A technical question about the Joule-Thomson Experiment

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It's written in Kubo's textbook:

we obtain
$$(1) \ \ \ \ \ \bigg( \frac{\partial T}{\partial p} \bigg)_H = \bigg[ T (\frac{\partial V}{\partial T})_p - V \bigg] / C_p$$

When the equation of state ##pV = RT(1+Bp)##, eq (1) becomes

$$ (2) \ \ \ \ (\partial T / \partial p)_H = (TdB/dT-B)/C_p$$

I tried getting (2) from (1), but I get something different, I get:
##T\partial V / \partial T - V = TR/p+TRB+RT^2dB/dT-RT/p-RTB = RT^2dB/dT##, how to resolve this conundrum?

Thanks.
 
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Hi,

Your steps are correct.

It seems that you must prove ##RT=1-\frac {B} {T\frac{dB} {dT}}##.
 
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MathematicalPhysicist said:
Seems so.

The equation pV=RT(1+Bp) is false.

The general equation is ##pV=RT+B(T)p+C(T)p^2+D(T)p^3+.##.
If you consider C,D,...=0, then you can end up to the following equation:
## \left( \frac {\partial T} {\partial p} \right)_H=\frac {\left( T\left( \frac {\partial V} {\partial T} \right)_p-V \right)} {C_p}=\frac {\left( T \frac {\partial \left(\frac {RT+Bp} {p} \right)_p} {\partial T}-\frac {RT+Bp} {p} \right)} {C_p} \Rightarrow##
##\left( \frac {\partial T} {\partial p} \right)_H = \frac { \left( \frac {RT} {p} +T \frac {dB} {dT} - \frac {RT}{p} - B\right)} {C_p}=\frac {T \frac {dB} {dT}-B} {C_p}##
 
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DoItForYourself said:
The equation pV=RT(1+Bp) is false.

The general equation is ##pV=RT+B(T)p+C(T)p^2+D(T)p^3+.##.
If you consider C,D,...=0, then you can end up to the following equation:
$$ \left( \frac {\partial T} {\partial p} \right)_H=\frac {T\frac {dB} {dT}-B} {C_p}$$
How do you get the last identity from ##pV = RT+B(T)p##?, I don't see it.
 
MathematicalPhysicist said:
How do you get the last identity from ##pV = RT+B(T)p##?, I don't see it.

I edited the post, so you can see the detailed process that I followed to reach the final result.