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Joule-Thomson Expansion of a van der Waals gas

  1. Mar 9, 2006 #1

    eep

    User Avatar

    Hi,
    Consider a pipe with thermally insulated walls. A thermally insulating porous plug in the pipe provides a constriction to the flow of gas. We model this as a sudden jump in pressure. A continous stream of gas flows from left to right, with the pressure [itex]p_1[/itex] upstream being larger than the pressure [itex]p_2[/itex] downstream. The gas is in thermal equilibrium on each of the two sides.

    For a van der Waals gas, what is the minimal starting temperature for cooling to occur in a Joule-Thomson expansion?

    I was able to derive that the enthalphy for a van Der Waals gas is

    [tex]
    H = \frac{5}{2}N\tau + N^2(\frac{b\tau}{V - Nb} - \frac{2a}{V})
    [/tex]

    and I know that enthalphy is conserved in the process.

    I set the enthalphy on both sides equal to one another, and then set [itex]\tau_2[/itex] equal to zero since that is the lowest possible temperature. Solving the equation for [itex]\tau_1[/itex] left me with

    [tex]
    \tau_1 = \frac{2Na(\frac{V_1 + V_2}{V_1V_2})}{\frac{5}{2} + \frac{Nb}{V_1 - Nb}}
    [/tex]

    I assumed that N would be the same on both sides of the barrier. I don't like this result, however, as it depends on [itex]V_1[/itex] and [itex]V_2[/itex]. Is this correct or should I be approaching this differently?
     
  2. jcsd
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