Answer to Question #130021 in Molecular Physics | Thermodynamics for Abhishek

Question

 The mean free path of the molecules of 

a gas at 27°C is 6.0 x 10-8 m. Calculate 

the pressure exerted by the gas. 

The radius of the molecule is 1.8 A. 

Take kB= 1.38x 10-23J K -1.

Answer 1

Mean free path equation $\lambda=\frac{kT}{\sqrt 2 \pi Pd^2_i}$

Where $\lambda$ = Mean free path of gas species

KB=Boltzmann constant

P= pressure

d= gas species diameter

T= temperature

substituting the values in the equation,

diameter of gas species=$1.8\times2=3.6A=3.6\times10^{-10}$

$6.0\times10^{-8}=\frac{1.38\times10^{-23}\times300}{\sqrt2 \pi P \times 3.6\times10^{-10}}$

making P subject of the formula,

$P=\frac{1.38\times10^{-23}\times300}{3.6\times10^{-10}\times\pi\times\sqrt2\times6.0\times10^{-8}}$

$P=\frac{4.14\times10^{-21}}{9.5966\times10^{-17}}$

$P=4.314\times10^{-5}Pa$