Question

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

• A. (a) $1.281\times {10}^{23}{m}^{-3}$ (b) $5.306\times {10}^{2}Pa$
• B. (a) $1.281\times {10}^{22}{m}^{-3}$ (b) $5.306\times {10}^{3}Pa$
• C. (a) $12.81\times {10}^{23}{m}^{-3}$ (b) $53.06\times {10}^{2}Pa$
• D. (a) $2.56\times {10}^{23}{m}^{-3}$ (b) $10.612\times {10}^{2}Pa$

Answer 1

The correct option is A (a) $1.281\times {10}^{23}{m}^{-3}$ (b) $5.306\times {10}^{2}Pa$

$\lambda =2.6\times {10}^{-5}m,\sigma =0.26nm=2.6\times {10}^{-10}m$
$T=300K$
$\lambda =\frac{1}{\sqrt{2}\pi {\sigma }^2{N}^{\ast }}$
$2.6\times {10}^{-5}=\frac{1}{\sqrt{2}\times 3.14\times (2.6\times {10}^{-10}{)}^2\times N^{\ast }}$
$N^{\ast }=1.281\times {10}^{23}{m}^{-3}$
$N^{\ast }=\frac{P}{KT}$
$P=1.281\times {10}^{23}\times 1.38\times {10}^{-23}\times 300$
$P=530.3Pa$