# Heat Capacities Given Equation of State

1. Jan 13, 2016

### Yoni V

1. The problem statement, all variables and given/known data
Given the equation of state $V(P,T)=V_1\cdot exp(\frac{T}{T_1}-\frac{P}{P_1})$ where $V_1\;,T_1\;,P_1$ are constants:
a. derive an equivalent equation $P(V,T)$;
b. given $C_V=DT^3$ where D is a const, calculate the entropy of the system $s(V,T)$ up to a const;
c. find heat capacity $C_P$.

2. Relevant equations
Definitions of heat and heat capacities.
First law? Enthalpy?

3. The attempt at a solution
Starting with a, we get via a simple calculation $P(V,T)=P_1\cdot (\frac{T}{T_1}-ln(\frac{V}{V_1}))$.
Then, using heat and heat cap. definitions we get
$$ds=dQ/T=CdT/T$$$$\Rightarrow (\frac{\partial S}{\partial T})_V = \frac{C_V}{T} = DT^2$$$$\Rightarrow s = \frac{1}{3}DT^3 + S_0$$
Of this step I'm not entirely sure. It looks okay, but s is not a function of the volume, and I think I need that dependence to work out $C_P$. So this is where I'm stuck.

Thanks!

2. Jan 13, 2016

### Staff: Mentor

What is the partial derivative of entropy with respect to volume at constant temperature?

3. Jan 13, 2016

### Yoni V

Hmm... from Maxwell's relations $$(\frac{\partial S}{\partial V})_T = (\frac{\partial P}{\partial T})_V$$
which equals $\frac{P_1}{T_1}$ from the given equation of state, and then $$s = \frac{P_1}{T_1}V + s_0'$$
The heat cap. is given by $C_P = T(\frac{\partial s}{\partial T})_P$.
I could try rewriting $$(\frac{\partial s}{\partial T})_P = -\frac{(\frac{\partial P}{\partial T})_s}{(\frac{\partial P}{\partial s})_T} = -\frac{(\frac{\partial S}{\partial V})_P}{(\frac{\partial P}{\partial s})_T}$$
But I can't see how these changes in variables get me where I want to go.
Maybe I can somehow relate $s_0$ to $s_0'$, but it doesn't seem relevant...

4. Jan 13, 2016

### Staff: Mentor

What happened to the DT3/3?

5. Jan 13, 2016

### Yoni V

Ok, I think I got it while writing this reply. What I started typing is:
It's part of the constant of integration $s_0'$, or if we neglect a constant that has nothing to do with the temp. and volume, it is exactly $s_0'$.
This I understand, but even writing it explicitly, I'm not sure on how to differentiate the entropy with respect to the temp. while holding P const, as V is free to change to keep the given pressure constant.

But..! We are given V in terms of both T and P. So plugging it to s, we finally get $$s = \frac{1}{3}DT^3 + \frac{P_1}{T_1}( V_1⋅exp(\frac{T}{T_1}−\frac{P}{P_1}))+s_0$$ where $s_0$ is now a true constant that we can entirely ignore and differentiate happily to get $$C_P = T(\frac{\partial s}{\partial T})_P = DT^3 + \frac{P_1V_1T}{T_1^2}exp(\frac{T}{T_1}-\frac{P}{P_1})$$
Is that correct?!
And thank you very much!

6. Jan 13, 2016

### Staff: Mentor

Correct. Nice job.