Maxwell distribution of kinetic energies

[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?

 

[Gerenuk]

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

 

[denian]

The Maxwell distribution of kinetic energies describes the probability of finding a molecule with a certain kinetic energy in an ideal gas. According to this distribution, the majority of molecules will have a kinetic energy close to the average value of 3/2kT. However, there will also be a small percentage of molecules that have higher kinetic energies.

To find the percentage of molecules that have energy above the average value, we can use the cumulative distribution function (CDF) for the Maxwell distribution. This function gives the probability that a molecule has a kinetic energy less than or equal to a certain value.

By setting the CDF equal to a certain percentage, we can find the corresponding kinetic energy value. For example, if we set the CDF equal to 95%, we can find the kinetic energy value that 95% of molecules will have less than or equal to. This value will be higher than the average value of 3/2kT, indicating that only a small percentage of molecules will have energies above this value.

In summary, the Maxwell distribution allows us to determine the percentage of molecules that have kinetic energies above a certain value by using the cumulative distribution function. This information is useful in understanding the behavior of an ideal gas and can also be applied to other systems in which the distribution of kinetic energies is important.