Question

Use Maxwell's speed distribution law to derive the expression for the most probable speed of molecules in a sample of gas dN/N = 4/squareroot pi [M/2RT]^3/2 v^2 e^-Mv^2/2 RT dv

Answer 1

Given That

$\frac{dN}{N} = \frac{4}{\sqrt{\pi }}\left ( \frac{M}{2RT} \right )^{3/2}v^{2}e^{\frac{-Mv^{2}}{2RT}}dv$

$dN = \frac{4N}{\sqrt{\pi }}\left ( \frac{M}{2RT} \right )^{3/2}v^{2}e^{\frac{-Mv^{2}}{2RT}}dv$

To get most probable speed we have to find maxima i.e putting this derivative equal to zero

i.e.

$\frac{4N}{\sqrt{\pi }}\left ( \frac{M}{2RT} \right )^{3/2}v^{2}e^{\frac{-Mv^{2}}{2RT}}dv = 0$

Here constant can not be zero i.e

$\frac{d\left (v^{2}e^{\frac{-Mv^{2}}{2RT}} \right )}{dv} = 0$

Using chain rule

$0 = 2v.e^{\frac{-Mv^{2}}{2RT}}+v^{2}.\left (\frac{-2Mv}{2RT} \right )e^{\frac{-Mv^{2}}{2RT}}$

Cancelling common terms

$0 = 1-v^{2}.\left (\frac{M}{2RT} \right )$

$v^{2}.\left (\frac{M}{2RT} \right ) = 1$

$v^{2} = \frac{2RT}{M}$

Here

$v^{2} = \frac{2RT}{N.m}$ R/N = k_B

$v^{2} = \frac{2k_{B}T}{m}$

$v_{mp} =\sqrt{ \frac{2k_{B}T}{m}}$