Is \( c_P \) Pressure Independent if \( \alpha = \frac{1}{T} \)?

[fluidistic]

Homework Statement

I'm stuck on a relatively simple exercise which can be found in Callen's book: Show that if $\alpha = \frac{1}{T}$ then $c_P$ is pressure independent.

Homework Equations

What they ask me to show is: If $\frac{1}{v} \left ( \frac{\partial v }{\partial T} \right ) _P=\frac{1}{T} \Rightarrow \left ( \frac{\partial c_P }{\partial P} \right ) _T=0$.
Where the lower case v is the molar volume and for what will follow the lower case letters are molar quantities.

The Attempt at a Solution

$c_P=T \left ( \frac{\partial s }{\partial T} \right ) _P$. Since both alpha and the specific heat at constant pressure are given in terms of T and P, I will think about v and $c_p$ as functions of T and P.
So $\left ( \frac{\partial c_P }{\partial P} \right ) _T=T \frac{\partial }{\partial P} \left [\left ( \frac{\partial s }{\partial T} \right ) _P \right ] _T=T \frac{\partial }{\partial T} \left [ \left ( \frac{\partial s }{\partial P} \right ) _T \right ] _P$. Now I use a Maxwell relation, namely that $\left ( \frac{\partial s }{\partial P} \right ) _T=- \left ( \frac{\partial v }{\partial T} \right ) _P$.
So I get that $\left ( \frac{\partial c_P }{\partial P} \right ) _T=-T\frac{\partial }{\partial T} \left [ \left ( \frac{\partial v }{\partial T} \right ) _P \right ] _P=-T \frac{\partial }{\partial T} \left [ \left ( \frac{v(T,P)}{T} \right ) \right ] _P$.
So if I can show that $\frac{\partial }{\partial T} \left [ \left ( \frac{v(T,P)}{T} \right ) \right ] _P=0$ then the job would be done. However I'm stuck here, I haven't been able to find a way to show that... Am I going the wrong way? Or simply missing a simple thing?

Edit: Nevermind, problem solved. lol, I was missing a simple thing. I had to perform the partial derivative and I indeed reach that it's worth 0 and hence the proof is completed.

 

[Chestermiller]

The equation for the enthalpy change expressed in terms of dT and dP is given by
$$dH=C_pdT+V(1-\alpha T)dP$$

The form of this equation is
$$dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP$$

Therefore,

$$\frac{\partial H}{\partial T}=C_p$$
$$\frac{\partial H}{\partial P}=V(1-\alpha T)$$
So,$$\frac{\partial^2 H}{\partial T \partial P}=\frac{\partial C_p}{\partial P}=\frac{\partial (V(1-\alpha T))}{\partial T}$$