Consider e.g. the differential of the Helmholtz free energy: $$dA = -SdT-pdV+\mu dn, $$ where, for simplicity, we only consider one particle species. We can then derive, for example, the following Maxwell relation: $$\left(\frac{\partial S}{\partial V} \right)_{T,n} = \left(\frac{\partial p}{\partial T} \right)_{V,n} (1)$$

Sometimes we have occasion to consider instead the molar entropy $s = S/n$ and the molar volume $v = V/n$. Certainly it holds that $$\left(\frac{\partial s}{\partial v} \right)_{T,n} = \left(\frac{\partial p}{\partial T} \right)_{V,n} (2)$$ but is it also true that $$\left(\frac{\partial s}{\partial v} \right)_{T} = \left(\frac{\partial p}{\partial T} \right)_{v} (3),$$ i.e. may we drop the restriction that $n$ be constant and instead demand that $v$ be constant on the RHS?

I cannot really see why this is true, but it is a relation which is used e.g. in "Introduction to Modern Statistical Mechanics" by David Chandler.


Some elaboration to clarify my confusion. To derive relation (1), note that $-S = \left(\frac{\partial A}{\partial T} \right)_{V,n}$ and $-p = \left(\frac{\partial A}{\partial V} \right)_{T,n}$. By equality of mixed partial derivatives, we get $$\left(\frac{\partial S}{\partial V} \right)_{T,n} = -\left(\frac{\partial}{\partial V}\left(\frac{\partial A}{\partial T} \right)_{V,n}\right)_{T,n}=-\left(\frac{\partial}{\partial T}\left(\frac{\partial A}{\partial V} \right)_{T,n}\right)_{V,n} = \left(\frac{\partial p}{\partial T} \right)_{V,n}.$$

Let's try something similar for relation (2) and (3). Put $a = A/n$ and use $S = sn$ and $V = vn$. We get $$nda = -nsdT-npdv-pvdn+\mu dn \Rightarrow da = -sdT-pdv+\frac{\mu-vp}{n}dn.$$ Now we can show relation (2) in just the same was as we showed relation (1) above, i.e. where we keep the requirement that $n$ be constant. But if we don't have this requirement, as in relation (3), I don't see how we can complete the derivation, because of the term $\frac{\mu-vp}{n}dn$ in $da$.

  • $\begingroup$ Isn't simply following? Is the second last notation that seems redundant to me. $\endgroup$
    – Alchimista
    Commented Dec 28, 2018 at 20:11
  • $\begingroup$ $p$ and $T$ are both intensive. So, it should work. I think your question was whether we can write $v$ instead of $V$. $\endgroup$ Commented Dec 28, 2018 at 20:41
  • $\begingroup$ @Alchimista Thank you for your comment. I have updated my question to clarify my concerns. $\endgroup$ Commented Dec 28, 2018 at 20:54
  • $\begingroup$ @EashaanGodbole Thank you for your comment. I have updated my question to clarify my concerns. $\endgroup$ Commented Dec 28, 2018 at 20:54

1 Answer 1


One could intuitively guess that, since fixing $n$ and $V$ implies fixing the ration $v=V/n$, and since fixing $v$, $V$ and $n$ must scale in the same way, an intensive quantity like $\partial{p}/\partial{T}$, should be the same at fixed $n$ and $V$ or at fixed $v$.

However, a more formal proof may help in getting a better understanding of the key ingredients behind the identity of the RHS of eqn. (2) and (3).

The two starting points are: the expression of the differential of $A$: $$ dA = -S dT -P dV + \mu d n~~~~~~~~~~~~(1) $$ and the requirement that $A$ should be a homogeneous function of degree one with respect to its extensive variables (which in turn may be seen as consequence of additivity and continuity of the free energy). This last requirement implies that, for any $\lambda>0$ and for all the values of $T,V,n$ in the domain of $A$: $$ A(T,\lambda V, \lambda n) = \lambda A(T,V,n). $$ By choosing $\lambda=1/n$, we get $$ a(T,v)=A(T,v,1)=\frac{1}{n}A(T,V,n)~~~~~~~(2) $$ which is just the formal statement that extensiveness implies the free energy per mole is function of $T$ and $v$ only.

Therefore, $$ da = \left. \frac{\partial{a}}{\partial{T}} \right|_{v}dT + \left. \frac{\partial{a}}{\partial{v}} \right|_{T}dv. ~~~~~~~~~(3) $$ In order to identify the coefficients of $dT$ and $dv$ in eqn(3), we can start from the differential of eqn(2): $$ da = d\left(\frac{A(T,V,n) }{n}\right)= \frac{dA}{n} - \frac{Adn}{n^2}~~~~~~~(4) $$ by using eqn(1) in the first term of the RHS of eqn(4) and using the identity $$ A = -PV+\mu n $$ (which is consequence of the Euler's theorem for homogeneous functions of degree 1), we get $$ da = -sdT-P\left( \frac{dV}{n}-\frac{v}{n}dn \right)~~~~~~~~(5) $$ and the parenthesis on the RHS is nothing but $ d\left( \frac{V}{n} \right) = dv. $ At this point, $\left. \frac{\partial{a}}{\partial{v}} \right|_{T} = -P $ and the connection between Maxwell relations using the $(T,V,n)$ or $(T,v)$ sets of variables should be more clear

  • $\begingroup$ Could you please elaborate why $A = -PV + \mu n$ is the consequence of the Euler's theorem for homogeneous functions? I can see that it is indeed the homogeneous function of $V$ and $n$ of order 1, but cannot see the converse. I would appreciate a reference. Your answer is very enlightening! $\endgroup$
    – mikeonly
    Commented Nov 4, 2021 at 7:49
  • $\begingroup$ @mikeonly, $A$ is a homogeneous function of degree 1 of its extensive variables ($V$ and $n$). Therefore, $A(T,\lambda V,\lambda n)= \lambda A(T,V,n)$. Taking the derivative wrt $\lambda$ on both sides of the equation, for $\lambda=1$, and by using the definition of $-P$ as the partial derivative of $A$ wrt volume, and $\mu$ as partial derivative wrt to $n$, one gets the result. $\endgroup$ Commented Nov 4, 2021 at 14:45
  • $\begingroup$ Thanks, it works. But then when I plug $A = -PV+\mu n$ and the expression (1) into (4), I am left with an extra $d\mu-\mu\frac{dn}{n}$: $da = \frac{1}{n}(-SdT-PdV+nd\mu)-\frac{dn}{n^2}(-PV+\mu n) = -sdT-P\frac{dV}{n}+d\mu+P\frac{v}{n}dn - \mu \frac{dn}{n} = -sdT-P(\frac{dV}{n}-\frac{v}{n}dn)+d\mu-\mu\frac{dn}{n}$. Am I missing something or there is a mistake somewhere? $\endgroup$
    – mikeonly
    Commented Nov 7, 2021 at 22:28
  • $\begingroup$ @mikeonly Sorry, you just discovered the consequence of a typo in eqn. 1. The last term is $+\mu dn$, it is not $ + n d \mu$. I have corrected it. $\endgroup$ Commented Nov 7, 2021 at 23:15
  • $\begingroup$ Ah, exactly! Glad it is corrected now, that $d\mu$ term was my another question. :) $\endgroup$
    – mikeonly
    Commented Nov 8, 2021 at 8:59

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