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Maxwell speed distribution

  1. Apr 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the number [tex]N(0,v)[/tex], of molecules of an ideal gas with speeds between 0 and v is given by

    [tex]N(0,v) = N \left[ erf(\xi) - \frac{2}{\sqrt{\pi}} \xi e^{-\xi^2} \right][/tex]

    Where,

    [tex]erf(\xi) = \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{-x^2} dx [/tex]

    And,

    [tex]\xi^2 = \left(\frac{mv^2}{2kT} \right)[/tex]

    2. Relevant equations

    [tex]\frac{dN_v}{N} = \sqrt{\frac{2}{\pi}} \left( \frac{m}{kT}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2kT}} \ dv [/tex]

    3. The attempt at a solution

    Alright, so I managed to get to the following

    [tex]\frac{dN_v}{N} = \frac{4}{\sqrt{\pi}} x^2 e^{-x^2} dx [/tex]

    Where,

    [tex]\alpha = \frac{m}{2kT}, \ x = \sqrt{\alpha}v [/tex]

    =========================\\===========================

    So far so good, but now when checking the solution on the textbook it claims the following algebraic manipulation which I can't follow

    [tex]\frac{2N}{\sqrt{\pi}}\int_{0}^{\xi} x (2xe^{-x^2} dx) = \frac{2N}{\sqrt{\pi}} x e^{-x^2} |_{\xi}^{0} \ + \ N \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{-x^2} dx [/tex]

    What was done on the integral above, how can you "split" it like that?
     
  2. jcsd
  3. Apr 22, 2011 #2

    L-x

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    Integration by parts
     
  4. Apr 22, 2011 #3

    Dick

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    Re: I need a hand understanding this integral manipulation

    It's integration by parts. udv=d(uv)-vdu with u=x and dv=(2x)e^(-x^2)dx.
     
  5. Apr 22, 2011 #4
    But of course! Agh, how didn't I see that, I feel very stupid right now. :biggrin:

    Thanks for the highlight, L-x!
     
  6. Apr 22, 2011 #5

    L-x

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    No worries. It's not immidiately obvious, although the integral is written in a way which gives you a clue: x(2x) instead of 2x^2
     
  7. Apr 22, 2011 #6

    Dick

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