The mean free path of the molecules of a certain gas at `300 K` is `2.6xx10^(-5)m`. The collision diameter of the molecule is `0.26m`. Calculate (`a`) the pressure of the gas and (`b`) the number per unit volume of the gas.

Explanation

Given
`l` (mean free path) `=2.6xx10^(-5)m`, `sigma=0.26xx10^(-9)m`,
`T=300 K`
` :' l=(1)/(sqrt(2)pi sigma^(2)N^(*))`,
`N^(*)=(1)/(sqrt(2)sigma^(2)l)`
`=(1)/((1.414)(3.14)(0.26xx10^(-9)m)(2.6xx10^(-5)m))`
`=1.281xx10^(23)m^(-3)`
Now, pressure `P=N^(*)kT`
`=(1.281xx10^(23)m^(-3))(1.38xx10^(-23)JK^(-1))xx(300 K)`
`=5.306xx10^(2)Pa`