Thermodynamics: determining potentials

1. Mar 6, 2013

Chiborino

1. The problem statement, all variables and given/known data
Consider an imaginary substance which is characterized by thermal energy
U=$\frac{NS^2}{V^2}$

(a) Determine the Helmholtz free energy F(T, V).
(b) Determine the Gibbs free energy G(T, p).
(c) Determine the enthalpy H(S, p)

2. Relevant equations
F=U-TS (maybe dF = dU - sdT = -pdV - sdT?)
G=U-TS+pV = F+pV (dG = -sdT + Vdp
H=U+pV (dH = TdS + Vdp)

3. The attempt at a solution
I'm really lost when it comes to this. My professor hasn't done any examples with this sort of problem, the book doesn't have anything like this in it, and I can't find a problem similar online anywhere. So my best guess is to just throw the N(S/V)^2 into the potential formula for each problem and circle it. But I get the feeling that F(T,V) =N(S/V)^2 -TS isn't a valid answer. A quick explanation of how to proceed would be extremely helpful.

2. Mar 7, 2013

vela

Staff Emeritus
The energy U(S,V) is a function of S and V while the Helmholtz free energy F(T,V) is a function of T and V. To go from U to F, you're replacing the variable S with its conjugate, T. This is what's called a Legendre transform.

If you differentiate U, you get
$$dU = \frac{\partial U}{\partial S} dS + \frac{\partial U}{\partial V} dV.$$ Comparing this to the first law of thermodynamics, $dU = T\,dS-p\,dV$, you can see that $T = \frac{\partial U}{\partial S}$ and $p = -\frac{\partial U}{\partial V}$. The first one is the one you want because you want to replace S with T.

For part a, start by differentiating U with respect to S to find T in terms of S and V. Invert that to find S in terms of T and V. You should get $S = \frac{TV^2}{2N}$. Then use this to eliminate S from the expression
$$F = U - TS = \frac{NS^2}{V^2} - TS.$$ In the end, you should have $F(V,T)=-\frac{V^2T^2}{4N}$.

To find Gibbs free energy G, note that you're starting with F and replacing V by p. Similarly, to find the enthalpy H, you're starting with G and replacing T by S. You follow the same basic procedure each time.

Last edited: Mar 7, 2013
3. Mar 7, 2013

Chiborino

Thank you!
That's such an easy problem, I don't know why professor blew over the concept in class.
I don't know if you feel like checking me over, but for G I got $\frac{3Np^2}{T^2}$ and the algebra's giving me trouble for H. I got p= -(dU/dV)= $\frac{2NS^2}{V^3}$, then $V=(\frac{2NS^2}{p})^{\frac{1}{3}}$

Last edited: Mar 7, 2013
4. Mar 7, 2013

vela

Staff Emeritus
I got $G = \frac{Np^2}{T^2}$. I think you made a sign error somewhere.

I got $H = 3\left(\frac{N S^2 p^2}{4}\right)^{1/3}$. Your expressions for p and V match what I found.

5. Mar 7, 2013

Chiborino

I got two seperate cube root expressions for H, but I see how I can factor out a 2 to combine them into what you have. And yes, I spotted the sign error. Thank you again for the help. :)