How Can Maxwell's Relations Be Applied to Thermodynamic Equations?

[Dewgale]

Homework Statement

This is question 2.18 from Bowley and Sanchez, "Introductory Statistical Mechanics" .

Show with the help of Maxwell's Relations that
$$T dS = C_v dT + T (\frac{\partial P}{\partial T})_V dV$$
and
$$TdS = C_p dT - T( \frac{\partial V}{\partial T})_P dP.$$

Then, prove that
$$(\frac{\partial U}{\partial V})_T = T (\frac{\partial P}{\partial T})_V - P$$

Homework Equations

The four common Maxwell's Relations. They can be found here: https://en.wikipedia.org/wiki/Maxwell_relations.

The Attempt at a Solution

The issue for me is the first part. I have managed to prove that the second is true.
My attempt thus far:
Using the fact that $dU = TdS - PdV$, we can rewrite the equation as $$TdS = dU + PdV.$$
We can then multiply and divide dU by dT to get $$TdS = (\frac{\partial U}{\partial T})_V dT + PdV.$$ However, $\frac{\partial U}{\partial T}_V = C_v$, and so we have $$TdS = C_v dT + PdV.$$
This then leaves me with the issue of how to convert the term PdV $\to T (\frac{\partial P}{\partial T})_V dV$. I was considering using the relation $\frac{\partial P}{\partial T} = \frac{\partial S}{\partial V}$, but I'm not sure how to get that to work. I'll keep working at it, but any help is appreciated!

 

[Fightfish]

We can then multiply and divide dU by dT to get $$TdS = (\frac{\partial U}{\partial T})_V dT + PdV.$$

Well, this is certainly not true! Recall that given a multivariable function $f(x,y)$, we have
$$df(x,y) = \left(\frac{\partial f}{\partial x}\right)_{y} dx + \left(\frac{\partial f}{\partial y}\right)_{x} dy$$

 

[Dewgale]

Well, this is certainly not true! Recall that given a multivariable function $f(x,y)$, we have
$$df(x,y) = \left(\frac{\partial f}{\partial x}\right)_{y} dx + \left(\frac{\partial f}{\partial y}\right)_{x} dy$$

I solved it!
Consider the fact that $$ dU = (\frac{\partial U}{\partial S}) dS + (\frac{\partial U}P\partial V}) dV.$$ Then since $dU = TdS - PdV$, we can see that $\frac{\partial U}{\partial S} = T$, or $\frac{\partial S}{\partial U} = \frac{1}{T}$. Now, consider a function $S=S(U,V)$.
$$dS = (\frac{\partial S}{\partial U}) dU_V + (\frac{\partial S}{\partial V}) dV$$
Using one of the Maxwell relations, we know that $\frac{\partial S}{\partial V} = \frac{\partial P}{\partial T}$. We also know that $dU_V = C_V dT$. Therefore, we have
$$dS = \frac{1}{T} C_V dT + (\frac{\partial P}{\partial T}) dV$$
and multiplying through by T gives
$$TdS = C_V dT + T(\frac{\partial P}{\partial T}) dV$$

A similar process using S(U,P) will give the other. Thank you!