Avogadro’s Number

The Distribution of Molecular Speed

The Maxwell speed distribution $P (v)$ is a function such that $P (v) d v$ gives the
fraction of molecules with speeds in the interval $d v$ at speed $v$ :

P (v) = 4 π v^{2} {(\frac{M}{2 π R T})}^{3 / 2} e^{\frac{- M v^{2}}{2 R T}}

Due to Bolzmann distribution :

f (\vec{v}) d v_{x} d v_{y} d v_{z} \propto e^{\frac{- M v^{2}}{2 R T}} d v_{x} d v_{y} d v_{z}

\frac{m}{k} = \frac{M}{R}

1 = \int_{0}^{\infty} P (v) d v = 4 π A \int_{0}^{\infty} e^{\frac{- m v^{2}}{2 k T}} v^{2} d v

As we know :

\int_{0}^{\infty} x^{2 n} e^{- a x^{2}} d x = \frac{(2 n - 1)!!}{2^{n + 1} a^{n}} \sqrt{\frac{π}{a}}

So due to $n = 1$ :

A = {(\frac{M}{2 π R T})}^{\frac{3}{2}}

The Distribution of Molecular Speed

v_{avg} = \int_{0}^{\infty} v P (v) d v = \sqrt{\frac{8 R T}{π M}} (average speed)

v_{rms} = \sqrt{\int_{0}^{\infty} v^{2} P (v) d v} = \sqrt{\frac{3 R T}{M}} (rms speed)

The most probable speed ( $v_{p}$ ) happens when $\frac{d P}{d v} = 0$ , so :

2 v e^{\frac{- m v^{2}}{2 R T}} + v^{2} e^{\frac{- m v^{2}}{2 R T}} (- \frac{m v}{R T}) = 0

v_{p} = \sqrt{\frac{2 R T}{M}}