some thermodynamics

Thanks Alna, for your answer! Yes, I'm sorry for the mistake, I just didn't notice it, it's actually :h=-T(dv/dT) at constant p.

Here's my answer for this question: 

$H=U+pv$(H is the enthalpy)

$H=Q+W+pv$(U=Q+W)

$dH=dQ+dW+d(pv)$

knowing that $dQ={C}_{v}dT+hdp$

$dH={C}_{v}dT+hdp-pdv+pdv+vdp$

$dH={C}_{v}dT+hdp+vdp$

$dH={C}_{v}dT+(h+v)dp$

H is a state function, that means it's a total  differential:

$\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T}=\begin{pmatrix} \frac{\partial({h+v})} {\partial T} \end{pmatrix}_{p}$

$\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} =\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p}$(1)

On the other hand we have:

$dQ={C}_{v}dT+hdp$$*(\frac{1}{T})$

$dS=\frac{{C}_{v}}{T}dT+\frac{h}{T}dp$

S is a state function, that means it's a total  differential:

$\begin{pmatrix} \frac{\partial({C}_{v}/T)} {\partial p} \end{pmatrix}_{T} = \begin{pmatrix} \frac{\partial({h/T})} {\partial T} \end{pmatrix}_{p}$

$\frac{1}{T}\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} = \frac{ {T} \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T²}}{T²}$

$\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} -\frac{h}{T}$(2)

(1)=(2)

$\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T}$

 

$\begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = - \frac{h}{T}$

 

$h= -T\begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p}$