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Thermodynamic Potentials

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    My question is how can one derive U(S(V,T),V) from the relation U(S,V,N).This is the beginning of a given solution which is not explained in more detail.

    2. Relevant equations


    3. The attempt at a solution
    I can understand U(S,V,N) as dU=TdS-pdV+μdN but not the transition: U(S,V,N)-->U(S(V,T),V).
    A full solution of this exercise is given at http://www.thphys.uni-heidelberg.de/~amendola/statphys/problems-students.pdf [Broken] on page 3.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 8, 2015 #2
    Are you referring to eqn 13?? If that's the case here's explanation:

    [itex]
    \frac{\partial U(S,V)}{\partial V}= \frac{\partial U}{\partial S} \frac{\partial S}{\partial V}+\frac{\partial U}{\partial V} \frac{\partial V}{\partial V}
    =\frac{\partial U}{\partial S} \frac{\partial S}{\partial V}+\frac{\partial U}{\partial V}
    [/itex]

    This is a simple chain rule. If you don't know chain rule, review your calculus course, I don't think you'll go anywhere without that.
     
  4. Apr 8, 2015 #3
    Thank you for your reply.I am not referring to equation 13, but before that.
    The exercise starts with U(S(V,T),V). My question is how one can derive that from U(S,V,N)?
     
  5. Apr 8, 2015 #4
    Entropy is a function of T,V or T,P.
    If you still can't recall,
    [itex]S=-\frac{\partial G(T,V)}{\partial T}[/itex] and,
    [itex]S=-\frac{\partial A(T,P)}{\partial T}[/itex]
     
    Last edited: Apr 8, 2015
  6. Apr 8, 2015 #5

    BvU

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    Hello Persephone, welcome to PF :smile: !

    Could it be as simple as N is constant, hence ##U(S, V, N) = U(S, V)## ?
     
  7. Apr 8, 2015 #6
    Thanks BvU!
     
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