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Suppose that the variables $E$, $T$, $V$, and $P$ are related by a pair of equations, $f(E,T,V,P)=0$ and $g(E,T,V,P)=0$, that can be solved for any two of the variables in terms of the other two, and suppose that the differential equation $\partial_VE-T\partial_TP+P=0$ is satisfied when $V$ and $T$ are taken as the independent variables. Show that $\partial_PE+T\partial_TV+P\partial_PV=0$ when $P$ and $T$ are taken as the independent variables.

If we put $\phi(V,T)=(E(V,T),T,V,P(V,T))$ and $F=(f,g)$, then $(F\circ\phi)(V,T)=0$ and hence $F'(\phi(V,T))\phi'(V,T)=0$.

How to proceed any further?

1 Answers1

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Taking $V$ and $T$ as independent variables means that $E$ and $P$ are determined as functions of $V$ and $T$.

Let $E=\eta(V, T) $ and $P=\zeta(V,T)$

Then given $\partial_VE-T\partial_TP+P=0$ becomes $\partial_1\eta -T\partial_2\zeta+P=0\quad \ldots (1) $

Let us take $P, T$ as independent variables.

Then differentiating $E=\eta(V, T) $ and $P=\zeta(V,T)$ w.r.to $P$ yields$$\partial_PE =(\partial_1\eta )(\partial_PV)$$

and $$ 1=(\partial_1\zeta )(\partial_PV) $$

Hence $$\partial_1\eta =\frac{\partial_PE}{\partial_PV}$$

and $$\partial_1\zeta =\frac{1}{\partial_PV}$$

Again differentiating $P=\zeta(V,T)$ w.r.to $T$ yields $$0 =(\partial_1\zeta )(\partial_TV)+\partial_2\zeta$$

Hence $\partial_2\zeta=-(\partial_1\zeta )(\partial_TV)=-\frac{\partial_TV}{\partial_PV}$

Now substituting the values of $\partial_1\eta, \partial_2\zeta$ in $(1) $ , we have

$\begin{align}0&=\frac{\partial_PE}{\partial_PV}-T ( -\frac{\partial_TV}{\partial_PV})+P\\&=\partial_PE +T \partial_TV+P\partial_TV\end{align}$

Sourav Ghosh
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