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A Partial differetial question

  1. Feb 1, 2008 #1
    PV = RT(1+B(T)/V)

    [tex]\beta[/tex] = (1/V)*([tex]\frac{dV}{dT}[/tex]) at constant P

    show [tex]\beta[/tex] =[tex]\frac{1}{T}[/tex]*[tex]\frac{V + B + T\frac{dB}{dT}}{V + 2B}[/tex]

    I got to

    [tex]\beta[/tex] =[tex]\frac{PV}{VRT+PRTB}[/tex]*([tex]\frac{R}{P}[/tex]+[tex]\frac{d}{dT}[/tex][tex]\frac{RTB}{V}[/tex])

    I need help with [tex]\frac{d}{dT}[/tex][tex]\frac{RTB}{V}[/tex])

    I dont know the latex format for pd
     
    Last edited: Feb 1, 2008
  2. jcsd
  3. Feb 3, 2008 #2
    Assuming this is the equation of state (click on any formula in a post to see the latex coding):

    [tex]pV=RT\left(1+\frac{B(T)}{V} \right)[/tex]

    The way to proceed is to calculate the derivative

    [tex]\frac{\partial V}{\partial T}[/tex]

    By taking the derivative of the equation of state. This gives you:

    [tex]p\frac{\partial V}{\partial T}= R\left(1+\frac{B(T)}{V}\right)+ RT\left(\frac{1}{V}\frac{\partial B(T)}{\partial T}- \frac{B(T)}{V^2}\frac{\partial V}{\partial T}\right)[/tex]

    Or:

    [tex]\left[ p+\frac{RTB(T)}{V^2}\right] \frac{\partial V}{\partial T}= R+ \frac{RB(T)}{V}+ \frac{RT}{V} \frac{\partial B(T)}{\partial T}[/tex]

    Now the question was to calculate:

    [tex]\beta=\frac{1}{V}\frac{\partial V}{\partial T}[/tex]

    Thus you have (after rewriting):

    [tex]\beta=\frac{RT\left(B+V+T\frac{\displaystyle \partial B(T)}{\displaystyle \partial T}\right)}{TV \left(pV+\frac{\displaystyle RTB(T)}{\displaystyle V}\right)}[/tex]

    Using now the given equation:

    [tex]\frac{pV}{RT}=1+\frac{B(T)}{V}[/tex]

    and substitute in the denominator, you obtain the result asked for.
     
  4. Feb 3, 2008 #3
    Cheers
     
  5. Feb 3, 2008 #4
    You're welcome. :smile:
     
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