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Maybe this could be case where a question, other than an answer, could be wrong, but whatever...

Let's start with the fundamental thermodynamic relation in differential form for a hydrostatic system, $dE=TdS-PdV+\mu dN$.

Via the Euler's homogeneous functions theorem we could write too $E=TS-PV+\mu N$.

So what could stop us from writing, i.e., $V$ as $V=-\frac{\partial E}{\partial P}$? Why we have to switch to enthalpy?

valerio
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Lo Scrondo
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1 Answers1

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It doesn't work because $T,p$ and $\mu$ are not extensive variables.

The natural variables for the internal energy $U$, $(S,N,V)$, are extensive, i.e. they are additive for subsystems: if I put together two systems of entropy $S_1$ and $S_2$, the total entropy is $S_1+S_2$. The same is valid for $N$ and $V$, but it is not valid for $T,P$ and $\mu$.

So it is true that $U$ is homogeneous of degree $1$ in $S,N$ and $V$:

$$U(\lambda S, \lambda N, \lambda V) = \lambda U(S,N,V)$$

and therefore we can use Euler's theorem to write*

$$U = \frac{\partial U}{\partial S}S+ \frac{\partial U}{\partial V}V + \frac{\partial U}{\partial N}N = TS-pV+\mu N$$

but $U$ is not homogeneous in $T,p$ and $\mu$. Therefore we cannot write

$$U = \frac{\partial U}{\partial T}T+ \frac{\partial U}{\partial p}p + \frac{\partial U}{\partial \mu}\mu$$

$^*$ More details here.

valerio
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