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Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India)

Copyright by Brij Bhooshan @ 2013

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Please welcome for any correction or misprint in the entire manuscript and your

valuable suggestions kindly mail us [email protected].

1986

Problem: Prove that

[IAS-1986]

Solution: Let entropy S be imagined as a function of T and V. Then S = S(T, V)

Since T(∂S/∂T)V = CV, heat capacity at constant volume, and Maxwell's third equation,

Then,

This is known as the first TdS equation. Finally, we get




2 Solution of UPSC Papers of Thermodynamic Relations

1987

Problem: Explain the salient features of Clapeyron equation. Calculate the change in

melting point of ice when it is subjected to a pressure of 100 atmospheres.

Density of ice = 0.917 gm/cm3

Latent heat of ice = 80.3 cal/gm.

[IAS-1987]

Solution: Clapeyron Equation: During the phase transitions like melting,

vapourization, and, sublimation, the temperature and pressure remain constant. A

phase change of the first order is known as any phase change that satisfies the following

requirements:

(a) There are changes of entropy and volume.

(b) The first-order derivatives of Gibbs function change discontinuously.

Now, phase change at fixed temperature and pressure and estimate changes in specific

entropy, internal energy and enthalpy during phase change.

Let us start with one of Maxwell relations;

For pure substances we have seen that during phase transformation at some

temperature the pressure is saturation pressure. Thus pressure is also independent of

specific volume and can be determined by temperature alone. Hence,

Here (∂p/∂T)sat is the slope of saturation curve on pressure-temperature (p–T) diagram

at some point determined by fixed constant temperature during phase transformation

and is independent of specific volume.

Substituting in the Maxwell relation.

Thus, during vaporization i.e. phase transformation from liquid to vapour state, above

relation can be given as,

P

Vapour

Solid

T

T

P

Slope

Liquid




3 Solution of UPSC Papers of Basic and Applied Thermodynamics By Brij Bhooshan

From differential form of specific enthalpy,

for phase change occurring at constant pressure and temperature,

for saturated liquid to dry vapour transformation,

Substituting hfg/T in place of entropy in (∂p/∂T)sat, then, we get

Above equation is termed as Clapeyron equation.

Numerical part: Given that: P = 100 atm, = 0.917 gm/cm3; Lice = 80.3 Cal/gm.

1989

Problem: Starting from fundamentals, show that the internal energy of an ideal gas is

a function of temperature only.

[IAS-1989]

Solution: Let properties of a substance are

Since dQ = T.ds; and dW = P.dv, then

T.ds = du + P.dv

Suppose internal energy u = f(T, v), then

Now, from Eqns. (1) and (2), then, we have

Suppose entropy s = f(T, v), then


Differentiating Eqn. (5) with respect to v, then, we get





Differentiating Eqn. (6) with respect to T, then, we get


For an ideal gas, Pv = RT, then


This means that u does not change when v changes at T as constant.

Similarly, if u = f (T, P), then, we have

Therefore, u does not change either when P changes at T as constant. So the internal

energy of an ideal gas is a function of temperature only.

1991

Problem: Define the Joule-Thomson coefficient and discuss its importance in low-

temperature applications.

[IAS-1991]

Solution: Joule-Thomson Co-efficient: Joule-Thomson coefficient is defined as the

rate of change of temperature with pressure during an isenthalpic process or throttling

process. So,

Liq.

T

Maximum inversion

temperature

Heating

region

Inversion curve ( )

Critical

point

Vap.

Cooling

region

Constant enthalpy

curve (isenthalpes)





Mathematically evaluating the consequence of μj we see,

for μj > 0, temperature decreases during the process.

for μj = 0, temperature remains constant during the process.

for μj < 0, temperature increases during the process.

The curve passing through the maxima of these isenthalpes is termed as the inversion

curve.

This graphical representation of isenthalpic curve gives the Joule-Thomson coefficient

by its slope at any point. Slope may be positive, negative or zero at different points on

the curve. The points at which slope has zero value or Joule-Thomson coefficient is zero

are termed as “inversion points” or “inversion states”. Temperature at these inversion

states is referred as “inversion temperature”. Locii of these inversion states is named as

“inversion line”. Temperature at the intersection of inversion line with zero pressure line

is termed as “maximum inversion temperature”.

The numerical value of the slope of an isenthalpe on a T-p diagram at any point is

named as the Joule-Kelvin coefficient or Joule-Thomson coefficient (μj). Thus the locus of

all points at which μj is zero is the inversion curve. The region inside the inversion curve

where μj > 0 is named as the cooling region and the region outside where μj < 0 is named

as the heating region.

1996

Problem: The specific heats of a gas are of the form Cp = a + kT and Cv = b + kT, where

a, b and k are constants and T is in K. Derive the formula Tbva–bekT = constant, for

adiabatic expansion of the gas.

[Engg. Services -1996]

Solution: Now, we know that – = R, then from given equation in the problem

– = a + kT – b – kT = a – b = R

Now, we know that change in entropy is

v

dvR

T

dTCds v

v

dvba

T

dTkTbds

v

dvbakdT

T

dTbds

For isentropic process, ds = 0, then

Const. = b ln T + kT + (a – b) ln v

kTbab evT = Const.

1998

Problem: Determine the pressure of saturated steam at 40ºC if the saturation pressure

at 35ºC is 5.628 kPa. The enthalpy of evaporation and the specific volume at 35°C are





2418.6 kJ/kg and 25.22 m3/kg respectively. Assume that enthalpy of evaporation is

essentially constant in this range and R = Pvg /T.

[IAS-1998]

Solution: Given that: T1 = 35°C, P1 = 5.628 kPa, T2 = 40°C, v1 = 25.22 m3/kg, h1 = 2418.6

kJ/kg, h → constant, and R = Pvg /T.

We know that

PRTT

h

Tv

h

dT

dP fg

fg

fg

/

2

1

2

2

1RT

dTh

P

dPfg

.211

2 11ln

sat

fg

TTR

h

P

P

Now,

35273

22.251000628.5

R = 0.46083 kJ/kg

Now,

313

1

308

1

46083.0

6.2418628.5lnln 2P

2lnP = 1.727 + 0.2722 = 1.992

Then, we get P2 = 7.3831 kPa.

Problem: Show using Maxwell's relations:

[IAS-1998]

Solution: Now, we know that first TdS equation is

Now, we know that second TdS equation is

Equating the first and second TdS equations

Again let temperature T be imagined as a function of V and p. Then T = T(V, p)

Comparing above two equations, then we get





Both these equations give

From cyclic relation

Problem: Using the Maxwell relation derive the following Tds equation

[IAS-1998]

Solution: Suppose S = S(T, P), then

Multiplying by T both side

Since

Then, we get

1999

Problem: Using Maxwell relation, derive Clapeyron equation. Hence derive Clapeyron-

Clausius equation.

[IAS-1999]

Solution: Clapeyron Equation: See the solution of IAS 1987.

Clapeyron - Clausius Equation: Therefore, Clapeyron equation can be modified in

the light of two approximations of “vf being negligible compared to vg at low pressures”

and ideal gas equation of state during vapour phase at low pressure, vg = RT/p.

From Clapeyron Equation

Clapeyron equation thus becomes, Clausius-Clapeyron equation as given here,





Above equation is termed as Clausius-Clapeyron equation.

Now integrating between initial state ‘i’ to final state ‘f’, then we have

Clausius-Clapeyron equation is thus a modified form of Clapeyron equation based upon

certain approximations and is valid for low pressure phase transformations of liquid-

vapour or solid-vapour type.

2001

Problem: Develop the Clapeyron equation for the pure substance changing the phase.

Hence find the enthalpy of evaporation for R-22 at -10°C and compare the same with the

tabulated value. What is the percentage error involved?

Properties of R-22

TS PS vf vg hfg

°C kPa liters/kg m3/kg kJ/kg

20 244.72 0.7409 0.0929 220.331

–10 354.16 0.7587 0.0654 213.136

0 497.41 0.7783 0.0472 205.369

[IAS-2001]

Solution: Clapeyron Equation: See the solution of IAS 1987.

Given that: R 22 at –10°C.

Now, from Clapeyron Equation

Now, 1st value from 0°C to –10°C

After solving, we get = 2711.30 kJ/kg.

Now, 2nd value from –10°C to 20°C

After solving, we get = 70.72 kJ/kg, which is not possible because temperature

difference is –30°C.

Now % is

Problem: IFoS 2001/1(c)





[IFoS 2001]

Solution: Now, we know that

Differentiating above Eqn. with pressure (p), then we get

Now, using Maxwell’s Eqn.

Differentiating above Eqn. with temperature (T), then we get

From Eqns. (i), and (ii)

For an ideal gas



Then from Eqn. (iii)

This means that is a function of T alone.

Problem: Using Maxwell's relations, show that for a pure substance

Where β is coefficient of cubical expansion, k is coefficient of compressibility and Cp and

Cv are specific heats at constant pressure and at constant volume respectively.

[Engg. Services-2001]

Solution: (i) Suppose entropy S be imagined as a function of T and p. Then S = S(T, p)





Since T(∂S/∂T)p = Cp, heat capacity at constant pressure, and Maxwell's fourth equation,

Then,

Now, we know that, volume expansivity

From, Eqns. (1) and (2)

(ii) Let entropy S be imagined as a function of T and V. Then S = S(T, V)


Then,

Now, we know that, volume expansivity, and compressibility


Now, utilizing Eqns. (4, 5, 6), we get

From, Eqns. (3) and (7)

(iii) Let entropy S be imagined as a function of p and V. Then S = S(p, V)





Since T(∂S/∂T)p = Cp, heat capacity at constant pressure, and T(∂S/∂T)V = CV, heat

capacity at constant volume, then we get

Now, using Eqns. (4), (5), and (8), we get

2002

Problem: By using Maxwell's relations of thermodynamics, show that Joule - Thomson

coefficient, μ of gas can be expressed as,


Solution: The difference in enthalpy between two neighbouring equilibrium states is

and the second TdS equation (per unit mass)

The second term in the above equation stands only for a real gas, because for an ideal

gas, dh = cp dT.

For an isenthalpic process dh = 0, then Eqn. [1] will be

Then we get

2003

Problem: Explain the terms (i) coefficient of cubical expansion, β and (ii) coefficient of

compressibility K. Hence, show that β/K = (∂P/∂T)V.


Solution: Coefficient of cubical expansion: Partial derivatives of V with respect to

temperature can be related to “volume expansivity” or “coefficient of volume expansion”

as below,





Coefficient of compressibility: Partial derivative of specific volume with respect to

pressure can be related to “isothermal compressibility”, K as below.

Using, Eqns. (1), and, (2) we get

Problem: Using Maxwell's and other equations, show that

Hence show that

Cp – CV = β2TV/K


Solution: From the first TdS equation,

and the second TdS equation,

Equating the first and second TdS equations

Again let temperature T be imagined as a function of V and P. Then T = T(V, P)

Comparing the Eqns. [1] and [2], then we get

Both these equations give

From 1st law of thermodynamics

dU + P.dV = dQ = T.dS

dU = T.dS – P.dV [4]

If U = U(T, V)





From Eqns. (4) and (5), then we get

From Eqns. (6) and 2nd T.dS Eqn., then we get

From Eqns. (2) and (7), then we get

From Eqns. 3, then we get


Now, we know that

Then, we get

Cp – CV = β2TV/K

2004

Problem: Derive Clausius-Clapeyron equation and calculate the change in freezing

temperature per bar change in pressure for water. Given that specific volume of water

at 0°C is 10−3 m3/kg and that of ice is 1.091 × 10−3 m3/kg. Latent heat of ice = 335 kJ/kg.

[IFoS-2004]

Solution: Clausius-Clapeyron Equation: See the solution of IAS 1999.

Numerical Part: Given that: T = 0°C, = 10−3 m3/kg, = 1.091 × 10−3 m3/kg, L = 335

kJ/kg.

From Clausius-Clapeyron equation





Problem: Using the Maxwell relation, derive the following equation:

dPT

VTVdTCdh

p

p

IFoS-2004]


H = U + PV

dH = dU + P.dV + V.dP [1]

and, dQ = dU + P.dV {dQ = T.dS}

T.dS = dU + P.dV [2]

From Eqn. 1 and 2

dH = T.dS + V.dP [3]

Now, we know that 2nd T.dS Eqn.,

From Eqn. 3 and 4, we get

dPT

VTVdTCdh

p

p

Problem: Derive the following Clapeyron and Clausius-Clapeyron equations:

Explain the physical significance of these equations.


Solution: See the solution of IAS 1999.

2005

Problem: Using Maxwell's relations, show that for a pure substance,

where β is the coefficient of cubical expansion, K is coefficient of compressibility and CP,

Cv are specific heats at constant pressure and constant volume respectively.

[IAS-2005]

Solution: See the solution of Engg. Services 2001.

Problem: Show that the slope on the h-s diagram is equal to:

(i) T for a reversible constant pressure process;

(ii) T − (1/β) for a reversible isothermal process;





(iii) v

vp

C

CCT

.

for a reversible constant volume process.

where β = coefficient of volume expansion. The following relations:

K

TvCC vp

2 and

v

vT

STC

can be used. K is compressibility.

[IFoS-2005]

Solution: Now, from the property relation

H = U + PV

Suppose P as a constant, then

dh = dU + P.dV [1]

from 1st law of thermodynamics

dQ = dU + P.dV {dQ = T.dS}


From Eqns. (1) and (2), we have

dh = T.dS

TS

h

p

This, means that T for a reversible constant pressure process.

(ii) Now, again we know that

dh = dU + V.dP [3]

from 2nd TdS Eqn.

Now, utilizing Eqns. 3 and 4, we have

dPT

VTVdTCdh

p

p

For isothermal process dT = 0, then

TP

h

= V − T

PT

V

Now, dividing by PT

V

both side

P

T

TV

Ph

/

/=

PTV

V

/− T

P

T

TV

Ph

/

/=

PT

V

V

1

1− T [5]

Now,

TP

h

=

TS

h

×

TP

S

[6]

From Maxwell’s Eqn.

TP

S

= −

PT

V

[7]

Now, utilizing Eqns. 6 and 7, we have





TP

h

=

TS

h

× (−1)

PT

V

P

T

TV

Ph

/

/= −

TS

h

[8]

From Eqns. (5) and (8), we have

−TS

h

=

PT

V

V

1

1− T

Now, we know that β = PT

V

V

1, then

TS

h

= T −

1

Problem: Given that

p

pT

STC

and

v

vT

STC

obtain the expression for

T

p

P

C

and

T

v

v

C

using Maxwell's and other relations.

Using the above derived equations, show that CP and CV for an ideal gas are functions of

temperature only.

[IFoS-2005]

Solution: Given that

p

pT

STC

Differentiating above Eqn. w.r.t. P, then

T

p

P

C

= T

PT

S2

From Maxwell’s Eqn

Now, from ideal gas Eqn.

Differentiating above Eqn. w.r.t. T, then





Then

This, means that CP for an ideal gas are functions of temperature only.

Also given that

V

VT

STC

Differentiating above Eqn. w.r.t. P, then

T

V

V

C

= T

VT

S2

From Maxwell’s Eqn

Now, from ideal gas Eqn.

Differentiating above Eqn. w.r.t. T, then

Then

This, means that CV for an ideal gas are functions of temperature only.

Problem: Derive the expression for (Δh)T for a substance that obeys the equation of

state given by:


Solution: Given that

Now, start from 1st law of thermodynamics



From 1st T.dS Eqn.





From property relation

dh = dU + V.dP [4]

from 2nd TdS Eqn.

From Eqn. (2), and (3), then we have

Suppose that U = f(T, V)

From Eqn. (6) and (7)

If

Differentiating above Eqn. w.r.t. T, V as a constant

From, Eqn. 1, 8, and 9 then

From Eqn. (2), and (4), then we get

dH = dU + P.dV + V.dP

From Eqn. (10), and (11), then

2006

Problem: With the help of Maxwell's relation of thermodynamics, prove that Joule-

Thomson coefficient, μJ of a gas is given by the following expression:

[IAS-2006]






Problem: Using appropriate T.dS relations, show that the slope of the constant volume

line is higher than that of a constant pressure line passing through a given state

represented on a temperature-entropy diagram for a perfect gas. Sketch the

temperature-entropy diagram.

[IFoS-2006]


T.dS = dU + P.dV

T.dS = CV.dT + P.dV

T.dS = dh − V.dP

T.dS = CP.dT + V.dP

Now, we know > , then

The slope is constant volume line passing through point A is steeper than that of the

constant pressure line passing through the same point.

Problem: If u = f (T, V) and h = f (T, P) prove that


Solution: Suppose that U = f(T, V), then

Now, we know that


From 1st T.dS Eqn.


S

T

A

P = C

V = C





We know, that


Suppose that h = f(T, P), then

Now, we know that

dh = T.dS + V.dP [2]

From 2nd T.dS Eqn.


We know, that


2007

Problem: Prove the Mayer relation

Cp – CV = β2TV/

where = isothermal compressibility and β = coefficient of volume expansion.

[IFoS-2007]

Solution: See the solution of Engg. Services-2003.

Problem: Derive four Maxwell relationships.

[IFoS-2007]

Solution: A pure substance existing in a single phase has only two independent

variables. Of the eight quantities p, V, T, S, U, H, F (Helmholtz function), and G (Gibbs

function) any one may be expressed as a function of any two others.

For a pure substance undergoing an infinitesimal reversible process

(a) dU = T.dS ‒ p.dV

(b) dH = dU + p.dV + V.dp = T.dS + V.dp

(c) dF = dU ‒ T.dS ‒ S.dT = ‒p.dV ‒ S.dT

(d) dG = dH ‒ T.dS ‒ S.dT = V.dp ‒ S.dT

Since U, H, F and G are thermodynamic properties and exact differentials of the type

Then





From above four equations for properties to be exact differentials, we can write

functions;

For differential of function ‘U’ to be exact;

For differential of function ‘H’ to be exact;

For differential of function ‘F’ to be exact;

For differential of function ‘G’ to be exact;

These four equations are known as Maxwell's equations. Maxwell relations have large

significance as these relations help in estimating the changes in entropy, internal

energy and enthalpy by knowing p, V and T.

Problem: Prove the following relations


Solution: Suppose that U = f(T, V), then

Now, we know that


From 1st T.dS Eqn.


Comparing Eqns. (1) and (4), we get

Suppose that h = f(T, P), then

Now, we know that

dh = T.dS + V.dP [2]

From 2nd T.dS Eqn.







2008

Problem: Show that the properties at the critical state for a gas obeying van der Waals

equation of state

are given by

Hence show that the coefficients 'a' and 'b' are expressed as

the critical coefficient for the van der Waals gas is 2.66.


Solution: For a van der Waals gas,

where a, b, and R are the characteristic constants of the particular gas.

or pv3 – (pb + RT)v2 + av − ab = 0

It is therefore a cubic in v and for given values of p and T has three roots of which only

one need be real. For low temperatures, three positive real roots exists for a certain

range of pressure. As the temperature increases the three real roots approach one

another and at the critical temperature they become equal. Above this temperature only

one real root exists for all values of p.

The critical isotherm Tc at the critical state on the p-v plane, where the three real roots

of the van der Waals equation coincide, not only has a zero slope, but also its slope





changes at the critical state (point of inflection), so that the first and second derivatives

of p with respect to v at T = Tc are each equal to zero. Therefore

From Eqns. (2), and (3)


From Eqns. (1), (5) and (4)



2009

Problem: Derive equations for the change in internal energy and entropy of a gas

which obeys the vander Waals equation of state.

[IAS -2009]

Solution: From Vander Wall’s Eqn.

Differentiating above Eqn. w.r.t. T, V as a constant

Again, differentiating above Eqn. w.r.t. T, V as a constant

Internal Energy: Suppose that U = f(T, V), then

Now, start from 1st law of thermodynamics



From 1st T.dS Eqn.







From, Eqn. (a), (b), and (5) then

Entropy: From Eqns. (3) and (5)

Problem: Define the Joule - Thomson coefficient and prove that for an ideal gas, the

value of Joule - Thomson coefficient tends to zero.

[IAS -2009]

Solution: Joule - Thomson coefficient: See the solution of IAS 1991, and, Engg.

Services 2002.

We know that

For an ideal gas pv = RT

There is no change in temperature when an idea' gas is made to undergo a Joule-Kelvin

expansion (i.e. throttling).

Problem: Write a short note on Redlich - Kwong equation of state.

[IAS -2009]

Solution: The Redlich-Kwong equation proposed in 1949 and given by





the constants a and b in terms of critical properties as follows:

Problem: With the help of Maxwell's relations, prove that Joule-Thomson coefficient μj

of a gas is given by

What does this equation signify?

[IFoS -2009]


Problem: Derive the equations:

(i) SP

PT

P

T

VTC

(ii) TV

CP

ST

P

(iii) 1/

/

V

S

TP

TP


Solution: (i) We know that

Now, from Maxwell’s Eqn.

Now, from Eqn. (1) and (2), then we have

(ii) Now, rearrange the Eqn. (3)

We know, that


(iii) We know that






Now

Now, we know that

Cp – CV = β2TV/ [8]


Problem: Show that for an ideal gas, the slope of the constant volume line on the T-s

diagram is more than that of the constant pressure line.


Solution: See the solution of IFoS 2006.

2010

Problem: For a gas, the equation of state is expressed as below over a certain range of

temperatures and pressures:

where a is constant. Prove that the change in enthalpy is given by

for isothermal process. Also find out the expression for change of entropy.

[IFoS – 2010]

Solution: We know that

Now, according to problem

From Eqn. 1 and 2, we have

Entropy: From 2nd T.dS Eqn.





Problem: Using Maxwell's relations and the thermodynamic definitions for Cp and Cv

in terms of gradients, show the following:

[Engg. Services – 2010]

Solution: Let entropy S be imagined as a function of T and V. Then S = S(T, V)


Then,

This is known as the first TdS equation.

Again let entropy S be imagined as a function of T and p. Then S = S(T, p)

Since T(∂S/∂T)p = Cp, heat capacity at constant pressure, and Maxwell's fourth equation,

Then,

This is known as the second TdS equation.

Problem: Joule-Thomson coefficient


Solution: See the solution of IAS 1991, Engg. Services 2002, and, IAS 2009.





Problem: There is no change in temperature when an ideal gas is made to undergo

Joule - Thomson expansion.


Solution: See the solution of IAS 2009.

2011

Problem: The following expressions for the equations of state and the specific heat Cp

are obeyed by a certain gas:

where α, A, B, C are constants. Obtain an expression for (i) the Joule-Thomson

coefficient and (ii) the specific heat CV.

[IFoS – 2011]

Solution: Given that:

(i) the Joule-Thomson coefficient

From Eqn. (1)

From Eqn. (3) and (4)

Specific heat CV: now from Eqn. (1)

Now we know that





2012

Problem: Derive

from the first principles. Explain any assumptions needed here.

[IAS-2012]

Solution: See the solution of IFoS-2007.

Problem: Given that

p

pT

STC

and

v

vT

STC

obtain the expression for

T

p

P

C

and

T

v

v

C

using Maxwell's and other relations.

Using the above derived equations, show that CP and CV for an ideal gas are functions of

temperature only.

[IFoS-2012]

Solution: See the solution of IFoS 2005.

Problem: For an isentropic expansion of a gas with CP = a + kT, CV = b + kT and CP −

CV = R, show that Tb va-b ekT = constant.


Solution: See the solution of Engg. Services -1996.

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