# Proving a thermodynamic relationship

1. Nov 30, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
Prove that $TdS = C_vdT + \alpha T / \kappa dV$

2. Relevant equations
$T dS = dU - pdV$
$\alpha = \frac{1}{v}\left(\frac{\partial v}{\partial T}\right )_P$
$\kappa = -\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_T$

3. The attempt at a solution

The $C_vdT$ part is quite easy since for a constant volume process $dU = C_vdT$ but I can't seem to figure out how to get the second part of the expression. After multiplying by forms of 1 I end up with $$-pdV = \frac{\alpha\left(\frac{\partial v}{\partial P}\right)_T}{\kappa \left(\frac{\partial v}{\partial T}\right)_P}PdV$$, now using the cyclical rule here doesn't seem logical since that would introduce a negative so it seems like I need to replace the pressure P with something although I'm not sure what relation I can use to do that.

2. Dec 3, 2016

### Mapes

You've probably figured it out over the past few days, but for one thing, you've got a sign problem: $T\,dS=dU+p\,dV$ because $p$ is compressive stress.

3. Dec 4, 2016

### Staff: Mentor

Your mistake is that dU is not equal to $C_vdT$. That is only correct for an ideal gas. In general, $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV=\frac{C_vdT}{T}+\left(\frac{\partial S}{\partial V}\right)_TdV$$
From one of the Maxwell relationships, $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$Therefore, $$dS=\frac{C_vdT}{T}+\left(\frac{\partial P}{\partial T}\right)_VdV$$So, $$TdS=C_vdT+T\left(\frac{\partial P}{\partial T}\right)_VdV$$

4. Dec 4, 2016

### Mapes

At constant volume, $dU=C_V\,dT$ for all materials, as Potatochip911 noted.

5. Dec 4, 2016

### Staff: Mentor

But the problem statement does not say anything about constant volume. In fact, it explicitly indicates that the volume is not considered constant.

Last edited: Dec 4, 2016