# Maxwell Boltzmann Distribution

1. Jun 11, 2014

### TeslaPow

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

Last edited by a moderator: Jun 11, 2014
2. Jun 11, 2014

### hilbert2

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.

3. Jun 11, 2014

### TeslaPow

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

4. Jun 11, 2014

### hilbert2

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

5. Jun 11, 2014

### TeslaPow

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
6. Jun 12, 2014

### hilbert2

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 $f(x)$.

7. Jun 12, 2014

### TeslaPow

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

8. Jun 12, 2014

### hilbert2

^ Yes, that seems to be the correct value.

9. Jun 12, 2014

### TeslaPow

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

10. Jun 12, 2014

### TeslaPow

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
11. Jun 12, 2014

### hilbert2

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...

12. Jun 12, 2014

### TeslaPow

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

Last edited: Jun 12, 2014