Question

Calculate the means free path of nitrogen moleculeat $27^{\circ }C$ when pressure is $1.0atm$. Given, diameter of nitrogen molecule $=1.5\mathring{A}$, $k_B=1.38\times 10^{-23}J{K}^{-1}$. If the average speed of nitrogen molecule is $675m{s}^{-1}$. The time taken by the molecule between two successive collisions is

• A. $0.6ns$
• B. $0.4ns$
• C. $0.8ns$
• D. $0.3ns$

Answer 1

Answer: (A)

Here, $T=27{}^{\circ }C=27+273=300K$.
$P=1atm=1.01\times 10^{5}N{m}^{-2}$, $d=1.5\mathring{A}=1.5\times 10^{-10}m$,
$k_B=1.38\times 10^{-23}J{K}^{-1}$, $\lambda =?$
From $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}=\frac{1.38\times 10^{-23}\times 300}{1.414\times 3.14{\left(1.5\times 10^{-10}\right)}^2\times 1.01\times 10^5}$
$=4.1\times 10^{-7}m$
Time interval between two successive collisions
$t=\frac{distance}{speed}=\frac{\lambda }{v}=\frac{4.1\times 10^{-7}}{675}\approx 0.6\times 10^{-9}s$