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Thermodynamics: integrating partial derivatives

  1. Sep 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Given that mechanical equation of state for a paramagnetic substance is
    [tex]m=\left(\frac{DH}{T}\right)[/tex]
    where D is a constant, H is the magnetic field, m is molar magnetization and the molar heat capacity
    [tex]c_{m}[/tex] is constant, find entropy and enthalpy

    2. Relevant equations
    [tex]\left(\frac{\partial S}{\partial T}\right)_{m}=\frac{C_{m}}{T}[/tex]
    [tex]\left(\frac{\partial S}{\partial m}\right)_{T}=-\left(\frac{\partial H}{\partial T}\right)_{m}[/tex]

    3. The attempt at a solution
    Integrating above equations I found that
    [tex]S=-\frac{m^{2}}{2D}+c_{m}\ln{\frac{T}{T_{0}}}.[/tex]
    Or alternatively
    [tex]T=T_{0}\exp{(2DS+m^{2})/(2DC)}.[/tex]
    Now to the entalphy h. By definition
    [tex]\left(\frac{\partial h}{\partial S}\right)_{H}=T=T_{0}\exp{(2DS+m^{2})/(2DC)}[/tex]
    and
    [tex]\left(\frac{\partial h}{\partial H}\right)_{S}=-m=DH/T.[/tex]
    Now, I'm a bit confused here. When integrating these kind of partial derivatives, I guess that one should express the integrand using only those variables that are used to perform the differentiation, i.e. S and H in this particular case. If this is the case, one should probably eliminate m from above equation using the equation of state. But this gives
    [tex]T=T_{0}\exp{(2DS+DH^{2}/(2CT^{2}))/(2DC)}[/tex]
    which makes no sense as T appears on both sides... What's going wrong?
     
    Last edited: Sep 12, 2007
  2. jcsd
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