# Maxwell relations with heat capacity.

1. Feb 28, 2008

### fraggedmemory

1. The problem statement, all variables and given/known data
Use the Maxwell relations and the Euler chain relation to express (ds/dt)p in terms of the heat capacity Cv = (du/dt)v. The expansion coefficient alpha = 1/v (dv/dt)p, and the isothermal compressibility Kt = -1/v (dV/dp)T. Hint. Assume that S= S(p,V)

2. Relevant equations
dQ(rev) = Tds
The maxwell relations
Euler Chain relation

3. The attempt at a solution

Alright, my attempts at this involved trying find common partial derivatives from the information already given. I couldn't find anything. But then looking at the hint I thought that there might be a way to express the change in entropy with respect to pressure and volume. I get this ds = (dU + PdV)/T assuming constant pressure. I am really not sure what I am suppose to do. I especially don't get what the expansion coefficient and thermal compressibility has to do with anything, but that might be because I can't see the big picture with this problem.

A step by step explanation would be greatly appreciated.

2. Mar 1, 2008

### Mapes

I'd rather consider S to be a function of T and V. Then I could differentiate S as

$$dS=\left(\frac{dS}{dT}\right)_V\,dT+\left(\frac{dS}{dV}\right)_T\,dV$$

Then I'd differentiate with respect to T at constant p:

$$\left(\frac{dS}{dT}\right)_p=\left(\frac{dS}{dT}\right)_V+\left(\frac{dS}{dV}\right)_T\left(\frac{dV}{dT}\right)_p$$

You should be able to figure out the rest. This is a useful trick for when you want to compare derivatives taken under different conditions.

3. Mar 3, 2008

### fraggedmemory

I am very thankful for your reply. However, I managed to solve the problem several hours after my post. Your method though is something that I didn't think of, so I do appreciate it.

4. Nov 15, 2009

### tetis

I'm not sure about the sentence "all variables and given/known data".

In fact I succeed in order to obtain a relation between:
$$\left(\frac{\partial s}{\partial T}\right)_p$$
and $$c_v, \alpha, k_T, T, v$$ just following the mapes's hint to consider S as a function of T and V. Where minuscle letter for extensive quantity means: "this quantity is molar", and all transformation are intended to involve a costant number of molecules.

In fact:

$$\left(\frac{\partial s}{\partial T}\right)_p = \left(\frac{\partial s}{\partial T}\right)_v + \left(\frac{\partial s}{\partial v}\right)_T \left(\frac{\partial v}{\partial T}\right)_T$$

The Maxwell's relation following from $$d(-p dv - sdT)=0$$ tell us:
$$\left(\frac{\partial s}{\partial v}\right)_p= \left(\frac{\partial p}{\partial T}\right)_v$$
Now the Euler's chain rule give us the link between the first derivative in second addend, the compressibility and the thermal expansion coefficient. In fact:

$$\left(\frac{\partial p}{\partial T}\right)_v \left(\frac{\partial T}{\partial v}\right)_p\left(\frac{\partial v}{\partial p}\right)_T = -1$$

and:

$$\left(\frac{\partial T}{\partial v}\right)_p = 1/\left(\frac{\partial v}{\partial T}\right)_p$$

so that:

$$\left(\frac{\partial p}{\partial T}\right)_v = \frac{\alpha}{k_T}$$

In order to complete the derivation we need to use the given alternative definition of specific heat:

$$c_v = T \left(\frac{\partial s}{\partial T}\right)_v = \left(\frac{\partial u}{\partial T}\right)_v$$

$$T dS = dU + p dV$$
$$\left(\frac{\partial s}{\partial T}\right)_p = \frac{c_v}{T} + \frac{\alpha^2 v}{k_T}$$