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Difference between isobar and isochor heat capacity

  1. Nov 21, 2011 #1

    I want to show:
    [itex] C_p - C_v = -T \big( \frac {\partial V}{\partial p} \big)_{T,n} \big( \frac {\partial p}{\partial T} \big)_{V,n}^2 [/itex]

    I started by doing this:
    [itex] C_p - C_v= \big( \frac {\partial H}{\partial T} \big)_{p,N} - \big( \frac {\partial U}{\partial T} \big)_{V,n} [/itex]

    Applying the definitions of enthalpy and energy:
    [itex] dH = TdS + V dp + \mu dn [/itex]
    [itex]dU = TdS - p dV + \mu dn[/itex]

    I can rewrite the equation like this:
    [itex]= V \big(\frac {\partial p}{\partial T} \big) + p \big( \frac {\partial V}{\partial T} \big) [/itex]
    (while TdS and µdn terms cancel out each other)

    Now I do not know how to continue. Can anyone help :)
  2. jcsd
  3. Nov 21, 2011 #2
    Start with:

    T dS = dU + p dV (Combined 1st and 2nd laws) (1)

    Write U as a function of T and V:

    T dS = [(∂U/∂T)V dT + (∂U/∂V)T dV] + p dV

    = [(∂U/∂T)V + p ] dV + (∂U/∂T)V dT

    The last term is CV dT, so:

    T dS = [(∂U/∂T)V + p ] dV + CV dT

    Now do an isobaric process, and divide by dT to get:

    CP - CV = [(∂U/∂T)V + p ] dV (∂V/∂T)P (2)

    which is a standard relation derived, for example, in Sears

    Now, starting again with the combined 1st and 2nd law (equ (1)):

    dU + p dV = T dS

    Divide by dV keeping T constant :

    ( ∂U/∂V)T + p = T (∂S/∂V)T (3)

    Substituting this for the term in square brackets in equ (2)

    CP - CV = T (∂S/∂V)T (∂V/∂T)P (4)

    Use one of Maxwell's equations (Ref: Sears) to substitute for the first partial derivative on the right of equ (4):

    CP - CV = T (∂P/∂T)V (∂V/∂T)P (5)

    Now use the standard relation for partial derivatives (also ref: Sears)

    (∂V/∂T)P (∂T/∂P)V (∂P/∂V)T = -1

    to substitute for the second partial derivative on the right of equ (5) and get what you need.

    Please let me know if you need clarification
  4. Nov 22, 2011 #3
    Thank you for your answer :)

    I have got a question on that:
    What does it exactly mean when you say 'Now do an isobaric process'. I don't see how the C_p drops in.
  5. Nov 22, 2011 #4
    So let us start with the equation before that line:

    T dS = [(∂U/∂T)V + p] dV + CV dT

    In an isobaric process, each of the changes dS, dV and dT will have certain values. So when we divide by that value of dT, we get:

    T (∂S/∂T)p = [(∂U/∂T)V + p] (∂V/∂T)p + CV

    The constant p on the partial derivatives signifying the isobaric process.

    The left hand side is precisely Cp. Now take the CV to the left and you get

    Cp - CV = [(∂U/∂T)V + p] (∂V/∂T)p

    This is equation (2) in what I wrote earlier. Incidentally, there was a typo in the earlier equation (2). There is a dV extra which should be erased.

    Let me know if you need any more help
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