Answer to Question #176744 in Molecular Physics | Thermodynamics for Arsalan

Question

The expression for the number of molecules in a Maxwellian gas having speeds in

the range v to v + dv is

dN_V=4(3.14)N((M/(2(3.14)k_BT))^3/2 V² exp[-(mv²/2k_BT)]dV)

Using this relation, obtain an expression for average speed. Also, plot Maxwellian

distribution function versus speed at three different temperatures.

Answer 1

$f(v)dv=(\frac{m}{2\pi k T})^{\frac 32}4\pi v^2 e^{-\frac{mv^2}{2kT}}dv,$

$\frac{df(v)}{dv}=-8\pi(\frac{m}{2\pi k T})^{\frac 32}v e^{-\frac{mv^2}{2kT}}(\frac{mv^2}{2kT}-1)=0,$

$\frac{mv^2}{2kT}-1=0,$

$v=\sqrt{\frac{2kT}{m}}=\sqrt{\frac{2RT}{M}}.$