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Van der Waals gas is not real gas?

  1. Nov 13, 2012 #1
    From van der Waals , (P+a/v^2)(v-b)=RT,
    At critical temperature, I get (∂P/∂V)at constant temperature =0
    and (∂^2P/∂V^2) at constant temperature ,T=0.
    then critical pressure,P = a/(27b^2)--------1
    critical volume,v=3b-----------2
    critical temperature=8a/(27Rb)----------3
    then simultaneous equation 1 and 3,
    I get b=(RT/8P), b=(v/3) ------------------4
    But from the experiment, we get T,P,v and then substitute into the two equation from 4,both b have different values. Why?

    Thank you
     
  2. jcsd
  3. Nov 14, 2012 #2

    mfb

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    (P+a/v^2)(v-b)=RT

    Using your equations:
    P = a/(27b^2)
    v=3b
    T=8a/(27Rb)

    I get
    (a/(27b^2)+a/(9b^2))(2b)=8Ra/(27Rb)
    8/(27b) = 8/(27b)

    Looks fine.

    Are you sure your real gas is a perfect van-der-Waals gas?
     
  4. Nov 14, 2012 #3
    I though all real gas is van der Waals ? Then what do you mean by perfect van der Waals?
     
  5. Nov 14, 2012 #4

    Philip Wood

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    On other threads which you've started, it's been made clear (I think) that the V der W equation is a theoretical equation based on some quite crude assumptions. No actual gas obeys the V der W equation perfectly. [The confusion may be caused because 'real gas' is sometimes used to mean non-ideal gas, even a theoretical non-ideal gas, and not necessarily an actual gas.]
     
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