# Partial derivatives (Maxwell relations) in thermodynamics

1. Sep 26, 2010

### AxiomOfChoice

My professor did this in lecture, and I can't figure out his logic. Can someone fill in the gaps?

He went from:

$$dS = \left( \frac{\partial S}{\partial P} \right)_T dP + \left( \frac{\partial S}{\partial T} \right)_P dT$$

(which I totally understand; it just follows from the fact that $S$ is an exact differential) to the following:

$$\left( \frac{\partial S}{\partial T} \right)_V = \left( \frac{\partial S}{\partial P} \right)_T \left( \frac{\partial P}{\partial T}\right)_V + \left( \frac{\partial S}{\partial T} \right)_P$$

Where the heck does THAT come from? Anyone have any ideas?

2. Sep 26, 2010

### gabbagabbahey

The first equation tells you that $S$ can be written as a function of $T$ and $P$ only... i.e. $S=S(P,T)$....the second equation is just a straight forward application of the chain rule to $S(P,T)$

3. Sep 26, 2010

### AxiomOfChoice

Thanks for your response, but I'm afraid I don't understand. How can a "straightforward application of the chain rule to $S(P,T)$" involve the variable $V$? I would think such an application could only contain the variables $S, T, P$.

Last edited: Sep 26, 2010
4. Sep 27, 2010

### gabbagabbahey

Both $T$ and $P$ may have some dependence on $V$.

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

simply means the partial derivative of $S$, with repect to $T$, taken at constant $V$. Whenever you calculate a partial derivative of a multivariable function, you're always holding at least one variable constant.

5. Sep 27, 2010

### George Jones

Staff Emeritus
Actually, all the other variables are constant.
When it come to the chain rule, often physicists are very sloppy. Start with $S=S\left(P,T\right)$ and suppose that pressure is a function of volume and temperature. Define

$$\tilde{S} \left(V,T\right) = S \left( P\left(V,T\right) , T \right).$$

Now, use the chain rule to calculate $$\partial \tilde{S} / \partial T$$.

Although $S$ and $$\tilde{S}$$ are very related, they actually are different functions.

6. Sep 27, 2010

### jambaugh

You beat me to it. But let me add. The real problem (sloppiness) is that we confuse the variables with the functions e.g. x = x(t). We should really begin with entropy as a function of T, P,N and V.
S = s(T,P,N,V) so
$$dS = \frac{\partial s}{\partial P} dP + \frac{\partial s}{\partial T} dT + \frac{\partial s}{\partial V}dV + \frac{\partial s}{\partial N}dN$$
But we also have a constraint on T,P, N, and V via the physical gas law. This we can express implicitly f(P,V,N,T)=0, or solve for any of the four quantities as a function of the other three.

$$P = p(V,T,N)$$
$$V = v(P,T,N)$$
$$T = t(P,V,N)$$
$$N = n(P,V,T)$$

Differentiate each of these to relate their partials to f's.

We then may impose a condition such as constancy of volume or temperature, etc.

It may be rather lengthy but I think worthwhile to go through the long derivation then practice converting to the "sloppy" form used.

[EDIT]: Actually "we begin with" the gas law and that is incorporated into the entropy function so I got overly inclusive with the variables.

Last edited: Sep 27, 2010
7. Oct 10, 2011

### nj79

Maybe he used Maxwell relations?
Its a difficult to calculate.
But you can find loads of ways to find similar relations :
G(T,P)