State and explain Graham's law of Diffusion.

Explanation

Graham.s Law of Diffusion : At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or) vapour pressure (or) molecular weight.
`r prop (1)/(sqrt(d)),r prop (1)/(sqrt(VD)), r prop (1)/(sqrt(m)) (or) r = K xx (1)/(sqrt(d)), r = K xx (1)/(sqrt(VD)), r = K xx (1)/(sqrt(m))`
If `r_(1) and r_(2)` are the rates of diffusion of two gses and `d_(1), d_(2)` are their densities then.
`(r_(1))/(r_(2)) = sqrt((d_(2))/(d_(1)))" "...(1)`
If `r_(1) and r_(2)` are the rates of diffusion of two gases and `VD_(1), VD_(2)` are their vapour pressures, then
`(r_(1))/(r_(2)) = sqrt((VD_(2))/(VD_(1)))" "...(2)`
If `r_(1) and r_(2)` are the rates of diffusion of two gases and `m_(1) m_(2)` are their molecular weights, then `(r_(1))/(r_(2)) = sqrt((m_(2))/(m_(1)))`
`therefore (r_(1))/(r_(2))= sqrt((d_(2))/(d_(1))) = sqrt((VD_(2))/(VD_(1))) = sqrt((m_(2))/(m_(1)))`
`therefore` But rate of diffusion, `r = ("V (volume)")/("t (time of diffusion)")`
Then `(r_(1))/(r_(2)) = (V_(1))/(t_(1)) xx (t_(2))/(V_(2))`
`therefore (r_(1))/(r_(2)) = (V_(1)t_(2))/(V_(2)t_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((VD_(2))/(VD_(1))) = sqrt((m_(2))/(m_(1)))`
Case - 1 : If the times of diffusions are equal i.e., `t_(1) = t_(2)`, then we can write
`(r_(1))/(r_(2)) = (V_(1))/(V_(2)) = sqrt((d_(2))/(d_(1))) = sqrt((VD_(2))/(VD_(1))) = sqrt((m_(2))/(m_(1)))`
Case - 2 : If the volumes of the two gases are the same (i.e.) `V_(1) = V_(2)`
then `(r_(1))/(r_(2)) = (t_(1))/(t_(2)) = sqrt((d_(2))/(d_(1))) = sqrt((VD_(2))/(VD_(1))) = sqrt((m_(2))/(m_(1)))`