1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    3c0a0cc895e45267dbaa601d4c29318d.png

    and

    ea5771ed8bd497e2dd089cfcf3e502bf.png

    and that this implies the following:

    32b7fb39a707cb5f2360d5c6d8b3d76e.png

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

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

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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
    $$C_p-C_v=T (\partial_T p)_V (\partial_T V)_p.$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Cp = 0 (by definition)?
  1. Cp and Cv (Replies: 2)

  2. Vector definition (Replies: 2)

  3. Current Definition (Replies: 2)

Loading...