Partial differentiation: thermodynamic relations

[unscientific]

Homework Statement

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

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 S_initial - S_final but why is this method wrong? Is it because this method is only an approximation?

 

[gabbagabbahey]

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 S_initial - S_final 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.

 

[unscientific]

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

 

[gabbagabbahey]

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?