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Finding the average height, z, of molecules in the box

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Let a box of height h be filled with classical ideal gas molecules of mass m in a constant gravitational filed g. As will be shown later, the distribution molecular height z obeys:

    f(z) = C exp (-mgz/KT)

    Where C is the normalization constant
    a) find the average height z, of molecules in the box
    b) find the limiting value of z when h->0
    c) find the limiting value of z when h->infinitiy

    3. The attempt at a solution

    I am a little confused as to where to begin, should i start with determining C?

    I started with:

    Area * Height * average_density = number of molecules
    = Area * integral (z = 0, H [constant*exp(-mgz/(kT))] dz
    = Area*constant*(kT/(mg))*integral(0,mgH)/(kT) [exp(-u)]du
    = Area*constant*(kT/(mg))*(1-exp(-mgH/(kT))

    Therefore,
    number = N = Area*constant*(kT/(mg))*(1-e^(-mgH/(kT))

    So:
    Constant = N*(mg/(kT))/(Area*(1-e^(-mgH/(kT))

    Am i on the right track? thanks in advance
     
  2. jcsd
  3. Jan 20, 2013 #2

    haruspex

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    It doesn't ask you to find C, so I think you can just use C as given and not bother with determining it. You can only find it by knowing N, which is not given.
     
  4. Jan 20, 2013 #3
    Thanks,

    I think i have figured out a) with the below:
    <z> = ∫ z f(z) dz / ∫ f(z) dz

    = ∫ z exp(-mgz/(kT)) dz / ∫ exp(-mgz/(kT)) dz

    where the integrations are from z = 0 to z = h.

    I am pretty stumped on b and c though
     
  5. Jan 20, 2013 #4

    haruspex

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    b) and c) as posted make no sense. It would make sense to ask for limiting values of the average height. Perhaps the original question has ##\bar z##?
     
  6. Jan 20, 2013 #5
    Yes sorry, all the z's in a,b,c are z¯
     
  7. Jan 20, 2013 #6

    haruspex

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    Ok, so did you solve the integrals and get an answer for (a)?
     
  8. Jan 20, 2013 #7
    Yes I have,

    <z> = ∫ z f(z) dz / ∫ f(z) dz

    = ∫ z exp(-mgz/(kT)) dz / ∫ exp(-mgz/(kT)) dz

    = (kT)^2/(mg)^2 * (e^(-mgh/kT)*(-mgh/kT - 1) - 1)) / -(kT/mg)[(e-mg/h/KT)-1]
     
  9. Jan 20, 2013 #8

    haruspex

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    In LaTex:
    ##\frac{\left(\frac{kT}{mg}\right)^2\left(e^{-\frac{mgh}{kT}}\left(-\frac{mgh}{kT} - 1\right) - 1\right) }{ -\frac{kT}{mg}\left(e^{-\frac{mgh}{kT}}-1\right)}##
    Right? But I think you have a sign wrong - pls double check.
    Some cancellation would be in order.
    So b asks for the limit of the above as h tends to 0.
     
  10. Jan 20, 2013 #9
    I cant seem to find the sign thats wrong, the only thing that I can reduce is the (KT/mg)^2 right? I cant figure out how to cancel anything else.

    Setting h=0 should give me the term i need and same with inifinity for c). Thanks. But i cant seem to figure out the above
     
  11. Jan 20, 2013 #10

    haruspex

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    I believe it should be ##\frac{\left(\frac{kT}{mg}\right)^2\left(e^{-\frac{mgh}{kT}}\left(-\frac{mgh}{kT} - 1\right) +1\right) }{ -\frac{kT}{mg}\left(e^{-\frac{mgh}{kT}}-1\right)}##, i.e. ##\frac{kT}{mg}\frac{1-e^{-\frac{mgh}{kT}}\left(1+\frac{mgh}{kT} \right) }{1 -e^{-\frac{mgh}{kT}}}##
    What's an approximation for e-x for small x?
     
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