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Thermo heat capacity proof: cp - cv

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that for a general (but simple) substance,

    c[itex]_{p}[/itex]-c[itex]_{v}[/itex]=-T([itex]\frac{∂v*}{∂T}[/itex])[itex]_{p}[/itex]([itex]\frac{∂p}{∂v*}[/itex])[itex]_{T}[/itex]

    where

    v* is the specific volume
    p is the pressure
    c[itex]_{p}[/itex] is the heat capacity when p is const
    c[itex]_{v}[/itex] is the heat capacity when v is const
    Q is heat
    T is temperature in K


    2. Relevant equations

    Standard Maxwell relations. Suppose to use jacobian to manipulate

    c[itex]_{p}[/itex] = ([itex]\frac{∂Q}{∂T}[/itex])[itex]_{p}[/itex]
    c[itex]_{v}[/itex] = ([itex]\frac{∂Q}{∂T}[/itex])[itex]_{v}[/itex]

    3. The attempt at a solution

    I started by inserting the above equations for specific heat in terms of heat (Q). Then I plugged the partial derivatives into the equation dQ = dE + p dv* which left me with

    c[itex]_{p}[/itex]-c[itex]_{v*}[/itex]=p([itex]\frac{∂v*}{∂T}[/itex])[itex]_{p}[/itex]+([itex]\frac{∂E}{∂T}[/itex])[itex]_{p}[/itex]-([itex]\frac{∂E}{∂T}[/itex])[itex]_{v*}[/itex]

    Next I rewrote the partial derivatives using dE = -TdS-pdV, but clearly at this point I'm just going around in circles. I think I'm missing some simple step to combine the partials, but I'm not sure what it is. Your help would be appreciated.
     
    Last edited by a moderator: Apr 20, 2012
  2. jcsd
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