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Stuck on a thermodynamics problem

  1. May 13, 2013 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    I'm stuck on a relatively simple exercise which can be found in Callen's book: Show that if ##\alpha = \frac{1}{T}## then ##c_P## is pressure independent.


    2. Relevant equations
    What they ask me to show is: If ##\frac{1}{v} \left ( \frac{\partial v }{\partial T} \right ) _P=\frac{1}{T} \Rightarrow \left ( \frac{\partial c_P }{\partial P} \right ) _T=0##.
    Where the lower case v is the molar volume and for what will follow the lower case letters are molar quantities.

    3. The attempt at a solution
    ##c_P=T \left ( \frac{\partial s }{\partial T} \right ) _P##. Since both alpha and the specific heat at constant pressure are given in terms of T and P, I will think about v and ##c_p## as functions of T and P.
    So ##\left ( \frac{\partial c_P }{\partial P} \right ) _T=T \frac{\partial }{\partial P} \left [\left ( \frac{\partial s }{\partial T} \right ) _P \right ] _T=T \frac{\partial }{\partial T} \left [ \left ( \frac{\partial s }{\partial P} \right ) _T \right ] _P##. Now I use a Maxwell relation, namely that ##\left ( \frac{\partial s }{\partial P} \right ) _T=- \left ( \frac{\partial v }{\partial T} \right ) _P##.
    So I get that ##\left ( \frac{\partial c_P }{\partial P} \right ) _T=-T\frac{\partial }{\partial T} \left [ \left ( \frac{\partial v }{\partial T} \right ) _P \right ] _P=-T \frac{\partial }{\partial T} \left [ \left ( \frac{v(T,P)}{T} \right ) \right ] _P##.
    So if I can show that ##\frac{\partial }{\partial T} \left [ \left ( \frac{v(T,P)}{T} \right ) \right ] _P=0## then the job would be done. However I'm stuck here, I haven't been able to find a way to show that... Am I going the wrong way? Or simply missing a simple thing?

    Edit: Nevermind, problem solved. lol, I was missing a simple thing. I had to perform the partial derivative and I indeed reach that it's worth 0 and hence the proof is completed.
     
    Last edited: May 13, 2013
  2. jcsd
  3. May 13, 2013 #2
    The equation for the enthalpy change expressed in terms of dT and dP is given by
    [tex]dH=C_pdT+V(1-\alpha T)dP[/tex]

    The form of this equation is
    [tex]dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP[/tex]

    Therefore,

    [tex]\frac{\partial H}{\partial T}=C_p[/tex]
    [tex]\frac{\partial H}{\partial P}=V(1-\alpha T)[/tex]
    So,[tex]\frac{\partial^2 H}{\partial T \partial P}=\frac{\partial C_p}{\partial P}=\frac{\partial (V(1-\alpha T))}{\partial T}[/tex]
     
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