Thermodynamic Relations Questions and Answers

Answer: D

The term \(u - Ts\) is the Helmholtz function, which is the Helmholtz free energy. The Gibbs function is \(h - Ts\), the Keenan function is the steady-flow availability \(h - T_0 s\), and so the key gives the Helmholtz function. Both Helmholtz and Gibbs functions are sometimes called free energy functions, but the specific one here is Helmholtz.

Answer: A

The correct relation is \(U = -T^{2}\left[\frac{\partial (F/T)}{\partial T}\right]_{V}\), the Gibbs-Helmholtz equation.

Since \(F = U - TS\) and \(S = -(\partial F/\partial T)_{V}\), we have \(U = F - T(\partial F/\partial T)_{V}\). Now differentiate \(F/T\):

\[ \begin{aligned} \frac{\partial (F/T)}{\partial T} &= \frac{1}{T}\frac{\partial F}{\partial T} - \frac{F}{T^{2}} \\ -T^{2}\frac{\partial (F/T)}{\partial T} &= F - T\frac{\partial F}{\partial T} = U \end{aligned} \]

The '+' version has the wrong sign. \(\pm T(\partial F/\partial T)_{V}\) equals \(\mp TS\), which is only the entropy term, not the internal energy.

Answer: B

Joule-Thomson expansion is isenthalpic.

In the throttling process gas is pushed steadily through a porous plug from high pressure \(P_{1}\) to low pressure \(P_{2}\), with no heat exchanged. The work done on the gas entering is \(P_{1}V_{1}\) and by the gas leaving is \(P_{2}V_{2}\), so

\[ \begin{aligned} U_{1} + P_{1}V_{1} &= U_{2} + P_{2}V_{2} \\ \Rightarrow H_{1} &= H_{2} \end{aligned} \]

It is not isentropic, because throttling is irreversible and the entropy rises. It is not isobaric, since the whole point is a pressure drop, nor isochoric, since the gas expands.

Answer: C

The derivative of \(c_{v}\) with respect to v at constant T is \(T(\partial^{2}p/\partial T^{2})_{v}\).

Start from \(c_{v} = T(\partial s/\partial T)_{v}\). Since s and its derivatives are continuous, the order of differentiation can be swapped:

\[ \begin{aligned} \left(\frac{\partial c_{v}}{\partial v}\right)_{T} &= T\frac{\partial}{\partial T}\left(\frac{\partial s}{\partial v}\right)_{T} \\ &= T\frac{\partial}{\partial T}\left(\frac{\partial p}{\partial T}\right)_{v} = T\left(\frac{\partial^{2}p}{\partial T^{2}}\right)_{v} \end{aligned} \]

The Maxwell relation turns the entropy derivative into a pressure derivative. The other options use the wrong prefactor or differentiate with respect to v, which the derivation never does. For an ideal gas p is linear in T, so \(c_{v}\) does not depend on v, as expected.

Answer: A

\(\left(\frac{\partial h}{\partial p}\right)_{T} = -\mu c_{p}\).

Apply the cyclic relation to h, p and T:

\[ \begin{aligned} &\left(\frac{\partial h}{\partial p}\right)_{T}\left(\frac{\partial p}{\partial T}\right)_{h} \\ &\times \left(\frac{\partial T}{\partial h}\right)_{p} = -1 \end{aligned} \]

Here \((\partial T/\partial h)_{p} = 1/c_{p}\) and \((\partial p/\partial T)_{h} = 1/\mu\), where \(\mu = (\partial T/\partial p)_{h}\) is the Joule-Thomson coefficient. Solving gives \((\partial h/\partial p)_{T} = -\mu c_{p}\).

\(\mu c_{p}\) loses the minus sign of the cyclic rule. \(c_{p}T\) and \(-c_{p}/T\) are dimensionally different from an enthalpy change per unit pressure. For an ideal gas \(\mu = 0\), so h does not vary with p at constant T, as expected.

Answer: D

The adiabatic bulk modulus is \(-v\left(\frac{\partial P}{\partial v}\right)_{s}\).

A bulk modulus measures the pressure increase needed per fractional decrease in volume, \(K = -v\frac{\partial P}{\partial v}\), with the minus sign making it positive because volume falls as pressure rises. 'Adiabatic' for a reversible process means constant entropy, so the derivative is taken at constant \(s\).

\(-v(\partial P/\partial v)_{T}\) is the isothermal bulk modulus. The forms \(-\frac{1}{v}(\partial v/\partial P)\) are compressibilities, the reciprocals of bulk moduli: isothermal at constant \(T\) and isentropic at constant \(s\).

Answer: C

The correct match is A1→B2, A2→B4, A3→B3, A4→B1.

• Helmholtz function: \(a = u - Ts\).
• Gibbs function: \(g = h - Ts\).
• T-ds equation: \(du = Tds - Pdv\), the first law for a simple compressible substance with reversible work and heat.
• Clapeyron-Clausius equation: \(\left(\frac{d\ln P}{dT}\right)_{sat} = \frac{h_{fg}}{RT^{2}}\), the Clapeyron equation for a liquid-vapour change with an ideal-gas vapour and negligible liquid volume.

The common confusion is swapping the two free-energy functions: Helmholtz uses internal energy and Gibbs uses enthalpy. Options pairing A3 with B1 also swap the T-ds relation and the vapour-pressure equation.

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.

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.

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.