# Why does the mean velocity of gas particles include $\pi$?

In chemistry, we are learning about kinetic molecular theory (KMT) of gasses, and I just couldn't help being surprised when I saw pi in the equation of mean velocity. I know that whenever $$\pi$$ is involved in an equation, it somehow involves circles, but KMT assumes that the molecules are points. I don't know how to add equations here from mobile, but here it is: $$\sqrt{\frac{8RT}{\pi M}}$$, where $$R$$ is the universal gas constant (8.3144 J/K mol), T stands for the temperature in Kelvin, and M is the molar mass of the molecule.

It just fascinates me that $$\pi$$ is everywhere, especially here and I would like to know why.

Thanks!

1) Number of microstates: $$W=\frac{N!}{\prod_{i}n_{i}!}$$ 2) Gibbs entropy: $$S=-k\sum_{i}p_{i}\ln p_{i}$$ 3) Boltzmann distribution: $$p_{i}=Ke^{-\frac{E_{i}}{kT}},$$ which implies the Maxwell–Boltzmann distribution: $$f(v)=4\pi \left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}v^2 e^{-\frac{mv^2}{2kT}}$$
Now the mean velocity is calculated by $$\langle v\rangle =\int_0^\infty vf(v)\, \mathrm dv,$$ so $$\langle v\rangle =4\pi \left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}\int_0^\infty v^3e^{-\frac{mv^2}{2kT}}\, \mathrm dv,$$ which evaluates to $$\sqrt{\frac{8kT}{\pi m}}=\sqrt{\frac{8RT}{\pi M}}$$.
To sum it up, $$\pi$$ comes from the fact that $$\int_0^\infty e^{-v^2}\, \mathrm dv =\frac{\sqrt{\pi}}{2}.$$