Question

The mean free path of molecules of a gas is ${10^{ - 8}}\,cm.$ if the number density of gas is ${10^9}\,c{m^{ - 3}}.$ Calculate the diameter of the molecule.

Answer 1

Hint: In order to solve this question, we need to understand the basic concept of the mean free path of a gas molecule. In kinetic theory of gases when gas molecules collide with each other kept in a closed container, then the average distance covered by a molecule of the gas between the collision with another gas molecule is known as the mean free path of a molecule of given gas. We will use the general relation between diameters of molecules of gas and mean free path to calculate the diameter of a molecule of the gas.

Formula used:
Diameter of a gas molecule is related as,
${d^2} = \dfrac{1}{{\sqrt 2 \pi n\lambda }}$
where, $n$ is the number density of the gas molecules, $d$ is the diameter of a molecule, $\lambda $ and is the mean free path of a gas molecule.

Complete step by step answer:
According to the question, we have given the following parameters.
$\lambda = {10^{ - 8}}cm = {10^{ - 10}}m$ Mean free path of the gas molecule.
$n = {10^9}c{m^{ - 3}} = {10^{15}}{m^{ - 3}}$ Number density of the gas molecule.
On putting these values in formula, ${d^2} = \dfrac{1}{{\sqrt 2 \pi n\lambda }}$ we get,
${d^2} = \dfrac{1}{{1.414 \times 3.14 \times {{10}^{15}} \times {{10}^{ - 10}}}}$
$\Rightarrow {d^2} = \dfrac{1}{{4.44 \times {{10}^5}}}$
$\Rightarrow {d^2} = 0.225 \times {10^{ - 5}}$
$\therefore d = 1.5 \times {10^{ - 3}}m$
Or we can write $d = 1.5mm$

Hence, the diameter of a molecule of the gas is $d = 1.5\,mm$.

Note: It should be noted that, the basic unit of conversions are used while solving question are $1cm = {10^{ - 3}}m$ and ${10^{ - 3}}m = 1mm$ Kinetic terror of gases is the study of behaviour of gas molecules at given pressure temperature and volume of certain gas. It was James Clerk Maxwell who first introduced the Kinetic theory of gases.