Showing that intensive parameters obtain by considering molar quantities

Question

In Callen's Thermodynamics textbook, he writes that $$\left(\frac{\partial u}{\partial s}\right)_v = \left(\frac{\partial U}{\partial S}\right)_{V,N}$$

where $u = U/N$, $s = s/N$, and $v = V/N$ and, moreover, $U$ is first order homogeneous so that $$ U (S,V,N) = U(SN/N,VN/N,N) = NU(S/N,V/N,1) \equiv Nu(s,v)$$ where the last equality represents another way to define $u$. I'm not able to see how to argue for the given equality of derivatives though. I get stalled out: $$\left(\frac{\partial u}{\partial s}\right)_v \equiv \left(\frac{\partial (U/N)}{\partial (S/N)}\right)_{V/N} \stackrel{?}{=} \left(\frac{\partial U}{\partial S}\right)_{V,N}$$

I want to be clear that I understand why (namely because the quantity so obtained on the RHS is intensive) this equality must hold, but I can't see how to show it mathematically as I get lost in the morass of partial derivatives and chain rules.

Answer 1

The key point here is the homogeneity of extensive thermodynamic properties. For $U(S,V,N)$ this means that if we increase $S$, $V$, $N$, simultaneously by the same factor, then $U$ increases by the same factor as well: $$ U(\lambda S, \lambda V, \lambda N) = \lambda U(S,V,N) $$ It follows that the partial derivatives of $U$ are intensive. This can be shown by calculus for any homogeneous function with degree 1 but in this case it is clear since these derivatives are known intensive properties: $$ \left(\frac{\partial U}{\partial S}\right)_{VN} = T;\quad \left(\frac{\partial U}{\partial V}\right)_{VN} = -P;\quad \left(\frac{\partial U}{\partial S}\right)_{VN} = \mu. $$ As intensive properties they do not depend on the absolute $U$, $V$ and $N$, but have the same value in all systems with the same ratios $S:V:N$. Therefore, they have the same values in a system with $s=S/N$, $v=V/N$, $n=N/N=1$: $$ \left(\frac{\partial U}{\partial S}\right)_{VN} = \left(\frac{\partial u}{\partial s}\right)_{v} = T. $$

Answer 2

Observe that $$ \left(\frac{\partial U}{\partial S}\right)_{V,N} \equiv \left(\frac{\partial (Nu(S/N,V/N))}{\partial S}\right)_{V,N} = N\left(\frac{\partial (u(S/N,V/N))}{\partial S}\right)_{V,N} $$ $$\stackrel{(1)}{=} N \left[\left(\frac{\partial (u(S/N,V/N))}{\partial (S/N)}\right)_{V/N}\left(\frac{\partial (S/N)}{\partial S}\right)_{V,N} + \left(\frac{\partial (u(S/N,V/N))}{\partial (V/N)}\right)_{S/N}\left(\frac{\partial (V/N)}{\partial S}\right)_{V,N} \right]$$ $$ N \left[\left(\frac{\partial (u(S/N,V/N))}{\partial (S/N)}\right)_{V/N}\frac{1}{N} + 0 \right] = \left(\frac{\partial (u(S/N,V/N))}{\partial (S/N)}\right)_{V/N}$$ $$\equiv \left(\frac{\partial u}{\partial s}\right)_v $$ where in (1) we have used the chain rule for partial derivatives and in the last step we have simply used the definitions of $s$ and $v$.

Answer 3

I would consider the very simple and immediate proof based on the definition of the partial derivative and the homogeneity of $U$: $$ \left(\frac{\partial U}{\partial S}\right)_{V,N} = \lim_{\Delta S \rightarrow 0} \frac{U(S+\Delta S,V,N)-U(S,V,N)}{\Delta S}=\lim_{\Delta s \rightarrow 0} \frac{U(N s+ N \Delta s,N v,N)-U(N s,N v,N)}{N\Delta s}= \lim_{\Delta s \rightarrow 0} \frac{u(s+\Delta s,v)-u(s,v)}{\Delta s}=\left(\frac{\partial u}{\partial u}\right)_{v} $$