Question

Calculate the means free path of nitrogen moleculeat $27^{\circ}C$ when pressure is $1.0 \,atm$. 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 $675 \,ms^{-1}$. The time taken by the molecule between two successive collisions is

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

Answer 1

Answer: (A)

Here, $T= 27\,{}^{\circ}C = 27 + 273 = 300\, K$.
$P = 1 \,atm = 1.01 \times 10^{5}\, N\, m^{-2}$, $d = 1.5 \,\mathring{A} = 1.5 \times 10^{-10}\, m$,
$k_{B}= 1.38 \times 10^{-23} \,J \,K^{-1}$, $\lambda = ?$
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
$t = \frac{distance}{speed} = \frac{\lambda}{v} = \frac{4.1 \times 10^{-7}}{675} \approx 0.6 \times 10^{-9}\,s$