Calculate the average volume available to a molecule in a sample of `N_(2)` at `NTP` Also report the average distance between two neighbouring molecules if a nitrogen molecule is assumed to be spherical Comment on the result if the radius of one `N_(2)` molecule ` =2 xx 10^(-8) cm^(3)` .

Explanation

Volume occupied by `N` molecules of `N_(2) = 22400 cm^(3)`
`:.` Volume occupied by 1 molecules of `N_(2)`
` = (22400)/(6.02 xx 10^(23)) = 3.72 xx 10^(-20) cm^(3)`
Also volume of 1 molecule of `N_(2) = (4)/(3) pi r^(3)`
`:. (4)/(3) xx (22)/(7) xx r^(3) = 3.72 xx 10^(-20)`
` :. r =20.7 xx 10^(-8) cm`
Thus average distnace in between two `N_(2)` molecules
` = 2xx r = 41.4 xx 10^(-8) cm`
Given radius of `N_(2) = 2 xx 10^(-8)cm` obtained radius is between two molecules is gaseous state is alomost 10 times of the diameter of each molecule This confirms the empty space in gaseous state and also a reasonable justification for their compression .