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Thermodynamics: DT/DV at constant entropy? (last maxwell relation I haven't figured)

  1. Jan 22, 2013 #1
    So, until now I know:
    (DV/DS)p=(DT/Dp)s=a*T/cp*(rho) (enthalpy)
    (Dp/DT)v=(DS/DV)t=-a/k (helmoltz)
    (DS/Dp)t=-(DV/DT)p=-Va (gibbs)

    a=expansion coefficient
    k=isothermal compression coefficent
    cp=heat capacity at constante pressure

    I want to deduce DT/DV at constant entropy=(DT/DV)s. BUT HOW?
    Let me try to write S(T,V), then,
    dS=Cv/T*dT-a/k*dV
    putting S=0, i get,
    a/k*dV=Cv/T*dT <=> (DT/DV)s=a*T/Cv*k

    am I right?
     
  2. jcsd
  3. Jan 22, 2013 #2
    Re: Thermodynamics: DT/DV at constant entropy? (last maxwell relation I haven't figur

    Is this the one you want?

    [tex]\begin{array}{l}
    T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_V} \\
    {\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} = \left[ {\frac{\partial }{{\partial V}}{{\left( {\frac{{\partial U}}{{\partial S}}} \right)}_V}} \right] = \frac{{{\partial ^2}U}}{{\partial V\partial S}} \\
    \end{array}[/tex]

    and

    [tex]\begin{array}{l}
    P = - {\left( {\frac{{\partial U}}{{\partial V}}} \right)_S} \\
    {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V} = - \left[ {\frac{\partial }{{\partial S}}{{\left( {\frac{{\partial U}}{{\partial S}}} \right)}_V}} \right] = - \frac{{{\partial ^2}U}}{{\partial S\partial V}} \\
    \end{array}[/tex]

    Therefore

    [tex]{\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} = - {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V}[/tex]
     
  4. Jan 22, 2013 #3
    Re: Thermodynamics: DT/DV at constant entropy? (last maxwell relation I haven't figur

    Hey, thanks for worring so much, but until there I knew...
    I want to evaluate that derivative further and write in terms of a,k,Cv,Cp,T,p,... as I did
    (∂T/∂p)s=a*T/cp*(rho)
     
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