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Homework Help: Thermodynamics: determining potentials

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider an imaginary substance which is characterized by thermal energy

    (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. jcsd
  3. Mar 7, 2013 #2


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    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
  4. Mar 7, 2013 #3
    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 [itex]\frac{3Np^2}{T^2}[/itex] and the algebra's giving me trouble for H. I got p= -(dU/dV)= [itex]\frac{2NS^2}{V^3}[/itex], then [itex]V=(\frac{2NS^2}{p})^{\frac{1}{3}}[/itex]
    Last edited: Mar 7, 2013
  5. Mar 7, 2013 #4


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    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.
  6. Mar 7, 2013 #5
    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. :)
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