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Homework Help: Partial differentiation: thermodynamic relations

  1. Aug 5, 2012 #1
    1. The problem statement, all variables and given/known data

    This question is about entropy of magnetic salts. I got up to the point of finding H1, the final applied field.


    3. The attempt at a solution



    But instead of doing integration I used this:

    dS = (∂S/∂H)*dH

    = (M0/4α)(ln 4)2


    I removed the negative sign because they wanted decrease in S.


    I know integration is the sure-fire way to get Sinitial - Sfinal but why is this method wrong? Is it because this method is only an approximation?
     

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  2. jcsd
  3. Aug 5, 2012 #2

    gabbagabbahey

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    I'm not sure exactly what you'e done here. [itex]dH[/itex] is a differential, so it makes no sense to say that [itex] \left( \frac{ \partial S}{ \partial H } \right)_{T} dH = \frac{M_0}{4\alpha}\ln(4)^2[/itex].

    If you Taylor expand [itex]S(H, T)[/itex] around the point H=0, holding T constant. then to first order you get

    [tex]S(H_1, T) \approx \left. \left. \left( \frac{ \partial S}{ \partial H } \right)_{T} \right. \right|_{H=H_1} \cdot (H_1-0)= \frac{M_0}{4\alpha}\ln(4)^2[/tex]

    but that is only a first order approximation.
     
  4. Aug 5, 2012 #3
    What I meant was:

    dS = (∂S/∂H)*dH + (∂S/∂T)dT

    But since dT = 0 since T is kept constant,

    dS = (∂S/∂H)*dH
     
  5. Aug 5, 2012 #4

    gabbagabbahey

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    That's completely correct, but how did you get from that to (M0/4α)(ln 4)2?
     
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