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Number Density in the Atmosphere

  1. Nov 30, 2006 #1
    Assume that the Earth's atmosphere has a uniform temperature of 20ºC and uniform composition, with an effective molar mass of 28.9 g/mol. a) show that the number density of molecules depends on height according to:

    nv (y) = (n0) e^ -(mgy)/kBT

    b) commercial jetliners typically cruise at an altitude of 11.0 km. Find the ratio of the atmospheric density there to the density at sea level.

    Ok, so I really am kind of stuck here, for the normal boltzmann distribution equation, where nv(E) = (n0) e^-E/kBT is it possible just to say that since the atmosphere is assumed to have uniform temperature and composition that (assuming no heat loss or gain through interaction with the ground or space) all the molecules have the same kinetic energy and therefore the only change in E would occur through a change in potential energy, therefore the E in the boltzmann distribution eq can be subsituted for U and U can be substituted for mgy?? Seems far too easy this way. Help anyone?

    Correct title should have been Bolztmann distribution law
    Last edited: Nov 30, 2006
  2. jcsd
  3. Nov 30, 2006 #2
    Anybody please??????
  4. Nov 30, 2006 #3


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    I think you're good, and here is a nice little paper on the subject

  5. Nov 30, 2006 #4
    Ok thanks, but my prof just emailed me and said to skip that part! haha all that was just for fun then. I am still having problems with the question though, I tried finding the ratio, but I am confused for what I should use for m in the equation, should I use the molar mass and find it on a per mole basis or should I find the mass of one molecule? I really don't understand it too much, but I'll read the link you sent, maybe it will help? Thanks!
  6. Nov 30, 2006 #5
    Alright thank you! That link was very very helpful in understanding whats going on! The equation given below from that link uses "psi" to represent the potential energy of one molecule, so I think from this that I should find the mass of one molecule and use that in the mgy of my equation. Awesome, thanks again!

    Attached Files:

  7. Nov 30, 2006 #6


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    I can't see your diagram yet, but the m in the derivation is the molecular mass.
  8. Dec 1, 2006 #7
    oh really? ok yeah that makes sense i didn't see in the equation that its n(psi) and not just nv(y). Well if I'm going to use the molecular mass should i use kg/mol or g/mol? I converted to kilograms and used that but i ended up with the ratio equal to e^-7.27X10^22 which my calculator automatically rounds to zero since it is such a small number.
  9. Dec 1, 2006 #8
    Should I leave the answer expressed as e^-7.27X10^22 or should i redo with grams per mole?
  10. Dec 1, 2006 #9
    Well I redid it with g/mol and it still gave me 0 so I think i'll leave it with kg
  11. Dec 1, 2006 #10


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    The argument of an exponential, like the argument of a trig function, must be dimensionless. The units you use for m must be consistent with the units of everything else that is there. When you multiply everything in the exponential together you must have a pure number. This is also true of logarithms. That is why you almost always see logarithms of ratios when they appear in formulas. It would be incorrect to write

    ln(I/Io) = ln(I) - ln(Io)

    where I and Io are quantities with dimensions. What would be correct is if represents the units of I and Io then

    ln(I/Io) = ln(I/) - ln(Io/)

    so the the arguments of the ln() are dimensionless.
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