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I Cp = 0 (by definition)?

  1. May 25, 2016 #1
    Hi all, I'm working through a derivation of the general relationship between Cp and Cv and there's one point which is confusing me.

    I understand that




    and that this implies the following:


    but isn't this equal to 0? Shouldn't the two partial derivatives on the right hand side, by the cyclic rule, multiply to -(∂S/∂T)V?

    I know that I'm missing something here but I can't work out what it is.

    Help appreciated!
  2. jcsd
  3. May 25, 2016 #2

    Simon Bridge

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    What you hold constant makes a difference.
  4. May 26, 2016 #3


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    Hm, I'm not sure what you want to derive, but let's start defining the heat capacities and see, where this leads to.

    Start from energy conservation employing the 1st and 2nd fundamental laws
    $$\mathrm{d} U=\mathrm{d} Q-p \mathrm{d} V=T \mathrm{d} S-p \mathrm{d} V.$$
    This implies
    $$C_V:=\left (\frac{\partial Q}{\partial T} \right)_{V}=\left (\frac{\partial U}{\partial T} \right)_V=T \left (\frac{\partial S}{\partial T} \right)_V.$$
    For ##C_p## we need the enthalpy, given by the Legendre transform
    $$H=U+p V ; \Rightarrow \; \mathrm{d} H=T \mathrm{d} S+V \mathrm{d} p,$$
    and thus
    $$C_p=\left (\frac{\partial H}{\partial T} \right )_p=T \left (\frac{\partial S}{\partial T} \right)_p.$$
    Now we can use
    $$\frac{C_p}{T}=\left (\frac{\partial S}{\partial T} \right)_p = \det \left (\frac{\partial(S,p)}{\partial(T,p)} \right) = \det \left (\frac{\partial(S,V)}{\partial(T,V)} \right) \det \left (\frac{\partial(T,V)}{\partial(T,p)} \right) = \left [ (\partial_T S)_V (\partial_V p)_T-(\partial_V S)_T (\partial_T p)_V \right ] (\partial_p V)_T = \frac{C_V}{T} - (\partial_p S)_T (\partial_T p)_V.$$
    Now we can use the Gibb's free energy
    $$G=U-TS+pV \; \Rightarrow \; \mathrm{d} G=-S \mathrm{d} T+V \mathrm{d} p,$$
    to derive
    $$\partial_p \partial_T G=-(\partial_p S)_T=\partial_T \partial_p G=(\partial_T V)_p$$
    to get
    $$C_p-C_v=T (\partial_T p)_V (\partial_T V)_p.$$
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