Question

The mean free path of the molecules of a certain gas at $300\text{K}$ is $2.6\times {10}^{-5}\text{m}$. The collision diameter of the molecule is $0.26\times {10}^{-9}\text{m}$. Calculate (a) the pressure of the gas and (b) the number per unit volume of the gas.

• A. $5.306\times {10}^2\text{Pa}$ and $2.798\times {10}^{23}{\text{m}}^{-3}$
• B. $5.306\times {10}^2\text{Pa}$ and $4.183\times {10}^{23}{\text{m}}^{-3}$
• C. $10.6\times {10}^2\text{Pa}$ and $1.281\times {10}^{23}{\text{m}}^{-3}$
• D. $5.306\times {10}^2\text{Pa}$ and $1.281\times {10}^{23}{\text{m}}^{-3}$

Answer 1

The correct option is D $5.306\times {10}^2\text{Pa}$ and $1.281\times {10}^{23}{\text{m}}^{-3}$

Given,
$l\text{(mean free path)}=2.6\times {10}^{-5}\text{m}$, $\sigma =0.26\times {10}^{-9}\text{m}$, $T=300\text{K}$
$\because l=\frac{1}{\sqrt{2}\pi {\sigma }^{2}N}$
$N=\frac{1}{\sqrt{2}\pi {\sigma }^{2}l}$
$=\frac{1}{\left(1.414\right)\left(3.14\right)(0.26\times {10}^{-9}\text{m}{)}^2(2.6\times {10}^{-5}\text{m})}$
$=1.281\times {10}^{23}{\text{m}}^{-3}$
Now, pressure $P=N.kT$
$=(1.281\times {10}^{23}{\text{m}}^{-3})(1.38\times {10}^{-23}{\text{J K}}^{-1})\times \left(300\text{K}\right)$
$=5.306\times {10}^2\text{Pa}$