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A A technical question about the Joule-Thomson Experiment

  1. Nov 20, 2017 #1

    MathematicalPhysicist

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    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.
     
  2. jcsd
  3. Nov 20, 2017 #2

    DoItForYourself

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    Hi,

    Your steps are correct.

    It seems that you must prove ##RT=1-\frac {B} {T\frac{dB} {dT}}##.
     
  4. Nov 20, 2017 #3

    MathematicalPhysicist

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    Seems so.
     
  5. Nov 20, 2017 #4

    DoItForYourself

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    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}##
     
    Last edited: Nov 20, 2017
  6. Nov 20, 2017 #5

    MathematicalPhysicist

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    How do you get the last identity from ##pV = RT+B(T)p##?, I don't see it.
     
  7. Nov 20, 2017 #6

    DoItForYourself

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    I edited the post, so you can see the detailed process that I followed to reach the final result.
     
  8. Nov 20, 2017 #7
    I confirm your result.
     
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