Majid

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UsernameMajid
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Questions 6
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 #3
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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)

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

 

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

 

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

.
Jan 15, 2017