2.50 dm3. Calculate the enthalpy change in this process. Cv.m.for argon is 12.48 JK–1 mol–1. [JEE 2000]
B
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Introduction and First Law of Thermodynamics
Page 36
“BOOST YOUR BASIC” 1.B 15.D 29.C 43.C
2.D 16.B 30.B 44.C
1. A,C 8. B,D
3.A 17.D 31.A 45.A
4.D 18.C 32.C 46.B
5.A 19.A 33.A 47.A
6.A 20.C 34.C 49.A
7.B 21.A 35.D
8.D 22.A 36.C
9.C 23.B 37.B
10.A 24.C 38.C
11.B 25.B 39.C
12.A 26.C 40.C
“CHOOSE YOUR FAVOURITE” 3. B 4. C,D 5.C,D 6. A,B 10.A,B,C,D 11. A,D “FACE THE CHALLENGE” (Subjectives) 2. [ E = 556.74J, H= 927.9 J] 3.[ 8]
2. A,B,D 9. B,C,D
1.[T2 = 662.2 K]
13.D 14.C 27.A 28.C 41.B. 42.D
7. A,C
4. 25% 5. [-8.3657 kJ mol-1] 6. (i)[ q=-w = 17.54 kJ,U (ii) [ q=0, U = w=-10.356 kJ, kJ] (iii) [ q= U = w= (iv) [q=-w= 10kJ, U =H=0]
R I S
7. [q = 3708.59 Jmol-1,w = -3583.88 J mol-1, U = 124.71 Jmol-1, H = 207.85J mol-1] 8. [ w = 0.00279 J]
E C IN
9. [(a) 1040.8 J, 1456.5 J, (b) 1043.3 J, 1459 J]
10. [-2271.1 J & zero] 11. [ w= -11.5 kJ] 12. [ 1350 J] 13. [U=501 J H = 99.5 kJ] 14. [w = -101.8 kJmol-1] 15. [ w = -8.9 kJ] -1 -1 16. [H=+41KJmol , U = +38kJmol ] 17. [ 2091.49 kJ] 18. [Ans-3]
C-(1) C-(2) C-(3) C-(4)
1. B 1.C 1.D 1.B
B
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“GO COMPREHENSIVE” 2. C 3.A 2.B 3.A 2. C 3.A 2.C 3.C
“MATCH YOUR SKILLS” Q.1 A(P,Q), BQ,CR Q.2 A (P,R), BS, C Q Q.4.AQ, BR,CS, D P Q.5.AS, BP,CQ, D R ASSERTIONS AND REASONS 1. B 2. C 3.A 4.D 5.D 6.C 7.A 8.D 9.D 10.B
Q1. A,D Q.7 C
Q.2 Q.8
“FEEL THE HEAT” (Previous years JEE questions) A Q.3 A Q.4 D B Q.9 C Q.10 A
Q.5
Q.3.AR, B(P,Q),CS
A
Q.6
C
Q.11. 2 Q.12. 9 kJ Q.13. –557 kJ/mol Q.14 U = 0.1 litre atm, H = 9.9 litre atm Q.15 (ii) –w = q = 620.77 J, (iii) H = 0, U = 0, S = 0 Q.16 H = –114.52 J
Q.9. Q.10.
Q.11.
Q.12.
Q.13.
Select the correct alternate about entropy : (A) Tlim S=0 (B) Tlim S= 0 Which is not true about G ? (A) G is a state function (C) If G is positive, reaction is spontaneous Which is not true about G ? (A) G is state function (C) If G is positive, reaction is spontaneous An exothermic reaction with S = – ve, is : (A) Spontaneous at all the temperature (C) nonspontaneous at low temperature For the equilibrium at 298 K 2H2O H3O+ + OH¯ G° is approximately : (A) 100 kJ (B) –80 kJ
(C) Tlim S=0 0
(D) S(liquid) > S(vapour)
(B) Decrease in G, (–G) is equal t net work done (D) G = G° if K = 1 (B) Decrease in G, (–G) is equal to net work done (D) G = G° if K = 1 (B) Spontaneous at high temperature (D) Spontaneous at low temperature
(C) 80 kJ
(D) –100 kJ
Column I gives the reaction of internal energy change dU and column II gives the gases with requisite condition. [Assume CV to be independent of T] Column I Column II (A)
dU = nCV dT
(B)
dU = nCVdT + n2 a
(C)
dU = n2 a
Q.26
(S)correct? Vander Waal's gas under isochoric condition. Which of the following statement(s) is/are (A) The quantities E, H and G have the same dimension (B) Gibb’s free energy of 10 gm ice at 0°C and 1.0 atm is less than the Gibb’s free energy of 10 gm water at 0°C and 1 atm. (C) Ssys = 0 for every adiabatic process in a closed system.
dV V2
dV V2
(P)
Ideal gas
(Q)
Vander Waal's gas
(R)
Vander Waal's gas under isothermal condition.
(D) For every reversible process, in a closed system, Ssys =
H sys
T Standard enthalpies of formation of O3, CO2, NH3 and HI are 142.2, –393.2, –46.2 and +25.9 kJ mol–1 respectively.The order of their increasing stabilities will be : (A) O3, CO2, NH3, HI (B) CO2, NH3, HI, O3 (C) O3, HI, NH3, CO2 (D) NH3, HI, CO2, O3 Q.1 Which among of the following represents the reaction of formation of the product ? (A) C(diamond) + O2(g) CO2(g) (B) S(monoclinic) + O2(g) ® SO2(g) (C) 2N2(g) + O2(g) 2N2O(g) (D) None of the above
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