Discussion
One of the Maxwell equations is expressed as
\[ (\partial s/\partial v)_{T} = (\partial p/\partial T)_{v} \]
where \(s\) is the entropy per unit mass, \(v\) is the mass specific volume, \(p\) is the pressure, and \(T\) is the temperature. In this expression, \(s\) is a continuous function of \(T\) and \(v\). The derivatives of \(s\) are also continuous. Let \(c_{v}\) be specific heat capacity at constant volume for a gas. Then, \(\left(\frac{\partial c_{v}}{\partial v}\right)_{T}\) can be written as
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.
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