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Velocity of molecules in a gas

  1. Apr 9, 2006 #1
    I am a little confused. I was talking to my teacher a couple days ago, and he asked me to put the following three into order of velocity (greatest first):

    1. Average velocity of the molecules in a gas (AV)
    2. Most probable velocity of the molecules in a gas (MPV)
    3. RMS (root mean square) velocity of the molecules in a gas (RMSV)

    Well, I thought that AV and MPV must be the same? Because surely the MPV is based upon the AV? And then the formula used to calculate RMSV is:

    [​IMG]

    So I figured this is greater than the other 2? So I said to him RMSV had the greatest velocity, and then AV and MPV had equal velocities, but he told me this was incorrect. Can anyone help me out? Are they all equal?
     
  2. jcsd
  3. Apr 9, 2006 #2
    Rms>av>mpv
     
  4. Apr 9, 2006 #3
    ah thanks, that makes good sense. i did some more research and that's the answer that made most sense to me as well. thanks.
     
  5. Apr 9, 2006 #4

    Astronuc

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    Staff: Mentor

  6. Apr 9, 2006 #5

    nrqed

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    Just a comment. For *symmetric* probability distributions, the average value and most probable value are equal. The MB distribution is asymmetric...From a graph it is clear that the average value is larger than the most probable value (which would corespond to the peak) because the distribution is ''stretched'' farther from the peak for speeds above the most probable value.

    Patrick
     
  7. Apr 9, 2006 #6

    lightgrav

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    The average velocity of a gas molecule = most probable velocity = 0.

    The typical Maxwell-Boltzmann distribution function is for molecule SPEEDs,
    which are never negative.

    It's not so much that the distribution is "stretched" toward large speeds,
    as that the negatives have been "folded over to +'ve" by Pythagoras.
     
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