6. HEAT & THERMODYNAMICS.pdf

October 22, 2017 | Author: Ashok Pradhan | Category: Heat Capacity, Heat, Heat Transfer, Gases, Temperature

Description

HEAT & THERMODYNAMICS

These topics are taken from our Book: ISBN : 9789386320063 Product Name : Heat & Thermodynamics for JEE Main & Advanced (Study Package for Physics) Product Description : Disha's Physics series by North India's popular faculty for IIT-JEE, Er. D. C. Gupta, have achieved a lot of acclaim by the IIT-JEE teachers and students for its quality and indepth coverage. To make it more accessible for the students Disha now re-launches its complete series in 12 books based on chapters/ units/ themes. These books would provide opportunity to students to pick a particular book in a particular topic. Laws of Motion and Circular Motion for JEE Main & Advanced (Study Package for Physics) is the 6th book of the 12 book set. • The chapters provide detailed theory which is followed by Important Formulae, Strategy to solve problems and Solved Examples.

• Each chapter covers 5 categories of New Pattern practice exercises for JEE - MCQ 1 correct, MCQ more than 1 correct, Assertion & Reason, Passage and Matching based Questions. • The book provides Previous years’ questions of JEE (Main and Advanced). Past years KVPY questions are also incorporated at their appropriate places. • The present format of the book would be useful for the students preparing for Boards and various competitive exams.

Contents 5. Thermometry, Expansion & Calorimetry

365-414

5.1 Introduction 366 5.2 Temperature 366 5.3 Thermal expansion of solids 370 5.4 Expansion of liquids 373 5.5 Expansion of gases 376 5.6 Heat and calorimetry 382 5.7 Special heat 382 5.8 Heat capacity or thermal capacity 385 5.9 Water equivalent 385 5.10 Change in phase 386 5.11 Latent heat 387 5.12 Law of mixture or law of calorimetry 389 Review of formulae & important points 392 Exercise 5.1 - Exercise 5.6 393-404 Hints & solutions 405-414

6. Kinetic Theory of Gases

415-456

6.1 Introduction 416 6.2 Gas laws 416 6.3 Ideal gas 418 6.4 Kinetic theory of gases 423 6.5 Translational kinetic energy 424 6.6 Graham’s law of diffusion 425 6.7 Dalton’s law of partial pressure 426 6.8 The distribution of molecular speed 426 6.9 Degrees of freedom 429 6.10 Law of equipartition of energy 431 6.11 Mean free path 432 6.12 Phases and phase diagrams 433 6.13 Vapour pressure 436 6.14 Dew point 437 6.15 Humidity 437 Review of formulae & important points 438 Exercise 6.1 - Exercise 6.6 439-448 Hints & solutions 449-456

7. Laws of Thermodynamics

Contents

7.1 7.2 7.3 7.4 7.5 7.6

Thermodynamical terms Internal energy Internal energy of an ideal gas Work in volume change Work done in cyclic process First law of thermodynamics

457-526 458 458 458 459 460 461

7.7 Reversible process 7.8 Thermodynamical processes 7.9 Change in internal energy 7.10 Polytropic process 7.11 Weakness of first law and need of second law 7.12 Second law of thermodynamics 7.13 Entropy 7.14 Heat engine 7.15 Carnot reversible heat engine 7.16 Refrigerator or heat - pump Review of formulae & important points Exercise 7.1 - Exercise 7.6

478 478 479 479 482 485 486-508

509-526

Hints & solutions

8. Heat Transfer

464 464 469 470 477

527-570

8.1 Modes of heat transfer 8.2 Thermal conduction 8.3 Steady state and temperature gradient 8.4 Rate of flow of heat : heat current 8.5 Thermal resistance 8.6 Determination of thermal conductivity 8.7 Combination of metallic rods 8.8 Redial flow of heat 8.9 Cylindrical flow of heat 8.10 Formation of ice on ponds 8.11 Application of conductivity in daily life 8.12 Convection 8.13 Phenomenon based on convection 8.14 Radiation 8.15 Spectral absorptive power 8.16 Spectral emissive power 8.17 Emissivity

528 528 528 529 529 530 532 533 534 534 539 539 539 541 541 541 542

8.18 Black body 542 8.19 Kirchhoff’s law 542 8.20 Prevost theory of heat exchange (1792) 543 8.21 Stefan -boltzmann law 543 8.22 Newton’s law of cooling 543 8.23 Wien’s displacement law 547 8.24 Solar constant 547 Review of formulae & important points 548 Exercise 8.1 - Exercise 8.6 549-560

Hints & solutions

561-570

Chapter 5

382

Thermometry, Expansion & Calorimetry

5.7 SPECIAL HEAT It is defined as the amount of heat required to raise the temperature of unit mass of a substance through 1 K (1 °C). Suppose Q amount of heat is supplied to m amount of substance, the rise in temperature of substance is DT , then specific heat is given by Q c = m DT Q = mcDT or we can write,

Units of specific heat

In CGS system, the unit of Q is calorie, m is gram and DT in °C. Therefore unit of c in this system becomes c = cal/g –°C (ii) In SI system, the unit of Q is Joule, m in kg and DT in kelvin. Therefore in this system unit of specific heat becomes J/kg-K. 1. Specific heat of water is 1cal/g-°C or 4200 J/kg-K. 2. Specific heat of ice is 0.5 cal/g-°C or 2100 J/kg-K. 3. The maximum value of specific heat is 3.5 cal/g°C for hydrogen. 4. The minimum value of specific heat is 0.022 cal/g°C for radon. The specific heat of a substance is not constant at all temperature. Therefore specific heat used in the above formula is the mean value of specific heats. When c varies considerably with temperature, then for small change in temperature dT, we can write dQ = mcdT (i)

T2

\

Q =

ò mcdT

T1

Here T1 and T2 are the initial and final temperatures.

Dulong and Petit’s Law (1819)

“At near about room temperature the molar specific heat of most of the solids is equal to 3R or 6 cal/mol-K at constant volume”. In case of solids, the significant motion of atoms are vibratory motion. During vibration, the kinetic energy (Ek) of an atom changes periodically into potential energy (EP) and vice-versa. So the average values of Ek and Ep are equal. For each form of energy there are three degrees of freedom. Therefore a molecule has six degrees of freedom (3 for kinetic + 3 for potential). According to law of equi-partition of kT energy, each degree of freedom possesses energy per atom. Therefore total energy 2 associated with one mole of a substance at a temperature T is given by; kT 3 = kT Ek = 3 ´ 2 2 kT 3 = kT EP = 3 ´ 2 2 \ Average vibrational energy per atom = Ek + E p = 3kT The internal energy (due to vibration) of one mole of an atom of the solid is given by

Also, we have

or Proof of formula CV =

U = (3kT ) ´ N = 3(kN)T = 3RT dU CV = dT d (3RT ) = dT CV = 3R dU , will be discussed in the next chapter.. dT

(kN = R)

THERMOMETRY, EXPANSION AND CALORIMETRY

383

Variation of specific heat of solid with temperature The figure shows the variation of molar specific heat (CV) as a function of temperature. It can be easily understand that at higher temperature the molar specific heat of all solids is close a value 3R.

Specific heat of gas Limit of specific heat of gas : Consider a gas of mass m and volume V at a pressure P. (i)

Suppose gas is compressed suddenly without supplying heat : Let the temperature of the gas rises by DT . \

c

=

DQ m DT

But DQ = 0 , \ c = 0. (ii)

Heat is supplied to the gas and it is allowed to expand in such a way that there is no rise in temperature. i.e., DT = 0 \

Fig. 5.24

c

=

DQ m DT

DQ =¥ m´ 0 Thus the specific heat of a gas may varies from zero to infinity. It may have any positive or negative value. The exact value depends on the conditions of pressure and volume when heat is being supplied. Out of the many specific heats of a gas, two are of prime significance.

=

1.

Molar specific heat at constant volume CV It is the amount of heat required to raise the temperature of 1 mole of a gas through 1 K (1 °C) at constant volume. If QV is the heat given to n moles of a gas at constant volume and change in temperature be DT , then

or 2.

CV

=

Qv n DT

QV

=

nCv DT

Molar specific heat at constant pressure CP It is the amount of heat required to raise the temperature of 1 mole of a gas through 1K (1°C) at constant pressure. If QP is the heat given to n moles of a gas at constant pressure and change in temperature be DT , then

or

CP

=

QP n DT

QP

=

nCP DT

Relation between CV and CP : Mayer’s formula Heat supplied to a gas at constant volume entirely used to raise its temperature. When a gas is heated at constant pressure, it expand to keep pressure constant and therefore some mechanical work is to be done in addition to raise the temperature of the gas. Hence more heat is required at constant pressure than that at constant volume. Thus for one mole of a gas, we have

384

MECHANICS, HEAT, THERMODYNAMICS & WAVES CP – CV = Work done =

PDV

… (1)

PV1

=

RT

… (2)

PV2

=

R (T + 1)

… (3)

At constant pressure, we have

and

where V1 is the volume of gas at temperature T and V2 is the volume of gas at temperature (T + 1). Subtracting equation (2) from (3), we get

or

P (V2 - V1 )

= R

PDV

= R

Substituting this value in equation (1), we get CP - CV

= R

Mayer’s formula

Note: 1.

Substance which expand on heating, PDV is positive and therefore, CP - CV = +ve or CP > CV . If any substance contracts on heating, PDV will be

negative and therefore, CP - CV = -ve or CP < CV . 2.

CP - CV = R holds good for all ideal gases.

3.

For one gram of a gas, we have cV and cPand we can write CV = McV and CP = McP . Also, cP - cV = r R Here r = which is different for different gases. M

Fig. 5.25

Chapter 7

Laws of Thermodynamics 7.8 THERMODYNAMICAL PROCESSES Any process may have own equation of state, but each thermodynamical process must obey PV = nRT.

1.

Isobaric process If a thermodynamic system undergoes physical change at constant pressure, then the process is called isobaric. (i) Isobaric process obeys Charle’s law, V µ T dP (ii) Slope of P ~ V curve, = 0. dV (ii) Specific heat at constant pressure

Fig. 7.21

5R 7R for monoatomic and CP = for diatomic 2 2 (iv) Bulk modulus of elasticity: As P is constant, DP = 0

CP =

and

B =

DP =0 æ DV ö ç ÷ è V ø

(v) Work done: W = PDV = nRDT (vi) First law of thermodynamics in isobaric process Q = DU + W = DU + PDV = DU + nRDT =

Fig. 7.22

nCV DT + nR DT = n(CV + R)DT

= nCP DT (vii) Examples: Boiling of water and freezing of water at constant pressure etc.

2.

Isochoric or isometric process A thermodynamical process in which volume of the system remain constant, is called isochoric process. (i)

An isochoric process obeys Gay - Lussac’s Law, P µ T

dP =¥ dV (ii) Specific heat at constant volume

(ii) Slope of P – V curve,

3R 5R for monoatomic and CV = for diatomic 2 2 (iv) Bulk modulus of elasticity : As V is constant, DV = 0

CV =

DP = ¥` æ -DV ö ç ÷ è V ø W = PDV = 0 (v) Work done : (vi) First law of thermodynamics in ischoric process Q = DU + W = DU+ 0 or Q = DU = nCVDT

\

3.

B =

Isothermal process A thermodynamical process in which pressure and volume of the system change at constant temperature, is called isothermal process. (i) An isothermal process obeys Boyle’s law PV = Constant. (ii) The wall of the container must be perfectly conducting so that free exchange of heat between the system and surroundings can take place. (iii) The process must be very slow, so as to provide sufficient time for the exchange of heat.

Fig. 7.23

LAWS OF THERMODYNAMICS

465

(iv) Slope of P – V curve: For isothermal process PV = Constant After differentiating w.r.t. volume, we get P +V

dP dV

= 0

dP -P -P = or tan q = dV V V (v) Specific heat at constant temperature: As DT = 0,

or

Fig. 7.24

DQ =¥ n DT (vi) Isothermal elasticity: Bulk modulus at constant temperature is called isothermal elasticity. It can be defined as DP dP = Eiso = B = -DV æ dV ö ç ÷ V è V ø dP From above = P æ - dV ö ç ÷ è V ø

\

C =

\

Eiso

= P Vf

(vi) Work done :

W =

ò PdV

Vi

nRT PV = nRT Þ P = V

By

Vf

\

W =

æ dV ö ÷ ø

ò nRT çè V

Vi

V

f = nRT lnV V i

or æ Vf Here çç V è i

W =

ö ÷÷ ø

ö ÷÷ is called expansion ratio. ø

Also PiVi = PfVf,

\

æVf nRT ln çç è Vi

\

Vf Vi

=

W =

Pi Pf æVf nRT ln çç è Vi

æ Pi ö ÷÷ = nRT ln çç ø è Pf

(viii) First law of thermodynamics in isothermal process. As DT = 0, \ DU = 0 Q = DU + W = 0 + W or Q = W

ö ÷ ÷ ø

466 4.

MECHANICS & THERMODYNAMICS

Adiabatic process An adiabatic process is one in which pressure, volume and temperature of the system change but heat will not exchange between system and surroundings. (i) Adiabatic process must be sudden, so that heat does not get time to exchange between system and surroundings. (ii) The walls of the container must be perfectly insulated. (iii) Adiabatic relation between P and V According to first law of thermodynamics dQ = dU + dW For adiabatic process, dQ = 0, \ dU + dW = 0 ...(1) For one mole of gas dU = CVdT and dW = PdV Substituting these values in equation (1), we have CVdT + PdV = 0 ...(2) For one mole of an idea gas, PV = RT ...(3) After differentiating equation (3), we get PdV + VdP = RdT or

dT =

PdV +VdP R

From equation (2)

æ PdV + VdP ö CV ç ÷ + PdV R è ø or CV PdV + CVVdP + RPdV or (CV + R) PdV + CVVdP or CPPdV + CVVdP After rearranging, we get CP dV dP + CV V P

Substituting

= 0 = 0 = 0 = 0

= 0

CP = g , we have CV

dV dP + = 0 V P Integrating equation (4), we get

or

g

dV dP + V ò P

= C

or

glnV + ln P

= C

or

lnV g + lnP

= C

or

ln( PV g )

= C

or

PV g

=

or

PV g

= k

...(4)

eC

Adiabatic relation between V and T & P and T For one mole of gas PV = RT, or P = Substituting in PVg = k, we get

RT V

LAWS OF THERMODYNAMICS æ RT ç è V

ö g ÷V ø

V g -1T

or Also

= k =

V =

æ RT ö Pç ÷ è P ø

\

k = new constant R RT P

g

= k

P1-gT g

or

467

=

k

= another constant

Rg

(iv) Slope of P – V curve : We have PV g = k . On differentiating, dP dV

= 0

or

dP dV

=

-g

P V

or

tan q

=

-g

P V

we have

P g V g-1 + V g

Fig. 7.25

As slope of isothermal curve = -

P V

\ Slope of adiabatic curve = g × slope of isothermal curve. Since g > 1, so slope of adiabatic always be greater than slope of isothermal curve. Q 0 = =0 n DT n DT (vi) Adiabatic elasticity: Bulk modulus of gas at constant heat is called adiabatic elasticity. If can be defined as

(v) Specific heat : Fig. 7.26

C =

dp æ dV ö ç÷ è V ø

As

\ As

= B=

=

DP dP = æ -DV ö æ - dV ö ç ÷ ç ÷ è V ø è V ø

gP

= gP = g Eiso Vf

(vii) Work done : For adiabatic process or

W = PV g

=

ò PdV

Vi

g g PV i i = Pf V f = k

P = kV – g

468

MECHANICS & THERMODYNAMICS Vf

\

W =

=

or

-g

dV

Vi

k

V 1-g (1 - g )

Vf Vi

=

1 [kV f 1-g - kVi1-g ] (1 - g )

=

1 g 1-g [( Pf V f g )V f 1-g - ( PV ] i i )Vi (1 - g )

=

1 é Pf V f - PV i i ùû (1 - g ) ë

W =

Also

ò kV

( PV i i - Pf V f )

(g - 1)

PiVi = nRTi and PfVf = nRTf

\

W =

nR [Ti - T f ] g -1

(viii) First law of thermodynamics in adiabatic process Q = DU + W \ DU = – W

As

Q = 0,

or \

Uf – Ui = – W Uf = Ui – W

Summary of four gas processes : Fig. 7.28 shows four different processes : Isobaric, isothermal, adiabatic and isochoric. Path

Constant

Process

quantity

type

Some results ΔU = Q – W and ΔU = nCV ΔT for all paths

1

P

Isobaric

2

T

Isothermal

Q = W = nRT ln(V f / Vi ); DU = 0

Q = nCP ΔT ; W = PΔV

3

PV g , TV g-1

Q = 0; W = -DU

4

V

Isochoric

Q = DU = nCV DT ; W = 0

Fig. 7.28 P – V diagram representing four different processes for an ideal gas