How Can Maxwell Relations Be Applied to This Thermodynamics Problem?

[Lucas Mayr]

Homework Statement

2. The attempt at a solution
I've tried using the relation Cp = T(dS/dT), isolating "T" for T = Cv²(dT/dS) and using the maxwell relations to reduce the derivatives, reaching, T = Cv²/D (dV/dS), but i don't think this is the right way to do solve this problem, i couldn't find a similar example on the chapter either.

 

[TSny]

$T$ is a state variable that can be thought of as a function of $v$ and $P$: $\;T(v,P)$.

Consider $dT$ which will be something times $dv$ plus something times $dP$. Can you express the coefficients of $dv$ and $dP$ in terms of $\alpha$ and $\kappa_T$?

 

[Lucas Mayr]

Ok, so I've tried looking at dT as a function of both P and v and reached dT = (∂T/∂P)_v dP + (∂T/∂v)_P dv
And after reducing the derivatives, dT = K_T/α dP + 1/vα dv , and using the problem's K_T and α.
dT = 1/D dv + Ev²/D dP
dT = 1/D dv + EP_av_a²/(P_bD) dP
T_b = T_a + (v_b - v_a)/D + EP_av_a² Ln(P_b/P_a)/D
which is close but still different from the answer given on the question and i can't find a reason why, what did i miss?

 

[TSny]

Ok, so I've tried looking at dT as a function of both P and v and reached dT = (∂T/∂P)_v dP + (∂T/∂v)_P dv
And after reducing the derivatives, dT = K_T/α dP + 1/vα dv ,...

Check the sign of the first term on the right. Otherwise, that looks good.

T_b = T_a + (v_b - v_a)/D + EP_av_a² Ln(P_b/P_a)/D

Can you express $\ln(P_b/P_a)$ in terms of $v_a$ and $v_b$?