Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0 \; atm$ and temperature $17\,^{\circ} C$. Take the radius of a nitrogen molecule to be roughly $1.0 \; \mathring{A}$. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $N_{2} = 28.0 \; u$).

(A) Given: Pressure $P = 2.0 \; atm = 2.026 \times 10^{5} \; Pa$,Temperature $T = 17^{\circ} C = 290 \; K$,Radius $r = 1.0 \; \mathring{A} = 1.0 \times 10^{-10} \; m$,Diameter $d = 2r = 2.0 \times 10^{-10} \; m$,Molecular mass $M = 28.0 \times 10^{-3} \; kg/mol$.
$1$. Root mean square speed $v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 290}{28 \times 10^{-3}}} \approx 508.26 \; m/s$.
$2$. Mean free path $l = \frac{kT}{\sqrt{2} \pi d^{2} P}$. Using Boltzmann constant $k = 1.38 \times 10^{-23} \; J/K$,$l = \frac{1.38 \times 10^{-23} \times 290}{\sqrt{2} \times 3.1416 \times (2.0 \times 10^{-10})^{2} \times 2.026 \times 10^{5}} \approx 1.11 \times 10^{-7} \; m$.
$3$. Collision frequency $f = \frac{v_{rms}}{l} = \frac{508.26}{1.11 \times 10^{-7}} \approx 4.58 \times 10^{9} \; s^{-1}$.
$4$. Collision time $t_{c} = \frac{d}{v_{rms}} = \frac{2.0 \times 10^{-10}}{508.26} \approx 3.93 \times 10^{-13} \; s$.
$5$. Time between collisions $t_{f} = \frac{l}{v_{rms}} = \frac{1.11 \times 10^{-7}}{508.26} \approx 2.18 \times 10^{-10} \; s$.
Ratio $\frac{t_{f}}{t_{c}} = \frac{2.18 \times 10^{-10}}{3.93 \times 10^{-13}} \approx 555 \approx 500$.