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Homework Help: Berthelot Equation

  1. Aug 2, 2007 #1
    1. The problem statement, all variabl
    The attempt at a solution

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Aug 2, 2007
  2. jcsd
  3. Aug 2, 2007 #2

    chemisttree

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    The answer is obviously 'c'.
     
  4. Aug 2, 2007 #3
    I'm sorry. I had some problems using the Latex editor. It's my first time using it. I did post earlier about proton affinity though. I haven't gotten a heads up from anyone about it but I think my reasoning is correct. I really wish that there was something like PF when I was in school.
    Oh, and the correct answer to the above problem is c.
     
  5. Aug 2, 2007 #4
    What I was trying to ask was how does the author Klotz get from a to b?
    a
    (P + a/Vm^2)( Vm - b) = RT

    b
    P Vm = RT [1 + 9/128 P/Pc * Tc / T (1 - 6 T^2c / T^2)
     
  6. Aug 7, 2007 #5
    See below please
     
    Last edited: Aug 7, 2007
  7. Aug 7, 2007 #6
    (P + a/V2m)(Vm - b) = RT

    P = RT/( Vm - b) - a/Vm^2

    ∂Vm/∂ Pc = 0 = -RTc/( Vmc - b)^2 + 2a/TcVmc^3
    ∂Vm2/∂2 Pc = 0 = 2RTc/( Vmc - b)^3 - 6a/TcVmc^4


    b = Vmc /3 ; 2a = 9RTc/4Vmc^2 * TcVmc^3

    a = 9/8 RTc^2 Vmc = 3PcVmc^2 Tc


    PcTcVmc^2 = 3/2R Tc^2 Vmc - a

    4 PcTcVmc^2 = 3/2R Tc^2 Vmc

    3 R Tc/8Pc = Vmc = 3b
    b = R Tc/8Pc
    4/3 a = 3/2 RTc^2 Vmc
    a = 3/4 * 3/2 * 3/8 R2 Tc^3 / Pc
    a = 27/64 R^2 Tc^3 / Pc ; b = R Tc/8Pc ; R = 8Pc Vmc/3 Tc
    PVmc = RT [1+1/8 PTc /PcT - 9/8 Tc Vmc /T^2 Vm + 3/8 Tc^2 Vmc^2 /T^2 Vm^2 ]

    How do I get Vm and Vmc in terms of Pc & Tc? Solve a cubic in terms of Vm?

    P Vm = RT[1 + 9/128 P/Pc * Tc/T (1 - 6 Tc^2 /T^2)]
     
  8. Aug 7, 2007 #7

    Gokul43201

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    (a) is the van der Waals equation and (b) is the Berthelot equation. They arise from two different models, and I don't believe they are functionally identical (I could be wrong here, I haven't tried to check). Is there a reason that you believe one is derivable from the other?
     
  9. Aug 8, 2007 #8
    Hello sir. Thank you for your time and attention on these matters. Weell, a should actually be

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

    which is a form of the Berthelot equation. The development and derivations for a, b, R and the form of the eq. which contains Vmc and Vm are the same as the original post.

    P = RT/(Vm - b) - a/TVm^2

    ∂Vm/∂ Pc = 0 = -RTc/( Vmc - b)^2 + 2a/TcVmc^3
    ∂Vm2/∂2 Pc = 0 = 2RTc/( Vmc - b)^3 - 6a/TcVmc^4


    b = Vmc /3 ;
    2a = 9RTc/4Vmc^2 * TcVmc^3

    a = 9/8 RTc^2 Vmc = 3PcVmc^2 Tc

    The constant a differs from the van der Waals constant value by a factor of Tc.

    PcTcVmc^2 = 3/2R Tc^2 Vmc - a

    4 PcTcVmc^2 = 3/2R Tc^2 Vmc
    3 R Tc/8Pc = Vmc = 3b

    b = R Tc/8Pc
    which is the same as the van der Waals value.

    4/3 a = 3/2 RTc^2 Vmc
    a = 3/4 * 3/2 * 3/8 R2 Tc^3 / Pc

    a = 27/64 R^2 Tc^3 / Pc ; b = R Tc/8Pc ; R = 8Pc Vmc/3 Tc

    R and b are the same as for the van der Waals eq. The constant a varies only by a factor of Tc.


    PVmc = RT [1+1/8 PTc /PcT - 9/8 Tc Vmc /T^2 Vm + 3/8 Tc^2 Vmc^2 /T^2 Vm^2 ]

    So how do I get Vm and Vmc in terms of Pc & Tc? Solve a cubic in terms of Vm?

    P Vm = RT[1 + 9/128 P/Pc * Tc/T (1 - 6 Tc^2 /T^2)]
     
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