## partial differentiation: thermodynamic relations

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 Thumbnails

 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study

Recognitions:
Homework Help
 Quote by unscientific 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?
I'm not sure exactly what you'e done here. $dH$ is a differential, so it makes no sense to say that $\left( \frac{ \partial S}{ \partial H } \right)_{T} dH = \frac{M_0}{4\alpha}\ln(4)^2$.

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

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

but that is only a first order approximation.

 Quote by gabbagabbahey I'm not sure exactly what you'e done here. $dH$ is a differential, so it makes no sense to say that $\left( \frac{ \partial S}{ \partial H } \right)_{T} dH = \frac{M_0}{4\alpha}\ln(4)^2$. If you Taylor expand $S(H, T)$ around the point H=0, holding T constant. then to first order you get $$S(H_1, T) \approx \left. \left. \left( \frac{ \partial S}{ \partial H } \right)_{T} \right. \right|_{H=H_1} \cdot (H_1-0)= \frac{M_0}{4\alpha}\ln(4)^2$$ but that is only a first order approximation.
What I meant was:

dS = (∂S/∂H)*dH + (∂S/∂T)dT

But since dT = 0 since T is kept constant,

dS = (∂S/∂H)*dH

Recognitions:
Homework Help

## partial differentiation: thermodynamic relations

 Quote by unscientific What I meant was: dS = (∂S/∂H)*dH + (∂S/∂T)dT But since dT = 0 since T is kept constant, dS = (∂S/∂H)*dH
That's completely correct, but how did you get from that to (M0/4α)(ln 4)2?

 Tags partial differential