Thermodynamic Relations Questions and Answers

Q: In a graph with Helmholtz function on the \(y\)-axis and volume on the \(x\)-axis, the slope of the isothermal curves for a finite volume system containing an ideal gas is

The slope is finite, negative, and non-zero.From \(dA = -S\,dT - P\,dV\), the slope of an isotherm on an A-V diagram is\[ \left(\frac{\partial A}{\partial V}\right)_T = -P \]For an ideal gas at finite volume, \(P = nRT/V\) is positive and finite, so the slope is negative, finite and non-zero, and it flattens as V grows.A zero slope would need zero pressure, an infinite one would need infinite pressure, and a positive slope would need negative pressure, none of which applies to an ideal gas.

Q: Match each quantity in Column M with the appropriate relation from Column N. Here, \(\psi\) is Helmholtz function, \(P\) is pressure, \(v\) is specific volume, \(T\) is temperature, \(h\) is specific enthalpy, and \(s\) is specific entropy.Column MColumn N(M1) \(v\)(N1) \(-\left.\frac{\partial \psi}{\partial v}\right|_T\)(N5) \(\left.\frac{\partial h}{\partial P}\right|_s\)(M2) \(P\)(N2) \(\left.\frac{\partial \psi}{\partial v}\right|_T\)(N6) \(-\left.\frac{\partial h}{\partial s}\right|_P\)(M3) \(s\)(N3) \(\left.\frac{\partial \psi}{\partial T}\right|_v\)(N7) \(-\left.\frac{\partial h}{\partial P}\right|_s\)(M4) \(T\)(N4) \(-\left.\frac{\partial \psi}{\partial T}\right|_v\)(N8) \(\left.\frac{\partial h}{\partial s}\right|_P\)

The correct matching is M1-N5, M2-N1, M3-N4, M4-N8.The Helmholtz function satisfies \(d\psi = -s\,dT - P\,dv\), so\[ \begin{aligned} P &= -\left.\frac{\partial \psi}{\partial v}\right|_T \\ s &= -\left.\frac{\partial \psi}{\partial T}\right|_v \end{aligned} \]giving M2-N1 and M3-N4. Enthalpy satisfies \(dh = T\,ds + v\,dP\), so\[ \begin{aligned} T &= \left.\frac{\partial h}{\partial s}\right|_P \\ v &= \left.\frac{\partial h}{\partial P}\right|_s \end{aligned} \]giving M4-N8 and M1-N5. The first option gives T the wrong sign (N7), the second gives the wrong signs throughout, and the last swaps the relations for v and T.

Q: The equation of state for a non-ideal gas is\[ \frac{Pv}{RT} = 1 + BP \]where \(P\) is pressure, \(v\) is specific volume, \(R\) is the specific gas constant, \(T\) is temperature, and \(B\) is a temperature dependent parameter. For this gas, the partial derivative of enthalpy with respect to pressure at constant temperature is

The derivative is \(-RT^{2}\left(\frac{dB}{dT}\right)\).The general relation is \((\partial h/\partial P)_T = v - T(\partial v/\partial T)_P\). From the equation of state, \(v = RT/P + BRT\), so\[ \begin{aligned} \left(\frac{\partial v}{\partial T}\right)_P &= \frac{R}{P} + BR + RT\frac{dB}{dT} \\ v - T\left(\frac{\partial v}{\partial T}\right)_P &= -RT^{2}\frac{dB}{dT} \end{aligned} \]The ideal-gas part \(RT/P\) and the \(BRT\) term both cancel. BRT is what remains if B is treated as constant in only one of the two terms, and 0 is the ideal-gas result, which holds here only if B does not depend on T.

Q: Consider any substance whose mass specific heat capacity at constant pressure and mass specific heat capacity at constant volume are \(c_{P}\) and \(c_{V}\), respectively. If the gas constant is R, which ONE of the following relations is CORRECT for the substance at all conditions?

The relation true for every substance is \(c_{P} \ge c_{V}\).The general thermodynamic relation is:\[ c_{P}-c_{V} = \frac{Tv\beta^{2}}{\kappa_{T}} \]Temperature, specific volume and the isothermal compressibility \(\kappa_{T}\) are always positive, and \(\beta^{2}\) cannot be negative, so the difference is never negative. It is zero when the expansivity \(\beta\) is zero, for example liquid water at 4 °C, or for an idealised incompressible substance.So \(c_{P} \gt c_{V}\) fails in those cases, \(c_{P} - c_{V} = R\) holds only for an ideal gas, and \(c_{P} = c_{V}\) holds only in the special cases just named.

Q: Consider a real gas, which obeys the following equation-of-state:\[ v = \frac{RT}{P} + C_{1} - \frac{C_{2}}{Pv}, \]where v is the mass specific volume, P is the pressure, T is the temperature and R is the gas constant. \(C_{1}\) and \(C_{2}\) are constants. If s is the mass specific entropy, the quantity \(\left(\frac{\partial s}{\partial v}\right)_{T}\) for the gas is given by

The quantity is \(\frac{R}{v-C_{1}}\).By the Maxwell relation \((\partial s/\partial v)_{T} = (\partial P/\partial T)_{v}\). Solve the equation of state for P:\[ \begin{aligned} v - C_{1} &= \frac{1}{P}\left(RT - \frac{C_{2}}{v}\right) \\ P &= \frac{RT - C_{2}/v}{v-C_{1}} \end{aligned} \]\[ (\partial P/\partial T)_{v} = \frac{R}{v-C_{1}} \]R/v is the ideal-gas result (C1 = C2 = 0), which this gas does not obey. The other two forms come from differentiating without first making P explicit.

Q: A fluid undergoes a process where its pressure (P), temperature (T) and volume (V) changes from \((P_{1}, T_{1}, V_{1})\) to \((P_{2}, T_{2}, V_{2})\). During the process, volume expansivity \((\beta)\) and isothermal compressibility \((\kappa_{T})\) remains constant. Given that \(\beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\) and \(\kappa_{T} = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\), the ratio \(\left(\frac{V_{2}}{V_{1}}\right)\) is

The ratio is \(\frac{\exp[\beta(T_{2}-T_{1})]}{\exp[\kappa_{T}(P_{2}-P_{1})]}\).Write the total differential of V(T, P) using the two definitions:\[ \begin{aligned} \frac{dV}{V} &= \beta\,dT - \kappa_{T}\,dP \\ \ln\frac{V_{2}}{V_{1}} &= \beta(T_{2}-T_{1}) - \kappa_{T}(P_{2}-P_{1}) \end{aligned} \]with \(\beta\) and \(\kappa_{T}\) constant. Exponentiating gives the keyed expression.The difference \(\beta\Delta T - \kappa_{T}\Delta P\) is ln(V2/V1), not the ratio itself, and the product and quotient forms do not come from integrating the differential.

Q: The entropy of black-body radiation is given by \(S = \frac{4}{3}\sigma V^{1/4}U^{3/4}\) where \(\sigma\) is a constant V is the volume and U is internal energy of the system. The temperature of the radiation is :

The temperature is \(T = U^{1/4}/(\sigma V^{1/4})\).Temperature follows from the thermodynamic relation \(1/T = (\partial S/\partial U)_{V}\):\[ \begin{aligned} \frac{1}{T} &= \frac{4}{3}\sigma V^{1/4}\cdot\frac{3}{4}U^{-1/4} \\ &= \sigma V^{1/4}U^{-1/4} \end{aligned} \]Inverting gives \(T = U^{1/4}/(\sigma V^{1/4})\), equivalent to \(U \propto VT^{4}\), the Stefan-Boltzmann law for the energy of cavity radiation. Option B drops the volume, and options C and D have the wrong powers.

Q: Helmholtz free energy function is defined by :

The Helmholtz free energy is F = U - TS.It is the Legendre transform of the internal energy that replaces entropy by temperature as a natural variable: \(dF = -S\,dT - P\,dV\). At constant temperature and volume a system moves towards minimum F, and F is the work obtainable from a system at constant temperature.U + PV is the enthalpy H, and U + PV - TS is the Gibbs free energy G, the potential minimised at constant temperature and pressure. U + TS is not a standard thermodynamic potential.

Q: The thermodynamic identity for the ideal rubber band is \(dU = Tds + \tau dL\) where T is the temperature, \(\tau\) is the tension, U is the internal energy of the rubber band, S is its entropy, and L is its length. Which of the following Maxwell's relation is correct for this rubber band system is ....

The correct relation is \(-(\partial S/\partial L)_{T} = (\partial\tau/\partial T)_{L}\).Start from the Helmholtz free energy \(F = U - TS\). Using \(dU = TdS + \tau dL\):\[ dF = -S\,dT + \tau\,dL \]so \(S = -(\partial F/\partial T)_{L}\) and \(\tau = (\partial F/\partial L)_{T}\). Equating the mixed second derivatives of \(F\) gives the keyed Maxwell relation. It is the rubber-band analogue of \((\partial S/\partial V)_{T} = (\partial p/\partial T)_{V}\), with the sign changed because the tension term enters \(dU\) as \(+\tau dL\) where the gas has \(-p\,dV\).The option without the minus sign misses exactly that sign. The other two pair variables wrongly: from \(dU\) itself the correct relation is \((\partial T/\partial L)_{S} = (\partial\tau/\partial S)_{L}\), with no minus sign, and the Gibbs-type function \(U - TS - \tau L\) gives \((\partial S/\partial\tau)_{T} = (\partial L/\partial T)_{\tau}\), again without the minus sign of the last option.

11.

In a graph with Helmholtz function on the \(y\)-axis and volume on the \(x\)-axis, the slope of the isothermal curves for a finite volume system containing an ideal gas is

A.always zero
B.infinite
C.finite, positive, and non-zero
D.finite, negative, and non-zero

Exam Year(s)

GATE XE - 2024,

Answer: D

The slope is finite, negative, and non-zero.

From \(dA = -S\,dT - P\,dV\), the slope of an isotherm on an A-V diagram is

\[ \left(\frac{\partial A}{\partial V}\right)_T = -P \]

For an ideal gas at finite volume, \(P = nRT/V\) is positive and finite, so the slope is negative, finite and non-zero, and it flattens as V grows.

A zero slope would need zero pressure, an infinite one would need infinite pressure, and a positive slope would need negative pressure, none of which applies to an ideal gas.

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12.

Match each quantity in Column M with the appropriate relation from Column N. Here, \(\psi\) is Helmholtz function, \(P\) is pressure, \(v\) is specific volume, \(T\) is temperature, \(h\) is specific enthalpy, and \(s\) is specific entropy.

Column M	Column N
(M1) \(v\)	(N1) \(-\left.\frac{\partial \psi}{\partial v}\right\|_T\)	(N5) \(\left.\frac{\partial h}{\partial P}\right\|_s\)
(M2) \(P\)	(N2) \(\left.\frac{\partial \psi}{\partial v}\right\|_T\)	(N6) \(-\left.\frac{\partial h}{\partial s}\right\|_P\)
(M3) \(s\)	(N3) \(\left.\frac{\partial \psi}{\partial T}\right\|_v\)	(N7) \(-\left.\frac{\partial h}{\partial P}\right\|_s\)
(M4) \(T\)	(N4) \(-\left.\frac{\partial \psi}{\partial T}\right\|_v\)	(N8) \(\left.\frac{\partial h}{\partial s}\right\|_P\)

A.M1-N5, M2-N1, M3-N4, M4-N7
B.M1-N6, M2-N2, M3-N3, M4-N8
C.M1-N5, M2-N1, M3-N4, M4-N8
D.M1-N8, M2-N1, M3-N4, M4-N5

Exam Year(s)

GATE XE - 2024,

Answer: C

The correct matching is M1-N5, M2-N1, M3-N4, M4-N8.

The Helmholtz function satisfies \(d\psi = -s\,dT - P\,dv\), so

\[ \begin{aligned} P &= -\left.\frac{\partial \psi}{\partial v}\right|_T \\ s &= -\left.\frac{\partial \psi}{\partial T}\right|_v \end{aligned} \]

giving M2-N1 and M3-N4. Enthalpy satisfies \(dh = T\,ds + v\,dP\), so

\[ \begin{aligned} T &= \left.\frac{\partial h}{\partial s}\right|_P \\ v &= \left.\frac{\partial h}{\partial P}\right|_s \end{aligned} \]

giving M4-N8 and M1-N5. The first option gives T the wrong sign (N7), the second gives the wrong signs throughout, and the last swaps the relations for v and T.

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13.

The equation of state for a non-ideal gas is

\[ \frac{Pv}{RT} = 1 + BP \]

where \(P\) is pressure, \(v\) is specific volume, \(R\) is the specific gas constant, \(T\) is temperature, and \(B\) is a temperature dependent parameter. For this gas, the partial derivative of enthalpy with respect to pressure at constant temperature is

A.\(BRT\)
B.\(-RT^{2}\left(\frac{dB}{dT}\right)\)
C.\(BRT - RT^{2}\left(\frac{dB}{dT}\right)\)
D.0

Exam Year(s)

GATE XE - 2024,

Answer: B

The derivative is \(-RT^{2}\left(\frac{dB}{dT}\right)\).

The general relation is \((\partial h/\partial P)_T = v - T(\partial v/\partial T)_P\). From the equation of state, \(v = RT/P + BRT\), so

\[ \begin{aligned} \left(\frac{\partial v}{\partial T}\right)_P &= \frac{R}{P} + BR + RT\frac{dB}{dT} \\ v - T\left(\frac{\partial v}{\partial T}\right)_P &= -RT^{2}\frac{dB}{dT} \end{aligned} \]

The ideal-gas part \(RT/P\) and the \(BRT\) term both cancel. BRT is what remains if B is treated as constant in only one of the two terms, and 0 is the ideal-gas result, which holds here only if B does not depend on T.

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14.

Consider any substance whose mass specific heat capacity at constant pressure and mass specific heat capacity at constant volume are \(c_{P}\) and \(c_{V}\), respectively. If the gas constant is R, which ONE of the following relations is CORRECT for the substance at all conditions?

A.\(c_{P} = c_{V}\)
B.\(c_{P} - c_{V} = R\)
C.\(c_{P} \ge c_{V}\)
D.\(c_{P} \gt c_{V}\)

Exam Year(s)

GATE XE - 2026,

Answer: C

The relation true for every substance is \(c_{P} \ge c_{V}\).

The general thermodynamic relation is:

\[ c_{P}-c_{V} = \frac{Tv\beta^{2}}{\kappa_{T}} \]

Temperature, specific volume and the isothermal compressibility \(\kappa_{T}\) are always positive, and \(\beta^{2}\) cannot be negative, so the difference is never negative. It is zero when the expansivity \(\beta\) is zero, for example liquid water at 4 °C, or for an idealised incompressible substance.

So \(c_{P} \gt c_{V}\) fails in those cases, \(c_{P} - c_{V} = R\) holds only for an ideal gas, and \(c_{P} = c_{V}\) holds only in the special cases just named.

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15.

Consider a real gas, which obeys the following equation-of-state:

\[ v = \frac{RT}{P} + C_{1} - \frac{C_{2}}{Pv}, \]

where v is the mass specific volume, P is the pressure, T is the temperature and R is the gas constant. \(C_{1}\) and \(C_{2}\) are constants. If s is the mass specific entropy, the quantity \(\left(\frac{\partial s}{\partial v}\right)_{T}\) for the gas is given by

A.\(\frac{R}{v}\)
B.\(\frac{R^{2}T}{Pv^{2}}\)
C.\(\frac{R}{v-C_{1}}\)
D.\(\frac{C_{1}C_{2}}{Tv}\)

Exam Year(s)

GATE XE - 2026,

Answer: C

The quantity is \(\frac{R}{v-C_{1}}\).

By the Maxwell relation \((\partial s/\partial v)_{T} = (\partial P/\partial T)_{v}\). Solve the equation of state for P:

\[ \begin{aligned} v - C_{1} &= \frac{1}{P}\left(RT - \frac{C_{2}}{v}\right) \\ P &= \frac{RT - C_{2}/v}{v-C_{1}} \end{aligned} \]

\[ (\partial P/\partial T)_{v} = \frac{R}{v-C_{1}} \]

R/v is the ideal-gas result (C₁ = C₂ = 0), which this gas does not obey. The other two forms come from differentiating without first making P explicit.

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16.

A fluid undergoes a process where its pressure (P), temperature (T) and volume (V) changes from \((P_{1}, T_{1}, V_{1})\) to \((P_{2}, T_{2}, V_{2})\). During the process, volume expansivity \((\beta)\) and isothermal compressibility \((\kappa_{T})\) remains constant. Given that \(\beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\) and \(\kappa_{T} = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\), the ratio \(\left(\frac{V_{2}}{V_{1}}\right)\) is

A.\(\frac{\beta(T_{2}-T_{1})}{\kappa_{T}(P_{2}-P_{1})}\)
B.\([\beta(T_{2}-T_{1})][\kappa_{T}(P_{2}-P_{1})]\)
C.\(\frac{\exp[\beta(T_{2}-T_{1})]}{\exp[\kappa_{T}(P_{2}-P_{1})]}\)
D.\(\beta(T_{2}-T_{1})-\kappa_{T}(P_{2}-P_{1})\)

Exam Year(s)

GATE XE - 2026,

Answer: C

The ratio is \(\frac{\exp[\beta(T_{2}-T_{1})]}{\exp[\kappa_{T}(P_{2}-P_{1})]}\).

Write the total differential of V(T, P) using the two definitions:

\[ \begin{aligned} \frac{dV}{V} &= \beta\,dT - \kappa_{T}\,dP \\ \ln\frac{V_{2}}{V_{1}} &= \beta(T_{2}-T_{1}) - \kappa_{T}(P_{2}-P_{1}) \end{aligned} \]

with \(\beta\) and \(\kappa_{T}\) constant. Exponentiating gives the keyed expression.

The difference \(\beta\Delta T - \kappa_{T}\Delta P\) is ln(V₂/V₁), not the ratio itself, and the product and quotient forms do not come from integrating the differential.

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17.

Let f(T, l) be the tension in a rubber band of length l, T being the absolute temperature. Let F denote Helmholtz free energy and S the entropy. Then :

A.\(\left.\frac{\partial F}{\partial l}\right|_{T} = S\), \(\left.\frac{\partial F}{\partial T}\right|_{l} = f\)
B.\(\left.\frac{\partial F}{\partial l}\right|_{T} = f\), \(\left.\frac{\partial F}{\partial T}\right|_{l} = -S\)
C.\(\left.\frac{\partial F}{\partial l}\right|_{T} = Sl\), \(\left.\frac{\partial F}{\partial T}\right|_{l} = fl\)
D.\(\left.\frac{\partial F}{\partial l}\right|_{T} = 0\), \(\left.\frac{\partial F}{\partial T}\right|_{l} = f\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The correct pair is \(\left.\partial F/\partial l\right|_{T} = f\) and \(\left.\partial F/\partial T\right|_{l} = -S\).

For a rubber band, the work done on it in stretching by dl is \(f\,dl\), playing the role that \(-p\,dV\) plays for a gas. So \(dU = T\,dS + f\,dl\), and the Helmholtz free energy \(F = U - TS\) gives

\[ dF = -S\,dT + f\,dl \]

Reading off the partial derivatives: at constant T, \(\partial F/\partial l = f\); at constant l, \(\partial F/\partial T = -S\). The first option swaps the roles of S and f, the third multiplies them by l, and the fourth sets the tension derivative to zero; none matches the differential.

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18.

The entropy of black-body radiation is given by \(S = \frac{4}{3}\sigma V^{1/4}U^{3/4}\) where \(\sigma\) is a constant V is the volume and U is internal energy of the system. The temperature of the radiation is :

A.\(T = \frac{U^{1/4}}{\sigma V^{1/4}}\)
B.\(T = \frac{U^{1/4}}{\sigma}\)
C.\(T = \frac{1}{\sigma^{4}}\left(\frac{U}{V}\right)\)
D.\(T = \sigma\left(\frac{V}{U}\right)^{1/4}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The temperature is \(T = U^{1/4}/(\sigma V^{1/4})\).

Temperature follows from the thermodynamic relation \(1/T = (\partial S/\partial U)_{V}\):

\[ \begin{aligned} \frac{1}{T} &= \frac{4}{3}\sigma V^{1/4}\cdot\frac{3}{4}U^{-1/4} \\ &= \sigma V^{1/4}U^{-1/4} \end{aligned} \]

Inverting gives \(T = U^{1/4}/(\sigma V^{1/4})\), equivalent to \(U \propto VT^{4}\), the Stefan-Boltzmann law for the energy of cavity radiation. Option B drops the volume, and options C and D have the wrong powers.

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19.

Helmholtz free energy function is defined by :

A.F = U + TS
B.F = U - TS
C.F = U + PV
D.F = U + PV - TS

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The Helmholtz free energy is F = U - TS.

It is the Legendre transform of the internal energy that replaces entropy by temperature as a natural variable: \(dF = -S\,dT - P\,dV\). At constant temperature and volume a system moves towards minimum F, and F is the work obtainable from a system at constant temperature.

U + PV is the enthalpy H, and U + PV - TS is the Gibbs free energy G, the potential minimised at constant temperature and pressure. U + TS is not a standard thermodynamic potential.

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20.

The thermodynamic identity for the ideal rubber band is \(dU = Tds + \tau dL\) where T is the temperature, \(\tau\) is the tension, U is the internal energy of the rubber band, S is its entropy, and L is its length. Which of the following Maxwell's relation is correct for this rubber band system is ....

A.\(-(\frac{\delta S}{\delta L})_{T} = (\frac{\delta\tau}{\delta T})_{L}\)
B.\((\frac{\delta S}{\delta L})_{T} = (\frac{\delta\tau}{\delta T})_{L}\)
C.\(-(\frac{\delta T}{\delta L})_{S} = (\frac{\delta\tau}{\delta S})_{L}\)
D.\(-(\frac{\delta S}{\delta\tau})_{T} = (\frac{\delta L}{\delta T})_{\tau}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The correct relation is \(-(\partial S/\partial L)_{T} = (\partial\tau/\partial T)_{L}\).

Start from the Helmholtz free energy \(F = U - TS\). Using \(dU = TdS + \tau dL\):

\[ dF = -S\,dT + \tau\,dL \]

so \(S = -(\partial F/\partial T)_{L}\) and \(\tau = (\partial F/\partial L)_{T}\). Equating the mixed second derivatives of \(F\) gives the keyed Maxwell relation. It is the rubber-band analogue of \((\partial S/\partial V)_{T} = (\partial p/\partial T)_{V}\), with the sign changed because the tension term enters \(dU\) as \(+\tau dL\) where the gas has \(-p\,dV\).

The option without the minus sign misses exactly that sign. The other two pair variables wrongly: from \(dU\) itself the correct relation is \((\partial T/\partial L)_{S} = (\partial\tau/\partial S)_{L}\), with no minus sign, and the Gibbs-type function \(U - TS - \tau L\) gives \((\partial S/\partial\tau)_{T} = (\partial L/\partial T)_{\tau}\), again without the minus sign of the last option.

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