# Maxwell distribution of kinetic energies

1. Jan 7, 2008

### bobwell

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. Jan 7, 2008

### Gerenuk

I tried to solve the integral
- see http://en.wikipedia.org/wiki/Maxwell_distribution Equation 10
- you integrate from $$v=\sqrt{\frac{3kT}{m}}$$ to infinity
- my result was
$$P(E > 3/2\,kT)=\sqrt{\frac{6}{\pi e^3}}+\operatorname{erfc}\left(\sqrt{\frac{3}{2}}\right)=39.2\%$$
Feel free to check. (for integration use Mathematica, Matlab or for example "Tables of indefinite integrals" by Brychekov)

Last edited: Jan 7, 2008