1. Not finding help here? Sign up for a free 30min 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!

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?
     

    Attached Files:

  2. jcsd
  3. Aug 5, 2012 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    That's completely correct, but how did you get from that to (M0/4α)(ln 4)2?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial differentiation: thermodynamic relations
  1. Partial Differentiation (Replies: 11)

  2. Partial differentiation (Replies: 10)

Loading...