# Homework Help: Free energy Helmholtz using only the equation of state

Tags:
1. Jul 15, 2017

### Dario SLC

1. The problem statement, all variables and given/known data
This is a state ecuation of a gas:
PV=AT+B/V, where A and B there are constants.
First: Demonstrate that $c_V$ depends only of T
Second: Find U(T,V) and S(T,V)

2. Relevant equations
$\left(\frac{\partial U}{\partial S}\right)_V=T\text{ (1)}$
$\left(\frac{\partial U}{\partial V}\right)_S=-p\text{ (2)}$
$F=U-TS\text{ (3)}$
$\left(\frac{\partial F}{\partial V}\right)_T=-p\text{ (4)}$
$\left(\frac{\partial F}{\partial T}\right)_V=-S\text{ (5)}$

3. The attempt at a solution
For second item, I think use first (2) and then integrate respect to V using the state ecuation, I got:
$U(S,V)=-AT\ln V+B/V+u(S)$
when u(S) is a constant integration, then I use the (1) for find u(S) and the complete expression for U(S,V):
$U(S,V)=-AT\ln V+B/V+TS+U_0$
now using (4) I would like to obtain the free energy of Helmholtz:

F(T,V)=-ATlnV+B/V+f(T)

when f(T) it is a new constant integration, then I use the (5) I got:
-S(T,V)=-AlnV+f'(T) then S(T,V)=AlnV-f'(T)
if F=U(S,V)-TS(T,V) and derivate respect to T, this it is the same
-AlnV=-AlnV+f'(T)
therefore f(T)=c with c constant, then
S(T,V)=AlnV-c
and
U(T,V)=-ATlnV+B/V-cT+U0

Well, I doubt that the entropy don't depends of T, I haven't see the error, it should be something like to
$S(T,V)=S_0+c_V\ln\frac{T}{T_0}+R\ln\frac{V}{V_0}$ for a van der Waals gases
(Sears, Thermodynamics, chapter 9)

2. Jul 15, 2017

### Staff: Mentor

What is the equation for dU in terms of dT, Cv, P, and dV?

3. Jul 15, 2017

### Dario SLC

Hi, do you mean at this?
$$dU=TdS-pdV$$ or another form
$$dU=c_vdT-pdV$$
but I can't the real form for $c_v$, except of course $$c_V=T\left(\frac{dS}{dT}\right)_V$$ but I haven't if the expression for S(T,V) found is correct.

4. Jul 15, 2017

### Staff: Mentor

This equation is incorrect. It should read $$dU=c_vdT+\left(T\frac{\partial S}{\partial V}-P\right)dV$$From a Maxwell relation involving dF=-SdT-PdV, what is $\partial S/\partial V$ at constant T in terms of P, V, and T?

Last edited: Jul 16, 2017
5. Jul 16, 2017

### Dario SLC

Yeah! thanks a lot, reviewing a lot of writings, I see the same but I do not complete de entropy. My doubt now is the $c_V$ because except that the problem say it is constant, good! but if $c_V=c_V(T)$ there will be a integral in the expression for energy and entropy.

For the first issue, I think this:
for the first principle $dU=TdS-pdV$ and $dU$ and $dS$ there are exact differential, rewriting:
$$\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=T\left(\frac{\partial S}{\partial T}\right)_VdT+T\left(\frac{\partial S}{\partial V}\right)_TdV-pdV$$
gathering differentials for $dT$
$$\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V$$
when the term $T\left(\frac{\partial S}{\partial T}\right)_V$ is the specific heat at V constant $c_v$, therefore only depend of T

This is correct?

6. Jul 16, 2017

### Staff: Mentor

Well, your math is correct. But you still haven't proven that Cv depends only on T for this particular equation of state. Do you know how to determine $\left(\frac{\partial S}{\partial V}\right)_T$ in terms of P, V, and T? Do you know how to show that
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$

7. Jul 16, 2017

### Dario SLC

I think I understand what you say:
I must prove that:
$$\frac{\partial c_v}{\partial V}=0$$
how
$$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial S}{\partial T}\right)_V\right]_T$$
and how the Maxwell relation say:
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
then
$$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial P}{\partial T}\right)_V\right]_T$$
using the equation of state:
$$\left(\frac{\partial P}{\partial T}\right)_V=A/V$$
therefore deriving again respect to T it is null, then $c_V$ it does not depend on $V$.

8. Jul 16, 2017

### Staff: Mentor

Correct.

9. Jul 16, 2017

### Staff: Mentor

Actually, you have $$dU=C_vdT-\frac{B}{V^2}dV$$. This also shows that Cv is dependent only on temperature.

Last edited: Jul 16, 2017