Ncert Exercises & Solutions

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 $\overset{o}{A}$ $\overset{o}{A}$ and 2 $\overset{o}{A}$ $\overset{o}{A}$ . The gases may be considered under identical conditions of temperature, pressure and volume.

Hint: The mean free path depends on the diameters of the molecules.

Step 1: Write down the formula of the mean free path.

Mean free path of a molecule is given by,

l = \frac{1}{\sqrt{2} d^{2} n}

$l = \frac{1}{\sqrt{2} d^{2} n}$

where n = number of molecules/ volume
d = diameter of the molecule

Step 2: Find the ratio of mean free paths of the gases.
Now, we can write,

l \propto \frac{1}{d^{2}}

$l \propto \frac{1}{d^{2}}$

Given,

d_{1} = 1 \overset{o}{A}, d_{2} = 2 \overset{o}{A}

$d_{1} = 1 \overset{o}{A}, d_{2} = 2 \overset{o}{A}$

l_{1} \propto \frac{1}{d_{1}^{2}} a n d l_{2} \propto \frac{1}{d_{2}^{2}}

$l_{1} \propto \frac{1}{d_{1}^{2}} a n d l_{2} \propto \frac{1}{d_{2}^{2}}$

\Rightarrow

$\Rightarrow$ So,

\frac{l_{1}}{l_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{2}{1})}^{2} = \frac{4}{1}

$\frac{l_{1}}{l_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{2}{1})}^{2} = \frac{4}{1}$

Hence,

l_{1} : l_{2} = 4 : 1

$l_{1} : l_{2} = 4 : 1$

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