0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

When studying the KMT, you should be familiar with these equations:

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 $$\int_0^\infty e^{-v^2}\, \mathrm dv =\frac{\sqrt{\pi}}{2}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.