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Proof that Heat Capacity is independent of Pressure for a new Equation of State

  1. Oct 12, 2009 #1
    For a gas that obeys the equation of State P(v-b) = RT, where b is a constant, show that Cp is independent of Pressure, i.e., ([tex]\delta[/tex]Cp/[tex]\delta[/tex]p) at constant T is equal to zero



    2. Relevant equations
    Maxwell Relations
    H = U+PV
    dh = TdS + PdV
    dh/dT at constant P is defined as Cp


    3. The attempt at a solution
    Unfortunately I can't simply say that since Cp is defined as existing at a constant pressure state, that it's independent of pressure; would have made the problem much simpler.

    I figure that I'm supposed to prove that ([tex]\delta[/tex](dH/dT)/[tex]\delta[/tex]p) at constant T is equal to zero, but I'm having trouble figuring out which maxwell relation is the best fit.

    d/dP (dH/dT) = d[((TdS)/dT + (VdP)/dT) at constant P]/dP at constant T

    My problem is the dS portion of the equation. It's defined in terms of Cp, among other things, and that doesn't really help me in any way.... I think.

    I have a similar problem when I attempt it with dH = d(U+PV). dU is defined in terms of Cv, or if I break it apart with U = Q+W, I get U = TdS- PdV. Again, not too helpful.

    Is it an issue of which relation I'm using to start? Or can the Cv and Cp actually help me?

    Thanks for the help
     
  2. jcsd
  3. Oct 12, 2009 #2

    Mapes

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    This is a good start. Now how about writing [itex]\partial/\partial P[(\partial S/\partial T)_P]_T[/itex] in a way that lets you use a Maxwell relation?
     
    Last edited: Oct 12, 2009
  4. Oct 12, 2009 #3
    Oh.... Is it possible to switch the order of differentiation here? I suppose this is derived from an exact differential? If that's so, then (dS/dP) at constant T is equal to (dv/dT) at constant P...
     
  5. Oct 12, 2009 #4

    Mapes

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    Yep! Should be no problem now. (Don't forget any sign changes!)
     
  6. Oct 12, 2009 #5
    Right! thanks a bunch
     
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