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Deriving heat capacity using thermodynamics identity

  1. Dec 14, 2006 #1
    1. The problem statement, all variables and given/known data
    Use the thermodynamic identity to derive the heat capacity formula [tex]C_V=T\frac{\partial{S}}{\partial{T}}_V[/tex]


    2. Relevant equations
    [tex]C_V=\frac{\partial{U}}{\partial{T}}[/tex]
    [tex] T=\frac{\partial{U}}{\partial{S}}[/tex]
    dU=TdS-PdV+[tex]\mu[/tex]dN

    3. The attempt at a solution

    I used [tex]C_V=\frac{\partial{U}}{\partial{S}}\frac{\partial{S}}{\partial{T}}=T\frac{\partial{S}}{\partial{T}}_V=\frac{\partial{U}}{\partial{T}}=C_V[/tex].
    This is the only solution I can think of, but I don't think I used the thermodynamics identity, did I?
     
    Last edited: Dec 14, 2006
  2. jcsd
  3. Dec 14, 2006 #2

    Kurdt

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    There are a distinct lack of thermal physicists on this website and with good reason as it is one of the most boring subjects ever. I think if you know that:

    [tex] dS=(\frac{\partial{S}}{\partial{U}})_{PV}dU[/tex]

    Then you can substitute the following into the equation for dU,

    [tex] dU=(\frac{\partial{U}}{\partial{T}})_VdT[/tex]

    and have a fiddle about, that should set you off on the right lines.
     
    Last edited: Dec 15, 2006
  4. Dec 14, 2006 #3
    This may be a dumb question, but what's the difference between E and U?
     
  5. Dec 15, 2006 #4

    Kurdt

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    There is none. I apologise. I normally write E and I was trying to conform to your notation but obviously forgot half way through I will change it.
     
  6. Dec 15, 2006 #5

    dextercioby

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    The solution provided by the OP is correct.

    Daniel.
     
  7. Dec 15, 2006 #6

    Kurdt

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    I'd have thought if they wanted a derivation from the thermodynamic identity then they'd want you to start from that. However the OP proof might be accepted.
     
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