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Maxwell Boltzmann Distribution

  1. Jun 11, 2014 #1
    I don't know how to integrate the Maxwell-Boltzmann distribution without approximation or help from Maple.

    Given the Maxwell-Boltzmann distribution:

    [tex]f(v) = 4\pi\left[\frac{m}{2\pi kT}\right]^{3/2}v^2\textrm{exp}\left[\frac{-mv^2}{2kT}\right][/tex]

    Observe the appearance of the Boltzmann factor ##\textrm{exp}\left[\frac{-mv^2}{2kT}\right]## with ##E = \frac{mv^2}{2}##.

    Assuming a fixed temperature and mass, one can simplify this equation:

    [tex]f(v) = av^2\textrm{exp}[-bv^2][/tex]
    [tex]a = 4\pi \left[\frac{m}{2\pi k T}\right]^{3/2}[/tex]
    [tex]b = \frac{m}{2kT}[/tex]

    In order to calculate the fraction of particles between two speeds ##v_1## and ##v_2##, one should evaluate the definite integral. It's possible to use this formula directly with low speeds, but for higher speeds between let's say 400-500 m/s an integration is needed.

    [tex]\int f(v)dx[/tex]

    Here is an link to integral-tables, http://integral-table.com/
    How would I solve this problem for let's say a certain amount of moles with hydrogen between two different velocities? Best regards, Tor
     
    Last edited by a moderator: Jun 11, 2014
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  3. Jun 11, 2014 #2

    hilbert2

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    First note that
    ##\int_a^b e^{-x^2}dx=\frac{\sqrt{\pi}}{2}(Erf(b)-Erf(a))##,
    because this is the definition of the error function. Now you can integrate
    ##\int_a^b e^{-\lambda x^2}dx##
    using the substitution ##u=\sqrt{\lambda}x##. Finally, we see that
    ##\int_a^b x^2 e^{-\lambda x^2}dx=-\int_a^b \frac{d}{d\lambda}e^{-\lambda x^2}dx=-\frac{d}{d\lambda}\int_a^b e^{-\lambda x^2}dx##.
    That way, you can do the required integration.
     
  4. Jun 11, 2014 #3
    By looking in the integral table at number 70, is this the right solution?
     
  5. Jun 11, 2014 #4

    hilbert2

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    Yes, usually when relatively difficult integrals like this are needed in physics calculations, they are just looked from tables of integrals.
     
  6. Jun 11, 2014 #5
    Thanks for the short reply, I've never used the error function in integrals before. How do I go on about from here when I want to integrate between 400-500m/s from the page I just copied from my physics books. I can't find an derivation example in either of my books, it would seem a rather difficult integration. http://s29.postimg.org/6kpwe4kc7/molecular.jpg
     
    Last edited: Jun 11, 2014
  7. Jun 12, 2014 #6

    hilbert2

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    The calculation on that page uses linear approximation instead of actual integration. Now that you know the integration formula for ##\int x^2e^{-\lambda x^2}dx##, you just use ##\int_a^bf(x)dx=F(b)-F(a)## to calculate the result (here ##F(x)## is the indefinite integral of ##f(x)##.
     
  8. Jun 12, 2014 #7
    Do I put in -0.0645 for the lambda as calculated from the copy?
     
  9. Jun 12, 2014 #8

    hilbert2

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    ^ Yes, that seems to be the correct value.
     
  10. Jun 12, 2014 #9
    Can you use the same formula for calculating between 400-401 m/s to check yourself that
    it's the correct answer?
     
  11. Jun 12, 2014 #10
    I had to leave 1.74e14 outside the integral and for the lambda part it was necessary to calculate it like this: 2(1.67e-27)(x^2)/2(1.38e-23)(300K) => 4.03382e-7x^2 and integrate x^2*e-4.03382*10-7*x^2. Thanks for your help, now I know how to integrate the MBD. Mucho appreciato!
     
    Last edited: Jun 12, 2014
  12. Jun 12, 2014 #11

    hilbert2

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    Wait a minute, the value of lambda should not include the factor ##v^2##... The correct value of ##\lambda## in the calculation is ##4.03\times 10^{-7}\frac{s^2}{m^2}##.

    I calculated the number of molecules that have speed between 400 and 500 m/s and I got the result ##3.26\times 10^{21}##. Now you can calculate and see if you get the same result.

    EDIT: oh, you noted the problem with lambda value...
     
  13. Jun 12, 2014 #12
    Yeah I got the exact same value. Now I can use this for all kinds of gases.
     
    Last edited: Jun 12, 2014
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