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Homework Help: The temperature change of an ideal gas, Joule Kelvin expansion (const. enthalpy)

  1. May 4, 2012 #1
    1. The problem statement, all variables and given/known data

    This is the last part of the question. So far have been made to derive:

    ## \mu _{\text{JK}}=\left(\frac{\partial T}{\partial P}\right)_H=-\frac{1}{C_P}\left(\frac{\partial H}{\partial P}\right)_T ##

    Then

    ##\left(\frac{\partial H}{\partial P}\right)_T=V - T \left(\frac{\partial V}{\partial T}\right)_P ##

    It says you need to derive an expression for the temperature change as an integral over pressure.


    3. The attempt at a solution

    ##dT=\left(\frac{\partial T}{\partial P}\right)_HdP+\left(\frac{\partial T}{\partial H}\right)_PdH ##

    At constant enthalpy ## dH = 0 ##.

    ## dT=\left(\frac{\partial T}{\partial P}\right)_HdP\text{=}-\frac{1}{C_P}\left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right) ##

    So I think the change in temperature will be:

    ## \text{$\Delta $T} =-\frac{1}{C_P} \int_{P_1}^{P_2} \left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right) \, dP ##

    Then it says derive an expression for the temperature change for an ideal gas.

    ## p V = n R T ##

    ## V = \frac{ n R T}{P} ##

    ## T\left(\frac{\partial V}{\partial T}\right)_P = \frac{ n R T}{p} ##

    So it would seem the integral vanishes, and ## \Delta T = 0##

    I don't think this is right.
     
  2. jcsd
  3. May 6, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Oh, but it is. The Joule-Kelvin coefficient μJK for an ideal gas is identically zero.
     
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