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Dieterici's equation

  1. Aug 2, 2007 #1
    Dieterici's equation ( an alternative to Van der waal's equation) states that the pressure p, volume v and absolute temperature T of a mass of gas are connected by the equation
    [tex] p=\frac{RT}{(v-b)}\exp{\frac{-a}{vRT}} \\ [/tex],
    where a, b and R are constants. Verify that both [tex] \frac{{\partial p}}{{\partial v}} = 0 \\[/tex]
    and [tex] \frac{{\partial^2 p}}{{\partial^2 v}} =0 \\[/tex], for the critical volume and temperature [tex] v_c [/tex] and [tex] T_c [/tex] respectively,where [tex] v_c = 2b [/tex]and [tex] T_c = \frac{a}{4bR} \\ [/tex]. What is the value of [tex] p_c [/tex] the critical pressure in terms of a,b and e?
    I have a question in solving this: namely is [tex] \frac{d\exp{\frac{-a}{vRT}}}{dv} = \frac{d \exp{\frac{-a}{vRT}}}{d v^{-1}} \frac{d v^{-1}}{dv} \\ [/tex] Because I don't think so: could someone explain what the l.h.s. is equal to. Thanks for the help.
     
    Last edited: Aug 2, 2007
  2. jcsd
  3. Aug 3, 2007 #2

    HallsofIvy

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    [tex]\frac{d exp(\frac{-\alpha}{vRT})}{dv}= exp(\frac{-\alpha}{vRT})\frac{d\frac{-\alpha}{vRT}}{dv}= exp(\frac{-\alpha}{vRT})\frac{\alpha}{v^2RT}[/tex]

    That's because
    [tex]\frac{d \frac{A}{v}}{dv}= \frac{d Av^{-1}}{dv}= -Av^{-2}[/tex]
    for any constant A.
     
  4. Sep 17, 2007 #3
    Ok, I finally got to this step. But how do I take the 2nd derivative of this last result? It's gnarly.
     
  5. Sep 17, 2007 #4
    Ok,

    I think I got the 2nd derivative, and then I set both 1st and 2nd derivative to zero.
    Now I have 3 equations (original, 1st derivative, 2nd derivative), but how many unknowns? I know that V is an unknown, but isn't T also an unknown? I treated it as a constant.

    Do I solve for V in the 1st derivative and plug it into the 2nd derivative?

    This is all very confusing . . .
     
  6. Sep 27, 2007 #5
    [tex] \frac{{\partial p}}{{\partial v}} = \frac{{\partial}}{{\partial v}} (\frac{RT}{v-b}\exp^{\frac{-a}{vRT}}) \\ [/tex]
    which [tex] = \exp^{\frac{-a}{vRT}} \frac{{\partial }}{{\partial v}}(\frac{RT}{v-b}) + \frac{RT}{v-b} \frac{{\partial }}{{ \partial v}}(\exp^{\frac{-a}{vRT}}) \\ [/tex].
    Now use HallsofIvy's equation to evaluate the second term of the product rule expression to get the following:
    [tex]\frac{RT}{(v-b)^2}\exp^{\frac{-a}{vRT}} - \frac{a}{(v-b)v^2} \exp^{\frac{-a}{vRT}} \\ = \exp^{\frac{-a}{vRT}}(\frac{RT}{(v-b)^2} - \frac{a}{(v-b)v^2}) \\ [/tex] [tex]\mbox{ For the critical volume } \ v_c \ \mbox{ and the critical temperature } \ T_c \\ \ \frac{{\partial p}}{{\partial v}}= \exp^{-2}( \frac{a}{4b(b)^2} - \frac{a}{b4b^2})=0 [/tex]
     
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