Question

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 ${}^{o}C$. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N${}_2$ = 28.0 u)

Answer 1

Mean free path $=1.11\times {10}^{-7}m$

Collision frequency $=4.58\times {10}^9{s}^{-1}$
Successive collision time $\simeq 500\times$ (Collision time)

Pressure inside the cylinder containing nitrogen, $P=2.0atm=2.026\times {10}^{5}Pa$

Temperature inside the cylinder, $T={17}^{o}C=290K$

Radius of a nitrogen molecule, $r=1.0\mathring{A}=1\times {10}^{10}m$

Diameter, $d=2\times 1\times {10}^{10}=2\times {10}^{10}m$
Molecular mass of nitrogen, $M=28.0g=28\times {10}^{-3}kg$

The root mean square speed of nitrogen is given by the relation:

${\upsilon }_{rms}=\sqrt{\frac{3RT}{M}}$

where,

R is the universal gas constant $=8.314Jmo{l}^{-1}{K}^{-1}$

$\therefore {\upsilon }_{rms}=\frac{3\times 8.314\times 290}{28\times {10}^{-3}}=508.26m/s$

The mean free path $\left(l\right)$ is given by relation:

$l=\frac{kT}{\sqrt{2}\times d^2\times P}$

Where,

k is the boltzmann constant $=1.38\times {10}^{-23}kg{m}^2{s}^{-2}{K}^{-1}$

$\therefore l=\frac{1.38\times {10}^{-23}\times 290}{\sqrt{2}\times 3.14\times (2\times {10}^{-10}{)}^2\times 2.026\times {10}^5}$ $=1.11\times {10}^{-7}m$

Collision frequency $=\frac{{\upsilon }_{rms}}{l}$ $=\frac{508.26}{1.11\times {10}^{-7}}=4.58\times {10}^9{s}^{-1}$

Collision time is given as:

$T=\frac{d}{{\upsilon }_{rms}}$ $=\frac{2\times {10}^{-10}}{508.26}=3.93\times {10}^{-13}s$

Time taken between successive collisions:

$T^{\prime }=\frac{l}{{\upsilon }_{rms}}$ $=\frac{1.11\times {10}^{-7}}{508.26}=2.18\times {10}^{-10}$

$\therefore \frac{T^{\prime }}{T}=\frac{2.18\times {10}^{-10}}{3.93\times {10}^{-13}}=500$

Hence, the time taken between successive collisions is 500 times the time taken for a collision.