Question

Collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2\text{atm}$ and temperature of ${17}^{\circ }\text{C}$
(Assume radius of a nitrogen molecule to be $1.0\stackrel{\circ }{\text{A}}$)

• A. $6\times {10}^9{\text{s}}^{-1}$
• B. $8\times {10}^9{\text{s}}^{-1}$
• C. $4.6\times {10}^9{\text{s}}^{-1}$
• D. $7\times {10}^9{\text{s}}^{-1}$

Answer 1

The correct option is C $4.6\times {10}^9{\text{s}}^{-1}$

Given,
Pressure, $P=2.0\text{atm}=2\times {10}^{-5}{\text{N/m}}^2$
Temperature, $T={17}^{\circ }\text{C}=290\text{K}$
Radius of a nitrogen molecule, $r=1.0\stackrel{\circ }{\text{A}}=1\times {10}^{-10}\text{m}$
So, diameter, $d=2\times {10}^{-10}\text{m}$
We know that mean free path,
$\lambda =\frac{k_{b}T}{\sqrt{2}\pi d^{2}P}$
On putting all data,
$\lambda =\frac{1.4\times {10}^{-23}\times 290}{\sqrt{2}\times 3.14\times (2\times {10}^{-10}{)}^2\times 2\times {10}^5}\Rightarrow \lambda =1.1\times {10}^{-7}$
Also, rms speed speed, $V_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.3\times 290}{28\times {10}^{-3}}}\approx 508\text{m/s}$
$\therefore \text{Collision frequency,}f=\frac{V_{rms}}{\lambda }=\frac{508}{1.1\times {10}^{-7}}=4.6\times {10}^9{\text{s}}^{-1}$