Pressure turning out to be zero?

Question

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

$$dU=TdS-pdV$$

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

$$dU=TdS-PdV$$

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

$$P=0\,\,\,\,\,\,\text{(?!)}$$

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.

Answer 1

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:

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

Answer 2

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

If you write

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

then

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

Next use

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

to obtain

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

or

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

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. $\endgroup$