91Ó°ÊÓ

The maximum velocity of the nitrogen molecules in the air is $300\mathrm{m}/\mathrm{s}$.$300\mathrm{m}/\mathrm{s}$

Maxwell speed distribution function is given by

$D\left(v\right)={\left(\frac{m}{2\mathrm{Ï€}kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4\mathrm{Ï€}{v}^2{e}^{-\raisebox{1ex}{$m{v}^2$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}.........................\left(1\right)$

The formula to calculate the probability that the speed of the nitrogen molecules lies in the range of $0<v<300$is given by

role="math" localid="1646974039312">P=∫0300m2πkT324πv2e-mv22kTdv=4πm2πkT32∫0300v2e-mv22kTdv...................(2)

Change the variable $v$in equation (2) to $x=v\sqrt{\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$v_{\mathrm{mp}}$}\right.$, where $v_{\mathrm{mp}}$is the most probable speed.

Substitute $300\mathrm{m}/\mathrm{s}$for $v$and $422\mathrm{m}/\mathrm{s}$for $v_{\mathrm{mp}}$to calculate the upper limit for the variable role="math" $x$.

role="math" $\begin{array}{rcl}x& =& \frac{300\mathrm{m}/\mathrm{s}}{422\mathrm{m}/\mathrm{s}}\\ & ≈& 0.71\end{array}$

Substitute the required parameters into equation (2) and evaluate the integral.

role="math" localid="1646974577626" $\begin{array}{rcl}P& =& 4\mathrm{π}{\left(\frac{m}{2\mathrm{π}kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left(\frac{2kT}{m}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{∫}_0^{0.71}{x}^2{e}^{-x^2}dx\\ & =& \frac{4}{\sqrt{\mathrm{π}}}{∫}_0^{0.71}{x}^2{e}^{-x^2}dx\\ & ≈& 0.20\end{array}$