# Maxwell Boltzmann Distribution

I don't know how to integrate the Maxwell-Boltzmann distribution without approximation or help from Maple.

Given the Maxwell-Boltzmann distribution:

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

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:

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

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.

$$\int f(v)dx$$

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

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hilbert2
Gold Member
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.

Uriel and TeslaPow
By looking in the integral table at number 70, is this the right solution?

hilbert2
Gold Member
Yes, usually when relatively difficult integrals like this are needed in physics calculations, they are just looked from tables of integrals.

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

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hilbert2
Gold Member
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

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)##.

1 person
Do I put in -0.0645 for the lambda as calculated from the copy?

hilbert2
Gold Member
^ Yes, that seems to be the correct value.

^ Yes, that seems to be the correct value.

Can you use the same formula for calculating between 400-401 m/s to check yourself that

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!

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hilbert2
Gold Member
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...

Yeah I got the exact same value. Now I can use this for all kinds of gases.

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