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.
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 $$ T = \left(\frac{\partial u}{\partial s}\right)_v$$
