Thermodynamic Relations Questions and Answers

Answer: B

The coefficient is \(-4\times10^{-6}\) K/Pa.

The Joule-Thomson coefficient follows from the equation of state through

\[ \begin{aligned} \mu &= \frac{1}{C_{p}}\Big[T\left(\frac{\partial\bar{v}}{\partial T}\right)_{P} \\ &\quad - \bar{v}\Big] \end{aligned} \]

Solve the equation of state for the molar volume and differentiate at constant pressure:

\[ \begin{aligned} \bar{v} &= b + \frac{RT}{P} + \frac{aP}{T} \\ \left(\frac{\partial\bar{v}}{\partial T}\right)_{P} &= \frac{R}{P} - \frac{aP}{T^{2}} \end{aligned} \]

\[ \begin{aligned} T\left(\frac{\partial\bar{v}}{\partial T}\right)_{P} &= \frac{RT}{P} - \frac{aP}{T} \\ T\left(\frac{\partial\bar{v}}{\partial T}\right)_{P} - \bar{v} &= -b - \frac{2aP}{T} \end{aligned} \]

With \(a = 10^{-5}\), \(P = 10^{6}\) Pa, \(T = 500\) K and \(b = 0.08\) m³/kmol:

\[ \begin{aligned} \mu &= -\frac{0.08 + 0.04}{30000} \\ &= -4\times10^{-6} \text{ K/Pa} \end{aligned} \]

The coefficient is negative, so this gas warms on throttling. The two positive options have the wrong sign, and dropping the factor 2 on the \(aP/T\) term would give \(-3.3\times10^{-6}\), which is not offered.

Answer: A

The correct pair is \(\dfrac{1}{v}\left(\dfrac{\partial v}{\partial T}\right)_{P}\) and \(\dfrac{1}{T}\).

Volume expansivity \(\beta\) is the fractional change in volume per unit temperature rise at constant pressure. For an ideal gas \(v = RT/P\), so

\[ \begin{aligned} \beta &= \frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{P} \\ &= \frac{1}{v}\cdot\frac{R}{P} = \frac{1}{T} \end{aligned} \]

\(-\frac{1}{v}(\partial v/\partial P)_{T}\) is the isothermal compressibility, which equals \(1/P\) for an ideal gas; the options built on it describe compressibility, not expansivity. The option with \(1/P\) as the ideal-gas expansivity mixes the two.

Answer: C

The Joule-Thomson coefficient is \(\mu_{j} = -\dfrac{b}{c_{p}}\).

In general

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

From \(v = RT/p + b\), \((\partial v/\partial T)_{p} = R/p\), so \(T(R/p) - v = (v - b) - v = -b\), and \(\mu_{j} = -b/c_{p}\).

The negative value means such a gas always warms on throttling. The positive \(b/c_{p}\) has the wrong sign, and \(\pm c_{p}/b\) inverts the expression and does not even have units of K/Pa.

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.