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Joule Coefficient at Constant U

  1. Jun 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Using basic thermodynamic relations, show that the Joule Coefficient is given by:
    2. Relevant equations
    [tex] {\left(\partial T \over \partial V \right)_U}=-{T^2 \over C _v}{\partial \over \partial T} \left(P \over T \right)_V [/tex]
    3. The attempt at a solution
    I started with the cyclic thermodynamic relation:
    [tex] {\left( \partial T \over \partial V \right)_U \left( \partial V \over \partial U \right)_T \left (\partial U \over \partial T \right)_V}=-1 [/tex]
    Rearranging the equation:
    [tex] {\left( \partial T \over \partial V \right)_U}={-\left( \partial U \over \partial V \right)_T \left( \partial T \over \partial U \right)_V} [/tex]
    Knowing that: [itex] C_V= \left( \partial U \over \partial T \right)_V [/itex] → [itex] {1 \over C_V}= \left( \partial T \over \partial U \right)_V [/itex] I get:
    [tex] {\left( \partial T \over \partial V \right)_U}={-{1 \over C_V} \left( \partial U \over \partial V \right)_T} [/tex]
    And from there I don't know how to continue.
    I hope someone here can help me, thanks in advance.
     
  2. jcsd
  3. Jun 11, 2014 #2
    You need to get (∂U/∂V)T. Start out with

    dU = TdS-PdV

    so

    [tex]dU=T\left(\left(\frac{∂S}{∂T}\right)_VdT+\left(\frac{∂S}{∂V}\right)_TdV\right)-PdV=C_vdT+\left(T\left(\frac{∂S}{∂V}\right)_T-P\right)dV[/tex]
    So,
    [tex]\left(\frac{∂U}{∂V}\right)_T=\left(T\left(\frac{∂S}{∂V}\right)_T-P\right)dV[/tex]

    Now, you need to make use of an appropriate Maxwell relation to get the partial of S with respect to V at constant T.

    Chet
     
  4. Jun 12, 2014 #3
    Thanks for your answer Chet, I used [tex] {\left( \partial S \over \partial V \right)_T}= {\left( \partial P \over \partial T \right)_V} [/tex]

    But now I'm stuck here: [itex] {\left(\partial T \over \partial V \right)_U}=-{1 \over C_V}{\left[ T {\left( \partial P \over \partial T \right)_V}-P \right]dV} [/itex]

    I suppose that somehow [tex] {\left[ T {\left( \partial P \over \partial T \right)_V}-P \right]dV}={T^2{\partial \over \partial T}\left( P \over T \right)} [/tex] But I don't know how to get there.
     
  5. Jun 12, 2014 #4

    DrClaude

    User Avatar

    Staff: Mentor

    Note that there is a typo in Chet's post, such that there is no ##dV## here.

    Correcting, you have to show that
    [tex] \left[ T {\left( \partial P \over \partial T \right)_V}-P \right] = T^2{\partial \over \partial T}\left( P \over T \right)_V [/tex]
    Just apply the product rule on the right-hand-side.
     
  6. Jun 12, 2014 #5
    Thanks for spotting that typo Dr. C.

    Chet
     
  7. Jun 12, 2014 #6
    Thanks a lot, DrClaude and Chet your answers were of great help.
     
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