Answer to Question #301806 in Molecular Physics | Thermodynamics for Dhanush

Question

The expression for the number of molecules in a Maxwellian gas having speeds in the range vto v+ dv is

dNv= 4πN(m÷2πkbT)^3/2 v^2exp[-(mv^2÷2kbT)]dv using this relation, obtain an expression for most probable speed

Answer 1

$dN=4πN\left(\frac{m}{2πkT}\right)^{3/2} v^2\exp[-(mv^2/2kT)]dv$

$( v^2\exp[-(mv^2/2kT)])'=2v\exp[-(mv^2/2kT)]\\ -mv^3/kT\exp[-(mv^2/2kT)]=0$

$2=mv^2/kT$

$v_{mp}=\sqrt{2kT/m}$