+101 Prove these relations:
\((\frac{\partial T}{\partial p})_{v}.(\frac{\partial v}{\partial T})_{p}=-(\frac{\partial v}{\partial p})_{T}\)
\(h=-T(\frac{\partial v}{\partial p})_{p}\)
h=-T[dv/dp]p
Solve the separable equation h = -p T(( dv(p))/( dp)):
Solve for ( dv(p))/( dp):
( dv(p))/( dp) = T^(-1) (-h/p)
Integrate both sides with respect to p:
Answer: |v(p) = integral T^(-1) (-h/p) dp = integral T^(-1) (-h/p) dp + c_1, where c_1 is an arbitrary constant.
+33661 Here's the first part:
The second part isn't correct: dv/dp at constant p would be zero!
+101 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)
\(\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T}\)
\( \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = - \frac{h}{T}\)
\(h= -T\begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} \)
