A technical question about the Joule-Thomson Experiment

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In summary: 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.You must use the Leib
  • #1
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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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  • #2
Hi,

Your steps are correct.

It seems that you must prove ##RT=1-\frac {B} {T\frac{dB} {dT}}##.
 
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  • #3
Seems so.
 
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  • #4
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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  • #5
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.
 
  • #6
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.
 
  • #7
MathematicalPhysicist said:
It's written in Kubo's textbook:
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.
I confirm your result.
 

1. What is the Joule-Thomson Experiment?

The Joule-Thomson Experiment is an experiment that involves the expansion of a gas through a porous plug or valve. It was first conducted by James Prescott Joule and William Thomson (Lord Kelvin) in the 1850s and is used to study the cooling or heating effects of gases under different conditions.

2. How does the Joule-Thomson Experiment work?

In the Joule-Thomson Experiment, a gas is compressed and then allowed to expand through a small opening in a porous plug or valve. This expansion causes the gas to cool or heat up depending on the pressure and temperature conditions. The temperature change is measured using a thermometer.

3. What is the significance of the Joule-Thomson coefficient?

The Joule-Thomson coefficient is a measure of how much a gas will cool or heat up under specific pressure and temperature conditions during the Joule-Thomson Experiment. It is an important factor in understanding the behavior of gases and is used in various industrial and scientific applications.

4. How is the Joule-Thomson coefficient calculated?

The Joule-Thomson coefficient is calculated by dividing the change in enthalpy (heat content) of the gas by the change in pressure, while keeping the temperature constant. It is denoted by the symbol μ and has units of Kelvin per megapascal (K/MPa).

5. What are the applications of the Joule-Thomson Experiment?

The Joule-Thomson Experiment has many applications in industries such as refrigeration and natural gas processing. It is also used in research to study the behavior of gases at different temperatures and pressures. The experiment has also been used to develop the Joule-Thomson effect, which is the basis for the operation of many refrigeration systems today.

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