Question

Calculate the means free path of nitrogen molecule at ${27}^o$C when pressure is $1.0$ atm. Given, diameter of nitrogen molecule $=1.5\stackrel{o}{A}$, $k_B=1.38\times {10}^{-23}$J $K^{-1}$. If the average speed of nitrogen molecule is $675$ $m{s}^{-1}$. 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

The correct option is A $0.6$ns

Here, $T={27}^{o}C=27+273=300$K.

$P=1$atm $=1.01\times {10}^5$N $m^{-2}$, d$=1.5\stackrel{o}{A}=1.5\times {10}^{-10}$m,

$k_B=1.38\times {10}^{-23}$J $K^{-1},\lambda =?$

From $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}p}=\frac{1.38\times {10}^{-23}\times 300}{1.414\times 3.14(1.5\times {10}^{-10}{)}^2\times 1.01\times {10}^5}$

$=4.1\times {10}^{-7}$m

Time interval between two successive collisions

$t=\frac{distance}{speed}=\frac{\lambda }{v}=\frac{4.1\times {10}^{-7}}{675}=0.6\times {10}^{-9}s$