If the mean free-path of gaseous molecule is `60cm` at a pressure of `1xx10^(4)mm` mercury, what will be its mean free-path when the pressure is increased to `1 xx10^(-2)` mm mercury ? .

Options

A. `6.0 xx 10^(-1) cm`
B. `6.0 cm`
C. `6.0 xx 10^(-2) cm`
D. `6.0 xx 10^(-3) cm`

Explanation

Mean free path `prop (1)/("Pressure")`
`(l_(1))/(l_(2)) = (P_(2))/(P_(1)) implies l_(2) = (l_(1)P_(1))/(P_(2))`
` = (60 cm xx 1 xx 10^(4)mm)/(1xx10^(-2)mm)`
` = 6 xx 10^(-1) cm` .