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PHYSICS & PHYSICAL SCIENCES

Kinetic theory, Thermodynamics

(IIT-JAM/JEST/TIFR/M.Sc Entrance)


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Thermal & Statistical Mechanics

1. Kinetic theory of gases…………………………………………………………………..(1-29) 1.1 Basic assumption of kinetic theory

1.1.1 Pressure exerted by a gas

1.2 Gas Law for Ideal gases:

1.2.1 Boyle’s Law

1.2.2 Charle’s Law

1.2.3 Avogadro’s Law

1.2.4 Graham’s Law of Diffusion

1.2.5 Ideal Gas Equation:

1.3 Kinetic Interpretation of Temperature

1.4 Maxwell-Boltzmann Distribution Law

1.4.1 The Distribution in term of Magnitude

1.4.2 To Determine Value of β in term of Temperature T

1.4.2 Average Velocity

1.4.3 Root Mean Square Velocity

1.4.4 Most Probable Velocity

Questions and Solutions

2. Real Gases……………………………………………………………………….(30-43)

2.1 Andrew’s Experiment on Carbon Dioxide

2.2 van der Waals Equation of State.

2.3 Correction in Ideal Gas Equation to Achieve van der Waals Gas Equation of State.

2.3.1 Correction for Finite Size

2.3.2 Correction for Intermolecular Attraction

2.3.3 Maxwell Equal Area

2.3.4 Critical Point

2.3.5 van der Waals Equation of State and Virial Coefficient



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3. Basics of Thermodynamics and Laws of Thermodynamics…………………(44-79)

3.1 Mathematical Formulations of thermodynamics.

3.1.1 Some important Formulas

3.2 Fundamental Concept

3.2.1 System

3.2.2 Isolated System

3.2.3 Thermodynamical State

3.2.4 State Function

3.2.5 Intensive and Extensive Properties

3.3 The Ideal Gas:

3.4 Laws of Thermodynamics

3.4.1 Zeroth law of Thermodynamics:

3.4.2 First law of Thermodynamics:

3.4.3 Work Done during Different Process.

3.4.4 Specific Heat

3.4.5 Heat Capacity of Ideal Gas:

3.4.6 Molar Heat Capacity

3.4.7 Coefficient of Volume Expansion or Expansivity 3.4.8 Isothermal Elasticity and Isothermal Compressibility

3.5 Different Types of Thermo Dynamical Process and use of First Law of

Thermodynamics

3.5.1 Isochoric Process:

3.5.2 Isobaric Process

3.5.3 Isothermal Process

3.5.4 Adiabatic Process



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4. Second Law of Thermodynamics and Entropy……………………………...(80-110)

4.1 Second Law of Thermodynamics

4.2 Heat Engines

4.2.1 Heat Reservoir

4.2.2 Efficiency of Heat Engine (η)

4.2.3 Carnot Cycle

4.3 Entropy

4.3.1 Inequality of Clausius


5. Maxwell relation and Thermodynamic Potential………………………..(111-142) 5.1 Maxwell relations

5.2 Different types of thermodynamic potential and Maxwell relation

5.2.1 Internal Energy

5.2.2 Enthalpy 5.2.3 Helmholtz Free Energy

5.2.4 Gibbs Energy

5.3 Application of Maxwell Relation

5.3.1 First dST − Equation

5.3.2 Second dST − Equation

5.3.3 Third T-dS Equation:

5.3.4 First Energy Equation

5.3.5 Second Energy Equation



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6. Phase Transition and Low Temperature Physics………………………...(143-167)

6.1 Third Law of Thermodynamics and Attainable of Low Temperature

6.2 Production of Low Temperature and Joule – Kelvin Expansion:

6.3 Phase Transition

6.3.1 First Order Phase Transition

6.3.2 Equilibrium Between Two Phases

6.3.3 Clapeyron-Clausius Equation

6.3.4 Liquid-Vapour Phase Transition

6.3.5 Properties of First Order Phase Transition

6.3.6 Second Order Phase Transition:



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Chapter - 1

Kinetic Theory of Gases

1.1 Basic Assumption of Kinetic Theory:

1. Any infinitely small volume of a gas contains a large number of molecule.

2. A gas is made up identical molecule which behaves as rigid, perfectly elastic, hard

sphere.

3. The molecules continuously move about in random directions. All directions of motion

are equally probable.

4. The size of the molecules is much less than the average distance between them.

5. The molecule of a gas exert no force on each other except when they collide.

6. The collision between molecules and with walls are perfectly elastic.

7. The direction of molecular velocities are assumed to be distribute uniformly.

8. The molecules move with all speeds ranging from 0 to ∞.

9. The time of collision is much less than the time between collisions.

1.1.1 Pressure Exerted by a Gas

Suppose there are n molecules per cubic meter each of mass m, and its is assumed that ni

no. of molecule have velocity iv .

Mathematically

nni =∑ and 2222iziyixi vvvv ++=

where ixv iyv and izv are x, y, z component of velocity of gases.

From assume of kinetic theory of gases 222iziyix vvv ==

2

3iv

=

suppose molecules are kept in the cubic container of parameter L .

A molecule moving in the x direction will have momentum ixmv normal to face of the

cube before collision

( ) ixixixix mvmvmvP 2=−−=Δ

Force acting on the wall by molecule is Lmvn

Lmvn

tmvnf ixiixiixi

ix

22

222

===Δ

=





Pressure exert on the wall of container by molecule 2

3i ix

ixmn vP

L=

so that pressure in the x direction expected by all group

23 ixiixx vn

LmPP ∑=∑=

Average value of v2 is given by

∑∑

=i

iixi

xn

vnv

2

2

2

1i ix

i

n v

n==∑

For three dimensional system 2 2 2 2x y zv v v v+ + = and

for isotropic system 2

2 2 2

3x y z

vv v v= = =

So Px can be written as

23 xx vn

LmP = , 2

331 vn

LmPP x ==

Vvmn

P2

31

=

2

31 vmNPV =

where V is volume of the container and 2v is average value of square of velocity.

1.2 Gas Law for Ideal Gases:

1.2.1 Boyle’s Law

At constant temperature ( )T , the pressure ( )P of a given mass a gas is inversely

proportional to its volume (V)

V

P 1∝

1.2.2 Charle’s Law

At constant pressure ( )P the volume of a given mass of a gas is proportional to its

temperature (T)

V T∝





1.2.3 Avogadro’s Law

At the same temperature and pressure, equal volume of all gases contain equal number of

molecules (N).

N1 = N2

1.2.4 Graham’s Law of Diffusion

When two gases at the same pressure and temperature are allowed to diffuse into each

other, the rate of diffusion (r) at each gas is inversely proportional to square root at

density of gas (ρ)

1 2

2 1

rr

ρρ

=

Dalton’s Law of Partial Pressure: The sum of pressure exerted (P) by each gas occupying

the same volume as that of the mixture (P1, P2, P3,….)

P = P1 + P2 + P3 +….

1.2.5 Ideal Gas Equation:

Consider a sample of an Ideal gas at pressure P, volume V and temperature T the gas

follows the equation

PV nRT=

Where n is number of molecules and R is proportionality constant known as gas constant

314.8=R J/mol/K

Boltzmann constant K is ratio between R to Avogadro number NA 23

8.3146.03 10B

A

RkN

= =×

KJkB /103.1 23−×= Example: Find the maximum attainable temperature of ideal gas in each process given

by ;20 Vpp α−= where α,0p and β are positive constants, and V is the volume of one

mole of gas.

Solution: 20 VPP α−= (i)

Number of mole of gas = 1





We know nRTPV = VRTP =⇒ put in (i)

20 VP

VRT α−=

RV

RVPT

30 α

−=⇒ (ii)

For T maximum, 0=dVdT 03 2

0 =−⇒RV

RP α

α30P

V = put in (ii) one will get α33

2 00max

PPT =

Example: Two thermally insulated vessel 1 and 2 are filled with air. They are connected by a

short tube with a value. The volume of vessels and the pressure and temperate of air in

them are 1 1 1(V , P , T ) and ( )2 2 2V , P , T respectively. Calculate the air temperate and

pressure established after opening of value if air follow Ideal gas equation.

Solution: For vessel (1) 1111 RTnVP = 1

111 RT

VPn =

For vessel (2) 2222 RTnVP = 2

222 RT

VPn =

After opening the value let pressure volume and temperature is P, V, T

nRTPV =

21 VVV +=

2

22

1

1121 RT

VPRT

VPnnn +=+=

Hence system is isolated then

Energy of (1) + energy of (2) = energy of composite

( )KTnnKTnKTn 212211 2

323

23

+=+

( ) ./212211 TnnTnTn +=+

21

2211

nnTnTnT

++

=





( )

2

22

1

11

22

221

1

11

RTVP

RTVP

TRT

VPTRT

VP

+

+=

( )1 1 2 21 2

1 1 2 2 2 1

PV PVT T T

PV T PV T+

⇒ =+

nRTPV = V

nRTP = 21

2211

VVVPVPP

++

=

Example: A horizontal cylinder closed from one end is rotated with a constant angular velocity

ω about a vertical axis passing through the open end of the cylinder. The outside air

pressure is equal to 0p , the temperature to T , and the molar mass of air to M . Find the air

pressure as a function of the distance r from the rotation axis. The molar mass is assumed

to be independent of r .

Solution: Force equation of dr element.

( ) 2ωrdmdF = if S is cross section area then

2ωrS

dmS

dFdP ⎟

⎠⎞

⎜⎝⎛== dP

rSdm ⎟

⎠⎞

⎜⎝⎛= 2ω

Also we know

( ) RTMdmSdrP ⎟

⎠⎞

⎜⎝⎛=

( ) dPrS

MRTdrPS ⎟

⎠⎞

⎜⎝⎛= 2ω

∫∫ =P

P

r

PdPRTrdrM

00

2ω

0

22

ln2 P

PRTrM=

ω

RTrM

ePP 20

22ω

=

TM,S

ω

0Pr

This end is open in air





Example: Prove that 2

21 vmNPA = and 2

2 B BE k T k T= = in two dimension.

Solution: A molecule moving in the x direction will have momentum ixmv normal to face of the

cube before collision

( ) ixixixix mvmvmvP 2=−−=Δ

Force acting on the wall by molecule is Lmvn

Lmvn

tmvnf ixiixiixi

ix

22

222

===Δ

=

Pressure exert on the wall of container by molecule 2

3i ix

ixmn vP

L=

So that pressure in the x direction expected by all group

23 ixiixx vn

LmPP ∑=∑=

Average value of 2v is given by

∑∑

=i

iixi

xn

vnv

2

2

2

1i ix

in v

n==∑

For two dimensional system 2 2 2x yv v v+ = and

22 2

2x y

vv v= =

So xP can be written as

22 xx vn

LmP = , 2

221 vn

LmPP x ==

A

vmnP

2

21

=

2

21 vmNPA =





1.3 Kinetic Interpretation of Temperature

According to assumption of Kinetic theory of gases, there is only translation motion of

the molecule and there is not any potential acting between them, so

Average energy E of gases are equivalent to Average translation energy of a molecule

2

21 vmE =

Pressure at P as 2

31 vmnP = ⎟

⎠⎞

⎜⎝⎛= 2

21

32 vmn En

32

=

23

PV Vn E= 23

PV N E= where NnV

= number density

32 A

RTEN

= and TNRE

A⎟⎠⎞

⎜⎝⎛=

23

32 BE K T= where Bk is Boltzman constant

So average kinetic energy is given by

TkE B23

= where T is absolute temperature.

Example: It is possible to treat electromagnetic radiation in container whose wall is mirrors, as a

gas of particle (photons) with a constant speed c and whose energy is related to their

momentum p which is directed parallel to their velocity by E pc= .Show that if

container is full of radiation the equation of state is 13

PV E=

Solution: Pressure 2

31 vnmP = vmvn ⋅=

31 vpn ⋅=

31

For Photon v c= and velocity is parallel to momentum, so

PcnP31

= pcVNP

31

=⇒

13

PV Npc= 13

PV E⇒ =





1.4 Maxwell-Boltzmann Distribution Law:

Distribution of Molecular velocity in perfect gas.

Maxwell-Boltzmann distribution law is applicable for Ideal

gas where molecules have no vibrational or rotational

energies.

In the equilibrium state of the molecules have complete

random motion and probability that a molecule has a given velocity component is

independent of other two components.

In given figure dv is volume element in velocity space for a molecule at velocity

( )zyx vvvv ,,≡ .

2222zyx vvvv ++=

We need to calculate number of molecules simultaneously having component in the range

xv to yxx vdvv ,+ to yy dvv + and zv to zz dvv +

It is assumptions in Maxwell-Boltzmann distribution law is that probability that molecule

selected at random has velocities in a given range is a function purely at the magnitude of

velocity and the width of the interval.

So fraction of molecule having velocity component in the range xv to yxx vdvv ,+ to

yy dvv + and zv to zz dvv + is ( ) ( ) yyxx dvvfdvvf , and ( ) zz dvvf respectively.

( ) ( ) ( ) zyxzyx dvdvdvvfvfvfN

dN=

where dN is number of molecule having between velocity v to dvv + and N is total

number of molecules.

( ) ( ) ( ) zyxzyx dvdvdvvfvfvfNdN =

Number of molecule having velocity vx to vx + dvx, vy to vy + dvy and vz to vz + dvz is same

as number of molecule having velocity v to v + dv.

So ( ) ( ) ( ) ( ) zyxzyxzyx dvdvdvvFNdvdvdvvfvfvfN 2=

z

x

y

dv

v





F is some function of 2v (magnitude of velocity) and for fixed value of ,v ( )2F v is

constant.

So ( ) 02 =vdF is equivalent to ( ) ( ) ( )[ ] 0=zyx vfvfvfd

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0x x y z y y x z z z x yf v dv f v f v f v dv f v f v f v dv f v f v′ ′ ′+ + =

Dividing both side with f(vx) f(vy) f(vz)

( )( )

( )( )

( )( ) 0=′

+′

+′

zz

zy

y

yx

x

x dvvfvfdv

vfvfdv

vfvf (i)

2v = constant 2222 vvvv zyx =++

vxdvx + vydvy + vzdvz = 0 (ii)

by method of Lagrange’s method of undetermined multiplies multiply by β2 in equation

(ii) and add in equation (i)

( )( )

( )( )

( )( ) 0222 =⎟⎟

⎠

⎞⎜⎜⎝

⎛+

′+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

′+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

′zz

z

zyy

y

yxx

x

x dvvvfvfdvv

vfvfdvv

vfvf βββ

hence yx vv , and zv are independent

( )( )

2 0xx

x

f vv

f vβ

′+ =

( )( ) 2 0y

yy

f vv

f vβ

′⇒ + =

( )( )

2 0zz

z

f vv

f vβ

′⇒ + =

( ) 2xv

xx eAvf β−= ( ) 2yv

yy eAvf β−= ( ) 2zv

zz eAvf β−=

( ) ( ) ( )zyx vfvfvf ,, are probability density, so

( ) ( ) ( )∫∫∫

∞

∞−

∞

∞−

∞

∞−

=== ,1,1,1 zzyyxx dvvfdvvfdvvf

Use the integration

2

1( )0 2

1 12

2

v nn

ne v dvβ

β

∞−

+

+=∫

2

1xvx xA e dvβ∞

−

−∞

=∫ =2

0

2 1xvx xA e dvβ

∞−⋅ =∫





1/ 2

xA βπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

Similarly, 1/ 2 1/ 2

y zA Aβ βπ π

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) 2

2/1xv

x evf β

πβ −⎟⎠⎞

⎜⎝⎛= , ( ) 2

2/1yv

y evf β

πβ −⎟⎠⎞

⎜⎝⎛= , ( ) 2

2/1zv

z evf β

πβ −⎟⎠⎞

⎜⎝⎛=

( )

zyxvvv dvdvdve

NdN

zyx222

2/3++−⎟

⎠⎞

⎜⎝⎛= β

πβ

where ∞<<∞− xv , ∞<<∞− yv , ∞<<∞− zv 1.4.1 The Distribution in Term of Magnitude

2222zyx vvvv ++= which is equation of sphere and zyx dvdvdv can be replace by dvv 24π

( ) ∞<<⎟⎠⎞

⎜⎝⎛== − vdvve

NdNdvvf v 04 2

2/32βπ

πβ

1.4.2 To Determine Value of β in Term of Temperature T.

Mean square velocity ( )2v can be calculated by

( )∫∞

0

22 dvvfv

∫∞

−⎟⎠⎞

⎜⎝⎛

0

42/3

24 dvev vβ

πβπ

3/ 2

5/ 2

14 5 / 22

βππ β

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

πβπ

βπ21

23

214

2/5

2/3

⋅⎟⎠⎞

⎜⎝⎛

β1

232 ⋅=⇒ v

Now average energy of temperature T equivalent to

23 12 2Bk T m v=

3 1 3 12 2 2Bk T m

β= =

2 B

mk T

β =

So ( )( )2 2 23/ 2

2, ,2

x y z

B

m v v v

k Tx y z x y z

B

mf v v v e dv dv dvk Tπ

+ +−⎛ ⎞

= ⎜ ⎟⎝ ⎠

( )vf

v

2T1T

11 TT <





( )23/ 2

2242

B

mvk T

B

mf v dv v e dvk T

ππ

−⎛ ⎞= ⎜ ⎟

⎝ ⎠

1.4.2 Average Velocity

( )dvvvfv ∫∞

=0

23/ 2

2 342

B

mvk T

B

m e v dvk T

ππ

−⎛ ⎞= ⎜ ⎟

⎝ ⎠∫ 8 Bk T

mπ=

1.4.3 Root Mean Square Velocity

[ ] ( )2/1

0

22/12⎥⎦

⎤⎢⎣

⎡= ∫

∞

dvvfvv

2 1/ 21/ 23/ 2

2 2

0

42

B

mvk T

B

m e v dvk T

ππ

∞ −⎡ ⎤⎡ ⎤⎛ ⎞⎢ ⎥⎢ ⎥= ⎜ ⎟⎢ ⎥⎢ ⎥⎝ ⎠⎣ ⎦ ⎣ ⎦∫

3 Bk Tm

=

1.4.4 Most Probable Velocity pv : 0=dvdf

2 Bp

k Tv

m⇒ =





Example: For Maxwellian gas find the v

v 1×

Solution: 8 Bk Tvmπ

= ( )3/ 2

0

1 142 B

m f v dvv k T v

ππ

∞⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠∫

1/ 22

B

mk Tπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

1 4vv π

⇒ × =

Example: If vx and vy are x and y component of velocity then find the average value of

( )2yx bvav +

( ) yxyxyx vvabvbvabvav ⋅++=+ 222222

yxyx vvabvbva 22222 ++=

( )2 2 23/ 22

2

x y z

B

m v v v

k Tx x x y z

B

mv v e dv dv dvk Tπ

+ +∞ ∞ ∞ −

−∞ −∞ −∞

⎛ ⎞= ⎜ ⎟⎝ ⎠

∫ ∫ ∫ = 0

( )2 2 23/ 222 2

2

x y z

B

m v v v

k Tx x x y z

B

mv v e dv dv dvk Tπ

+ +∞ ∞ ∞ −

−∞ −∞ −∞

⎛ ⎞= ⎜ ⎟⎝ ⎠

∫ ∫ ∫ Bk Tm

=

Similarly, 0=yv 2 By

k Tvm

=

( ) yxyxyx vvabvbvabvav 222222 ++=+

2 2 0B Bk T k Ta bm m

= + + ( )2 2Bk T a bm

= +

Example: Write down expression of energy distribution function for Maxwellian gas between E

and E dE+ . Hence find E down 2E .

Solution: 2

21 mvE = ,

( ) 2/12mEdEdv =

( )23/ 2

2 242

B

mvk T

B

mf v dv e v dvk T

ππ

−⎛ ⎞= ⎜ ⎟

⎝ ⎠ put value of v and dv

( )( )

1/ 23/ 2

2 1 0B

Ek T

B

f E dE e E dE Ek Tπ

−

= < < ∞





( )∫∞

=0

dEEEfE 32 BE k T=

( )2 2

0

E E f E dE∞

= ∫ ,

( )2 2 1/ 2

3/ 20

2 1B

Ek T

B

E E e E dEk Tπ

∞ −

= ∫ = ( ) ( )2 22 5 3 15. . . .

2 2 2B Bk T k Tππ

=

Example: Write down expression of energy distribution function for Maxwellian gas between E

and E dE+ in two dimensional system . Hence find E .

2

21 mvE = ,

( ) 2/12mEdEdv =

( )22 / 2

222

B

mvk T

B

mf v dv e vdvk T

ππ

−⎛ ⎞= ⎜ ⎟

⎝ ⎠ put value of v and dv

( ) ( ) ∞<<=−

EdEeTk

dEEf TKE

B

B 01

( )∫

∞

=0

dEEEfE TkE B=⇒

Example: Using the Maxwell distribution function, calculate the mean velocity projection xv

the mean value of the modulus of the modulus of this projection xv if the mass of each molecule is equal to m and the gas temperature isT .

Solution: We know Mean Velocity ∫∞

∞−==

NdNvv x

x

21/ 22

2x

B

m vk T

x xB

mv N e dvk T

Nπ

−∞

−∞

⎛ ⎞⎜ ⎟⎝ ⎠=

∫= 0

Mean speed N

dveTk

mNvv

x

vTk

m

Bx

x

xB∫

∞

∞−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

2

22/1

2π

N

dveTk

mNvv

x

vTk

m

Bx

x

xB∫

∞ −

⎟⎟⎠

⎞⎜⎜⎝

⎛

=0

22/1 2

22

π 2 B

xk Tvmπ

⇒ =





MCQ (Multiple Choice Questions)

Q1. Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas. Let

vmp and vrms denote the most probable velocity and the root mean square velocity,

respectively. The magnitude of the ratio rms

mp

vv

is

(a) 32

(b) 2/3 (c) 3/2 (d) 3 / 2

Q2. For temperature 21 TT > , the qualitative temperature dependence of the probability

distribution ( )vF of the speed v of a molecule in three dimensions is correctly

represented by the following figure

(a) (b)

(c) (d)

Q3. The speed v of the molecules of mass m of an ideal gas obeys Maxwell’s velocity

distribution law at an equilibrium temperatureT . Let ( )zyx vvv , denote the components of

the velocity an Bk the Boltzmann constant. The average value of ( )2

x yv vα β− , where α

and β are constants, is

(a) ( ) mTkB /22 βα − (b) ( ) mTkB /22 βα +

(c) ( ) mTkB /2βα + (d) ( ) mTkB /2βα −

F(v)

2T

1T

v

F(v)

2T

1T

v

F(v)

2T 1T

v

F(v)

1T

2T

v





Q4. The speed v of the molecules of mass m of an ideal gas obeys Maxwell’s velocity

distribution law at an equilibrium temperatureT . Let ( )zyx vvv , denote the components of

the velocity and Bk the Boltzmann constant. The average value of ( )2

x yv vα , where α

and β are constants, is

(a) 0 (b) 2

2 ⎟⎠⎞

⎜⎝⎛

mTkBα

(c) 2

2

2Bk Tm

α ⎛ ⎞⎜ ⎟⎝ ⎠

(d) 2

2 2 Bk Tm

α ⎛ ⎞⎜ ⎟⎝ ⎠

Q5. The statistical energy distribution underlying an ideal gas law gives the number of

molecules with kinetic energies between E and dEE + as ( )dEEN . Which one of the

following expressions can be used to obtain average kinetic energy over the collection of

N molecules?

(a) ( )∫∞

0

1 dEEENN

(b) ( )∫∞+

∞−dEEN

N1

(c) ( )∫∞

0

1 dEENN

(d) ∫∞

∞−dEE

N1

Q6. The plots of Maxwell’s distribution fraction ⎟⎠⎞

⎜⎝⎛

dvdN

versus speed (c) for a given sample of a gas at three

different temperatures 21 ,TT and 3T respectively,

are shown in the above diagram. If the areas on the

c-axis under three curve I, II and III be denoted by

III , AA and IIIA respectively, then which one of the

following is correct?

(a) IIIIII AAA >> (b) IIIIII AAA ==

(c) IIIIII AAA << (d) IIIIII AAA <<

dvdN

I

IIIII

1T

2T

3T

v





Q7. Temperature of an ideal gas is increased such that the most probable velocity of

molecules increase by a factor of 4. By what factor will the rms velocity increase?

(a) 23 (b)2 (c)4 (d)16

Q8. A gas at thermal equilibrium satisfying Maxwell’s velocity distribution

Given =v average speed of the molecules

=pv most probable speed

=rmsv root mean square speed

Select the correct sequence for :,, rmsp vvv

(a) prms vvv >> (b) vvv prms >>

(c) rmsp vvv >> (d) prms vvv >>

Q9. A hypothetical speed distribution for a sample of N gas particles is shown below.

Here ( ) 0=vP for 02vv > . How many particles have speeds between 02.1 v and 09.1 v ?

(a)5N (b)

157N (c)

212N (d) None of these

Q10. A parallel beam of nitrogen molecules moving with velocity v m/s impinges on a wall at

an angle θ to its normal. The concentration of molecules in the beam n cm3. The pressure

exerted by the beam on the wall assuming the molecules to scatter in accordance with the

perfectly elastic collision law is given by

(a) θcos2 2nmv (b) θcos2nmv

(c) θsin2 2nmv (d) θsin2nmv

V 02V0V0

a( )vP

Speed





Q11. If The mass of each molecule is equal to m then The temperature of a gas will the number

of molecules, whose velocities fall within the given interval from v to dvv + be the

greatest

(a) 2

B

mvTk

= (b) 2

2 B

mvTk

= (c) 2

3 B

mvTk

= (d) 2

4 B

mvTk

=

Q12. Using the Maxwell distribution function, the mean value of the modulus of the modulus

of this projection in x direction .e. xv if the mass of each molecule is equal to m and

the gas temperature T is given by

(a) 0 (b)mTkB

π (c)

mTkB (d)

mTkB

π2

Q13. Making use of the Maxwell distribution function, if v1 the mean value of the reciprocal

of the velocity of molecules in an ideal gas and v is the average velocity at a

temperatureT ,if the mass of each molecule is equal to m .then which one of the following

is correct.

(a) 1 1v v

= (b) 1 4v vπ

= (c) 1 2v vπ

= (d) 1 4v v

π=

Q14. If the root mean square velocity of hydrogen molecules exceeds their most probable

velocity by vΔ m/s then temperature is given by

(a) ( )

2

23 2B

m vTk

Δ=

− (b)

( )23 2B

m vTk

Δ=

−

(c) ( )2

3 2B

m vTk

Δ=

− (d)

( )3 2B

m vTk

Δ=

−





Q15. In the case of ideal gaseous in three dimensional the temperature at which the velocities

of the molecules smv /1 and smv /2 are associated with equal values of the Maxwell

distribution function ( )vF

(a) ( )2 2

2 1

2 14 ln /B

m v vT

k v v

−= (b) ( )

12

21

22

/ln4 vvkvvmT

B

+=

(c) ( )21

21

22

/ln4 vvkvvmT

B

−= (d) ( )

21

21

22

/ln4 vvkvvmT

B

+=

Q16. A gas consists of molecules of mass m and is at a temperatureT in three dimension.

Making use of the Maxwell velocity distribution function, the corresponding distribution

of the molecules over the kinetic energies E is given by .

(a) dEEeTk

dEEf TkE

B

B .1)(2/3

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ππ

(b) dEEe

TkdEEf Tk

E

B

B .12)(2/3

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ππ

(c) dEEeTk

dEEf TkE

B

B .1)(2/3

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ππ (d) dEEe

TkdEEf Tk

E

B

B .12)(2/3

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ππ

Q17. A gas consists of molecules of mass m and is at a temperature T in two dimensions.

Making use of the Maxwell velocity distribution function, the corresponding distribution

of the molecules over the momentum p is given by

(a)

pdpeTmk

pf Tmkp

B

B2

2

21)(

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

π (b) pdpe

Tmkpf Tmk

p

B

B2

2

1)(−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

(c) dpeTmk

pf Tmkp

B

B2

2

21)(

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

π (d)

dpe

Tmkpf Tmk

p

B

B2

2

1)(−

⎟⎟⎠

⎞⎜⎜⎝

⎛=





MSQ (Multiple Select Questions)

Q18. Consider the following statements related to kinetic theory of gases are correct:

(a) The molecules of a gas are all alike in size and shape and are hard, smooth, spherical

particles.

(b) The size of the molecules is very small compared to the volume occupied by the gas.

(c) The molecules exert no appreciable force on one another except during a collision.

(d) The collisions of the molecules with the walls of the containing vessel are inelastic.

Q19. Consider the following statements. Which of the following is correct?

(a) The root mean square velocity of molecules of a gas having Maxwellian distribution

of velocities, is higher than their most probable velocity, at any temperature.

(b) A very small number of molecules of a gas which posses very large velocities

increase the root mean square velocity without affecting the most probable velocity

(c) Most probable velocity is lowest among the most probable velocity, average velocity

and root mean square velocity.

(d) Mean square velocity is equal to square of mean velocity

Q20. Consider a collision between an oxygen molecule and a hydrogen molecule in a mixture

of oxygen and hydrogen kept at room temperature. Which of the following are possible?

(a) The kinetic energies of both the molecules increase.

(b) The kinetic energies of both the molecules decrease.

(c) The kinetic energy of the oxygen molecule increases and that of the hydrogen

molecule decreases.

(d) The kinetic energy of the hydrogen molecule increases and that of the oxygen

molecule decreases.

Q21. Consider a mixture of oxygen and hydrogen kept at room temperature. As compared to a

hydrogen molecule an oxygen molecule an oxygen molecule hits the wall

(a) with greater average speed (b) with smaller average speed

(c) with greater average kinetic energy (d) with smaller average kinetic energy





Q22. Which of the following quantities is zero on an average for the molecules of an ideal gas

in equilibrium?

(a) any component of momentum (b) magnitude of momentum

(c) x component of velocity (d) speed

Q23. The average momentum of a molecule in a sample of an ideal gas depends on

(a) temperature (b) number of moles

(c) volume (d) mass of molecule

Q24. Which of the following quantities is the same for all ideal gases at the same temperature?

(a) The kinetic energy of 1 mole (b) The kinetic energy of 1g

(c) The number of molecules in 1 mole (d) The number of molecules in 1 g

Q25. Which of the following is correct for ideal gas in two dimensional system

(a) The energy distribution is ( ) exp n

B

Ef E Ek T

⎛ ⎞∝ −⎜ ⎟

⎝ ⎠ the value of 1n = for two

dimensional function. (b) The average kinetic energy is equal to Bk T

(c) The rms velocity of the gas is 3 Bk Tm

(d) The most probable velocity of the gas is Bk Tm

Q26. Keeping the number of moles, volume and temperature the same, which of the following

are not the same for all ideal gases?

(a) rms speed of a molecule (b) density

(c) pressure (d) average magnitude of momentum





NAT (Numerical Answer Type)

Q27. At temperature ……………… 0C the rms speed of gaseous hydrogen molecules equal to

that of oxygen molecules at C047 .

Q28. Temperature of an ideal gas is increased such that the most probable velocity of

molecules increase by a factor of 4. The rms velocity increase by the factor ……………

Q29. If density of hydrogen gas is 3kg/m1.0 and atmospheric pressure is ,N/m 100.1 25× then

root mean square speed of hydrogen molecule is…………m/sec

Q30. The temperature in Kelvin, at which the average speed of 2H molecules will be same as

that of 2N molecules at Co35 , will be……………

Q31. Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas three

dimension. Let mpv and rmsv denote the most probable velocity and the root mean square

velocity, respectively. The magnitude of the ratio rms

mp

vv

is………….(Answer must be up to

one decimal point)

Q32. Consider a Maxwellian distribution of the energy of the molecules of an ideal gas in three

dimensions. Let avE and rmsE denote the average energy and the root mean square

energy, respectively. The magnitude of the ratio rms

av

EE


one decimal point).

Q33. Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas in two

dimension. Let avv and rmsv denote the average velocity and the root mean square

velocity, respectively. The magnitude of the ratio rms

av

vv


two decimal point)





Solutions


Ans. 1: (a)

Solution: For Maxwellian distribution 2 Bmp

k Tvm

= , m

Tkv Brms

3=

32

rms

mp

vv

⇒ =

Ans. 2: (a)

Solution: Area under the ( )vf curve is conserved and the mean velocity shift towards right for

higher temperature.

Ans. 3: (b)

Solution: Ideal gas obeys Maxwell velocity distribution law at equilibrium temperature. Then

average value of ( )2

x yv vα β−

Now ( )2yxv βα − yxyx vvvv αββα 22222 −+=

0,0 == yx vv and 2 2 2Bx y z

k Tv v vm

= = =

Then ( )2

x yv vα β− yxyx vvvv αββα 22222 −+=

( )2

x yv vα β−mTk

mTk BB 22 βα += ( )

mTkB22 βα +=

Ans. 4: (b)

Solution: Ideal gas obeys Maxwell velocity distribution law at equilibrium temperature. Then

average value of ( )2

x yv vα

Now ( )2

x yv vα 2 2 2x yv vα=

a 222zy

Bx vv

mTkv === ,Then ( )2

x yv vα2

2 Bk Tm

α ⎛ ⎞= ⎜ ⎟⎝ ⎠

Ans. 5: (a)

Ans. 6: (b)





Solution: By Maxwell’s distribution law the area, of graph between ⎟⎠⎞

⎜⎝⎛

dvdN versus velocity v is

same at all temperature. Hence, IIIIIIA AA == Ans. 7: (c)

Solution: mRTvrms

3= (i)

Most probable velocity mRTv p

2=

From Equation (i) and (ii)

rmsv⇒ and pv both proportional T

Ans. 8: (d)

Ans. 9: (b)

Solution: Since, total probability is one, hence area of the figure should be one

( ) 1221

000 =−+⇒ vvaav

0

0 321

23

vaav =⇒=⇒

now area between 02.1 vv = to 09.1 vv = ( )1572.19.1

32

000

Nvvv

=−×=

Ans. 10: (a)

Solution: Momentum transfer in one collision θcos2mv=

Number of molecules collision per second ( )vAn=

( )Amvnvdtdp

mt

θcos2=⎟⎠⎞

⎜⎝⎛

θcos2 2nmvF =

θcos2/ 2nmvPAF ==

V 02V0V0

a( )vP

Speed





Ans. 11: (c)

Solution: ( ) 222/3

42

2

veTk

mvf Tkmv

B

B ππ

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Here constant=v but T is variable.

Then for ( )vf maximum. ( ) 0=dT

vdf

( ) 222/32/3

42

2

veTk

mvf Tkmv

B

B ππ ⎟⎟

⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

( ) 0=dT

vdf , 2/52

2/7

23

2−− =⇒ T

kmvT

B

Bk

mvT3

2

=⇒

Ans. 12: (d)

Solution:

2 21/ 2 1/ 20 2 2

02 2x x

B B

m mv vk T k T

x x x xB B

x

m mv N e dv v N e dvk T k T

vN

π π

− −∞

−∞

⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=∫ ∫

=mTkv B

x π2

=

Ans. 13: (b)

Solution: ( )

N

dNv

v∫∞

=0

11

∫∞

⎟⎠⎞

⎜⎝⎛=

0

1N

dNv

∫∞ −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

0

222/3

42

1 2

dvveTk

mv

vTk

m

B

B ππ

vTk

mv B ππ

421==

Ans. 14: (a)

Solution: m

Tkv Bp

2= and

mTkv B

rms3

=m

Tkm

Tkvvv BBprms

23−=Δ=−

( ) Tkmv

B

=−Δ

⇒23

( )22

23 −

Δ=⇒

Bk

vmT





Ans. 15: (a)

Solution: ( )3/ 2

/ 2 242

Bmv k T

B

mF v e vk T

ππ

−⎛ ⎞= ⎜ ⎟⎝ ⎠

Let temperature T at which for 1v and 2v , ( )F v are same.

2 21 2

3/ 2 3/ 2/ 2 / 22 2

1 24 42 2

B Bmv k T mv k T

B B

m me v e vk T k T

π ππ π

− −⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Here range will be taken same for both

( )( )

2 22 1

2 14 ln /B

m v vT

k v v−

=

Ans. 16: (b)

Solution: dvveTk

mvf Tkmv

B

B 222/3

42

)(2

ππ

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

21.. mvEEK ==

mEv 2

=⇒ Differentiate: mvdvdE = mdEvdv =⇒

mdE

mEe

TkmdEEf Tk

E

B

B .242

)(2/3

ππ

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

3/ 21( ) 2 .B

Ek T

B

f E dE e E dEk T

ππ

−⎛ ⎞= ⎜ ⎟

⎝ ⎠

Ans. 17: (b)

Solution: In two dimensional vdveTk

mvf Tkmv

B

B ππ

22

)( 2

2−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

mpmvEK22

1..2

2 ==

mdvdp = putting the value in vdveTk

mvf Tkmv

B

B ππ

22

)( 2

2−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

mdp

mpe

Tkmpf Tmk

p

B

B ππ

22

)( 2

2−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pdpe

Tmkpf Tmk

p

B

B2

2

1)(−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒






Ans. 18: (b) and (d)

Solution: Kinetic theory of gases are based on the following assumptions:

(i) All gases are made up of tiny elastic particles known as molecules.

(ii) the volume of the molecule is negligible.

(iii) The collision between the molecules is elastic.

(iv) They exert no force on each other.

Ans. 19: (a), (b) and (c)

Solution:

Ans. 20: (c) and (d)

Solution: Momentum will transferred from one molecule to other as higher momentum will

change to lower momentum and vice versa .

Ans. 21: (b)

Ans. 22: (a) and (c)

Ans. 23: (a) and (d)



Solution: For two dimension ( ) expB

Ef Ek T

⎛ ⎞∝ −⎜ ⎟

⎝ ⎠ 0n =

average energy is 0

0

( )

( )B

Ef E dEk T

f E dE

∞

∞ =∫

∫

for two dimensional system 2 Brms

k Tvm

= Bmp

k Tvm

=

dCdN

avC

rmsC

mpC





Ans. 26: (a), (b) and (d)

Solution: Only pressure is not function of mass.


Ans. 27: 0253 C−

Solution: mRTVrms

3=

H

H

mRT

mRT 33

0

0 =⇒ = ( )00

Tmm

T HH =

amumamumH 32,2 0 ==

and CT 00 32027347 =+=

so, ( )320322

=HT

KTH 20= = ( ) CTH027320 −= C0253−=

Ans. 28: 4

Solution: mRTvrms

3= and

mRTv p

2=

rmsv⇒ and pv are both proportional to T

rmsv⇒ increases by 4 times

Ans. 29: 1710 m/s

Solution: By kinetic theory of gases the pressure exerted by the gas on the wall of container is

given as

Pressure 213

P d v=

Here, .kg/m1.0,N/m101 325 =×= dP





So, by Eq. (i) dPv 3

=

53 1 100.1rmsv × ×

= m/s 1710m/s103 6 =×=

Ans. 30: KTH 22=

The average speed of a molecule of gas at constant temperature T is given as

MRTvav π

8= H NH

N H N

T Mvv M T

⇒ =

given, NH vv =

NHHN MTMT =⇒

N

H

N

H

MM

TT

=⇒282

=

NH TT ×=⇒141 ( )35273

141

+×=

KTH 22=

Ans. 31: 0.8

Solution: For Maxwellian distribution Bmp

2k Tvm

= , 3 Brms

k Tvm

=23

mp

rms

vv

⇒ =

Ans. 32: 1.8

Solution: 32B

avk TE = 15.

2rms BE k T=

152.72 1.83 1.5

2

rms

av

EE

= = =

Ans. 33: 2.25





Solution: 2

2

0 0

( ) . 22 8

B

mvk T B

avB

k Tmv v vf v dv v e vdvk T m

πππ

∞ ∞ −

= = = =∫ ∫

2

22 2 2

0 0

2( ) . 22

B

mvk T B

B

k Tmv v f v dv v e vdvk T m

ππ

∞ ∞ −

= = =∫ ∫

2 Brms

k Tvm

=

In two dimension system 2 B

rmsk Tvm

=

21.41 2.25.62

8

rms

av

vv π

= = =





Chapter - 2

Real Gases

2.1 Andrew’s Experiment on Carbon Dioxide

Andrew’s experiment investigated the behaviour of 2CO and analyse the Pressure

( )P versus volume ( )V at different temperature T .

The observations are following:

1. Above a temperature of about ( )048T C= the 2CO resemble that of Ideal gas.

2. As temperature lowered, the isotherms exhibit distortion which gradually

increases, which is indication from the ideal gas character.

3. At 031.4 C a Kink is observed which suggests the gas can be liquified under

compression.

4. As temperature is lower further the kink spread into a horizontal line, i.e.

compression produces liquification.

From A to B , 2CO behave as a gas. At the point

B the liquification of the gas just starts. The gas

condenses at constant pressure from B to C so

that liquid and vapour coexist. At C , the gas is

completely in the liquid phase.

From C to D the slop is very steep since a liquid

is almost incompressible.

Conclusion: The temperature at which it becomes possible to liquefy a gas under

compression is known as critical temperature ( )CT [In Andrew experiment

( ) 048CT C= ], corresponding pressure and volume is known as critical pressure ( )CP

and critical volume ( )CV .

A gas can be liquified only if it cooled upto or below its characteristic critical

temperature.

048 C031.4 C021.5 C013.1 C

VQ

C

P gasJK

DP

liqui

d

AFcondensation

B





There exist a continuity of liquid and gaseous states, i.e. they are two distinct stages at a

continuous physical phenomenon.

2.2 van der Waals equation of state.

The van der Waals equation for real gases are given by

( ) RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + 2 for 1 mole of gas and

( )2

2

n aP V nb nRTV

⎛ ⎞+ − =⎜ ⎟

⎝ ⎠ for n mole of gas.

Assumption for real gas:

1. Gas molecules have finite size

2. There are weak interaction force, which depends only upon distance between them.

3. The molecular density is small and the number of collisions with the walls of the

container are exactly the same point and finite size molecular.

2.3 Correction in Ideal Gas equation to achieve van der Waals gas equation of state.

2.3.1 Correction for finite size: if V is volume available for one mole of gas (volume of

container). If size of molecule take into account then ( )V b− is volume available for real

gas which is less thanV . b is popularly known as covolume which is dependent

on the nature of gas.

Example: If mV is molecular volume of real gas then prove that mb = 4NV if N is total number

of molecule in container.

The volume available to first molecule = V

The volume available to second molecule = sV V−

Where sV volume of exclusion i.e. around any

molecule, a spherical volume is 34

3sdV π⎛ ⎞

= ⎜ ⎟⎝ ⎠

will

be denied to every other molecule.

Volume of exclusion ( )34 23s

rV

π=

Volume of exclusion ( )

324

3rVs π=

ms VV 8=

rd 2=

r r

mV





Similarly volume available to N th molecule = sV - (N - 1)V

Average volume available for each molecule

( )1

1 1N

si

V V i VN =

= − −∑ ( )12

sN N VV

N−

= −2 sNV V= − 8

2 mNV V= − 4 mV NV= −

m(V - b) = V - 4 N V so mb = 4NV

2.3.2 Correction for intermolecular attraction:

A molecule in the equally in all direction so that there is no resultant force on it.

But for outermost layer close to surface there will be net inward force. So whenever a

molecule strikes the walls of container, the momentum exchange will be less than for

Ideal gas.

There forces are cohesive in nature and proportional to number of molecule.

So for real gas change in pressure is 2

aV

. So for real gas pressure will be 2

aPV

⎛ ⎞+⎜ ⎟⎝ ⎠

So gas equation reduce to ( )2

aP V b RTV

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

Then 2

RT aPV b V

= −−





2.3.3 Maxwell Equal Area

James Clerk Maxwell replaced the isotherm

between a and c with a horizontal line positioned so that

the areas of the two hatched regions are equal (means

area of adb and bec are equal). The flat line portion of

the isotherm now corresponds to liquid-vapor

equilibrium. As shown in figure.

The portions a d− and c e− are interpreted as metastable states of super-heated liquid

and super-cooled vapor respectively. The equal area rule can be expressed as:

( )

G

L

V

V G LV

P V V PdV− = ∫

where VP is the vapor pressure (flat portion of the curve), LV is the volume of the pure

liquid phase at point a on the diagram, and GV is the volume of the pure gas phase at

point c on the diagram. The sum of these two volumes will equal the total volumeV .

Example One mole of a certain gas is contained in a vessel of volume V . At a temperature

1T the gas pressure is 1p atm and at a temperature 2T the pressure is 2p atm. Find the Van

der Waals parameters for this gas.

Solution: it is given no of mole 1n =

( ) 121 RTbV

VaP =−⎟⎠⎞

⎜⎝⎛ + (i)

( ) 222 RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + (ii)

from (i) and (ii)

( )( )12

12212

TTPTPTVa

−−

=

( )( )12

12

PPTTRVb

−−

−=

2

1

0LV GV 4321

VP

P

db

eca





Example: Under what pressure will carbon dioxide of molar mass M have the density ρ at the

temperature T . If given gas is obeying for a Van der Waals gas.

Solution: Assume M is molar mass of the carbon dioxide and V is the volume so VM

=ρ

Van der wall equation (for one mole gas):


⎜⎝⎛ + 2

2

2

a MP b RTMρ

ρ⎛ ⎞⎛ ⎞

+ − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2

2

Ma

bMRTP ρ

ρ

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⇒ 2

2

Ma

bMRTP ρρ

ρ−

−=⇒

2.3.4 Critical Point

The van der Waals equation of state for a gas is given by


⎜⎝⎛ + 2

where VP, and T represent the pressure, volume and temperature respectively, and a

and b are constant parameters. At the critical point, where all the roots of the above

cubic equation are degenerate means all roots are equal.

In another way mathematically For the critical isotherm is the point of inflection point

On basis of above definition one can find the critical volume cV , critical pressure cP and

critical temperature cT for van der waal gas.

For Van der Waals equation

( )2

aP V b RTV

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

2

RT aPV b V

= −−

(i)

0

T

PV∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

for extremum point

( )2 2

2 0T

P RT aV VV b∂⎛ ⎞ = − + =⎜ ⎟∂⎝ ⎠ −

at cV V= , cT T= (ii)





2

2 0T

PV

⎛ ⎞∂=⎜ ⎟∂⎝ ⎠

for inflection point

( )3 3

2 6 0RT aVV b

− =−

at cV V= , cT T= (iii)

solving (ii) and (iii)

3cV b= RbaTc 27

8=

Put value of cV and cT one can get 227baPc =

ccc

c cvP

RT==

38 which is popular known as critical coefficient for van der Waals gas.

2.3.5 van der Waals Equation of State and Virial Coefficient

According to virial theorem the equation of state is given by

2 .....pVV Vβ γα= + + + (i)

Where ,α β and γ are first second and third virial coefficient .

For the Ideal gas RTα = and other coefficient are zero.

Virial coefficient for Van der Waals gas

To put van der Waals equation in virial form we first rewrite it as

1

1 b apV RTV V

−⎛ ⎞= − −⎜ ⎟⎝ ⎠

Using binomial theorem, we have

1 2

21 1 ....b b bV V V

−⎛ ⎞− = + + +⎜ ⎟⎝ ⎠

Hence

2

2 .....RTb a RTbpV RTV V−

= + + + …. (ii)

As will be noted, van der Waals equation has only three virial coefficients and a

comparison with equation (i) yields





2, and RT RTb a RTbα β γ= = − =

At the Boyle temperature, the second virial coefficient is zero. Hence,

0BRT b a− =

or BaT

Rb=

From the preceding, section we recall that the critical temperature of a gas obeying van

der Waals equation of state is

827C

aTRb

=

on comparing these expressions, we get

27 3.3758B C CT T T= =

that is, the Boyle temperature, on the basis of Van der Waals equation, is 3.375 times the

critical temperature.






Q1. A gas behaves as an ideal gas at:

(a) very low pressure and high temperature

(b) high pressure and low temperature

(c) high temperature and high pressure

(d) low pressure and low temperature

Q2. In the van der Waals equation, the terms ⎟⎠⎞

⎜⎝⎛

2Va and ( )b are introduced to account for the:

(a) inter-molecular attraction and the total volume occupied by the gas

(b) molecular size and the size of the containing vessel

(c) inter-molecular attraction and the volume of the molecules

(d) inter-molecular attraction and the force exerted by the molecules on the walls of the

container

Q3. ‘Critical temperature’ is defined as the:

(a) lowest temperature at which the gas can be liquefied at constant pressure

(b) lowest temperature at which the gas can be liquified by increase of pressure alone

(c) highest temperature at which the gas can beliquified by increase of pressure alone

(d) highest temperature at which the gas can be liquified at constant pressure

Q4. The work performed by one mole of a Van Der Waals gas during its isothermal

expansion from the volume 1V to 2V at a temperature T is given by .

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−−

121

2 11lnVV

abVbV

RT (b) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−−

211

2 11lnVV

abVbV

RT

(c) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−−

212

1 11lnVV

abVbV

RT (d) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−−

122

1 11lnVV

abVbV

RT





Q5.

Consider the above graph in respect of van der Waals equation of state. Which portion of

the graph cannot be explained?

(a) AB (b) BC (c) DE (d) BCD


Q6. Which of the following are important in case of a van der Waals gas?

(a) Short range attraction (b) Long range repulsion

(c) Short range repulsion (d) Long range attraction

Q7. Which of the important results of Andrews’ experiment are correct

(a) There exists a temperature called critical temperature, above which a gas cannot be

liquefied, however great the applied pressure is.

(b) There exists a temperature called critical temperature, below which a gas cannot be

liquefied, however great the applied pressure is.

(c) For van der Waal gases critical temperature is 27

aRb

.

(d) Oxygen, nitrogen and hydrogen are permanent gases and they cannot be liquefied.

Q8. In the equation of state for real gases ( ) RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + 2

(a) The critical points are point of inflection.

(b) Critical volume is given by 3cV b=

(c) Critical pressure is given by 227caPb

=

(d) critical temperature is given by 827c

aTRb

=

PA

B

C

D

E

Co1.13V





NAT (Numerical Answer Type Questions)

Q9. If the value of van der Waals constant b for a real gas is mol,/cm32 3 then the

approximate volume of one molecule of the gas i ……… 2310−× 3cm (Avogadro constant 231002.6 ×= ) (Answer must be upto one decimal)

Q10. van der Waals equation predicts that the critical coefficient of a gas ⎟⎟⎠

⎞⎜⎜⎝

⎛

cc

c

VPRT

has the

value………..(Answer must upto one decimal point )

Q11. If equation of state is given by expRT aPV b RTV

⎛ ⎞= −⎜ ⎟− ⎝ ⎠ then critical volume

cV =………..b .


⎛ ⎞= −⎜ ⎟− ⎝ ⎠ then critical volume

cT =……….. aRb

.


⎛ ⎞= −⎜ ⎟− ⎝ ⎠ then critical coefficient

........c

c c

RTPV

=





Solutions


Ans. 1: (a)

Solution: The relation between PV and P is given as

++++= 22 DPCPBPAPV (i)

…CBA ,, are virial constant .…DCBA >>> If P is very high then ≠PV constant.

So, for ideal gas P should be small and T should be large.

Ans. 2: (a)

Solution: The ideal gas equation is RTPV = whereas gas equation, for real gases is given by

van der Waals which is ( ) RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + 2 where, the factor b is due to volume

occupied by the molecules itself and ⎟⎠⎞

⎜⎝⎛

2Va is due to molecular attractive force.

Ans. 3: (c)

Ans. 4: (a)

Solution: We know van der Waals gas equation:


⎜⎝⎛ + 2 (i)

∫ ∫==Δ pdVdWW

from (i): 2Va

bVRTP −−

=

then ∫ ⎟⎠⎞

⎜⎝⎛ −

−=

2

1

2

V

V

dVVa

bVRTW

2

1 2 1

1 1ln V bW RT aV b V V

⎛ ⎞−= + −⎜ ⎟− ⎝ ⎠





Ans. 5: (d)

Solution: The Van der Waals equation is given as

( ) RTbV

VaP =−⎟⎠⎞

⎜⎝⎛ + 2

where ,a b are constant.

The graph drawn is shown as the curve ABCDE . This does not agree with the

experimental isothermals for 2CO as obtained by Andrews However the portion DE has

been explained as due to super cooling of vapours and the portion AB due to super

heating of the liquid. But the portion BCD cannot be explained because it shows

decrease in volume with decrease in pressure.


Ans. 6: (c) and (d)

Solution: van der Waals equation ( ) RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + 2 , where ba, are constant. In this case

short range of repulsion and long range of attraction is important.

Ans. 7: (a) and (d)

Solution: From the result of Andrew’s experiment we can define critical temperature as a

temperature above which a gas cannot be liquefied however great the applied pressure is

All gases are real gas so they can be liquefy

For van der Waals gases, critical temperature is given by 827c

aTRb

=

Ans. 8: (a), (b), (c) and (d)

Solution: The van der Waals equation is given as

( ) RTbV

VaP =−⎟⎠⎞

⎜⎝⎛ + 2

0dand0 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛∂∂

TT dVP

VP (point of inflection ) thus, we get bVc 3=

and RbaTc 27

8= ⇒ 227b

aPc = Thus, bVc 3=

PA

B

C

D

E

Co1.13V






Ans. 9: 1.3

Solution: The van der Waals constant b is given as b = four times the actual volume of the

molecules.

volume of one molecule Nb

4=

volume of one molecule 32323 cm103.1

106432 −×=××

=

Ans. 10: 2.6

Solution: For van der Waal Gases 827c

aTRb

= , 3cV b= , 227caPb

=

38

=cc

c

VPRT

Ans. 11: 2

Solution: expRT aPV b RTV

⎛ ⎞= −⎜ ⎟− ⎝ ⎠ critical point are point of inflection so at critical points

0dand0 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛∂∂

TT dVP

VP (point of inflection)

So 2cV b=

Ans. 12: 0.25



0dand0 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛∂∂

TT dVP

VP (point of inflection )

2 ,4c caV b TbR

= =





Ans. 13: 3.7



0dand0 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛∂∂

TT dVP

VP (point of inflection)

2 ,4c caV b TbR

= = and 2 24caP

e b=





Chapter – 3

Basics of Thermodynamics and Laws of Thermodynamics

3.1 Mathematical Formulations of Thermodynamics.

( )1 2, ,...... ny y x x x=

then differential dy is said to be exact and one can write

i ii

dy c dx= ∑

1 1 2 .......dy c dx cdx= + + =

1 21 2

....... ii i

y y ydy dx dx dxx x x

⎛ ⎞⎛ ⎞ ⎛ ⎞∂ ∂ ∂= + + = ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑

and its corresponding ic ix are said to be conjugate to each other.

3.1.1 Some Important Formulas

1. kllk xx

yxxy

∂∂∂

=∂∂

∂ 22

2. zz x

yyx

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

3. xzy z

yyx

zx

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

4. xyz y

zzx

yx

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

5. Zz z

x x wy w y

⎛ ⎞ ⎛ ⎞∂ ∂ ∂⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

6. yw

x xdx dy dwy w

⎛ ⎞∂ ∂⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

yz w z

x x x wy y w y

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠





3.2 Fundamental Concept

3.2.1 System: A system can be any object, region of space etc, selected for study and set apart

(mentally) from everything else. Which then become surrounding.

The system of interest in thermodynamics are finite and macroscopic rather than

microscopic. The imaginary envelope which encloses a system and separated it from its

surrounding is called the boundary of the system.

3.2.2 Isolated system It can not exchange either matter or energy with the surroundings.

If exchange of matter is allowed the system is said to be open; if only energy and not

matter is closed (but not isolated).

3.2.3 Thermodynamical state

A thermodynamic state is a set of values of properties of a thermodynamic system that

must be specified to reproduce the system.

Thermodynamic state is the macroscopic condition of a thermodynamic system as

described by its particular thermodynamic parameter. Such as temperature (T) pressure

(P) volume (V) density ( )ρ .

3.2.4 State function state function also called “State variable” thermodynamic variables

describe the momentary condition of thermodynamic system

For a continuous process, such variable are exact different also fully determined by their

initial and final thermodynamic states.

Example includes entropy, pressure, temperature, volume, etc.

3.2.5 Intensive and extensive properties

Intensive properties:-it is a physical property of a system that does not depend on the

system size or the amount of material in system. It is scale invariant.





Example: Chemical potential, density, viscosity, resistivity, specific heat capacity, pressure,

elasticity, magnetization, velocity, acceleration, temperatures, etc.

Solution: Extensive properties:- It is one that is additive for independent non-interacting

subsystem. It is directly proportional to the amount of material in the system.

Example energy, Entropy, Gibbs energy, mass momentum, volume, change, weight,

Note: If f and g are arbitrary intensive variable then dgdf

gffg ,, and gf + will also

intensive variable.

If F and G are also two arbitrary extensive variable then GF + will an extensive variable

but GF and

dGdF will be intensive.

If F is extensive variable and f is intensive variable the Ff and fF

and dfdF

is extensive

variable.

3.3 The Ideal Gas:

The Ideal gas law is the equation of hypothetical Ideal gas. It is derived from kinetic

theory and satisfied Boyle’s and Charles’s law.

The state of an amount of ideal gas is determined by its pressure (P) volume (V) and

temperature (T).

The Ideal gas equation of state for n mole is given by PV = nRT

Where R is gas constant and given by 8.314 J. K-1 mole-1

3.4 Laws of Thermodynamics

3.4.1 Zeroth Law of Thermodynamics: If two systems 1 and 2 are separately in thermal

equilibrium with third 3 they must be in thermal equilibrium with one another.

3.4.2 First Law of Thermodynamics: Energy is conserved when heat is

taken into account.

Mathematically If Qδ amount of heated to the system and if system

will do Wδ amount of work then change in internal energy is

dU given by

dU

Wδ

Wδ





dU Q Wδ δ= −

obviously heat exchange and work is dependent on path and internal energy is state

function.

U is a mathematical abstraction that keeps account of the exchange of energy that be fall

the system.

The term Qδ means the amount of that amount of energy added to or

remove by conduction of heat or by thermal radiation.

Wδ is amount of energy gained or lost as result of work.

3.4.3 Work Done during Different Process.

Work done when process is occur to between A to B

B

A

W PdV= ∫

work done is area under the PV diagram.

If work is done by the system it has positive sign, and if work is done on the system it has

negative sign.

3.4.4 Specific Heat:

Heat capacity of a body is numerically equal to quantity of heat required to raise its

temperature by 1 unit.

T

CΔΔ

=θ

The specific heat of a material is numerically equal to quantity of heat required to raise

the temperature of unit mass of that materials through 1unit

mdTdC θ

=

3.4.5 Heat Capacity of Ideal Gas: if f is degree of freedom of Ideal gas then from

equipartition of energy total sum of energy is equivalent to sum of kinetic energy

associated with each degree of freedom

P

A

V

B





3.4.6 Molar Heat Capacity: heat capacity defined as the energy required to raise the

temperature of one mole of Ideal gas by one Kelvin at constant volume.

KTNfU A

2= where NA is Avogadro number

22fRKNf

dTdU

A ==

2RfCV = , for Ideal gas

⎟⎟⎠

⎞⎜⎜⎝

⎛+=+= 1

2f

RRCC VP

γ is defined as ratio of heat capacity at constant pressure to constant volume.

21p

v

CC f

γ⎛ ⎞

= = +⎜ ⎟⎝ ⎠

3.4.7 Coefficient of Volume Expansion ( )α or expansivity 1

P

VV T

α ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

3.4.8 Isothermal Elasticity TT

PE VV∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

and Isothermal compressibility T

TE1

=β

Example: Find the isothermal compressibilityα of a Van der Waals gas as a function of volume

V at temperature T .

Note: By definition, Tp

VV ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

−=1α .

Solution: We know bulk modulus of a gas is given by

Van der Wall equation: ( ) RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + 2 (i)

If process is isothermal: =T constant

( ) ( ) 023 =⎟⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ −

+ dVVaPbVdV

VadP





( ) 023 =⎟⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ −

VaPbV

Va

dVdP

( )( )

( ) ( )( )bVV

bVaaVPVVa

bVVaP

dVdP

−−++−

=+−

+−= 3

3

3

2/

( )( ) ( )

( )( ) ( )[ ]VbaaVPV

bVVbVaaVPV

bVVV −+−

−=⎥

⎦

⎤⎢⎣

⎡−++−

−−= 3

2

3

31α (ii)

Put values of P from (i) in (ii)

( )( )[ ]23

2

2 bVaRTVbVV−−

−=α

3.5 Different Type of Thermodynamical Process and Use of First Law of

Thermodynamics.

3.5.1 Isochoric Process: When volume remain constant during the process the process is said

to be isochoric process.

dW = PdV dV = 0 dW = 0

In an isochoric process work done during the process is zero.

From first law of thermodynamics.

0dU Q W Wδ δ δ= − =

VdU Q nC dTδ= =

In isochoric process change in interval is equal to heat exchange.

3.5.2 Isobaric Process: When pressure of the system remain constant during the process.

Work done during the process is given by

( )∫ −=

B

AAB VVPPdV

( )AB TTnR −=

TnRΔ=

where TΔ is change in temperature during the process.

Q dU Wδ δ= + VnC T nR T= Δ + Δ ( )Vn C R T= + Δ

PB

VA

P B

V

A





PQ nC Tδ = Δ V PC R C+ = (for the ideal gas)

Heat exchange during the process for n mole of gas is equal to pnC TΔ where pC is

specific heat capacity at constant process.

3.5.3 Isothermal Process: when temperature remains constant during the process then system

is said isothermal process.

In isothermal process change in internal energy is zero

because 0=ΔT

W PdVδ = = ∫B

A

V

V VdVnRT ln B

A

VW nRTV

δ⇒ =

during the isothermal process heat exchange

Q Wδ δ= A

B

VVnRT ln=

3.5.4 Adiabatic Process: When there is not any heat exchange during the process then process

is said to be adiabatic process.

Adiabatic process is defined by

γPV = constant

For Ideal gas PV = RT for one mole of gas.

So 1TV γ − = constant

γγTP −1 = constant

where V

P

CC

=γ

work done during adiabatic process W PdVδ =

PV kγ = kdW dVV γ=

B

A

V

V

k V dVγ−= ∫

1

1

B

A

V

V

Vkγ

γ

− +

− +=

1 1

1B AV Vk

γ γ

γ

− −⎛ ⎞−⎜ ⎟−⎝ ⎠

1 1

1B B B A A AP V V P V Vγ γ γ γ

γ

− −−−

=1

B B A AP V P Vγ−−

P

V

P

A

B

V





For n mole Ideal gas

( )1

B AnR T TW

γ−

=−

For Adiabatic process Q dU Wδ δ= +

Hence 0Qδ = dU Wδ= −

So in adiabatic process change in internal energy is equal to minus of work done i.e.

( )

1A BnR T T

dUγ

−=

−

Sign convention: Work done by the system is positive, work done on the system is

negative, heat absorbed by system dQ is positive. Heat reject by the system is negative.

Example: n mole of certain ideal gas at temperature 0T were cooled isochorically so that the gas

pressure reduced n times. Then as a result at the isobaric process the gas expanded till its

temperature get back to initial value. Find the total amount of heat absorbed by the gas in

the process.

Solution: Let at state A, the pressure, volume, temperature is nTVP ,,, 000 is number of mole.

According to question:

n

TTnT

VPTVP

BB

000

0

00 ==

nVVTV

nP

TV

nP

cc

B===

0

000

In process A to B

vdQ nC T= Δ 00v

TnC Tn

⎛ ⎞= −⎜ ⎟⎝ ⎠

In process B to C

isochoric isochoricBT

VnP 00

0

00

TVP

0

0

T

VnP

c

)A( )B( )C(





vdQ nC T PdV= Δ + 0 00 0v

T TnC T nR Tn n

⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Total heat absorbed by gas

( ) ( )dQ A B dQ B C→ + →

0 0 00 0 0v v

T T TnC T nC T nR Tn n n

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

011nRTn

⎛ ⎞= −⎜ ⎟⎝ ⎠

Example: A sample of an ideal gas is taken though the cyclic process abca as shown in figure. It

absorbed 50 J of heat during part ab, no heat during bc and reject 70 J of heat during ca,

40 Jule of work is done on the gas during part bc.

(a) Find the internal energy of the gas at b and c if it is 1500 J at a.

(b) Calculate the work done by the gas during the part ca.

Solution: (a) during ba → V = constant

50Q dU JΔ = =

aU = 1500J so at bU = 1550J

Work b = - 40 J c→ 15901550 == cb UJU (no heat at bc)

(b) during path c → a

JU 9015901500 −=−=Δ

70Q JΔ = −

W Q UΔ = Δ −Δ

J209070 =+−=

Example: A sample of an Ideal gas has pressure P0, volume V0 and temperature T0. It is

isothermally expand to twice its original volume. It is then compressed at constant

pressure to have the original volume V0. Finally, the gas is heated at constant volume to

get the original temperature.

(a) Show the process in VT diagram.

(b) Calculate the heat absorbed in the process.

P

c a

b

V





Solution: (a)

(b) Hence the process is cyclic then change in internal energy in cycle is zero. So heat

supplied in the cycle is equal to work done.

Work done during A → B is isothermal, so

00

2lnABVW nRTV

= 2ln0nRT=

b c→ isobaric, so work done is PdV.

bbaa VPVP =

000 2VPVP ba =

20PPb =

( )00 02

2b cPW V V→ = −

200VP

−=

Wc a → is isochoric Wc a → = 0

Total work done is given by

⎟⎠⎞

⎜⎝⎛−+

22ln 00

0VPnRT

⎟⎠⎞

⎜⎝⎛ −=

212ln0nRT Ans.

c a

b

T

0VV

02V






Q1. The first law of thermodynamics is a statement of

(a) conservation of heat (b) conservation of work

(c) conservation of momentum (d) conservation of energy

Q2. If heat is supplied to an ideal gas in an isothermal process,

(a) the internal energy of the gas will increase

(b) the gas will do positive work

(c) the gas will do negative work

(d) the said process is not possible

Q3. Figure given below shows two processes A and B on a system. Let 1QΔ and 2QΔ be the

heat given to the system in processes A and B respectively. Then

(a) 1 2Q QΔ > Δ (b) 1 2Q QΔ = Δ (c) 1 2Q QΔ < Δ (d) 1 2Q QΔ ≤ Δ

Q4. Refer to figure in Let 1UΔ and 2UΔ be the changes in internal energy of the

system in the processes A and B . Then

(a) 1 2U UΔ > Δ (b) 1 2U UΔ = Δ (c) 1 2U UΔ < Δ (d) 1 2U UΔ ≠ Δ

P A

BV

P A

BV





Q5. Consider the process on a system shown in figure given below. During the process, the

work done by the system

(a) continuously increases (b) continuously decreases

(c) first increases then decreases (d) first decreases then increases

Q6. Consider the following two statements.

(A) If heat is added to a system, its temperature must increase.

(B) If positive work is done by a system in a thermodynamic process, its volume must

increase.

(a) Both A and B are correct (b) A is correct but B is wrong

(c) B is correct but A is wrong (d) Both A and B are wrong

Q7. An ideal gas goes from the state i to the state f as shown in figure below. The work

done by the gas during the process

(a) is positive (b) is negative

(c) is zero (d) cannot be obtained from this information

P

V

P

T

f

i





Q8. Consider two processes on a system as shown in figure below.

The volumes in the initial states are the same in the two processes and the volumes in the

final states are also the same. Let 1WΔ and 2WΔ be the work done by the system in the

processes A and B respectively.

(a) 1 2W WΔ > Δ (b) 1 2W WΔ = Δ (c) 1 2W WΔ < Δ

(d) Nothing can be said about the relation between 1WΔ and 2WΔ .

Q9. A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly

moved in to compress the gas and is maintained at this position. As time passes the

pressure of the gas in the cylinder

(a) increases (b) decreases (c) remains constant

(d) increases or decreases depending on the nature of the gas.

Q10. The pressure and density of a gas ⎟⎠⎞

⎜⎝⎛ =

57γ change adiabatically from ( )11,dP to ( )22 ,dP .

If 2

1

32dd

= , then the value of ⎟⎟⎠

⎞⎜⎜⎝

⎛

1

2

PP

(a) 32 (b) 128 (c) 321 (d)

1281

Q11. The pressure P volume V and temperature T for a certain material are related

by ( )V

BTATP2−

= where BA, are constants. The work done by the materials if the

temperature changes from T to T2 while the pressure remains constant?

(a) 2BTAT − (b) 22BTAT −

(c) 23BTAT − (d) 232 BTAT −

P

T

BA





Q12. The specific heat ( )c of a substance is found to vary with temperature ( )T as 2Tc βα += .

Where T is Celsius temperature. At what temperature does the specific heat of the

substance becomes equal to the mean specific heat of the substance in a temperature

range between 0 and 0T ?

(a) 0

2T (b) 0

2T (c) 0

3T (d) 0

3T

Q13. The equation of state of a gas is given as ( ) nRTbVP =− where b is a constant, n is the

number of moles and R is the universal gas constant. When 2 moles of this gas undergo

reversible isothermal expansion from volume V to volume 2V , what is the work done

by the gas?

(a) ( ) ( )[ ]bVbVInRT −− 2/2 (b) ( ) ( )[ ]bVbVInRT −− /22

(c) ( ) ( )[ ]VbVInRT 2/2 − (d) ( ) ( )[ ]bVVInRT −/22

Q14. If (1) represents isothermal and (2) represents adiabatic, which of the graphs given above

in respect of an ideal gas are correct?

(a) I and II (b) II and III (c) I and III (d) I, II and III

Q15. What is the minimum attainable pressure of an ideal gas in the process given by

,2bVaT += where a, b are constants and V is the volume of one mole of ideal gas?

(a) ab (b) abR (c) abR2 (d) ba /

(R is the universal gas constant)

( )1

( )2

V

T

I

( )1( )2

T

PII

( )1 ( )2

P

VIII





Q16. One mole of an ideal gas is takes from an initial state ( )TVP ,, to a final state

( )TVP 4,2,2 by two different paths as shown in the Fig. 1 and 2 given above. If the

changes in internal energy between the final and the initial states of the gas along the

paths I and II are denoted by IUΔ and IIUΔ respectively, then:

(a) III UU Δ=Δ (b) III UU Δ>Δ (c) III UU Δ<Δ (d) III 66.0 UU Δ=Δ

Q17. A thermodynamic system is taken from an initial state A

to another state B and back to state A via state C as

shown by the path ACBA →→→ in the P V−

diagram given then The work done by the system in going

from the state A to the state B is:

(a) J0.335× (b) J5.335×

(c) J5.350× (d) J0.550×

Q18. A system absorbs J3105.1 × of energy as heat and produces 500J of work. The change

in the internal energy of the system will be:

(a) 1500 J (b) 100 J (c) - 1500 J (d) 1000 J

( )TVPA ,,

⎟⎠⎞

⎜⎝⎛ TVPB ,2,

2

( )TVPC 4,2,2

P

VFig. 1 (Path I)

( )TVPB ,,

⎟⎠⎞

⎜⎝⎛ TVPC 4,

2,2

( )TVPC 4,2,2P

V

C

A

B

Fig. 2 (Path II)

⎟⎠⎞

⎜⎝⎛

2mN

P

A

BC

( )3mV

20

50

40

30

10

1 2 4 5 63





Q19. A given amount of heat cannot be completely converted into work. However it is possible

to convert a given amount of work completely into heat. This apparently contradictory

statement results from the:

(a) zeroth law of thermodynamics (b) first law of thermodynamics

(c) second law of thermodynamics (d) third law of thermodynamics

Q20. If the number of degrees of freedom of a molecule in a gas is n then the ratio of specific

heats is given by:

(a) n11+ (b)

n211+

(c)n21+ (d)

122−nn

Q21. The ratiocurve adiabatic of Slopecurve isothermal of Slope is equal to:

(a) 1 (b)γ (c)γ1 (d)2

Q22. One mole of a perfect gas expands adiabatically. As a result of this, its pressure,

temperature and volume change from ,,, 111 VTP to 22 ,TP and 2V respectively. if molar

specific heat at constant volume is VC , then the work done by the as is:

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛

1

211 log303.2

VVVP (b) ⎟⎟

⎠

⎞⎜⎜⎝

⎛

1

21 log

VVRT

(c) ( )12

22

22

21

21

TTRVPVP

−− (d) ( )21 TTCV −

Q23. For a diatomic gas having 3 translational and 2 rotational degrees of freedom, the energy

is given by:

(a) 52 Bk T (b) 3

2 Bk T (c) 12 Bk T (d) Bk T





Q24. The maximum attainable temperature of ideal gas in each of the following processes

0Vp p e β−= is

(a) 0max

PTe Rβ

= (b) 0max 2

PTe Rβ

= (c) 0max

2PTe Rβ

= (d) 0max

23

PTe Rβ

=

Q25. A horizontal cylinder closed from one end is rotated with a constant angular velocity ω

about a vertical axis passing through the open end of the cylinder. The outside air

pressure is equal to 0p , the temperature to T , and the molar mass of air to M . Find the air

pressure as a function of the distance r from the rotation axis. The molar mass is

assumed to be independent of r .

(a) RTrM

ePP 20

22ω−

= (b) RTrM

ePP 20

22ω

= (c) RTrM

ePP22

0

ω−

= (d) RTrM

ePP22

0

ω

=

Q26. A thermally insulated vessel containing a gas whose molar mass is equal to M and the

ratio of specific heats γ=Vp CC / moves with a velocity v . If vessel is suddenly

stopped then increment of temperature is given by

(a) ( ) 21

T MvR

γ −Δ = (b) ( ) 21

2T Mv

Rγ −

Δ =

(c) ( ) 22 1

T MvRγ −

Δ = (d) ( ) 22 1

T MvRγ −

Δ =

Q27. Gaseous hydrogen contained initially under standard conditions in a sealed vessel of

volume 0V was cooled by .TΔ The amount of heat will be lost by the gas if initial

pressure is 0P and temperature 0T

(a) ( )10

00

−Δ

=ΔγT

TVPQ (b) ( )10

00

−Δ

=Δγ

γT

TVPQ

(c) ( )0

0

1PV TQ

Tγ − Δ

Δ = (d)

( )0

00 1T

TVPQ

γγ Δ−

=Δ





Q28. A given quantity of gas is taken from the state A C→ reversibly, by two paths, A C→

directly and A B C→ → as shown in the figure. During the process

A C→ the work done by the gas is 100 J and the heat absorbed is

150 J . If during the process A B C→ → , the work done by the gas

is 30 J , then which one of following is correct?

(a) heat absorbed in 20A B C J→ →

(b) heat absorbed in 80A B C J→ →

(c) heat reject in 20A B C J→ →

(d) heat reject in 80A B C J→ →

Q29. Let QΔ be the heat exchange in a quasistatic reversible thermodynamic process. Then

which of the following is correct?

(a) QΔ is a perfect differential if the process is isothermal

(b) QΔ is a perfect differential if the process is isochoric

(c) QΔ is always a perfect differential

(d) QΔ cannot be a perfect differential

Q30. Let WΔ be the work done in a quasistatic reversible thermodynamic process. Which of

the following statements about WΔ is correct?

(a) WΔ is a perfect differential if the process is isothermal

(b) WΔ is a perfect differential if the process is adiabatic

(c) WΔ is always a perfect differential

(d) WΔ cannot be a perfect differential

P A

B

V

C






Q31. The pressure p and volume V of an ideal gas both increase in a process.

(a) Such a process is not possible

(b) The work done by the system is positive

(c) The temperature of the system must increase.

(d) Heat supplied to the gas is equal to the change in internal energy.

Q32. In a process on a system, the initial pressure and volume are equal to the final pressure

and volume.

(a) The initial temperature must be equal to the final temperature.

(b) The initial internal energy must be equal to the final internal energy.

(c) The net heat given to the system in the process must be zero.

(d) The net work done by the system in the process must be zero.

Q33. Refer to figure below. Let 1UΔ and 2UΔ be the change in internal energy in processes

A and B respectively, QΔ be the net heat given to the system in process A B+ and

WΔ be the net work done by the system in the process A B+ .

(a) 1 2 0U UΔ + Δ = (b) 1 2 0U UΔ − Δ =

(c) 0Q WΔ − Δ = (d) 0Q WΔ + Δ =

Q34. The internal energy of an ideal gas decreases by the same amount as the work done by

the system.

(a) The process must be adiabatic. (b) The process must be isothermal.

(c) The process must be isobaric. (d) The temperature must decrease.

PA

B

V





Q35. Which of the following is correct statement statements?

(a) The value of ( )Vp

p

CCC+

for helium is 38 for ⎟

⎠⎞

⎜⎝⎛ =

35γ

(b) value of specific heat capacity at constant volume for helium is 32R

(c) The value of ( )( )Vp

Vp

CCCC

+

− for 2H is

61 for ⎟

⎠⎞

⎜⎝⎛ =

57γ

(d) value of heat capacity at constant pressure is 72R

Q36. The first law of thermodynamics, WQU Δ−Δ=Δ , indicates that when a system goes

from its initial state to a final state which of following is correct statement

(a) UΔ is the same for every path

(b) QΔ the same for every path in isochoric process

(c) WΔ is the same for every path in adiabatic process

(d) WΔ and QΔ are the same for every path

Q37. Two thermally insulated vessels 1 and 2 are filled with air and connected by a short tube

equipped with a valve. The volumes of the vessels, the pressures and temperatures of air

in them are known ( )222111 ,, and,, TpVTpV . The pressure established after the opening of

the valve is given by

(a) 21

2211

VVVPVP

P++

= (b)

11

2211

11//

VV

VPVPP

+

+=

(c) 21

2211

VVVTVT

P++

= (d)( )

122211

112221

TVPTVPVPVPTT

T++

=





Q38. Two moles of certain ideal gas at a temperature 0T cooled isochorically so that the gas

pressure reduced n times. Then, as a result of the isobaric process, the gas expanded till

the temperature got back to the initial value.

(a) the work is given by 0RT

(b) the work is given by 02RT

(c) the heat exchange is given by 0RT

(d) the heat exchange is given by 02RT


Q39. A diatomic gas at S T P is expanded to thirty-two times its volume under adiabatic

conditions. the resulting temperature in ………….. 0K ( answer must be in two decimal

points)

Q40. The quantity of heat required to raise the temperature of one gram molecule through one

degree for a mono atomic gas at constant volume is …………….. R (where R is gas

constant )

Q41. A gas expands adiabatically at constant pressure such that its temperature V

T 1∝ . The

value of ⎟⎟⎠

⎞⎜⎜⎝

⎛=

V

p

CC

γ of the gas is ………………

Q42. One mole of diatomic gas ⎟⎠⎞

⎜⎝⎛ = RCV 2

5 and one mole of a monatomic gas ⎟⎠⎞

⎜⎝⎛ = RCV 2

3

are mixed. The value of γ for the mixture is, (where γ is the ratio of two specific heats

of the gas…………..

Q43. For a ideal gas having 3 translational and 2 rotational degrees of freedom at constant

temperature T , the internal energy is ……… Bk T





Q44. The minimum attainable pressure of ideal gas in the process 20 VTT α+= where 0T and

are positive constants, and V is the volume of one mole of gas is 0R Tβ α the value of β

is given by ……….

Q45. The volume of one mole of an ideal gas with the adiabatic exponent γ is varied

according to the law ,/TaV = where a is a constant. Find the amount of heat obtained

by the gas in this process if the gas temperature increased by TΔ .

Q46. Ten grams of ice at 0°C is added to a beaker containing 30 grams of water at 25°C. What

is the final temperature of the system is …………. 0C when it comes to thermal

equilibrium? (The specific heat of water is 1 cal/gm/°C and latent heat of melting of ice is

80 cal/gm)





Solutions


Ans. 1: (d)

Ans. 2: (b)

Solution: dQ dW= sign dQ is positive so dW is positive the gas will do positive work .

Ans. 3: (a)

Solution: Change in internal energy in both the path is same and

area under the path A is more than path B so work done in

the path A is more than path B .From first law of

thermodynamics 1 2Q QΔ > Δ

Ans. 4: (b)

Solution: Internal energy is point function

Ans. 5: (a)

Solution: work done is area under the curve so from the figure continuously increases

Ans. 6: (c)

Solution: From first law of thermodynamics sign of dQ dU dW= + is not dependent only on

change on temperature rather it can be compensated by sign of internal energy and work

done .but work done is positive if volume expand .

Ans. 7: (c)

Solution: This is isochoric process so work done is zero

(d) Nothing can be said about the relation between 1WΔ and 2WΔ .

Ans. 8: (c)

Solution: The process is isobaric, so work done is nRdT so change in temperature in B is more

than A so 1 2W WΔ < Δ

Ans. 9: (c)

Solution: After some time the system will be in mechanical equilibrium, so pressure will remain

constant.

P A

BV





Ans. 10: (b)

Solution: If there is no exchange of heat between a system and its surrounding the process is

known as adiabatic. The gas equation for adiabatic change is given as

constant=γPV since, 1volumedensity

∝ , ( )γdensity∝P

γ11 kdP =⇒ γ

22 kdP = γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2

1

2

dd

PP

since, 321

2 =dd

( ) ( ) 128232 5/755/7

1

2 ===pp

Ans. 11: (c)

Solution: Here,

VBTATP

2−=

where, BA, are constants.

The work done dW is given as

PdVdW = (i)

Now, 2BTATPV −=

( )2BTATdPdV −=⇒

dTBTAdTPdV .2−=⇒ (ii)

by equation (i) and (ii)

dTBTAdTdW .2−=

work done ( )dTBTAT

T∫ −=2

2 [ ] TTBTAT 22−= ( ) ( )[ ]2222 TTBTTA −−−=

23BTAT −=





Ans. 12: (c)

Solution: The amount of heat required to increase the temperature of unit mass is called specific

heat.

Q mc T= Δ

where =m mass of substance

TΔ = increment in temperature

and 2Tc βα += (i)

The mean specific heat 0

0 0

1 T

c cdTT

= ∫ ( )0

2

0 0

1 T

T dTT

α β= +∫3

00

0

13TT

Tβα

⎡ ⎤= +⎢ ⎥

⎣ ⎦

20

3BTc α= +

By question cc = 2

20

3BT BTα α⇒ + = +

220 0

3 3BT BT T Τ

⇒ = ⇒ =

Ans. 13: (b)

Solution: ( ) nRTbVP =−

Work done = PdVdW =

∫= PdVW where bV

nRTP−

= ∫ −=

V

VdV

bVnRTW

2

( ) ( )2

log 2 logV

V

dVnRT nRT V b V bV b

⎡ ⎤= = − − −⎣ ⎦−∫

Since, 2=n ( )⎟⎟⎠⎞

⎜⎜⎝

⎛−−

=bVbVRTW 2log2

Ans. 14: (d)

Solution: Slope of adiabatic curve ×= γ slope of isothermal curve. This is shown by every

curve given in question. Hence, all curve represent the ideal gas.





Ans. 15: (c)

Solution: We have 2bVaT += 2bVaR

PV+=⇒

⇒ VRbV

VaRP

2

+= , VbRVaRP +=

To find minimum pressure put 0=dVdP

⇒ 0=⎥⎦⎤

⎢⎣⎡ +VbR

VaR

dVd

⇒ 02 =+− RbVaR

baV =⇒ 2

Again differential Eq. (i) twice, we get 2

2

dVPd is +ve for ,2V

⇒ P is minimum at .2V

By Eqs. (i) and (ii), we get

⇒ babRba

aRP //min += abR2=

Ans. 16: (a)

Solution: The change in internal energy dU is independent of the path, i.e., if initial and final

states of change are same then dU will be same.

Ans. 17: (a)

Solution: The work done is given as

PdV=

= area of the triangle + area of the rectangle

( ) ( ) ( )32025205021

×+⎥⎦⎤

⎢⎣⎡ −×−×=

32033021

×+××= 105 J=

p

A

BC

V

20

50

2 5





Ans. 18: (d)

Solution: The first law of thermodynamics is given as

Q dU Wδ δ= +

Here, JQ 1500105.1 3 =×=δ

JW 500=δ

so, dU Q Wδ δ= − = J1000 Ans. 19: (b)

Ans. 20: (c)

Solution: We know that energy associated with each degree of freedom of one mole gas is given

RT21

=

Hence if there is n degree of freedom, then RTnU2

= where →R universal gas

constant.

We know that specific heat at constant volume is given as

dTdUCV = ⇒ RnCV 2

=

(i)

By Mayor’s formula RCC VP =− RnR2

+= ⇒ RnCP ⎟⎠⎞

⎜⎝⎛ +=

21

v

P

CC

=γ ⇒n21+=γ

Ans. 21: (c)

Solution:

The ideal gas equation is written as RTPV =

For isothermal T is constant differentiate above equation w.r.t.V we get

P

Visothermal curve

P

Vadiabatic curve





( ) 0=PVdVd

⇒ ⎟⎠⎞

⎜⎝⎛−=VP

dVdP ⇒ slope of isothermal curve

VP

−=

In case adiabatic expansion gas equation is constant=γPV .

differentiating this w.r.t. V we get 01 =+ − dVPVdPV γγ γ

⎟⎠⎞

⎜⎝⎛−=VP

dVdP γ ⇒ slope of adiabatic curve ⎟

⎠⎞

⎜⎝⎛−=VPγ (ii)

dividing equation (ii), we get γ1

curve adiabatic of Slopecurve isothermal of Slope

=

Ans. 22: (d)

Solution: Specific heat: The rate of change of internal energy w.r.t. temperature at constant

volume is called specific heat at constant volume.

⇒ V

V dTdUC ⎟

⎠⎞

⎜⎝⎛= (i)

by first law of thermodynamics WdUQ Δ+=Δ (ii)

For adiabatic 0=ΔQ so above equation (ii) becomes

0=Δ+ WdU

⇒ ( ) 0=Δ+ WdTCV ⇒ ( ) 012 =Δ+− WTTCV ⇒ ( )21 TTCW V −=Δ Ans. 23: (a)

Solution: The energy associated with each degree of freedom at temperature T is given as

1 ,2 B BE k T k= is Boltzmann’s constant

Here, total number of degrees of freedom= number of degree of rotational freedom +

translational freedom 3 + 2 = 5 thus, total energy 152 Bk T= ×





Ans. 24: (a)

Solution: 0VP P e β−= 0

VRT P eV

β−⇒ =

0 VPT VeR

β−⇒ = ….(i)

For maximum temperature, 0=dVdT ( )0 0V VP

e VeR

β ββ− −⇒ − =

This gives 1Vβ

= . Putting this value of V in equation (i), gives 0PTe Rβ

=

Ans. 25: (b)

Solution: Force equation of dr element.

( ) 2dF dm rω= , 2ωrA

dmA

dFdP ⎟⎠⎞

⎜⎝⎛== and dP

rAdm ⎟

⎠⎞

⎜⎝⎛= 2ω

where A is area of cross section

Also we know

( ) RTMdmAdrP ⎟

⎠⎞

⎜⎝⎛=

( ) dPr

AMRTdrPA ⎟

⎠⎞

⎜⎝⎛= 2ω

∫∫ =P

P

r

PdPRTrdrM

00

2ω

0

22

ln2 P

PRTrM=

ω 2 2

20

M rRTP P eω

⇒ =

Ans. 26: (b)

Solution: Suppose number of moles of gas = n

Directional kinetic energy of gas = ( ) 2

21 vnM

When vessel sudden stop, then after long time this directional kinetic energy of gas is

converted into random kinetic energy when thermodynamic equilibrium will be achieved

and then ( ) TnCvnM V Δ=2

21 ⇒ ( ) 2

21 Mv

RT −=Δ

γ

TM,S

ω

0Pr

This end is open in air





Ans. 27: (a)

Solution: Suppose initial temperature is 1T and final is ( )TT Δ−1 then

1

;1

21

−=

−=

γγVPUVPU fi

( )120

1PP

VU −

−=Δγ

1

1 1 1nRTPV nRT P

V= ⇒ = also

1

01

TVP

nR =

0

22 V

nRTP =

( )TnRV

nRTV

nRTVU Δ

−=⎟

⎠⎞

⎜⎝⎛ −

−=Δ

11120

γγ

( )11

01

−Δ

=ΔγT

TVPU here

oTTPP

==

1

01

( )10

00

−Δ

=ΔγT

TVPU

UΔ = increase in potential energy

WUQ Δ+Δ=Δ since vessel is sealed then 0=ΔW

( )0 0

0 1PV TQ U

T γΔ

Δ = Δ =−

Ans. 28: (b)

Solution: During path AC 150 100 50dU Q W Jδ δ= − = − =

Hence internal energy is point function dU will same in all path

In path ABC , 50 30 80Q dU W Jδ δ= + = + = .

Ans. 29: (b)

Solution: Heat exchange is perfect differential in isochoric process.

Q dUΔ =





Ans. 30: (b)

Solution: WΔ is a perfect differential if the process is adiabatic

W dΔ = − U


Ans. 31: (b) and (c)

Solution: If volume increase then workdone is positive and from PV nRT= if volume and

pressure increase then temperature will increase.

Ans. 32: (a) and (b)

Solution: from PV nRT= if pressure and volume of initial point and final point is same then

temperature is same ,so internal energy will also same


Solution: 1 2 0U UΔ = Δ = because internal energy is path independent so 0Q WΔ − Δ =


Solution: Q dU Wδ δ= + and VdU nC dT= , dU is negative so temperature decreases.

If dU Wδ− = so 0Qδ =

Ans. 35: (b), (c) and (d)

Solution: The ratio of pC and VC is known as coefficient of adiabatic expansion. Hence,

V

p

CC

=γ

(a) so, for helium 35

=γ

351

1

+= ,

83

=+ Vp

p

CCC

(b) 31 2V

R RCγ

= =−

for ⎟⎠⎞

⎜⎝⎛ =

35γ

(c) for hydrogen 75

p V

p v

C CC C

γ−

= =+

dividing by VC the numerator and denominator, we get





1/1/

+

−=

+

−

Vp

Vp

Vp

Vp

CCCC

CCCC

11

+−

=γγ

157

157

+

−=

61

=+

−⇒

Vp

Vp

CCCC

(d) 71 2P

R RC γγ

= =−

Ans. 36: (a), (b) and (c)

Solution: The internal energy UΔ is a state function. Hence, it does not depend on path whereas

both Q and W depend on path. Thus, QΔ is same for every path.


Solution:

When value is opened and thermodynamics equilibrium is attained then, number of moles

will be constant. Then

( )

RTVVP

RTVP

RTVP 21

2

22

1

11 +=+

( )

TVVP

TVP

TVP 21

2

22

1

11 +=+ (i)

Also we know, in whole system:

WUQ Δ+Δ==Δ 0

0=ΔQ because vessel is insulated. And also

0=ΔW because gas does not work on atmosphere because vessel closed

then UΔ system = 0 then

22112211 TnTnTCnTCn VV Δ=Δ⇒Δ=Δ

( ) ( ) 022

221

1

11 =−+− TTTVPTT

TVP

(ii)

( )

122211

112221

TVPTVPVPVPTTT

++

= put in (i) 21

2211

VVVPVPP

++

=

22 VV +PT

2

?111 ,, TPV 222 ,, TPV

1vessel






Solution:

Initial ( )VTP ;; 00 isochoric→ ⎟

⎠⎞

⎜⎝⎛→⎟

⎠⎞

⎜⎝⎛ VT

PVT

P;;

2;;

2 00

isobaric

10

Here nRTPV =

00 2RTVP =

0

02PRTV =

Isochoric process:

2

1

2

1

TT

PP

′′

=

22/0

11

0

0

0 TT

TT

PP

=⇒=

Isobaric process:

2

1

2

1

TT

VV

′′

=′′

VVT

TVV 2

2/1

0

0

2

=⇒=

In whole system: from initial to final position:

WUQ Δ+Δ=Δ

here 0=ΔU final temperature is zero.

[ ]22

0 00 VPV

PW =Δ+=Δ

0

00 22 P

RTP=

0RTW =Δ and 0RTQ =Δ

0

0

2

PTn = constant=V

20PPf =






Ans. 39: K25.68

Solution: For diatomic gas 25

=γ

∴Adiabatic equation =−1γTV constant 1

2

112

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒

γ

VVTT

here, KTVVVV 273,32, 121 ===

( ) 12 32273

−= γT ( ) ( )2

31252

32

273

32

273==⇒

−T K25.68=

Ans. 40: (a)

Solution: The quantity of heat required to increase the temperature at constant volume through

C°0 for per degree of freedom

( ) VVQ nC dT=

1 1, 12VC R n dT= = =

So, ( ) RRQ V 5.01211 =×⎟

⎠⎞

⎜⎝⎛×=

Ans. 41: 1.5

Solution: We know that for adiabatic expansion constant1 =−γTV

Given that V

T 1∝ 1/ 2TV constant⇒ =

by equation (i) and (ii),

211 =−γ ⇒

211+=γ , 50.1

23==γ





Ans. 42: 1.5

Solution: 1 21 2

1 2

V VV

n c n cc

n n+

=+

and 1 21 2

1 2

P PP

n c n cc

n n+

=+

p

V

cc

γ =

1 115 71, ,2 2V PR Rn c c= = =

1 113 51, ,2 2V PR Rn c c= = =

Ans. 43: 2.5 Solution: The energy associated with each degree of freedom at temperature T is given as

TkE B21

= is Bk Boltzmann’s constant

Here, total number of degrees of freedom= number of degree of rotational freedom +

translational freedom 3 + 2 = 5thus, total energy TkB215×=

Ans. 44: 2

Solution: 20 VTT α+=

We know RTnRTPV ==

If V is increasing, T will increase and hence P will increase. Hence calculation of minP :

R

PVT = put in (i)

20 VT

RPV α+= , [ ]VVTRP α+= −1

0

0dP

dV= , 2 0

0 0 TR T V Vαα

−⎡ ⎤− + = ⇒ =⎣ ⎦ put in (ii)

Put the value of α

0TV = in [ ]VVTRP α+= −10

0min 2 TRP α=





Ans. 45: 2

Solution: TaV = number of mole 1n =

VaT =⇒

we know Q dU Wδ δ= +

∫ Δ−

=Δ==Δ TRTCdUU V 1γ (i)

W PdV W PdVδ = ⇒ Δ = ∫ (ii)

RTnRTPV ==

2VaR

VRTP == put in (ii)

⎥⎦

⎤⎢⎣

⎡−=−==Δ ∫

212

111 2

1

2

1 VVaR

VaRdV

VaRW

V

V

V

V

[ ]TRRVa

VaW Δ−=⎥

⎦

⎤⎢⎣

⎡−=Δ

21

Q U WΔ = Δ +Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

Δ=Δ−−Δ

=1

111 γ

γγ

TRTRTR

( )2

1R T

Qγ

γΔ −

Δ =−

Ans. 46: 0

Solution: The amount of heat required to melt the ice of mass 10gm at 00C is

10 80 800Q m L Cal= × = × = . Where L is the latent heat of melting of ice and m is the

mass of the ice. The amount of heat available in water of mass 30gm at 250C is

30 1 25 750vQ m C T Cal= × × = × × =

Since the heat available is less than the heat required to melt the ice therefore ice will not

melt as a result the temperature of the system will be at 00C only





Chapter – 4

Second Law of Thermodynamics and Entropy

4.1 Second Law of Thermodynamics

Statement: It is impossible to transfer total heat into work in a cyclic process in the

absence of other effect (by Lord Kelvin).

In another way it is also stated that it is impossible for heat to be transferred by a cyclic

process from a body to one warmer than itself without producing other changes at same

time.

4.2 Heat Engines

A machine that can convert heat into work is said to be heat engine.

It is a system that performs the conservation of heat or thermal energy to mechanical

work.

Schematic of operation at heat engine

4.2.1 Heat Reservoir

It is an effectively infinite pool of thermal energy of a given, constant temperature. Ideal

its heat capacity is large enough that when it is in thermal contact with another system, its

temperature remains constant.

All heat engines have mainly three essential components.

1. A source: This is a hot region which is a part of the surrounding from which

energy flows by heat transfer. Popularly it is known as hot reservoir. Example is a

nuclear reactor, furnace.

Hot reservoir T1

Cold reservoir T2

W = Q1 – Q2

1Q

2Q





2. The working agent: This under goes change at state as part of a continuous cycle

e.g. steam-water.

3. A sink: This is a respectively cold region which is part or the surrounding into

which heat is rejected by heat transfer e.g. cooler.

4.2.2 Efficiency of Heat Engine (η):

work done by the systemheat given into the system

η =

Hence the process is cyclic dU = 0.

dQ dW=

Heat given to system is Q1 and heat rejected by system is Q2.

1

21

QQQ −

=η

4.2.3 Carnot Cycle: It is theoretical thermodynamics cycle proposed by Nicolas, Leonard Sadi

Carnot.

It can be shown that it is most efficient cycle for converting a given amount of thermal

energy into work.

A system undergoing a Carnot cycle is called a Carnot heat engine.

Stages of the Carnot cycle:

1. Reversible isothermal expansion of the gas at hot temperature, T1 (isothermal heat

addition). During this step (1 to 2) the gas is allowed to expand and it does work

on the surrounding. The gas expansion is propelled by absorption of quantity Q1

at heat from the high temperature T1.

P

( )44 ,VP

( )11 ,VP

( )33 ,VP

( )22 ,VP

1T

2T

V

4

1

2

3





2. Reversible adiabatic expansion of the gas. For this step (2 to 3) on figure, the gas

continues to expand, working on surrounding. The gas expansion causes it to cool

to the cold temperature T2.

3. Reversible isothermal compression of the gas at the “cold” temperature T2

(isothermal heat rejection). The process is shown by 3 → 4. Now the

surroundings do work on the gas, causing quantity Q2 of heat to flow out of the

gas to the low temperature reservoir.

4. Adiabatic compression of the gas (4 to 1). During this step, the surrounding do

work on the gas, compressing it and cause the temperature to rise to T1. At this

point the gas in the same state as at start at step 1.

Efficiency of Carnot engine:

1

WQ

η =

W is the work done during cycle and Q1 is heat given to system.

4321 WWWWW +++=

1W = work done during process 1 to 2 for isothermal process = 1

21 ln

VV

nRT

change in internal energy during the process 0dU =

21 1 1

1

ln VQ W nRTV

⎛ ⎞= = ⎜ ⎟

⎝ ⎠ (From first law of thermodynamics)

2W is work done during process 2 to 3 in adiabatic process.

0dQ = ( )2 1

2 1nR T T

Wγ−

=−

3W = work done during process 3 to 4 isothermal compression.

43 2

3

ln VW nRTV⎛ ⎞

= ⎜ ⎟⎝ ⎠

2 3 0Q W dU= = (isothermal process)





4W , the work done during the adiabatic process 4 to1, is given by

( )1 2

4 1nR T T

Wγ−

=−

2 4W W= −

2 41 2

1 3

ln lnV VW nRT nRTV V

⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

211 ln

VVnRTQ

4

32

1 1 2

1

ln1

ln

VVTW

Q T VV

η

⎛ ⎞⎜ ⎟⎝ ⎠= = +⎛ ⎞⎜ ⎟⎝ ⎠

….(i)

From the figure, 1 1

1 2 2 3TV T Vγ γ− −= ….(ii)

1 12 4 1 1T V TVγ γ− −= ….(iii)

Dividing equation (ii) by (iii) gives 1 1

32

1 4

VVV V

γ γ− −⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

32

1 4

VVV V

⇒ = ….(iv)

From equations (i) and (iv)

2

1

1 TT

η = − ….(v)

For Carnot cycle, 1

2

1

2

TT

QQ

=

This gives 2

1

1 QQ

η = − ….(vi)

P

( )44 ,VP

( )11 ,VP

( )33 ,VP

( )22 ,VP

1T

2T

V

4

1

2

3





Example: An ideal gas engine operator in a cycle which when represent on a P-V diagram is a

rectangle. If we call 1 2,P P and the lower and higher pressures respectively and 1 2,V V as

lower and higher volume respectively.

(a) Calculate the work done in complete cycle

(b) indicate in which parts of the cycle heat is absorbed and in which part librated.

(c) Calculate the quantity of heat following into the gas in one cycle

(d) show that efficiency of the engine is

2 1

2 1 2 1

1P V

P P V V

γη γ−

=+

− −

Solution: work done drawing ( )122 VVPAB −=

Heat absorbed = pnC TΔ

2 1( )1

nR T Tγγ

= −− 2 2 2 1( )

1PV PVγ

γ= −

−

( )2 2 11P V Vγ

γ= −

−

12 VV > so heat absorbed in the process.

In process B-C isochoric process

0=−cBW

dUdQ =

vnC TΔ

( )21 21 1

Vn R T P Pγ γ

Δ = −− −

21 PP < heat rejected

In process C-D isobaric process

( )211 VVPW DC −=−

P ( )2 1,P VA

D( )1 1,P VC( )1 2,P V

B( )2 2,P V

V





Heat exchange during the process

( )1 1 21C DQ P V Vγγ→ = −−

21 VV < so heat is rejected

In process D → A isochoric process

0=→ADW

Heat exchange during the process

vnC TΔ =1

nR TγΔ−

=( )1 2 1

1V P P

γ−−

heat absorbed.

(a) Work done = P2 (V2-V1) + P1(V1 – V2)

W = (P2 – P1) (V2 –V1)

(b) Heat absorbed ( ) ( )12 2 1 2 11 1

VP V V P Pγγ γ

− + −− −

(c) Heat flowing into the cycle

( ) ( ) ( ) ( )2 2 1 2

1 2 1 1 2 1 2 11 1 1 1P V V V P P P V V V P P

γ γ γγ γ γ γ

−+ − + − + −

− − − −

( )( ) ( ) ( )2 1 2 1 1 2 2 11

1 1V V P P P P V Vγ

γ γ= − − + − ⋅ −

− −

( )( ) ( )( )2 1 2 1 2 1 2 11

1 1V V P P P P V Vγ

γ γ= − − − − −

− −

( )( )2 1 2 11

1 1P P V V γ

γ γ⎡ ⎤

− − −⎢ ⎥− −⎣ ⎦( )( )1212 VVPP −−=

dWdQ =

(d) Efficiency = absorbed

WQ

( )( )( ) ( )

2 1 2 1

2 2 1 1 2 1

1 1

P P V VP V V V P Pγγ γ

− −=

− −+

− −

Therefore, 2 1

2 1 2 1

1P V

P P V V

γηγ

−=

+− −





4.3 Entropy

Entropy is an extensive thermodynamic property that is measure of a system’s thermal

energy per unit temperature that is unavailable for doing useful work. Thermodynamic

entropy is a non-conserved state function. For Isolated systems entropy never decreases.

In statistical mechanics, entropy is a measure of the number of ways in which a system

may be arranged, often taken to be a measure of ‘disorder’ (the higher the entropy the

higher the disorder).

The infinitesimal change in the entropy ( dS ) of a system is the infinitesimal transfer of

heat energy (δQ) to a closed system driving a reversible process, divides by temperature

(T) of the system.

QSTδ

Δ = ∫

It has unit Joule/Kelvin or QdSTδ

= ∫

Law of thermodynamics and entropy:

According to first law of thermodynamics

Q dU PdVδ = +

From definition of Entropy

QS Q T S TdSTδ δΔ = ⇒ = Δ ⇒

pdVdUTdS +=





4.3.1 Inequality of Clausius

Consider an irreversible cyclic engine working between T1 and T2. If reversible engine is

operating between same temperature then from Carnot theorem.

Efficiency of irreversible (ηir) will always smaller than efficiency of reversible

engine( rη ).

.. revirr ηη <

rev

revrev

irr

irrirr

QQQ

QQQ

1

21

1

21 −<

−

rev

rev

irr

irr

QQ

QQ

1

2

1

2 11 −<−

1

2

1

2 11TT

QQ

irr

irr

−<−

1

1

2

2

TQ

TQ irrirr

>

F0 irreversible cyclic Engine

02

2

1

1 <−T

QT

Q irrirr

or ∫ < 0TQδ

This relation is known as inequality of Clausius.





Example: Two Carnot Engines A and B are operated in series. The first one A receives heat at

900 K and rejects to a reservoir at temperature T K. The second engine, B, receives the

heat rejected by the first Engine and in then rejects to a heat reservoir at 400 K.

Calculate the temperature T for the situation.

(a) The work outputs of the two engines are equal

(b) The efficiency of the two engines are equal.

For Engine A take in heat Q1 at temperature T1 and rejected heat Q at temperature T; and

the engine B taken in heat Q at temperature T and reject heat Q2 at temperature T2.

Solution: (a) 1AW Q Q= − 2BW Q Q= −

1 2A BW W Q Q Q Q= − = −

221 =−Q

QQ

TT

QQ 11 = and

TT

QQ 22 =

T

TTQ

QQ 2121 +=

+ 1 2 2 650T T T KT+

⇒ = =

Solution: (b) A Bη η=

TT

TT 2

1

11 −=−

( )12

1 2 600T T T T K= ⋅ =

Example: Calculate the charge in isothermal expansion from an initial volume iV to volume fV

Solution: For reversible process

PdVdUTdS += For isothermal process dU = 0

TPdVdS =

VnRTP =

lnf

i

Vf

iV

VdVdS nR nRV V

= =∫





Example: A mass of liquid at a temperature T1 is mined with an equal mass of the same liquid at

a lower temperature T2. the system is thermally insulated.

(a) compute the entropy-change

(b) show that it is necessarily positive.

Solution: Let c be the specific heat of the liquid. On the mining equal mass m of the same liquid

at temperature T1 and T2 let (T1 > T2).

Let T be the equilibrium temperature at minute.

( ) ( )21 TTmcTTmc −=−

2

21 TTT

+=

∫=ΔT

T TQS

1

1δ Entropy change of hotter liquid to cool from T1 to T

1

11

lnT

T

dT TS mc mcT T

Δ = =∫

⇒Δ 2S Entropy change of hotter liquid to heat from T2 to T

22 ln

TTmcS =Δ

SΔ = Entropy change of the system.

21 SSS Δ+Δ=Δ

21

2

lnTT

Tmc=21

21

2

21

21 2ln22

lnTT

TT

mc

TT

TTmc

+

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛+

=

(b) we know arithmetic mean is greater than geometric mean.

..MGAM >

So 2121

2TT

TT>

+ So 0>ΔS





Example: Compute the change in entropy when ice melt into steam. It is given that L1 is latent

heat of fusion, c is specific heat at water and L2 latent heat at vaporization.

Solution: Assume T1 be the Kelvin temperature at which ice melts into water and T2 the Kelvin

temperature at which water is boiled to steam.

1SΔ is entropy change when ice is converted into water

1

11 T

mLS =Δ

2SΔ Entropy change when water is heated from T1 to T2

∫=Δ2

1

2

T

T TdTmcS

1

22 ln

TT

mcS =Δ

3SΔ Entropy change when water change into vapors

23

2

mLST

Δ =

Total change in entropy 1 2 3S S S SΔ = Δ + Δ + Δ = 1

1

mLT

2

1

ln TmcT

+ 2

2

mLT

+






Q1. Two Carnot engines A and B are operating between the same source and the same sink.

Engine A uses an ideal gas as the working fluid while engine B uses Van der Waals’ gas

as the working fluid. Which one of the following is correct?

(a) The efficiency of engine A is less than that of engine B

(b) The efficiency of engine A is equal to that of engine B

(c) The efficiency of engine A is more than that of engine B

(d) No comparison can be made

Q2. A heat engine converts a given quantity of heat into work with maximum efficiency

during which one of the following processes?

(a) Isobaric process (b) Isochoric process

(c) Isoenthalpic process (d) Isothermal process

Q3. If heat Q is added reversibly to a system at temperature T and heat Q′ is taken away from

it reversibly at temperature ,T ′ then which one of the following is correct?

(a) 0=′′

−TQ

TQ

(b) 0>′′

−TQ

TQ

(c) 0<′′

−TQ

TQ

(d) =′′

−TQ

TQ change in internal energy of the system

Q4. The temperature of water (mass, m ) increases from 1T to 2T . If c is the specific heat

capacity of water, then the total increase in entropy of water is given by:

(a) ( )12 TTmc − (b) ⎟⎟⎠

⎞⎜⎜⎝

⎛

2

1logTTmc e

(c) ( )21 TTmc − (d) ⎟⎟⎠

⎞⎜⎜⎝

⎛

1

2logTT

mc e





Q5. Consider the following that take place in Carnot cycle:

1. Adiabatic expansion

2. Adiabatic compression

3. Isothermal expansion

4. Isothermal compression

The correct sequence of the above processes is:

(a) 1, 3, 4, 2 (b) 3, 1, 2, 4

(c) 3, 1, 4, 2 (d) 1, 3, 2, 4

Q6. The change in entropy of the melting of 1kg of ice at C00 is

(a) KJ /66.3 (b) KJ /31.15

(b) KJ /103.12 2× (d) KJ /1014.1 6×

Q7. Which one of the following reversible cycles, represented by right angled triangles in a

T-S diagram, is the least efficient?

(a) (b)

(c) (d)

03T

0T

0S 02SS

B

CA

T

02T

0T

0S 03SS

B

CA

T

C03T

0T

0S 02SS

B

A

T

02T

0T

0S 03SS

B C

A

T





Q8. One mole of an ideal gas is carried from temperature 1T and molar volume 1V to 2 2,T V .

Then the change in entropy is given by

(a) 2

1

ln VS RV

Δ = (b) 2

1

lnVTS CT

Δ =

(c) 2 2

1 1

ln lnvT VS C RT V

Δ = − (d) 2 2

1 1

ln lnvT VS C RT V

Δ = +

Q9. Consider an engine working in a reversible cycle and using an ideal gas with constant

heat capacity pC as the working substance. The cycle consists of two processes at

constant pressure, joined by two adiabatic as shown in fig .

Then the efficiency of this engine in terms of 1 2,p p is given by

(a)

1

2

1

1 pp

γγ

η

−

⎛ ⎞= − ⎜ ⎟

⎝ ⎠ (b)

1

2

1

1 pp

γη

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

(c)

1

2

1

1 pp

γγ

η

−

⎛ ⎞= − ⎜ ⎟

⎝ ⎠ (d)

12

1

1 pp

γγ

η−⎛ ⎞

= − ⎜ ⎟⎝ ⎠

Q10. An insulated chamber is divided into two halves of volumes. The left

half contains an ideal gas at temperature 0T and the right half is

evacuated. A small hole is opened between the two halves, allowing the

gas to flow through and the system comes to equilibrium. No heat is

exchanged with the walls.

(a) During the process work done is zero but change in entropy of gas as well as universe

0SΔ =

(b) During the process work done is not zero but change in entropy of gas 0SΔ = and

change in entropy of universe 0SΔ >

(c) During the process work done is zero but change in entropy of gas as well as

universe 0SΔ >

(d) During the process work done is not zero but change in entropy of gas 0SΔ > as

well as universe 0SΔ >





Q11. Each of the two isolated vessels, A and B of fixed volumes, contains N molecules of a

perfect monatomic gas at a pressure P . The temperatures of A and B are 1T and 2T ,

respectively. The two vessels are brought into thermal contact. At equilibrium, the

change in entropy is

(a) ⎥⎦

⎤⎢⎣

⎡ΤΤΤ+Τ

Ν21

22

21

4ln

23

Bk (b) ⎟⎟⎠

⎞⎜⎜⎝

⎛ΤΤ

Ν1

2ln23

Bk

(c) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡

ΤΤΤ+Τ

Ν21

221

4ln

23

Bk (d) 2NkB


Q12. In diagram 1 Carnot cycle is represented in PV diagram while in Diagram II 1 Carnot

cycle is represented in TS diagram

Which one of the following is correct?

(a) 1 and A is isothermal expansion and heat is given into the system

(b) B is adiabatic compression

(c) In process 3 heat is rejected by system

(d) Work done during 4 and B are same in magnitude and opposite to sign

( )VP −

P

V

4

1

3

2

1T

2T

Diagram I ( )ST −

Diagram II

T

S

D

A

C

B





Q13. Consider the following statements regarding the characteristics of entropy are correct

(a) Entropy is a measure of disorder.

(b) Entropy changes during a reversible adiabatic process.

(c) Entropy of a system decreases in all irreversible processes.

(d) Change in entropy for a complete reversible thermodynamic cycle is zero

Q14. Consider the following statements regarding transition of a system from one

thermodynamic state another which of the following is correct

(a) The heat absorbed by it along any reversible path independent of the path.

(b) The change of entropy of the system in a reversible : process is independent of the

path.

(c) The change of entropy of the system in a irreversible process is also independent of

the path.

(d) The heat absorbed by it along any irrereversible path independent of the path.

Q15. The temperature entropy diagrams of two engines A and B working between the same

temperature 1T and 2T of the source and the sink respectively are shown in the given

figures.

The efficiency of A :

(a) is less than that of B

(b) is equal to that of B

(c) is greater than that of B

(d) and B cannot be compared on the basis of data given in the diagrams

1T

2T1S 2S

B

Entropy

Tem

pera

ture

2S1S

1TA

2T

Entropy

Tem

pera

ture





Q16. Which of the the following statements are correct

(a) The entropy change during a reversible adiabatic process is zero

(b) Entropy is a state function

(c) The entropy of a thermally isolated system never decreases

(d) The entropy change during an reversible adiabatic process is zero.

Q17. An ideal gas is expanded adiabatically from ( )1 1,P V to ( )2 2,p V . Then it is compressed

isobarically to ( )2 1,P V . Finally the pressure is increased to 1P at constant volume 1V then

which of the following is correct?

(a) the P V− indicator diagram is given by fig

(b) the work done in the cycle is ( ) ( )2 2 1 1 2 2 11

1W PV PV P V V

γ= − + −

−

(c) heat will absorb in isochoric process

(d) the efficiency of the cycle is

2

1

1

2

11

1

VVpp

η γ−

= −−

Q18. Consider an arbitrary heat engine which operates between reservoirs, each of which has

the same finite temperature-independent heat capacity c . The reservoirs have initial

temperatures 1T and 2T , where 2 1T T> , and the engine operates until both reservoirs have

the same final temperature 3T . Then which of the following statements are correct

(a) The change of entropy is given by 2

3

1 2

ln TTT

.

(b) In general 3 1 2 T T T≤

(c) In general 3 1 2 T T T≥

(d) The maximum work done is given by ( )2

max 1 2W c T T= −

P

C

1P

2P

1V 2V

B

Aadiabatic

V





Q19. Which of the statements are correct?

(a) A mole of an ideal gas undergoes a reversible isothermal expansion from volume 1V

to 12V .Then change in entropy of the gas is ln 2RT and there is no change in

entropy of the universe .

(b) A mole of an ideal gas undergoes a reversible isothermal expansion from volume 1V

to 12V .Then change in entropy of the gas is ln 2RT and change in entropy of the

universe is ln 2RT

(c) A mole of an ideal gas undergoes free isothermal expansion from volume 1V to

12V .Then change in entropy of the gas is ln 2RT and there is no change in entropy

of the universe

(d) A mole of an ideal gas undergoes free isothermal expansion from volume 1V to

12V .Then change in entropy of the gas is ln 2RT and change in entropy of the

universe is ln 2RT

Q20. A body of constant heat capacity PC and a temperature iT is put into contact with a

reservoir at temperature fT . Equilibrium between the body and the reservoir is

established at established at constant pressure. Assume f iT T> . Then

(a) Change of entropy of the body is ln fp

i

TC

T

(b) The change of entropy of the heat source is ( )p i f

f

C T TT−

(c) If f iT T> then the change in entropy of universe 0SΔ >

(d) If f iT T< then change of entropy of universe is 0SΔ <





Q21. n mole of an ideal gas is originally confined to a volume 1V in an insulated container of

volume 1 2V V+ . The remainder of the container is evacuated. The partition is then

removed and the gas expands to fill the entire container. If the initial temperature of the

gas was T

(a) Then which of the following statement is correct

(b) The temperature remain constant T

(c) The work done during the process is 1 2

1

ln V VnRTV

⎛ ⎞+⎜ ⎟⎝ ⎠

(d) The change entropy of gas in the process is 1 2

1

ln V VnRV

⎛ ⎞+⎜ ⎟⎝ ⎠

The change in entropy of universe during the process is 1 2

1

ln V VnRV

⎛ ⎞+⎜ ⎟⎝ ⎠


Q22. A Carnot engine has an efficiency of 16

. On reducing the sink temperature by ,65 C° the

efficiency becomes 13

. The source temperature is given by …………. 0K

Q23. A Carnot engine whose low-temperature reservoir is at C027 has an efficiency

37.5%.The high-temperature reservoir is …………. 0C

Q24. In the given T S− diagram, the efficiency is given by …….. % (Answer must be in two

decimal point).

200

100

500 1000

A C

BTemperature (K)

1V 2V

Insulated container





Q25. An ideal gas is confined to a cylinder by a piston. The piston is slowly pushed in so that

the gas temperature remains at C020 . During the compression, J730 work is done on

the gas. the entropy change by the gas is …….. /J K

Q26. 10 g of ice at C°0 is slowly melted to water at .0 C° The latent heat of melting is 80 cal/g.

The change in entropy is nearly ………. /cal K

Q27. If a capacitor of Fμ1 charged to a potential of V300 is charged, a resistor kept at room

temperature, then the entropy change of the universe in…….. 410−× /J K

Q28. One kg of 2H O at o0 C is brought in contact with a heat reservoir at o100 C . When the

water has reached o100 C Then the change in entropy of the universe is ………… /J K

(specific heat of water 2

4.18 /H OC J g= )

Q29. A reversible engine cycle is shown in the following T -S diagram. The efficiency of the

engine is ……………….. %

1T2

1TT

1S 12S 13SS





Solution


Ans. 1: (b)

Solution: The efficiency of a Carnot working between temperature limits 1T and 2T is given as

1

21TT

−=η

→1T absolute temperature of source

→2T absolute temperature of sink

Since, efficiency does not depend on -working substance hence the efficiency of both

engines A and B are same

Ans. 2: (c)

Solution: The first law of thermodynamics

PdVdUdQ +=

The maximum efficiency can be obtained, if process is isoenthalpic.

Ans. 3: (a)

Solution: For a reversible process the change in entropy is zero.

⇒ 0=dS

⇒ 1

1

0Q QT T

′− =

′

Ans. 4: (d)

Solution: The change in entropy of a system is given as ∫=2

1

T

T TQdS δ

If Qδ amount of heat is given to water then 1T change in temperature dT is given as

mcdTQ =δ

∫=2

1

T

T TmcdTdS

1

2logTT

mc e=





Ans. 5: (c)

Solution:

AB → isothermal expansion

BC → adiabatic expansion

CD →isothermal compression

DA → adiabatic compression

Ans. 6: (c)

Solution: The change in entropy is given as

TQdS δ

=

here, ,Q mLδ = L is latent heat

801000×= 80000Q calδ⇒ =

⇒ JQ 2.480000 ×=δ (ii)

KT 273=

so, KJdS /273

2.480000×= KJ /103.12 2×=

Ans. 7: (d)

Solution: Efficiency 1

WQ

η =

In a)

0 0 0 0 0 01 (3 )(2 )2

W T T S S T S= − − =

1 0 0 0 0 01 (3 )(2 ) 1.52

Q T S S T S= − =

1

66%WQ

η = =

V

P

1

3 C

BA

D

24

03T

0T

0S 02SS

B

CA

T

)a(





In (b)

0 0 0 0 0 01 (2 )(3 )2

W T T S S T S= − − =

1 0 0 0 0 01 (2 )(3 ) 22

Q T S S T S= − =

1

50%WQ

η = =

In (c)

0 0 0 0 0 01 (3 )(2 )2

W T T S S T S= − − =

1 0 0 0 0 0(3 )(2 ) 3Q T S S T S= − =

1

33%WQ

η = =

In (d)

0 0 0 0 0 01 (2 )(3 )2

W T T S S T S= − − =

1 0 0 0 0 0(2 )(3 ) 4Q T S S T S= − =

1

25%WQ

η = =

Ans. 8: (d)

Solution: From ( ) ( )1 1 and vdS dU PdV C dT PdV PV RTT T

= + = + =

We obtain 2 2

1 1

ln lnvT VS C RT V

Δ = +

02T

0T

0S 03SS

B

CA

T

)b(

C03T

0T

0S 02SS

B

A

T

)c(

02T

0T

0S 03SS

B C

A

T

)d(





Ans. 9: (a)

Solution: In the cycle, the energy the working substance absorbs from the source of higher

temperature is ( )ab p b aQ C T T= − .The energy it gives to the source of lower temperature

is ( )reject p c dQ C T T= − . Thus

1 1reject c d

ab b a

Q T TQ T T

η −= − = −

−

form the equation of state pV nRT= and the adiabatic equations

2 1 2 1,d a c bPV PV PV PVγ γ γ γ= = we have

1

2

1

1 PP

γγ

η

−

⎛ ⎞= − ⎜ ⎟

⎝ ⎠ Ans. 10: (c)

Solution: After a hole has been opened, the gas flows continuously to the right side and reaches

equilibrium finally. During the process, internal energy of the system E is unchanged.

Since E depends on the temperature T only for an ideal gas, the equilibrium

temperature is still 0T so from first law of thermodynamics work done is zero but process

is irreversible so change in entropy of gas as well as universe is 0SΔ > Ans. 11: (c)

Solution: Final temperature of each vessel at equilibrium is 1 2

2T TT +

=

1 2

T TV V

T T

C dT C dTST T

Δ = +∫ ∫ = ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

ΤΤΤ+Τ

Ν21

221

4ln

23

Bk where 32

BV

KC = for monatomic gas

P

V

c

ba

d

1P

2P

adiabatics






Ans. 12: (a), (c) and (d)

Solution:

Path 1: Expansion of constant temperature and heat is absorbed because vedQ += hence

S will increase Corresponding to A.

Path 2: Adiabatic expansion means S is constant Corresponding to B.

Path 3: Isothermal compression vedQ −= so heat will reject S is decrease corresponding

to C.

Path A and path B are isoentropic process so work done is dependent on points but

direction of both the case will opposite .


Solution: Entropy is function of no of microstate which will measurement of disorderness

For thermodynamic process 0dS ≥ and for reversible process 0dS = .


Solution: The entropy is point function and perfectly differential so it is path independent

Ans. 15: (a)

Solution: The efficiency is defined as

1

21

1

21

TTT

HHH −

=−

=η

Ans. 16: (a), (b), (c) and (d)

Solution: Entropy:

where →Qδ amount of used heat in reversible adiabatic process 0Qδ =

It is a state function and is defined as TQdS δ

= is exact differential so it is state function

4

1

3

2

1T

2T

P

V

T

S

D

A

C

B





If this taken in by a system then change in entropy is positive and so entropy of the

system increases.

The entropy change during an reversible adiabatic process is zero.

Ans. 17: (a), (b) and (d)

Solution: (a) The cycle is shown in the figure

(b)The work the system does in the cycle is

( )2 1 2AB

W PdV PdV P V V= = + −∫ ∫

Because AB is adiabatic and an ideal gas has the equations pV nkT= and p vC C R= + ,

we get

( ) ( )2 1 2 2 1 11

1v vAB ABPdV C dT C T T PV PV

γ= − = − − = −

−∫ ∫

(c) During the CA part of the cycle the gas absorbs heat which is isochoric

( ) ( )1 2 1 2 11

1v vCA CAQ TdS C dT C T T V P P

γ= = = − = −

−∫ ∫

(d) Hence, the efficiency of the engine is

2

1

1

2

11

1

VVWPQP

η γ−

= = −−

Ans. 18: (a), (c) and (d)

Solution: (a) The increase in entropy of the total system is

3 3

1 2

23

1 2

ln 0T T

T T

TcdT cdTS cT T TT

Δ = + = ≥∫ ∫

(c) Thus 23 1 2 3 1 2, or T T T T T T≥ ≥

(d) The maximum amount of work can be obtained using a reversible heat engine, for

which 0SΔ = .

( ) ( ) ( )2

max 1 2 3min 1 2 1 2 1 22 2W c T T T c T T TT c T T= + − = + − = −

C

1p

2p

1V 2V

B

Aadiabatic






Solution: In the process of isothermal expansion, the external work done by the system is

1 1

1 1

2 2ln 2

V V

V V

dVW PdV RT RTV

= = =∫ ∫

Because the internal energy does not change in this process the work is supplied by the

heat absorbed from the external world. Thus the increase of entropy of the gas is

1 ln 2Q WS RT TΔ

Δ = = =

The change in entropy of the heat source 2 1S SΔ = −Δ thus the total change in entropy of

the universe is 1 2 0S S SΔ = Δ + Δ = .

If it is a free expansion, the internal energy of the system is constant. As its final state is

the same as for the isothermal process, the change in entropy of the system is also the

same. In this case, the state of the heat source does not change, neither does its entropy.

Therefore the change in entropy of the universe is ln 2S RΔ = .

Ans. 20: (a), (b) and (c)

Solution: We assume i fT T≠ (because the change of entropy must be zero when i fT T= ). The

change of entropy of the body is

1 lnf

i

T p fpT

i

C dT TS C

T TΔ = =∫

The change of entropy of the heat source is

( )

2p i f

f f

C T TQST T

−ΔΔ = =

Therefore the total entropy change is

1 2 1 ln fip

f i

TTS S S CT T

⎛ ⎞Δ = Δ + Δ = − +⎜ ⎟⎜ ⎟

⎝ ⎠

when 0x > and 1x ≠ , the function ( ) 1 ln 0f x x x= − − > .





The change in entropy of universe during the process is 1 2

1

ln V VnRV

⎛ ⎞+⎜ ⎟⎝ ⎠

Ans. 21: (a), (c) and (d)

Solution: This is a process of adiabatic free expansion of an ideal gas. The internal energy does

not change; thus the temperature does not change, that is, the final temperature is still T .

From first law of thermodynamics the work done is zero . hence the process is

irreversible and entropy is state function so one can choose path isothermal reversible

process . so change in entropy in isothermal is 1 2

1

ln V VnRV

⎛ ⎞+⎜ ⎟⎝ ⎠

which is equal to change

in entropy of gas as well as universe .


Ans. 22: 390

Solution: If 2T is Sink temperature and 1T Source temperature

Then, efficiency is given 1

21

TTT −

=η

Case 1: 1

2161

TT

−= 12 65 TT =⇒

Case 2: If 2T is reduced by ,65 C° then 6522 −=′ TT

1

21TT ′

−=′η

⇒ 1

2 651

31

TT −

−=11

2 65131

TTT

+−=⇒

1

65651

31

T+−= KT 0

1 390665 =×=⇒





Ans. 23: 207

Solution: If T is temperature of source, T is the temperature of the smk then efficiency

1

21TT

−=η 1

37.5 3001100 T

⇒ = −

100

5.371300

1

−=T 200

75200 −= 1

300 8 4805

T ×⇒ = =

273480−=⇒ t C0207=

Ans. 24: 33.33

Solution: 1

WQ

η =

( )( )1 . 200 100 1000 500 25000

2W = − − =

⎟⎠⎞

⎜⎝⎛ ××+= 100500

21000,501Q J000,75=

1

33.33%WQ

η = =

Ans. 25: 1.85−

Solution: If Q amount of heat is taken by or given to (a) system at temperature T then change

in entropy is given as

QdS

Tδ

=

By first law of thermodynamic

WdUQ δδ +=

According to question, since temperature is constant so no change in internal energy

0=⇒ dU

and work of J730 is done on the system JQ 730−=⇒ δ

From first law of thermodynamics, ( )7300 −+−=Qδ 730Q Jδ⇒ = −

and 20273+=T K293=

TQdS δ

= 293730−

= KJ /85.1−=

200

100

500 1000

A C

BTemperature





Ans. 26: 2.9

Solution: The entropy is defined as

TQdS δ

=

Here, cal8008010 =×== MLQδ and K2732730 =+=T

So, cal/K2.9cal/K273800

==dS

Ans. 27: 1.5

Solution: The energy stored in capacitor 2

21 CV=

So, change in entropy is given TQdS δ

=T

CVdS2

2

=⇒

Here, KTVoltVFC 30027327,300,10 6 =+=== −

So, change in entropy ( )300230010 26

××

=−

dS 3002

10910 46

×××

=−

KJ /105.1 4−×=

Ans. 28: 184 Solution: We assume the process is a reversible process of constant pressure. The change in

entropy of the water is 2 2 2

373

273

373ln273H O H O H O

dTS mC mCT

⎛ ⎞Δ = = ⎜ ⎟⎝ ⎠∫

we substitute 1m kg= , and 2

4.18 /H OC J g= into it, and find

2

1305 /H OS J KΔ =

The change in entropy of the heat source is

1001000 4.18 11121 /373ha

QS J K

TΔ = − = − × × = −

Therefore the change of entropy of the whole system is 2 184 /H O haS S S J KΔ = Δ + Δ =





Ans. 29: 33

Solution: ( )1 1 1 1 1 11 2 (3 )2

W T T S S T S= − − =

( )1 1 1 1 1 1 1 1 1 1 11 2 (3 ) (3 ) 22

Q T T S S T S S T S T S= − − + − = + = 1 13T S

1 1

1 1

.333T SW

Q T Sη = = = =33%





Chapter - 5

Maxwell relation and Thermodynamic Potential

5.1 Maxwell Relations

( , )F F x y= and if it is perfect differential then dF Mdx Ndy= +

Where y

FMx

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠and

x

FNy

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

then M and N will satisfy the condition

yx

M My x

⎛ ⎞∂ ∂⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

Maxwell relations are relationship between two derivatives of thermodynamic variables,

and energy due to the equivalence of potential second derivative under a change of

operation order dxdyFd

dydxFd 22

= where F is thermodynamic potential and x and y are two of

its natural independent variables.

Maxwell relations are extremely important for two reasons.

First they show us that derivative of thermodynamic parameters are not all independent.

This can serve as a consistency check in both experiments and in theoretical analysis.

Maxwell relations provide a method for expressing some derivative in other ways. This

enables as to connect difficult to measure quantities to those which are readily accessible

experimentally.

The measurement of entropy and chemical potential can not be directly measurable in lab

but with the help of Maxwell relation there thermodynamic property can be determine

theoretically.

For Maxwell relation.

Let us Legendre the independent variable as x , and y such that

U = U(x,y), S = S(x, y) V = V(x, y)

So dyyUdx

xUdU

xy⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

dyySdx

xSdS

xy⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=





dxyVdx

xVdV

xy⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

from first law of thermodynamic

PdVTdSdU −=

yyy x

VPxST

xU

⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

xxx y

VpyST

yU

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

Hence U, V,and S are perfect differential.

Then yxxy y

Uxx

Uy ⎥

⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

yxxy yV

xxV

y ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

yxxy yS

xxS

y ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

xyyxxy xS

yT

xS

yT

xU

y ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

xyyx xV

yP

xV

yP

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

− ---(1)

Similarly

yx x xy y

U T S STx y x y x y

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ yxxy yV

xP

yV

xP

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

− ---(2)

Equating equation (1) and (2)

yxxyyxxx x

SyT

yS

xT

xV

yP

yV

xP

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂ (A)

Maxwell first relation:- put ,x T y V= =

VT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂





Maxwell Second Relation:- put ,x T y P= =

PT T

VPS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

Maxwell Third Relation:- put ,x S y V= =

VS S

PVT

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

Maxwell Fourth Relation:- put ,x S y P= =

PS S

VPT

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Thermodynamic potential is a scalar function used to represent the thermodynamic state

of system. The concept of thermodynamic potentials was introduced by Pierre Duhem in

1886.

One main thermodynamic potential that has a physical interpretation is the internal

energy. It is energy of configuration of a given system of conservative forces. Expression

for all other thermodynamic energy potentials are drivable via Legendre transformation.

5.2 Different Types of Thermodynamic Potential and Maxwell Relation

Thermodynamic potentials are different form of energy which can be used in different

thermodynamic process .thermodynamic potentials are path independent variables so they

are perfect differential

If F is unique thermodynamic potential defined by variables x and y as ( , )F F x y= and

if it is perfect differential then dF Mdx Ndy= +

Where y

FMx

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠and

x

FNy

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

then M and N will satisfy the condition

yx

M My x

⎛ ⎞∂ ∂⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠





5.2.1 Internal Energy :- U and second from the first laws of thermodynamics

dU = TdS - PdV

from Legendre transformation

PVUT

SU

S

−=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ ,

V

from given relation one can derive Maxwell

relationS V

T PV S∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

5.2.2 Enthalpy (H) the enthalpy is defined as PVUH += VdPPdVdUdH ++=

from Laws of thermodynamics

PdVdUTdS +=

VdPTdSdH +=


PPHT

SH

SP

=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

The Enthalpy H is Extensive quantity, which can not be measured directly. Thus change

in enthalpy is more useful.

HΔ is positive in endothermic reaction and negative in exothermic reaction.

From above relation one can derive Maxwell relation S P

T VP S∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

HΔ of a system is equal to sum of non-mechanical work done on it and the heat supplied

to it.

5.2.3 Helmholtz Free Energy (F) the Helmholtz free energy is defined F U TS= −

SdTTdSdUdF −−=

From laws of thermodynamics PdVTdSdU −=

SdTTdSPdVTdSdF −−−=

dF PdV SdT= − −





From Legendre transformation

STFP

VF

VT

−=⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ ,

From above relation one can derive Maxwell relation V T

P ST V∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

The free Energy F, which is available energy for work in reversible isothermal process.

5.2.4 Gibbs Energy: ‘G’ is defined as G = H – TS.

G U PV TS= + −

dG dU PdV VdP TdS SdT= + − − −

TdS PdV PdV VdP TdS SdT− + + − −

dG VdP SdT= −


VPG

T

=⎟⎠⎞

⎜⎝⎛∂∂ and

P

G ST∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

From above relation one can derive Maxwell relation

P T

V ST P∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Gibbs free energy is popularly as free enthalpy.

The Gibbs free energy is Maximum amount of nonexpanding work that can be exacted

from a closed system.

The maximum will activated when the system is in reversible process.

Gibbs free energy is also treated as chemical potential.

In thermodynamics, chemical potential, as partial molar free energy, is a form of potential

energy that can be absorbed or relived during a chemical reaction.

The chemical potential of a species in the minute can be defined the slope of the energy

at system with respect to a change in the no of moles.

V

dGdN

μ =

where μ is chemical potential, G is Gibbs energy and N is no of molecules





Example: Prove that internal energy U is given by

(a) ( )( )

/1/

V

F TU

T⎛ ⎞∂

= ⎜ ⎟⎜ ⎟∂⎝ ⎠

(b) ( )( ) PT

TGH ⎥⎦

⎤⎢⎣

⎡∂∂

=/1/

Solution: (a) F = U –TS =U = F + TS

V

FST∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

VT

FTFU ⎟⎠⎞

⎜⎝⎛∂∂−

+= VT

FTF ⎟⎠⎞

⎜⎝⎛∂∂

−=( )

VTTFT ⎟

⎠⎞

⎜⎝⎛

∂∂

−=/2 ( )

VTTF

⎟⎠⎞

⎜⎝⎛∂∂

=/1/

Solution: (b) ( )( ) PT

TGH ⎥⎦

⎤⎢⎣

⎡∂∂

=/1/

G = H – TS

pT

GS ⎟⎠⎞

⎜⎝⎛∂∂

−=

PT

GTGH ⎟⎠⎞

⎜⎝⎛∂∂

−= ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

∂∂

−=PT

GT

T 2

( )( ) PT

TG⎥⎦

⎤⎢⎣

⎡∂∂

=/1/

5.3 Application of Maxwell Relation

5.3.1 First dST − equation

Let T, and V are independent variable S = S (T, V)

dVVSdT

TSdS

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

dVVSTdT

TSTTdS

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

= putvT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

.dV

TPTdTCdST

Vv ⎟

⎠⎞

⎜⎝⎛∂∂

+=⋅





5.3.2 Second dST − Equation

Let T and P are independent variable S = S (T, P).

dPPSTdT

TSTTdS

TP⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

From Maxwell relation

PT

VPS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

dPTVTdTCTdS

PP ⎟

⎠⎞

⎜⎝⎛∂∂

−=

5.3.3 Third T-dS Equation:

Let P,V are independent variable S = S (P,V)

dVVSdP

PSdS

VP⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

dVVSTdP

PSTTdS

VV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

dVVT

TSTdP

PT

TST

PPVV⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

=

dVVTCdP

PTC

PP

VV ⎟

⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

5.3.4 The First Energy Equation

Let T and V are independent variable U = U(T, V)

dVVUdT

TUdU

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

From first law of thermodynamics.

PdVTdSdU −=

PVST

VU

TT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂





Using Maxwell relation T V

S PT TV T∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

PTPT

VU

VT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

dVVUdT

TUdU

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

dVVUdTC

Tv ⎟

⎠⎞

⎜⎝⎛∂∂

+=

PTPT

VU

VT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

dVPTPTdTCdU

Vv ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛∂∂

+=

5.3.5 Second Energy Equation

PdVTdSdU −=

TTT P

VPPST

PU

⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Using Maxwell relation

PT T

VPS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

TPT P

VPTVT

PU

⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ This is popularly known as second energy

equation

Application of second energy equation

If U is function of independent variable of T and P.

( )PTUU ,=

dPPUdT

TUdU

TP⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=





dPPVP

TVTdT

TUdU

TPP⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

=

Example: From relation PdVTdSdU −=

Derive Maxwell relation VS S

PVT

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

Solution: PdVTdSdU −= TSUP

VU

VS

=⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ ,

Hence U is exact differential

SVvS S

UVV

US ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

SV V

TSP

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

−

Example: A real gas which obey van der Waal’s equation of state are kept in container which

has temperature T0 and volume V0. if volume of container changes to V such that

temperature of gas become T what is change in entropy?

Solution: Assume CV is specific heat of constant volume

For van der Waal’s gas


⎜⎝⎛ + 2

From first dST − equation

dVTPdTCTdS

VV ⎟

⎠⎞

⎜⎝⎛∂∂

+=

bV

RTP

V −=⎟

⎠⎞

⎜⎝⎛∂∂

( )dVbV

RTdTCdS

V

V

T

TV ∫∫ −

+=00

00 0

ln lnVT V bS C R ST V b⎛ ⎞ ⎛ ⎞−

= + +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠





where 0S is integration constant

Example: For Vander wall gases, prove that

2Va

VU

T

=⎟⎠⎞

⎜⎝⎛∂∂ where U is internal energy.

Solution: From first energy equation

PTPT

VU

VT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ -----(i)


⎜⎝⎛ + 2

2Va

bVRTP −−

=

( )bVR

TP

V −=⎟

⎠⎞

⎜⎝⎛∂∂ put the value of

VTP⎟⎠⎞

⎜⎝⎛∂∂ in equation (i)

2Va

VU

T

=⎟⎠⎞

⎜⎝⎛∂∂

Example: Prove that

(a) VT

V

TPT

VC

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

2

2

(b) PT

P

TVT

PC

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

2

2

Solution: (a) we know Vv

SC TT∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

TVT

V

TS

VT

VC

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ 2

VTVS

TT ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=2

using Maxwell relation

VT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ one can get

2

T V

STT V

⎡ ⎤∂ ∂⎛ ⎞⎢ ⎥⎜ ⎟∂ ∂⎝ ⎠⎣ ⎦

=VT

PT

T ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂ 2





VT

V

TPT

VC

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

2

2

(b)p

p TSTC ⎟⎠⎞

⎜⎝⎛∂∂

=

TPT

P

TS

PT

PC

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

PTPS

TT ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=

Use Maxwell relation

PT T

VPS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

PT

P

TVT

PC

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

2

2

Example: If pα is thermal expansivity at constant pressure and KT isothermal compressibility

then prove that

(i) pT

VPS α−=⎟⎠⎞

⎜⎝⎛∂∂

(ii) T

p

V KTP α

=⎟⎠⎞

⎜⎝⎛∂∂

(iii) T

VP KPTVCC

2α=−

Solution: From Maxwell relation

PT T

VPS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

PT

VV

⎟⎠⎞

⎜⎝⎛∂∂

=1α

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

PT TV

VV

PS 1 , V

PS

T

α−=⎟⎠⎞

⎜⎝⎛∂∂

(ii) V T

P ST V∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠





1−=⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

PTV VT

PV

TP

PT

V

VT

PVT

P

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟

⎠⎞

⎜⎝⎛∂∂ 1

T

P

PV

V

TV

V

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=

1

1

TV KT

P α−=⎟

⎠⎞

⎜⎝⎛∂∂

dVVSdT

TSdS

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

= dVVSdT

TC

dST

V ⎟⎠⎞

⎜⎝⎛∂∂

+=

PT

V

P TV

VS

TC

TS

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

+=⎟⎠⎞

⎜⎝⎛∂∂

TP

VP

VSV

TC

TC

⎟⎠⎞

⎜⎝⎛∂∂

+= α

Use Maxwell relation

V

PVP

TPV

TC

TC

⎟⎠⎞

⎜⎝⎛∂∂

+= α ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

T

PP

VP

KV

TC

TC α

α

T

pVP K

VTCC

2α=−

Example: Prove that

(a) P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(b) For the Vander Waal’s gas prove that

21P VaC C R

RTV⎛ ⎞− = +⎜ ⎟⎝ ⎠

Solution: (a) P VP V

Q QC CT T∂ ∂⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

VP T

STTST ⎟

⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

= ---- (A)

( )VTSS ,=





dVVSdT

TSdS

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

= PTVP T

VVS

TS

TS

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Put the value of PT

S⎟⎠⎞

⎜⎝⎛∂∂ in equation (A)

P VV T P V

S S V SC C T T TT V T T∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

P VT P

S VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞⇒ − = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

V P T V

P V S PTT T V T

⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∵

(b) For van der Waal’s gas

( )2

aP V b RTV

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

---- (B)

( )2

a RTPV V b

⎛ ⎞+ =⎜ ⎟ −⎝ ⎠ → differentiate w.r.t. to T

V

P RT V b∂⎛ ⎞ =⎜ ⎟∂ −⎝ ⎠

Differentiate (B) with respect to V

( )23

2

P P

a V RT V RV T T V bV b

∂ ∂⎛ ⎞ ⎛ ⎞− = − +⎜ ⎟ ⎜ ⎟∂ ∂ −⎝ ⎠ ⎝ ⎠−

( )23

2P

RV V b

RTTaV b

V

∂⎛ ⎞ −=⎜ ⎟∂⎝ ⎠

− −

Substituting the value PV T

VTP

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂ in equation

P VT P

S VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( )

( )2 32P V

R RV b V b

C C T RT aVV b

⎛ ⎞⎜ ⎟− −⎝ ⎠− =−

−





( )2

321

RV ba

V RT

=−⎛ ⎞− ⎜ ⎟

⎝ ⎠

2

321

P VRC Ca V

V RT

− =⎛ ⎞− ⎜ ⎟⎝ ⎠

121 aR

RTV

−⎛ ⎞= −⎜ ⎟⎝ ⎠

, 21P VaC C R

RTV⎛ ⎞− = +⎜ ⎟⎝ ⎠

Example: From P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Prove 2P VC C TE Vα− = where E is bulk modulus of elasticity and α is coefficient of

volume expansion.

Solution: Let P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( )VTPP ,=

dVVPdT

TPdP

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

For constant pressure dP = 0

dVVPdT

TP

TV⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

PTV T

VVP

TP

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

2

P VT P

P VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

TV

PVE ⎟⎠⎞

⎜⎝⎛∂∂

−= and PT

VV

⎟⎠⎞

⎜⎝⎛∂∂

=1α

22

P VT

V P V VC C TV V V T

∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ 22αV

VET ⎟⎠⎞

⎜⎝⎛−=

2

P VC C TVEα− = −





Example: If Helmholtz free energy for radiation is given by

VchTKF B33

445

458π

−=

(a) What is radiation pressure

(b) If S entropy of system prove that specific heat at constant volume is given by

VC = 3S

Solution: (a) PdVSdTdF −−=

TV

FP ⎟⎠⎞

⎜⎝⎛∂∂

−= 33

445

458

chTKBπ

=

(b) VT

FS ⎟⎠⎞

⎜⎝⎛∂∂

−=

333

45

4532 VT

chKBπ

=

V

V TSTC ⎟⎠⎞

⎜⎝⎛∂∂

= 233

45

4532

3 VTchK

T B⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅=

π 333

42

4532

3 VTchK B

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

π

SCV 3=

Example: The internal energy E of a system is given by VNbSE

3

= where b is constant and other

symbols have their used meaning.

(a) Find the temperature of system

(b) Find Pressure of system

Solution: From first law of thermodynamics

VNbSEUPdVdUTdS

3

==+= PdVTdSdU −=

(a) VS

UT ⎟⎠⎞

⎜⎝⎛∂∂

= VNbST

23=

(b) SV

UP ⎟⎠⎞

⎜⎝⎛∂∂

−= ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

NVbSP 2

3

NV

bSP 2

3

=

Example: Consider an Ideal gas where entropy is given by





⎥⎦⎤

⎢⎣⎡ ++=

nVR

nURnS ln2ln5

2σ

where n = number of moles, R = universal gas constant, U = internal energy

V = volume and σ = constant

(a) Calculate specific heat at constant pressure and volume

(b) Prove that internal energy is given by PVU25

=

Solution: (a) From first law of thermodynamics

PdVdUTdS −= , 1 PdS dU dVT T

= − TU

S

V

1=⎟

⎠⎞

⎜⎝⎛∂∂

TURn 15

2= nRTU

25

=

nR

TUCV 2

5=⎟

⎠⎞

⎜⎝⎛∂∂

= P VC C R⇒ = +72PC nR⇒ =

(b) V

nRTVUnRTU

25

25

=⇒=

nRTPV = P

nRTV =⇒

P

VU

25

= PVU25

=⇒





Example: Using the equation of state PV = nRT and the specific heat per mole 2

3RCv = for

monatomic ideal gas

(a) Find Entropy of given system.

(b) Find free energy of given system

3,2V V

nRT RdU nC dT p CV

= = =

PdVdUTdS +=

dU pdS dVT T

= + or VdT dVds nC nRT V

= +

03 ln ln2

S NR T N V S= + + where S0 is constant

(b) TSUF −=

03 3 ln ln

2 2nRT nRT T nRT V F⎛ ⎞= − − +⎜ ⎟

⎝ ⎠ where 00 STF ⋅= is again constant

Example: From electromagnetic theory Maxwell found that the pressure P from an isotropic

radiation equal to 31 the energy density i.e. 1

3UPV

= where V is volume of the cavity

using the first energy equation prove that

Energy densityu is proportional to 4T .

Solution: T T T

U S PT P T PV V T∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

uP31

= VUu =

13V V

P UT T∂ ∂⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

3 3T V

U T U UV V T V∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

uVU

T

=⎟⎠⎞

⎜⎝⎛∂∂ u

dTduTu

v 31

3−⎟

⎠⎞

⎜⎝⎛=

udTduT 4= =

TdT

udu 4=

4Tu ∝ = 4Tu α= where α is a constant.






Q1. Which of the following is not a Maxwell’s thermodynamic relation?

(a) VT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ (b)

VS SP

VT

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

(c) PV S

VPT

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ (d)

PT TS

PV

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Q2. Which one of the following thermo dynamical relations is used for certain adiabatic

changes, such as the sudden compression of a liquid or sudden -stretching of a rod?

(a) PT T

VPS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ (b)

TP VP

TS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

(c) TP V

PTS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ (d)

PT TV

PS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

Q3. Which one of the Maxwell’s thermodynamic relations given below leads to

Clausius-Clapeyron equation?

(a) VS S

PVT

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ (b)

PS SV

PT

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

(c) VT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ (d)

PT TV

PS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

Q4. Which of the following is correct if α is volume expansivity and other variables have

usual meaning in thermodynamics .

(a) 2P

T

C TVP

α∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (b) 2P

T

C TVP

α∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

(c) 2P

T

C TVP α

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (d) 2

P

T

C TVP α

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

Q5. Which of the following can be derived by ( , )S S T V= ?

(a) VV

PTdS C dT T dVT∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠

(b) PP

VTdS C dT T dVT∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

(c) VV

PTdS C dT T dVT∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

(d) PP

VTdS C dT T dVT∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠





Q6. Which of the following can be derived by ( , )S S T P= ?

(a) VV

PTdS C dT T dVT∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠

(b) PP

VTdS C dT T dVT∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

(c) VV

PTdS C dT T dVT∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

(d) PP

VTdS C dT T dVT∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠

Q7. For an isolated thermodynamically system, , , , ,P V T U S and F represent the pressure,

volume, temperature, internal energy, entropy, and Helmholtz free energy respectively.

Then the following relation is true

(a) STF

V

−=⎟⎠⎞

⎜⎝⎛∂∂ (b) S

TF

P

−=⎟⎠⎞

⎜⎝⎛∂∂

(c) TSU

V

=⎟⎠⎞

⎜⎝⎛∂∂ (d) P

VU

V

−=⎟⎠⎞

⎜⎝⎛∂∂

Q8. Which of the following thermodynamic relation will give the Maxwell relation

?T P

S VP T∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(a) dU TdS PdV= − (b) dH TdS VdP= +

(c) dF SdT PdV= − − (d) dG SdT VdP= − + )

Q9. Which of the following is not an exact differential?

(a) dQ where Q heat absorbed (b) dU where U is internal energy

(c) dS where S is entropy (d) dFwhere S is entropy

Q10. Which among the following sets of Maxwell relations is correct? (U-internal energy, H-

enthalpy, A-Helmholtz free energy and G-Gibbs free energy)?

(a) VS S

UPVUT ⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

= and (b) PS S

HTPHV ⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

= and

(c) ST P

GVVGP ⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

−= and (d) VT P

ASSAP ⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

−= and





Q11. When a system is held at constant temperature and pressure in a state of equilibrium, then

it attains a minimum value of:

(a) internal energy (b) enthalpy

(c) Helmholtz energy (d) Gibb’s free energy

Q12. Given that

=H the enthalpy of a system

=T absolute temperature and

=S entropy

TSHG −= is the Gibbs function for the system

In the case of a reversible, isotherm and isobaric process:

(a) =G constant

(b) 0>G and changes withT

(c) 0<G and changes with S

(d) G changes with both T and S

Q13. The Gibb’s function G in thermodynamics is defined is

TSHG −=

where H is the enthalpy, T is the temperature and S is the entropy. In an isothermal,

isobaric, reversible process, G :

(a) remains constant, but not zero (b) varies linearly

(c) varies non-linearly (d) is zero

Q14. The internal energy E of a system is given byVNbSE

3

= , where b is a constant and other

symbols have their usual meaning. The temperature of this system is equal to

(a) VNbS 2

(b) VNbS 23 (c)

NVbS

2

3

(d) 2

⎟⎠⎞

⎜⎝⎛

NS





Q15. The free energy of gas of N particles in a volume V and at a temperature T is

( )[ ]NTkVaTNkF BB /ln 2/50= , where 0a is a constant and Bk denotes the Boltzmann

constant. The internal energy of the gas is

(a) TNk B23 (b) TNkB2

5

(c) ( )[ ] TNkNTkVaTNk BBB 23/ln 2/5

0 − (d) ( )[ ]2/50 /ln TkVaTNk BB

Q16. The entropy S of a thermodynamic system as a function of energy E is given by the

following graph

The temperatures of the phases BA, and C , denoted by BA TT , and CT , respectively,

satisfy the following inequalities:

(a) ABC TTT >> (b) BCA TTT >> (c) ACB TTT >> (d) CAB TTT >>

Q17. The entropy of an ideal paramagnet in a magnetic field is given approximately by 2

0S S cU= − where U is energy of the spin system and c is constant then which one is

correct plot between internal energy and temperature T where T−∞ < < ∞

(a) (b)

(c) (d)

U

T

U

T

U

T

U

T

S

AB

C

E→





Q18. A certain system is found to have Gibbs free energy given by

5/2

aPG(p,T) RT ln(RT)

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Where a and R are constants then specific heat at constant pressure ( pc ) is given by

(a) 3 R2

(b) 5 R2

(c) 7 R2

(d) 9 R2

Q19. Helmholtz free energy is given by 4CTF −= where C is constant and T is temperature in

Kelvin then which one is correct relation between specific heat at constant volume Cv and

entropy S is given by

(a) 2VC S= (b) 4VC S= (c) 3VC S= (d) SCv23

=

Q20. If VC is the specific heat of the ideal gas then which of the following is correct of

Vander wall gases for same degree of freedom .

(a) VdU C dT= (b) 2

adU dVV

= −

(c) 2VadU C dT dV

V= − (d) 2V

adU C dT dVV

= +

Q21. For a Van der Waals gas the equation of the adiabatic curve in the variables ;,VT

(a) ( ) tconsbVT PCR tan/ =− (b) ( ) tconsbVT VCR tan/ =−

(c) ( ) tconsbVT PCR tan/ =− − (d) ( ) tconsbVT VCR tan/ =− −






Q22. If H is enthalpy and G is Gibbs free energy of the thermodynamic system then which of

the following is correct

(a) H G TS= − (b) H G TS= +

(c) 2

P

GH TT T

⎛ ⎞∂ ⎛ ⎞= − ⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠ (d) 2

P

GH TT T

⎛ ⎞∂ ⎛ ⎞= ⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

Q23. Which of the following statements are correct .

(a) The first energy equation is be given by PTPT

VU

VT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

(b) The value of 0=⎟⎠⎞

⎜⎝⎛∂∂

TVU for Ideal gas

(c) The value of 2Va

VU

T

=⎟⎠⎞

⎜⎝⎛∂∂ for Vander Waal’s gases .

(d) The second energy equation is given by PTVT

PU

PT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Q24. Which of following is correct for heat capacity at constant pressure PC and volume VC

(a) p VT P

U VC C pV T

⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− = +⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

(b) for van der waal’s gas ( )21 2 1 / /

p VRC C

a b V VRT− =

− −

(c) van der Waal’s gas behave like a ideal as high temperature

(d) for Ideal gas p Vc c R− = Q25. Which of the following is correct?

(a) P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(b) P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(c) 2

P VT P

P VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(d) 2

P VT P

P VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠





Q26. If α is volume expansivity of substance and Tβ is isothermal compressibility Sβ is

adiabatic compressibility and it is given that 2

P VT P

P VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

then which

one of the following is correct

(a) 2

P VT

T VC C αβ

− = −

(b) 2

P VT

T VC C αβ

− =

(c) 2

s TP

T VCαβ β− =

(d)

2

s TP

T VCαβ β− = −

Q27. Which of the following is correct if all variable have usual meaning in thermodynamics

(a) T V

S PV T∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(b) 2

2V

T V

C PTV T

⎛ ⎞∂ ∂⎛ ⎞ = ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(c) T V

S VP T∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(d) 2

2P

T V

C VTP T

⎛ ⎞∂ ∂⎛ ⎞ = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Q28. One mole of oxygen is expanded from a volume 1V to 2V at a constant temperature T . If

the The gas is assumed to be a Van der Waals gas

(a) the increment of the internal energy of the gas is zero

(b) the increment of the internal energy of the gas ⎟⎟⎠

⎞⎜⎜⎝

⎛−

21

11VV

a

(c) heat exchange during the process is ⎥⎦

⎤⎢⎣

⎡−+

−−

121

2 11lnVV

abVbV

RT

(d) heat exchange during the process is bVbV

RT−−

1

2ln

Q29. The free energy for a photon gas is given by 4

3VTaF ⎟⎠⎞

⎜⎝⎛−= , where a is a constant. The

entropy S and the pressure P of the photon gas are

(a) 3

34 aVTS = (b) 4

3, TaP =

(c) 3

34 aVTS −= (d) 4

34 TaP =





Solutions MCQ (Multiple Choice Questions) Solution

Ans. 1: (d)

Ans. 2: (a)

Solution: The sudden stretching of wire of compression of liquid is given by

PT TV

PS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

Ans. 3: (c)

Solution: The rate of change of temperature pressure is given by Clausius-Clapeyron equation

which is given as ( ) LVVT

LdTdP ,

12 −= is latern heat

This can be derived by Maxwell’s first thermodynamical relation given as

VT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Ans. 4: (a)

Solution: from Maxwell relation T V

S VP T∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

T PP P

S VT P T T

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ P PT P

T S VTP T T T

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

2

2P

T V

C VTP T

⎛ ⎞∂ ∂⎛ ⎞⇒ = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ and 1

P

VV T

α ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠ so 2P

T

C TVP

α∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ Ans. 5: (a)

Solution: V T

S SdS dT dVT V∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

from Maxwell relation T V

S PV T∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

and

VV

T S CT∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

VV

PTdS C dT T dVT∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠





Ans. 6: (b)

Solution: p T

S SdS dT dPT P∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

from Maxwell relation T P

S VP T∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

and

PV

T S CT∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

PP

VTdS C dT T dVT∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

Ans. 7: (a)

Solution: dF SdT PdV= − −

Ans. 8: (d)

Solution: dG SdT VdP= − +

P

G ST∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

and T

G VP∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

P TT p

G GP T T P

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

T P

S VP T∂ ∂⎛ ⎞ ⎛ ⎞⇒ = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

And. 9: (a)

Solution: Heat exchange is path dependent, so it is not perfect differential.

Ans. 10: (b)

Solution: dH TdS VdP= + PS S

HTPHV ⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

= and

Ans. 11: (d)

Solution: The change in Gibb’s free energy is given as

SdTVdPdG −=

at constant P, 0=dP at constant T, 0=dT ⇒G constant

Ans. 12: (a)

Solution: Enthalpy UPVH +=

From laws of thermodynamics is given as

PdVdUTdS +=

so, TSPVUG −+=





( ) ( )dG dU d PV d TS⇒ = + −

TdSSdTVdPPdVdU −−++= by equation (i)

dG VdP SdT= −

dTSVdPdG −= (ii)

for isobaric 0=dP , for isothermal 0=dT

so, equation (ii) becomes 0=dG

constantG⇒ =

Ans. 12: (d)

Ans. 13: (a)

Solution: Gibb’s function is given as

TSHG −= And The enthalpy PVUH +=

TSPVUG −+= ⇒ ( ) ( )TSdPVddUdG −+=

SdTTdSVdPPdVdUdG −−++=

SdTVdPdG −=

According to question the process is isobaric i.e., 0=SdP isothermal i.e., 0=dT . So

Equation (v) becomes as 0=dG G c⇒ =

Ans. 14: (b)

Solution: PdVdETdS += PdVTdSdE −=⇒ TSE

V

=⎟⎠⎞

⎜⎝⎛∂∂

⇒VNbST

23=⇒

Ans. 15: (b)

Solution: ( )[ ]NTkVaTNkF BB /ln 2/50−= , TSUF −= , TSFU +=

dF SdT PdV= − − STF

V

−=⎟⎠⎞

⎜⎝⎛∂∂

⇒ or VT

FS ⎟⎠⎞

⎜⎝⎛∂∂

−=VT

FTFU ⎟⎠⎞

⎜⎝⎛∂∂

−=⇒

( )2/5ln TCTNkF B−= where N

VkaC B

2/50=

( ) 2/32/5

2/5

25ln T

CTCTNkCTNk

TF

BBV

−−=⎟⎠⎞

⎜⎝⎛∂∂ ( ) TNkCTTNk

TFT BB

V 25ln 2/5 −−=⎟

⎠⎞

⎜⎝⎛∂∂

⇒





TNkFTFT B

V 25

+=⎟⎠⎞

⎜⎝⎛∂∂

VTFTFU ⎟⎠⎞

⎜⎝⎛∂∂

−=⇒ TNk B25

= .

Ans. 16: (c)

Solution: Now temperature of phase TCTBTA ,,

Now TdE

dS 1=⎟

⎠⎞

⎜⎝⎛

Now dEdS will be stops then it will be zero for B - phase

So ∞=BT

And in C and A phases external energy of C phase is more so AC TT >

Now ACB TTT >>

Ans. 17: (c)

Solution: V

U 1TS 2cU

∂⎛ ⎞= = −⎜ ⎟∂⎝ ⎠ i.e 1U

2cT= −

Ans. 18: (b)

Solution: 5/2p

G 5 apS R R lnT 2 (RT)

⎡ ⎤∂⎛ ⎞= − = −⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎣ ⎦

pS 5C T RT 2∂⎛ ⎞= =⎜ ⎟∂⎝ ⎠

Ans. 19: (c)

Solution: 34CTTFS

V

=⎟⎠⎞

⎜⎝⎛∂∂

−=

VTdS = C dT + PdV ⇒ V

V

C PdS dT dVT C

= +

3VV V

V V

CS SC T C ST T T∂ ∂⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Ans. 20: (d)

Solution: ( , )U U T V= V T

U UdU dT dVT V

∂ ∂⎛ ⎞ ⎛ ⎞⇒ = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ V

T

UdU C dT dVV∂⎛ ⎞⇒ = + ⎜ ⎟∂⎝ ⎠

S

A

B

C

E→





2T

U aV V∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

so 2VadU C dT dV

V= +

Ans. 21: (b)

Solution: ( ) RTbVVaP =−⎟⎠⎞

⎜⎝⎛ + 2 and 2V

adU C dT dVV

= +

For adiabatic process: pdVdUdQ +== 0

dV

Va

bVRTdV

VadTCV ⎟

⎠⎞

⎜⎝⎛ −

−=−− 22

⎥⎦⎤

⎢⎣⎡

−=−

bVRTdVdTCV

∫∫ −=−

bVdV

RTdTCV

( )bVTkR

CV −=− lnln

( ) RCVTkbV /−=−

RCRC VV kTbV // −− ×=−

( ) RCRC VV kTbV // −=−

( ) tconsbVT VCR tan/ =−



Solution: dG SdT VdP= − + P

G ST∂⎛ ⎞⇒ = −⎜ ⎟∂⎝ ⎠

G H TS= − H G TS= + so P

GH G TT∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

2

P

GH TT T

⎛ ⎞∂ ⎛ ⎞⇒ = − ⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

Ans. 23: (a), (b) and (c)

Solution: The first law of thermodynamics is given as

dU PdV TdS+ =





PVST

VU

TT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ From Maxwell’s second relation

VT TP

VS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

PTPT

VU

VT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

For real gas PV nRT= so 0=⎟⎠⎞

⎜⎝⎛∂∂

TVU

For Vander wall’s ( )2

aP V b nRTV

⎛ ⎞− − =⎜ ⎟⎝ ⎠

so 2Va

VU

T

=⎟⎠⎞

⎜⎝⎛∂∂

T T T

U S VT PP P P

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

VT TP

VS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Ans. 24: (a), (b), (c) and (d)

Solution: (a) From H U PV= + , we obtain,

P P P

H U VPT T T

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Let ( ), ,U U T V T P⎡ ⎤= ⎣ ⎦ .

Let ( ), ,U U T V T P⎡ ⎤= ⎣ ⎦ . The above expression becomes

V T

U UdU dT dVT V

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ P V T p

U U U VT T V T

∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

P V T P

H U U VPT T V T

⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

Hence P VT P

U Vc c PV T

⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− = +⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

(b) to find p vc c− for a Van der Waals gas ( )2

aP V b RTV

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

For the Van der Waals gas, we have

( )

3

2P

V RT a V bRT

V b V

∂⎛ ⎞ =⎜ ⎟∂ ⎡ ⎤−⎝ ⎠−⎢ ⎥−⎣ ⎦





Hence, ( )21 2 1 / /

p vRc c

a b V VRT− =

− −

(c) and d) When & , P VV T c c R→∞ − → , which is just the result for an ideal gas.


Solution: ( , )S S T V=

V T

S SdS dT dVT V∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ P V T P

S S S VT TT T V T∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( , )P P T V=

V T

P PdP dT dVT V∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

for constant pressure 0dP =

PutV T P

P P VT V T∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

in P VV P

P VC C TT T∂ ∂⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

one will

2

P VT P

P VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠


Solution: 2

P VT P

P VC C TV T∂ ∂⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

it is known 1

P

VV T

α ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠ and 1

TT

VV p

β⎛ ⎞∂

= − ⎜ ⎟∂⎝ ⎠

So 2

P VT

T VC C αβ

− = ,we know that P T

V S

CC

β γβ

= =

2

1V T

T VCαγβ

− = 2

Ss T

T V

T VC

β αβ ββ

⇒ − = 2

s TP

T VCαβ β⇒ − = −

Ans. 27: (a), (b) and (d)

Solution: from Maxwell relation T V

S PV T∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠





T VV V

S PT V T T

⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ V VT V

T S PTV T T T

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 2

2V

T V

C PTV T

⎛ ⎞∂ ∂⎛ ⎞⇒ = ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Similarly another Maxwell relation is given by T V

S VP T∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

T PP P

S VT P T T

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ P PT P

T S VTP T T T

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 2

2P

T V

C VTP T

⎛ ⎞∂ ∂⎛ ⎞⇒ = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠


Solution: (a) for Vander waal’s gas 2T

U aV V∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=Δ

2121

11VV

aVa

VaU

(b) We know


⎜⎝⎛ + 2

2Va

bVRTP −−

= and work done 2

1

V

V

W PdV= ∫

⎥⎦

⎤⎢⎣

⎡−+

−−

=−−

==Δ ∫∫121

22

11ln2

1

2

1VV

abVbV

RTdVVadV

bVRTdWW

V

V

V

V

Now ⎥⎦

⎤⎢⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡−+

−−

=Δ+Δ=Δ21121

2 1111lnVV

aVV

abVbV

RTUWQ bVbVRTQ

−−

=Δ1

2ln

Ans. 29: (a) and (b)

Solution: dF SdT PdV= − −

V

F ST∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

, T

F PV∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠





Chapter - 6

Phase Transition and Low Temperature Physics

6.1 Third Law of Thermodynamics and Attainable of Low Temperature

The third law of thermodynamics is some time stated as follows:

It is impossible for any process, no matter how idealized, to reduce the entropy of a

system to its zero point value in a finite number of

operations.

Properties of material at low temperature

At 0→T

CP = CV = 0

At T → 0

S → 0

At T → 0

Thermal expansion coefficient 01=⎟

⎠⎞

⎜⎝⎛∂∂

=PT

VV

α

0=α 6.2 Production of Low Temperature: The Joule – Kelvin Expansion:

The Joule – Kelvin Expansion is essentially a continuous steady – state flow process in

which a compressed gas is made to expand adiabatically irreversibly through a porous

plug and do work.

1p

2p

1pwoolplug

porous

Constant temperatur

Fig: A schematic diagram of the porous-plug experiment for Joule-Kelvin

T

VC

PC

CapacityHeat





Let us assume that we start off with a gas of internal energy 1U and volume 1V . After

passing through the porous plug let find internal energy and volume of the gas by 2U and

2V . No heat enters the system.

So this work has to performed at the expense of internal energy

222111 VPUVPU +=+

21 HH =

Joule – Kelvin expansion is isenthalpic process

H = H (T, P)

dP

PHdT

THdH

TP⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

= P

P THC ⎟⎠⎞

⎜⎝⎛∂∂

=

dPPHdTCdH

TP ⎟

⎠⎞

⎜⎝⎛∂∂

+= dH TdS VdP= +

VPST

PH

TT

+⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

dPTVTVdTCdH

PP ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛∂∂

−+=

Hence H does not change 0dH =

The μ is defined HP

H⎟⎠⎞

⎜⎝⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛∂∂

==⎟⎠⎞

⎜⎝⎛∂∂ V

TVT

CPT

PPH

1μ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

∂∂

=PP T

VTC

T 2

where μ is known as the Joule – Kelvin Coefficient.

The equation defines a curve in the (T, P) plane and is known as the inversion curve

when μ = 0.





A series of isoenthalphs, i.e. curve with ( ),H T P = constant.

At jTT = 0=μ known as inversion temperature.

iT T< and Positiveμ = . There is heating effect of gases i.e. temperature and pressure

move in same direction.

iT > T Negativeμ = . There is cooling effect i.e. temperature and pressure move in

opposite direction.

Example: (a) For van der Waals gas. Prove that inversion temperature Rb

aTi2

= where a and b

are parameter used in van der Waals gas.

(b) Why Hydrogen and helium shows heating effect as pressure increased at constant

enthalphy.

Solution: (a) ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= VTVT

CpT

PpH

1μ

For van der Waals gas


⎜⎝⎛ + 2 ( )bV

Va

VaP

RTV

P

−−⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛∂∂

3

2

2

T

T

1T

2T

2P 1P P

1HH =

2HH =3HH =4HH =

5HH =

B

A

Inversion curve

Figure: Curves of constant enthalpy. The bold curve is the inversion curve. Inside it, the gas is cooled on expansion. The temperature change 12 TTT −=Δ produced in a





( )( )

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

−=⎟

⎠⎞

⎜⎝⎛∂∂

3

221RTV

bVaT

bVTV

P

since Vb <<

1211

−

⎟⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛∂∂

RTVa

Vb

TV

TV

P⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ −=

RTVa

Vb

TV 211

.21 ⎟⎠⎞

⎜⎝⎛ +−=

RTVa

Vb

TV b

RTaV

TVT

P

−=−⎟⎠⎞

⎜⎝⎛∂∂ 2

⎟⎠⎞

⎜⎝⎛ −= b

RTa

Cp

21μ

For inversion temperature

0=μ ⇒ bRT

a

i

=2

RbaTi

2=

(b) Since inversion temperature of Hydrogen and Helium is very small. For cooling effect

initial temperature must be smaller than inversion temperature, but for Helium and

Hydrogen inversion temperature is very small

so it is not possible to achieve initial condition lower than inversion temperature so

Helium and Hydrogen give heated effect.

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ V

TVT

CPT

PPH

1 1

P

VdT T dP VdPC T

∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

1−=⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

TVP VP

PT

TV

TV

TV

P PV

TP

VP

PTT

V⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟

⎠⎞

⎜⎝⎛∂∂ 1

dVTPdP

TV

VP⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂





6.3 Phase Transition

A phase of a thermodynamic system and the state of matter has uniform physical

properties.

A phase transition is the transformation of thermodynamic system from one phase or

state of matter to another.

During a phase transition of a given medium certain properties of the medium change,

often discontinuously, as a result of some external condition, such as temperature,

pressure and others.

Gibbs phase rule: It is proposed by Josiah Willard Gibbs, which is given by

2+−= PCF

Where C is number of components, P is the number of phase in thermodynamic

equilibrium with each other and F is number of degree of freedom.

Phase: A phase is form of matter that is homogeneous in chemical composition and

physical state.

Typical phases are solid, liquid and gas. Two immiscible (or liquid mixture with different

compositions) separated by distinct boundary are countered as two different phase.

Components: The number of components is the number of chemically independent

constituents of the system, i.e. minimum number of independent species necessary to

define the composition of all phase of the system.

The number of degree of freedom ( )F in the context is the number of intensive variable

which are independent to each other.

6.3.1 First Order Phase Transition

Let us consider one component system in which system having only one kind of

constituent particles . for first order phase transition P T− diagram are shown in fig 1 .

Depending on the system, at some values for temperature and pressure, the three phases

of the system may be found in equilibrium. In the P-T diagram, the line OA represents

equilibrium between solid and liquid phases, the line OB represents equilibrium between

solid and gas phases, and the line OC represents equilibrium between liquid and gas

phases. The point O where all the three phases are in equilibrium, is known as a triple





point. The line OC terminates at the point C, called the critical point. Beyond this point,

the gas phase cannot be converted into the liquid phase. In figure 1, the point C is at the

apex of the P-V curve at the critical temperature Tc. For the temperature T > Tc, the gas

phase of the matter cannot be converted into the liquid phase, but for T < Tc, the gas

phase can in general be converted into the liquid phase.

6.3.2 Equilibrium Between Two Phases

Let us consider an isolated system having a matter which is existing in two phases,

denoted by 1 and 2, simultaneously in equilibrium with each other (Figure 2). Suppose V1

and V2 are volumes, N1 and N2 the number of particles, E1 and E2 the internal energies,

and S1 and S2 the entropies of the two phases, respectively. For each phase, entropy is a

function of its volume, number of particles (mass) and internal energy.

From these relations, it follows that

21 TT = thermal equilibrium

21 PP = mechanical equilibrium

21 μμ = chemical equilibrium

Hence, when two different phases of the matter are in

equilibrium, their temperatures, pressures and chemical

potentials must be equal. If the chemical potentials are expressed

as functions of pressure and temperature, we have

( ) ( )TPTP ,, 21 μμ =

A1E

B2E

Figure 2: Equilibrium of two phases of an isolated one component system

OP

B T

A

CCritical

Liqui

Melting Solid

Triple Saturation

Gas

Figure 1: P-T phase of one component





where P (= P1 = P2) and T (= T1 = T2) are the common pressure and temperature,

respectively, of the two phases in equilibrium. Thus, from above equation , we have

( ) ( ) 2211 ,, NTPGNTPG =

2

2

1

1

NG

NG

=

where G1(P, T) and G2(P, T) are the Gibbs free energies, and N1 and N2 the number of

particles in the two phases, respectively. Since during the phase transition, the number of

particles is not changing (i.e., N1 = N2), we have

( ) ( )TPGTPG ,, 21 =

Hence, during the phase transition, the Gibbs free energy does not change. Gibbs energies

G1 and G2 of the two phases 1 and 2, respectively, can be exhibited as shown in figure 4.

6.3.3 Clapeyron-Clausius equation

When the two phases, denoted by 1 and 2, of the given matter are in equilibrium, we have

( ) ( )TPGTPG ,, 21 =

where G1 and G2 are Gibbs free energies of the two phases, respectively, and P (= P1 =

P2) and T (= T1 = T2) are the common pressure and temperature, respectively, of the two

phases. In the P-T diagram, along the phase-transition line, let us consider a point, where

the pressure is P + dP and the temperature is T + dT so that we have

( ) ( )dTTdPPGdTTdPPG ++=++ ,, 21

P

21 GG < 21 GG >

21

T

21 GG =

Figure 3: Phase equilibrium curve (G1 – G2) separating two phases 1 and 2.





Using Taylor series expansion and neglecting the higher order terms, we have

( ) ( ) dTTGdP

PGTPGdT

TGdP

PGTPG

PTPT

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+=⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+ 222

111 ,,

Using above two equation we get

02121 =+−− dTSdTSdPVdPV

VS

VVSS

dTdP

ΔΔ

=−−

=12

12

TL

THS =

Δ=Δ

where ( )12 HHH −=Δ is the change in heat (enthalpy) which is the molar latent heat L.

thus, from equation an we have

VT

LdTdP

Δ=

for V2 > V1, we have

TT P

GP

G⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂ 12 (A)

further, for S2 > S1, we have

PP T

GTG

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂ 12 (B)

6.3.4 Liquid-Vapour Phase Transition

Let us consider a phase transition from a liquid state to a vapor one. If Vi and Vg,

respectively, denote the volume in the liquid and gas phases, and Lv is the heat of

vaporization (latent heat for the transition from liquid to vapour), the Clapeyro-Clausius

equation is

( )ig

v

VVTL

dTdP

−=

Since in the phase transition, Vg is always greater than Vi and the heat of vaporization Lv

is positive and we have

0>dTdP





It shows that the boiling point of a liquid increases with the increase in pressure.

Now, if the vapour pressure is low, i.e., Vg >> Vi, in comparison to Vg, and we have

g

v

TVL

dTdP

=

Using the ideal gas equation, PVg = RT, we have

2RTPL

dTdP v= 2T

dTRL

PdP v=

( )[ ] CTR

LTP v +−=

1ln

where C is a constant of integration. At the critical point, we have P = Pc, T = Tc and

equation is

( )[ ] CTR

LTP v

cc +−=1ln

( ) ( ) ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

TTRL

TPTPc

vcc

11exp

Here, we have assumed that the heat of vapourisation vL is independent of the

temperature. However, it depends on the temperature. Suppose it varies as

,bTaLv −= then for an ideal gas at low pressure, we have

( )2T

dTRbTa

PdP −

=

( )[ ] ( ) CTRb

TRaTP +−−= ln1ln

where C is a constant of integration. At the critical point, we have cc TTPP == , and

equation is

( )[ ] ( ) CTRb

TRaTP c

ccc +−−= ln1ln

On subtracting equation from and rearranging, we have

( )( ) ⎟

⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=⎥

⎦

⎤⎢⎣

⎡TT

Rb

TTRa

TPTP c

ccc

ln11ln





6.3.5 Properties of First Order Phase Transition

(1) Gibbs free energy is continuous

(2) First order derivative with respect to temperature and pressure have finite

discontinuity i.e. entropy ( )S and pressures ( )P have finite discontinuity.

(3) Second and more higher order differential is infinite

Example: For a two phase system in equilibrium, p is a function of T only, so that

V S

P PT T∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Show that

2

V

S

C dPTVdTβ

⎛ ⎞= ⎜ ⎟⎝ ⎠

Solution: Let us take T and V as independent variables and write

( )VTSS ,=

so that

dVVSdT

TSdS

TV⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

For an adiabatic process, it yields

STV T

VVS

TS

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

1 2

T

g1 2

T

1

2

T

Fig. 4: A schematic representation of first order phase transitions





Using first Maxwell relation, we obtain

V V S

S P VT T T∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Since VV

PC TT∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

we can write

VV S

P VC TT T∂ ∂⎛ ⎞ ⎛ ⎞= − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

V S S

P V PTT P T∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

SV S

P PTVT T

β ∂ ∂⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

2

SdPTVdT

β ⎛ ⎞= ⎜ ⎟⎝ ⎠

where Sβ is adiabatic compressibility.

Example: Calculate under what Pressure water would boil at 120° C. One gram of steam

occupies a volume of 1677 cm3. Latent heat of steam = 540 cal/g, J = 4.2 × 107 erg/cal.

atmospheric pressure = 1.0 × 106 dyne/cm3

Solution: ( )2 1

dP LdT T V V

=−

( )2 1

L dTdPV V×

=−

16772 =V cm3/g 12 =V cm3/g

L = 4.2 × 107 × 540 erg/g o20=dT k

725.0=dP P2 – P1 = .725

725.1725.01725.0 12 =+=+= PP

Example: Liquid helium – 4 has normal boiling point of 4.2 k. However at pressure at 1 mm of

mercury it boils at 1.2 k. Estimate the average latent heat of vaporization of helium in

this temperature range.

Solution: ( ) gg e

dP L LdT TVT V V

= =−

gPV RT= gRTVP

=





PRTT

LdTdP

= 2RTLP

dTdP

=

∫ ∫=P

P

T

T TdT

RL

PdP

0 0

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

TTRL

PP 11ln

00

TT

PPR

L11

ln

0

0

−=

P0 = 746 mm T0 = 4.2 k

P1 = 1 mm T = 1.2

L = 93 J/mol.

Example: Liquid helium boils at temperature T0 when its vapour pressure is equal to P0 we now

pump on the vapour and reduce the pressure to much smaller value P. Assume that the

Latent heat L is approximately independent at temperature and helium vapour density is

much smaller than that of liquid, calculate the approximate temperature Tm of the liquid

in equilibrium with its vapour at pressure P.

Express your answer in terms of L, T0, P0, Pm and any other required constants.

Solution: VT

LdTdP

Δ= gas liq gasV V V VΔ = − ≈

2RTLP

dTdP

= ∫ ∫=m mP

P

T

T TdT

RL

PdP

0 0

2

m

m

PP

LRT

TT

00

0

ln1+=





Example: In the phase transition from a liquid state to a vapour state. The heat of vapourisation

vL varies with temperature T as 2/1bTaLv −= . Considering the gas as an ideal one at

low pressure, show that the pressure ( )TP at temperature T in terms of the critical

pressure ( )cc TP at critical temperature cT is given by

( )( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=⎥

⎦

⎤⎢⎣

⎡2/12/1

11211lncccc TTR

bTTR

aTPTP

Solution: Clapeyron-Clausius equation for the phase transition from liquid to vapour is

( )ig

v

VVTL

dTdP

−=

where vL is the heat of vapourisation and iV and gV , respectively, denote the volume is

the liquid and gas phase. For low pressure, ig VV >>

g

v

TVL

dTdP

=

Using ,2/1bTaLv −= we have

gTV

bTadTdP 2/1−

=

For an ideal gas equation, ,RTPVg = and thus,

( )2

2/1

RTPbTa

dTdP −

=

( )2/322

2/1

TdT

Rb

TdT

RadT

RTPbTa

dTdP

−=−

=

( )[ ] CTR

bTR

aTP ++−= 2/1

11ln

On subtracting equation above and rearranging, we have

( )( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=⎥

⎦

⎤⎢⎣

⎡2/12/1

11211lncccc TTR

bTTR

aTPTP





Derive Glausius-Clapeyron equation from Maxwell relationVT T

PVS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂ .

VT TP

VS

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

VT TPT

VST ⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

VT TPT

VQ

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

LdmQ =δ

liqvapV VVL

TPT

−=⎟

⎠⎞

⎜⎝⎛∂∂ ( )liqvapV VVT

LTP

−=⎟

⎠⎞

⎜⎝⎛∂∂

(b) Draw Phase diagram for water and explain why water expand after freezing.

Phase diagram for water

The slope of solid liquid phase is negative.

So from Clausius:-

Clapeyron equation

( )solidliqsat VVTL

TP

−=⎟

⎠⎞

⎜⎝⎛∂∂

veTP

−=⎟⎠⎞

⎜⎝⎛∂∂ 0<− solidliq VV solidliq VV <

so water expand on freezing

6.3.6 Second Order Phase Transition:

In some cases the state of matter does not change but the arrangement of its constituent

particle changes. This kind of phase transition is known as second order phase transition.

In the case of second order phase transition, no heat is evolved or absorbed. In second

order phase transition

1. Gibbs free energy is continuous

g

1 2

T

g

1 2

p

P

T

solid liquid point critical

gas





2. First order differential of Gibbs energy with respect to temperature ie entropy are

changes smoothly

3. Second order differential of Gibbs energy with respect to temperature ie specific heat

and second order differential of Gibbs energy with respect to pressure ie isothermal and

isobaric expansivity have finite discontinuity at critical temperature.

T p

T p

T

Cp

p

α T

Figure 5: A schematic representation of a second order phase transition





4. The thermodynamic property which are determine by more than second order

derivative will be infinite at critical temperature,

P75

50

25

0 1 2 3 4 5 T

He

He

solid

Figure: Phase diagram of helium

vapour






Q1. In phase transition of first order, there is a finitely discontinuity at the transition point of

(a) Gibbs free energy G

(b) The first-order derivatives of G

(c) The second -order-derivatives of G

(d) The higher -order-derivatives more than second order of G

(where G is the Gibbs function)

Q2. In phase transition of second order, there is a infinitely discontinuity at the transition

point of G is the Gibbs function

(a) Gibbs free energy G

(b) The first-order derivatives of G

(c) The second -order-derivatives of G

(d) The higher -order-derivatives more than second order of G

(where G is the Gibbs function)

Q3. Which of the following is finitely discontinuous at transition temperature for first order

transition?

(a) Gibbs free energy (b) Entropy

(c) Specific Heat (d) Volume expansibility

Q4. The Clausius-Clapeyron equation indicates that the increase of pressure increases the

melting point:

(a) in the case of all substances

(b) in the case of substances which expand on solidification

(c) in the case of substances which contract on solidification

(d) in the case of substances which neither expand nor contract on solidification

Q5. Consider the following statements in respect of first-order phase transition:

1. Clausius-Clapeyron latent heat equation holds well in the first-order phase transition.

2. There is change in entropy and volume in the first-order phase transition.

Which of the above statements is/are correct?

(a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2





Q6. The Joule-Thomson expansion produced cooling:

(a) at all initial temperatures and pressures

(b) above certain initial temperature

(c) above certain initial pressure

(d) below certain initial temperature

Q7. After Joule-Thomson expansion, the gas is:

(a) always heated

(b) heated or cooled depending upon the initial temperature of the gas

(c) neither heated nor cooled at any temperature

(d) always cooled

Q8. Match List I and List II and select the correct answer using the codes given below the

Lists:

List I List II

A

.

Temperature of inversion in Joule-

Thomson effect is related to

1.0=⎟

⎠⎞

⎜⎝⎛∂∂

TVU

B

.

For perfect gases 2.⎥⎦⎤

⎢⎣⎡ −=⎟

⎠⎞

⎜⎝⎛∂∂ b

RTa

CPT

PH

21

C

.

For a perfect gas Joule-Thomson

effect vanishes because

3. ( ) 0≠∂∂

TPVP

D

.

Deviation from Boyle’s law implies 4.0=⎟

⎠⎞

⎜⎝⎛ −

∂∂ V

TVT

Codes:

A B C D

(a) 2 4 1 3

(b) 3 1 4 2

(c) 2 1 4 3

(d) 3 4 1 2

Q9. On the inversion curve, the Joule-Thomson coefficient is:

(a) positive (b) zero (c) negative (d) infinite






Q10. Which of the following is correct statement about first order phase transition.

(a) The Gibbs free energy is continuous at transition temperature

(b) The entropy is finitely discontinues at transition temperature

(c) Pressure is infinitely discontinues at transition temperature

(d) Specific heat is infinitely discontinuous at transition temperature

Q11. Which of the following is incorrect statement about second order phase transition?

(a) The Gibbs free energy is continuous at transition temperature

(b) The entropy is finitely discontinues at transition temperature

(c) Pressure is infinitely discontinues at transition temperature

(d) Specific heat is infinitely discontinuous at transition temperature

Q12. Which of the following is finitely discontinuous at transition temperature for second

order transition?

(a) Gibbs free energy (b) Entropy

(c) Specific Heat (d) Volume expansibility

Q13. Which of the following is correct statement for phase transition?

(a) Clausius-Clapeyron latent heat equation holds well in the first-order phase transition.

(b) There is not change in entropy and volume at critical temperature in the second-order

phase transition.

(c) The change of phase when water transformed in Ice is first order transition

(d) The concept of superconductor and super fluid can be explain by second order

transition

Q14. Which is correctly matched for van der Waal’s gas?

(a) Inversion temperature is given by inversionT = Rb

a2

(b) Boyle temperature is given by BoyleT =Rb

a2

(c) Critical temperature is given by criticalT = bRa

278

(d) Boyle temperature is given BoyleT =Rba





Q15. Which of the following pairs are correctly matched?

(a) Law of liquefaction of gases → triple point

(b) Law of liquefaction of gases → critical temperature

(c) Cooling effect of gas → inversion temperature

(d) Phase transition → Boyle temperature

Q16. The Joule-Thomson coefficient HP

T⎟⎠⎞

⎜⎝⎛∂∂

=μ :

(a) less than zero at all temperatures and pressures for ideal gas

(b) zero at all temperatures and pressures for ideal gas

(c) less than zero at all temperatures and pressures for real gas

(d) can have any value depending on pressure and temperature for real gases

Q17. Consider the following statements:

A gas can be liquefied by increasing the pressure

(a) Above the critical below critical temperature only.

(b) Only when the temperature of the enclosed gas in below the critical temperature.

(c) Only when the volume of the enclosed gas is below the critical volume.

(d) Ideal gas can not be liquefied.

Q18. Consider the following statements:

When a compressed real gas is allowed to pass through a narrow hole, the temperature

(a) Falls for some gases if iT T<

(b) Falls for some gases if iT T> .

(c) Rise for some gases if iT T> .

(d) Rise for some gases iT T<





Solutions


Ans. 1: (b)

Solution: In first order phase transition Entropy PT

GS ⎟⎠⎞

⎜⎝⎛∂∂

−= is discontinuous at transition

point

Phase transition and low temperature physics.

Ans. 2: (d)

Ans. 3: (b)

Solution: for first order phase transition G is continuous P

GST∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

and S

GVp

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

is

finitely discontinuous and 2

2PP

GC TT

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

is infinitely discontinuous

Ans. 4: (c)

Solution: Clausius-Clapeyron equation is given as

( )2 1

dP LdT T V V

=−

⇒ If dTdPVV ,12 > is positive

If volume increases the pressure increases with T .

Ans. 5: (c)

Ans. 6: (d)

Solution: Cooling effect is produced when temperature is below a temperature known as

temperature of inversion.

Ans. 7: (b)

Solution: The drop of temperature dT in Joule.

Thomson effect is given as

⎟⎠⎞

⎜⎝⎛ −= b

RTa

MCPdTP

2 (i)

⇒No change in temperature if Rb

aTbRT

ai

202=⇒=−





This temperature is known as temperature of inversion Equation (i)

⇒ 0>dT if 02>− b

RTa ⇒

RbaT 2

<

⇒ If temperature T of the gas is less than temperature of inversion then cooling effect is

observed otherwise heating effect.

Ans. 8 : (c)

Solution: (i) The temperature of inversion in Joule-Thomson effect is the temperature above of

which all gases show heating effect. Below of this all gases show colling effect and at the

temperature of inversion no effect. This is given as

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛∂∂ b

RTa

CPT

PH

21

(ii) The internal energy of a perfect gas does not depend on volume this depends only on

temperature. Thus,

0=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

TT PU

VU

For real gases 2Va

VU

T

=⎟⎠⎞

⎜⎝⎛∂∂

(iii) For a perfect gas Joule-Thomson effect vanishes because

0=⎟⎠⎞

⎜⎝⎛ −

∂∂ V

TVT

(iv) The pressure P and PV is related as

++++= 32 DPCPBPAPV

…DCBA >>> are virial constants.

The Boyle’s law is applied when P is very low hence when

APV =

( ) 0PV

P∂

⇒ =∂

So, deviation from Boyle’s law ( ) 0≠∂∂

⇒ TPVP





Ans. 9: (b)

Solution: An isenthalpic curve is the locus of all points

representing equilibrium of same enthalpy.

If temperature is not very high then such curves

pass through a maximum, called the inversion

point. The locus of all inversion points is called

the inversion curve. Since, towards the left of the

maximum of either curve of constant enthalpy, the

gas shows cooling effect, the region inside the

inversion curve is the cooling region, for this

region gas shows cooling effect.

On the other hand, the region outside the inversion curve the gases show heating effect

and region is called region of heating.

The Joule-Thomson coefficient is defined as

⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

= VTVT

CPT

PPH

1μ

On the inversion 0=μ

Thus, Joule-Thomson coefficient is zero.


Ans. 10: (a), (b) and (d)

Solution: For first order phase transition G is continuous P

GST∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

and S

GVp

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

is

finitely discontinuous and 2

2PP

GC TT

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

is infinitely discontinuous

P pressure

T

tem

pera

ture

inversioncurve

cooling heating





Ans. 11: (b), (c) and (d)

Solution: For second order phase transition G is continuous P

GST∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

and S

GVp

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

is

continuous and 2

2PP

GC TT

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

is finitely discontinuous .

Ans. 12: (c) and (d)

Solution: For second order phase transition G is continuous P

GST∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

and S

GVp

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

is

continuous and 2

2PP

GC TT

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

is finitely discontinuous

Ans. 13: (a), (b), (c) and (d).

Ans. 14: (a), (c) and (d)

Solution: inversionT = Rb

a2 , BoyleT =Rba and criticalT =

bRa

278



Solution: Joule-Thomson coefficient for ideal gas is zero

i.e., 0=⎟⎠⎞

⎜⎝⎛∂∂

HPT

whereas for real gas it is given as

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛∂∂

= bRT

aCP

T

PH

21μ

a, b are Vander Waal’s constant.


Solution: When temperature of a gas is above a fixed temperature, known as critical

temperature, the gas cannot be liquid whatever the pressure is applied.

If temperature of the gas is equal to or smaller than the critical temperature, then it can be

liquefied.






Solution: The fall of temperature is given as

⎟⎠⎞

⎜⎝⎛ −= b

RTa

CdPdT

P

2

⇒ Rb

aTi2

=

Thus, temperature iTT > then heating effect,

if temperature iTT < then cooling effect.

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