Answer to Question #169506 in Molecular Physics | Thermodynamics for Khushi

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

$dN_v=4\pi N(\frac{m}{2\pi k_bT})^{\frac 32} v^2 e^{-\frac{mv^2}{2k_b T}}dv,$

$F(v)=4\pi (\frac{m}{2\pi k_bT})^{\frac 32} v^2 e^{-\frac{mv^2}{2k_b T}}dv,$

$<v>=\int_0^{\infin}vF(v)dv=4\pi (\frac{m}{2\pi k_bT})^{\frac 32} \int_0^{\infin}v^3 e^{-\frac{mv^2}{2k_b T}}dv=\sqrt{\frac{8k_b T}{\pi m}}$