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