The van der Waals constant `b` of `Ar` is `3.22xx10^(-5) m^(3) mol^(-1)`. Calculate the molecular diameter of `Ar`.

Explanation

We know that,
`b = 4 xx` volume occupied by the molecules in one mole of a gas
`= 4 xx N_(0) xx ((4)/(3) pi r^(3))`
`3.22 xx 10^(-5) = 4 xx 6.02 xx 10^(23) xx (4)/(3) xx (22)/(7) xx r^(3)`
or `r = ((3.22 xx 10^(-5) xx 3 xx 7)/(4 xx 6.02 xx 10^(23) xx 4 xx 22))^(1//3)`
`= 0.1472 xx 10^(-9) m`
`:.` Diameter `= 2 r = 2 xx 0.1472 xx 10^(-9)`
`= 0.2944 xx 10^(-9) m`