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Maxwell distribution of kinetic energies

  1. Jan 7, 2008 #1
    Given that the average kinetic energy for molecules of an ideal gas is <E>=3/2kT, how can one find the percentage of molecules of the gas that contain energy above this value?
  2. jcsd
  3. Jan 7, 2008 #2
    I tried to solve the integral
    - see http://en.wikipedia.org/wiki/Maxwell_distribution Equation 10
    - you integrate from [tex]v=\sqrt{\frac{3kT}{m}}[/tex] to infinity
    - my result was
    [tex]P(E > 3/2\,kT)=\sqrt{\frac{6}{\pi e^3}}+\operatorname{erfc}\left(\sqrt{\frac{3}{2}}\right)=39.2\%[/tex]
    Feel free to check. (for integration use Mathematica, Matlab or for example "Tables of indefinite integrals" by Brychekov)
    Last edited: Jan 7, 2008
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