Maxwell distribution of kinetic energies

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SUMMARY

The discussion focuses on calculating the percentage of molecules in an ideal gas that possess kinetic energy exceeding the average value of =3/2kT, derived from the Maxwell distribution. The integral is evaluated from v=\sqrt{\frac{3kT}{m}} to infinity, yielding a result of P(E > 3/2\,kT)=\sqrt{\frac{6}{\pi e^3}}+\operatorname{erfc}\left(\sqrt{\frac{3}{2}}\right)=39.2%. Tools such as Mathematica and Matlab are recommended for performing the integration, alongside reference to "Tables of indefinite integrals" by Brychekov for additional support.

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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?
 
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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 &gt; 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)
 
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