I was studying for my freshman chemistry course, and ran into a mind-boggling contradiction.

We know, at equilibrium, entropy is at a maximum: $$\left(\frac{\partial S}{\partial V}\right)_U=0$$

From the first law


which leads to

$$\left(\frac{\partial U}{\partial S}\right)_V=T$$

From the cyclic rule

$$\left(\frac{\partial U}{\partial V}\right)_S\cdot\left(\frac{\partial V}{\partial S}\right)_U\cdot\left(\frac{\partial S}{\partial U}\right)_V=-1$$

$$\left(\frac{\partial U}{\partial V}\right)_S=-\left(\frac{\partial U}{\partial S}\right)_V\cdot\left(\frac{\partial S}{\partial V}\right)_U$$

Using the result ($\left(\frac{\partial U}{\partial S}\right)_V=T$) it gives

$$\left(\frac{\partial U}{\partial V}\right)_S=-T\cdot\left(\frac{\partial S}{\partial V}\right)_U$$

But we know that $\left(\frac{\partial S}{\partial V}\right)_U=0$, so

$$\left(\frac{\partial U}{\partial V}\right)_S=0 \tag{1}\label{eq1}$$

So far so good.

Writing the total differential of $U$

$$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV$$

and comparing it to


We get:

$$\left(\frac{\partial U}{\partial V}\right)_S=-P$$

Now I'm in real trouble, because equation $\eqref{eq1}$ says this quantity is zero, and even in this general series of thermodynamic equations, I have reached the conclusion that


I would appreciate if someone could help identify what the fundamental problem is? I've checked and rechecked my calculations and assumptions, but I'm sure I'm missing something.

  • $\begingroup$ I think the cyclic rule is not valid in this case. $\frac{\partial V}{\partial S}=\frac{1}{0}$ so I suspect some of the manipulations there are faulty. $\endgroup$
    – Tyberius
    Commented Jan 24, 2020 at 18:37
  • $\begingroup$ @Tyberius I see your point. But I see another problem here (you reminded me by disallowing $\frac{1}{0}$): the derivative $\left(\frac{\partial U}{\partial S}\right)_V$, which has to be $T$ at all costs, would also take a similar form, leading to supposedly infinte T. :( $\endgroup$ Commented Jan 24, 2020 at 18:58

2 Answers 2


The problem begins with your first statement:

$$\left(\frac{\partial S}{\partial V}\right)_U=0$$

Because in fact:

$$\left(\frac{\partial S}{\partial V}\right)_U=\frac{p}{T},$$

which is clearly non-zero (except in limit of zero pressure or infinite temperature).

More generally, being at equilibrium does not equate to a partial derivative being zero! Rather, being at equilibrium equates to a partial derviative being defined (i.e., they are only defined at equilibrium). Thus, whether or not a partial derivative equals zero has nothing to do with equilibrium, but rather with the specific nature of the system. And, most typically, partial derivatives are non-zero; it's only for certain specialized cases that a partial would be zero.

Consider, for instance, the total differential for $U=U(T, V,n)$:

$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{\!V,\, n_i}\,\mathrm{d}T+\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}\,\mathrm{d}V +\sum_i \left(\frac{\partial U}{\partial n_i}\right)_{\!T,\, V,\, n_{j\ne i}}\,\mathrm{d}n_i,$$

where the last term allows for chemical reactions and/or matter flow into or out of the system.

Note that state functions ($U, H, A, G, S...$) are only defined at equilibrium, and thus these partial derivatives are likewise only defined at equilibrium. What the above partial derivatives tell us, then, is how the state function (in this case, $U$) changes as we infinitesimally change the system from one equilibrium state to a new equilibrium state by varying $T$, $V$, or one of the $n_i$.

Thus, far from being zero at equilibrium, these partials typically need to be non-zero at equilibrium, unless the state function is independent of the change in the corresponding independent variable, which only applies in special cases.

One such special case is the internal energy of an ideal gas, which depends only on $T$ and $n$. Hence, for a single ideal gas, or a mixture of ideal gases:

$$\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}=0$$

  • $\begingroup$ If I understood you correctly, derivatives of $S_{sys}$ don't have to be zero at equilibrium(although $S_{sys}$ shouldn't be changing)? So there's no condition applicable to the $S_{sys}$ function to signify equilibrium? I will be very grateful if you could help clean up this mess for me. $\endgroup$ Commented Jan 25, 2020 at 7:20
  • $\begingroup$ Ok, I think I had a revelation. dS is zero, but dV could also be zero, so the p.d. might be a non-zero limit. So I think the only query I have now is "How do I apply the $dS_{univ, equil}=0$ to a given system?" $S_{univ}$ has always sounded like an abstract concept to me. Also say I'm not allowed to use conditions on Gibbs energy yet. $\endgroup$ Commented Jan 25, 2020 at 7:26
  • 2
    $\begingroup$ @WilliamR.Ebenezer I rewrote the secnd part of my answer in a way that I hope makes things clearer. As to the connection between equilibrium and entropy maximization, that requires a completely separate answer. I think you will find several questions and answers on this subject if you search this site. And, of course, as an experienced user of this site, you know that if you are unsatisfied with those answers, you can write a new question that specifically addresses your confusion. $\endgroup$
    – theorist
    Commented Jan 25, 2020 at 19:10
  • $\begingroup$ Thanks for your added clarifications. I was not intending to ask for help in comments that could only be met by entire answers, but I agree that it is a separate topic altogether. I'll dig for a bit and ask a new question if required. Thanks again for your time. $\endgroup$ Commented Jan 25, 2020 at 19:22

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

If you write

$$dS=\pd SVT dV + \pd STV dT$$


$$\pd SVU=\pd SVT + \pd STV \pd TVU \tag{1}$$

Next use

$$dT = \pd TUV dU + \pd TVU dV$$

to obtain

$$\pd TUS = \pd TUV + \pd TVU \pd VUS$$


$$\pd TVU = \pd UVS \left[ \pd TUS - \pd TUV \right] \tag{2}$$

Substituting Eq. (2) in Eq. (1) and solving for $\pd SVU$

$$\pd SVU = \pd SVT + \pd STV \pd UVS \left[ \pd TUS - \pd TUV \right] \tag{3}$$

Therefore if $$\pd SVU = 0$$ then

$$\pd UVS = -p = \frac{-\pd SVT \pd TSV}{ \left[ \pd TUS - \pd TUV \right]}$$

While this is not extremely illuminating (and I haven't hit on the reason for the discrepancy with the OPs approach) it does not result in an internal contradiction. $\endgroup$

  • $\begingroup$ Thanks to @orthocresol for the nice partial derivative notation shortcut. Saved a lot of time! $\endgroup$
    – Buck Thorn
    Commented Jan 24, 2020 at 21:06

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