Question

In a cubical vessel $1m\times 1m\times 1m$ the gas molecules of diameter $1.7\times {10}^{-8}\text{cm}$ are at a temperature 300 K and a pressure of ${10}^{-4}\text{mm}$ mercury. The mean free path of the gas molecule is

• A. 1 meter
• B. 4 meter
• C. 2.42 meter
• D. 1 cm

Answer 1

The correct option is C 2.42 meter

The mean free path or average distance between collison for a gas molecule may be estimated from KTG(Kinetic Theory of Gases).

From Serway's approach:-

$\left(Meanfreepath\right)\lambda =\frac{1}{\pi d^2{n}_v}$

$d\to$ diameter of molecule

$n_v\to$ No. of molecules per unit volume

From idle gas law,

$n_v=\frac{n{N}_A}{V}\Rightarrow n_v=\frac{N_{A}P}{RT}$

So, from $\left(i\right)$

$\lambda =\frac{RT}{\sqrt{2}\pi d^2{N}_{A}P}$

A/Q we have

$d=1.7\times {10}^{-8}cm=17\times {10}^{-10}mP={10}^{-4}mmHgT=300K\lambda =\frac{8.314\times 300}{\sqrt{2}\times 3.14\times (1.7\times {10}^{-10}{)}^2}(6.022\times {10}^{23}({10}^{-4}\left)\right)=\frac{8.314\times 3\times {10}^3}{1.414\times 3.14\times 2.89\times 6.022}=2.4201m$