Chain rule in thermodynamics

Question

The question is by considering the thermodynamics law $dU=T\,dS-p\,dV$, show that $$\left(\frac{\partial \ln p}{\partial \ln V}\right)_T - \left(\frac{\partial \ln p}{\partial \ln V}\right)_S = \left(\frac{\partial pV}{\partial T}\right)_V \left[\frac{p^{-1}(\partial U/\partial V)_T+1}{(\partial U/\partial T)_V}\right]$$, where the subscipt means keep that variable as constant.

What I have done is consider the chain rule of $p=p(V,S(V,T))$

$$\text{LHS}-\text{RHS} = \frac{V}{p} \left[ \left(\frac{\partial p}{\partial V}\right)_T - \left(\frac{\partial p}{\partial V}\right)_S-\frac{\partial U/\partial V|_T +p}{\partial U/\partial T |_V}\right]$$

Since $\frac{\partial p}{\partial V}|_T=\frac{\partial p}{\partial V}|_S+\frac{\partial p}{\partial S}|_V \frac{\partial S}{\partial V}|_T$, we have $(\frac{\partial p}{\partial V})_T -(\frac{\partial p}{\partial V})_S=\frac{\partial S/\partial V|_T}{\partial S/\partial p|_V}$

$$LHS-RHS=\frac{V}{p} \left[\frac{\partial S/\partial V|_T}{\partial S/\partial p|_V}-\frac{\partial U/\partial V|_T +p}{\partial U/\partial T |_V} \right]$$

But I cannot go on to show the RHS equals to $0$, what should I do next?

Any help is appreciated.

Answer 1

You were sort of on the right track. First, note that differential form of the internal energy $U$ is $$dU = T \, dS - p \, dV = \left( \frac{\partial U}{\partial S} \right)_V dS + \left( \frac{\partial U}{\partial V} \right)_S dV $$ Thus, $$ p = - \left( \frac{\partial U}{\partial V} \right)_S \tag 1 $$ Now, as you suggested, let's consider $p = p(S(T,V),V)$, so that $$ \begin{eqnarray} \left( \frac{\partial p}{\partial T} \right)_V &=& \left( \frac{\partial p}{\partial S} \right)_V \left( \frac{\partial S}{\partial T} \right)_V \tag 2\\ \left( \frac{\partial p}{\partial V} \right)_T &=& \left( \frac{\partial p}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T + \left( \frac{\partial p}{\partial V} \right)_S \tag 3 \end{eqnarray} $$ Similarly, we expand the internal energy $U = U(S(T,V),V)$, so that $$ \begin{eqnarray} \left( \frac{\partial U}{\partial T} \right)_V &=& \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial T} \right)_V \tag 4 \\ \left( \frac{\partial U}{\partial V} \right)_T &=& \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T + \left( \frac{\partial U}{\partial V} \right)_S \tag 5 \end{eqnarray} $$ So, beginning as you did, we rewrite (3) as $$ \left( \frac{\partial p}{\partial V} \right)_T - \left( \frac{\partial p}{\partial V} \right)_S = \left( \frac{\partial p}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T $$ Eliminate $\partial S/\partial V$ on the right using (5) $$ \left( \frac{\partial p}{\partial V} \right)_T - \left( \frac{\partial p}{\partial V} \right)_S = \frac{\displaystyle\left( \frac{\partial p}{\partial S} \right)_V}{\displaystyle\left( \frac{\partial U}{\partial S} \right)_V } \left[ p + \left( \frac{\partial U}{\partial V} \right)_T \right] $$ Eliminate $\partial U/\partial S$ using (4) $$ \left( \frac{\partial p}{\partial V} \right)_T - \left( \frac{\partial p}{\partial V} \right)_S = \frac{\displaystyle\left( \frac{\partial p}{\partial S} \right)_V \left( \frac{\partial S}{\partial T} \right)_V }{\displaystyle\left( \frac{\partial U}{\partial T} \right)_V } \left[ p + \left( \frac{\partial U}{\partial V} \right)_T \right] $$ Finally, use (2) to write $$ \left( \frac{\partial p}{\partial V} \right)_T - \left( \frac{\partial p}{\partial V} \right)_S = \frac{\displaystyle\left( \frac{\partial p}{\partial T} \right)_V}{\displaystyle\left( \frac{\partial U}{\partial T} \right)_V } \left[ p + \left( \frac{\partial U}{\partial V} \right)_T \right] $$ This is the desired form, written in a more appealing way IMO.