As stated in some of the exercises in chapter 5, the first law of thermodynamics can be expressed as dU = TdS − PdV. By calculating and equating ∂²U/∂Y∂X and ∂²U/∂X∂Y, where X and Y are an unspecified pair of variables (drawn from P, V, T and S), prove that ∂(S, T)/∂(X, Y) = ∂(V, P)/∂(X, Y).

Question

As stated in some of the exercises in chapter 5, the first law of thermodynamics can be expressed as

dU = T dS − P dV .

By calculating and equating ∂²U/∂Y∂X and ∂²U/∂X∂Y , where X and Y are an unspecified pair of variables (drawn from P, V, T and S), prove that

[latex]\frac {\partial \left( S,T \right) }{\partial \left( X,Y \right) } =\frac {\partial \left( V,P \right) }{\partial \left( X,Y \right) }[/latex].

Using the properties of Jacobians, deduce that

[latex]\frac {\partial \left( S,T \right) }{\partial \left( V,P \right) } =1[/latex].

Accepted Answer

Starting from

dU = T dS − P dV ,

the partial derivatives of U with respect to X and Y are

[latex]\frac{\partial U}{\partial X}=T \frac{\partial S}{\partial X}-P \frac{\partial V}{\partial X} \quad ext { and } \quad \frac{\partial U}{\partial Y}=T \frac{\partial S}{\partial Y}-P \frac{\partial V}{\partial Y}.[/latex]

We next differentiate these two expressions to obtain two (equal) second deriva-tives. Note that, since X and Y can be any pair drawn from P, V, T and S, we must differentiate all four terms on the RHS as products, giving rise to two terms each. The resulting equations are

[latex]\begin{aligned}\frac{\partial^2 U}{\partial Y \partial X} & =T \frac{\partial^2 S}{\partial Y \partial X}+\frac{\partial T}{\partial Y} \frac{\partial S}{\partial X}-P \frac{\partial^2 V}{\partial Y \partial X}-\frac{\partial P}{\partial Y} \frac{\partial V}{\partial X} ,\\frac{\partial^2 U}{\partial X \partial Y} & =T \frac{\partial^2 S}{\partial X \partial Y}+\frac{\partial T}{\partial X} \frac{\partial S}{\partial Y}-P \frac{\partial^2 V}{\partial X \partial Y}\frac{\partial P}{\partial X} \frac{\partial V}{\partial Y} .\end{aligned} [/latex]

Equating the two expressions, and then cancelling the terms that appear on both side of the equality, yields

[latex]\begin{aligned}\frac{\partial T}{\partial Y} \frac{\partial S}{\partial X}-\frac{\partial P}{\partial Y} \frac{\partial V}{\partial X} & =\frac{\partial T}{\partial X} \frac{\partial S}{\partial Y}-\frac{\partial P}{\partial X} \frac{\partial V}{\partial Y} ,\\Rightarrow \quad \frac{\partial T}{\partial Y} \frac{\partial S}{\partial X}-\frac{\partial T}{\partial X} \frac{\partial S}{\partial Y} & =\frac{\partial P}{\partial Y} \frac{\partial V}{\partial X}-\frac{\partial P}{\partial X} \frac{\partial V}{\partial Y} ,\\Rightarrow \quad \quad \frac{\partial(S, T)}{\partial(X, Y)} & =\frac{\partial(V, P)}{\partial(X, Y)}\end{aligned}[/latex]

Now, using this result and the properties of Jacobians [latex]\left(J_{p r}=J_{p q} J_{q r} ext { and } J_{p q}= ight. \left.\left[J_{q p} ight]^{-1} ight)[/latex], we can write

[latex]\begin{aligned}\frac{\partial(S, T)}{\partial(V, P)} & =\frac{\partial(S, T)}{\partial(X, Y)} \frac{\partial(X, Y)}{\partial(V, P)} \& =\frac{\partial(S, T)}{\partial(X, Y)}\left[\frac{\partial(V, P)}{\partial(X, Y)} ight]^{-1} \& =\frac{\partial(S, T)}{\partial(X, Y)}\left[\frac{\partial(S, T)}{\partial(X, Y)} ight]^{-1} \& =1 .\end{aligned}[/latex]

Question 6.21: As stated in some of the exercises in chapter 5, the first l...

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