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I have to prove the identity $$ \left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial p}{\partial T}\right)_V-p $$ I know how to do this using Maxwell's relations, but the assignment specified that I have to use Jacobian transformations. Reading many websites, pdf's, etc. has brought me no closer to the solution.

I think that I have to start from the equation for internal energy and then differentiate it using $dV$ - this produces almost the end result, but when I work with the Jacobian I'm always lost.

Any explanations on how to use the Jacobian would be great. I know the basic rules, but cannot solve this.

Could you help me please?

Kyle Kanos
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user44502
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2 Answers2

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In a reversible cyclic process the heat absorbed is the work done, hence $ \int_{C} TdS = \int_{C} pdV$ where $C$ is a closed contour. But it is also true that for the enclosed area $A$ within $C$ we have $ \int_{C} TdS = \iint_{A} dTdS$ and $ \int_{C} pdV = \iint_{A} dpdV$, so $ \iint_{A} dTdS = \iint_{A} dpdV$, and if we go from the variables $T,S$ to $p,V$ one has $ \iint_{A} dTdS = \iint_{A} dpdV \frac{\partial(T,S)}{\partial(p,V)}$. This implies that the Jacobian must be 1: $$\frac{\partial(T,S)}{\partial(p,V)}=1$$, in other words $$\frac{\partial T}{\partial p}\frac{\partial S}{\partial V}-\frac{\partial T}{\partial V}\frac{\partial S}{\partial p} = 1$$ This implies that in general $$\frac{\partial(T,S)}{\partial(x,y)}= \frac{\partial(p,V)}{\partial(x,y)}$$ Now if $x=T,y=V$ as new variables then you get $$ \frac {\partial S}{\partial V} = \frac {\partial p}{\partial T}$$ from which and from $dU=TdS-pdV$ your equality follows.

hyportnex
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The given proof entirely use Jacobian techniques. Maxwell relations are also used in Jacobian from

Jyothis V V
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