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Derive Internal Energy from Thermodynamic Identity

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    For a single molecule, derive the internal energy U = 3/2kBT
    In terms of the partition function Z, F = -kBTlnZ
    Where Z = V(aT)3/2

    2. Relevant equations
    Thermodynamic identity: δF = -SδT - pδV
    p = kBT/V
    S = kB[ln(Z) + 3/2]


    3. The attempt at a solution
    U = F + TS
    δU = δF + δT.S + δS.T
    = -SδT - pδV + δT.S + δS.T
    δU = -pδV + TδS

    (δS/δU)V = 1/T

    However, don't have a variable U in S to differentiate with respect to.
     
  2. jcsd
  3. Jan 10, 2015 #2

    gabbagabbahey

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    Hint: don't forget the basic rules of differentiation, namely [itex]\left(\frac{\partial}{\partial U}\right)_{V}S(Z) = \left(\frac{\partial Z}{\partial U}\right)_{V}\left(\frac{\partial}{\partial Z}\right)_{V}S(Z)[/itex]
     
  4. Jan 11, 2015 #3
    I still don't see how I can solve it given that I don't have a term U to use.
    The only thing I thought was solving p.dV and then substituting using U = -pdV but thing started getting messy.
    Can't go for (dS/dU)V because again, no U.

    Solved it, and I was just massively overcomplicating things. Just needed to use S and F known in U = F + TS
     
    Last edited: Jan 11, 2015
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