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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
    U=[itex]\frac{NS^2}{V^2}[/itex]

    (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

    vela

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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

    vela

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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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