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The following equation is taken from Callen's Thermodynamics, page 39:

$$(\frac{\partial u }{\partial s})_v=(\frac{\partial u }{\partial s})_{V,N}$$

where $u = U/N$, $s = S/N$ and $v = V/N$.

The notation is conventional.

My question is:

The author doesn't make any assumption of constant $N$ in this context.

Why would constant molar volume imply the equality?

I feel so confused.

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  • $\begingroup$ Just to be clear, are you asking why the assumption of constant $v=V/N$ is equated to constant $V$ and $N$ and doesn't allow the possibility of both $V$ and $N$ changing proportionately? $\endgroup$ – Chemomechanics Nov 8 '17 at 17:47
  • $\begingroup$ Yes. I don't know why he discarded that possibility $\endgroup$ – Math The Novice Nov 8 '17 at 17:51
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We know that $$ dU = TdS - pdV + \mu dN$$ If we assume that the internal energy $U$ is an extensive, homogeneous function of degree 1, then it follows that $$ U = TS - pV + \mu N$$


Now consider the quantity $u \equiv U/N$. We would have that $$ du = \frac{1}{N}dU - \frac{U}{N^2}dN= \frac{TdS}{N} - \frac{pdV}{N} + \frac{\mu}{N}dN -\frac{u}{N}dN$$

If we further define $v\equiv V/N$ and $s\equiv S/N$, note that $$ dS = d(sN)= Nds + s dN$$

so after a bit of algebra, $$du= Tds- pdv+\frac{sT-pv+\mu-u}{N}dN$$

However, we already said that $$ u =\frac{U}{N}=\frac{ST - pV + \mu N}{N}= sT - pv + \mu$$ so the last term vanishes, and we find that $$ du = Tds - pdv$$

so it follows that $$ T = \left(\frac{\partial u}{\partial s}\right)_v$$

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  • $\begingroup$ How is this related to (∂u/∂s)v=(∂u/∂s)V,N ? $\endgroup$ – Math The Novice Nov 11 '17 at 7:13
  • $\begingroup$ If we express $u$ as a function of $s,V,$ and $N$ (so we have the entropy per particle, but the total volume and particle number), you should be able to see that $du = Tds -\frac{p dV}{N} + \frac{TS+\mu-u}{N}dN$, so $T=(\frac{\partial u}{\partial s})_{V,N}$. $\endgroup$ – J. Murray Nov 11 '17 at 12:09

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