# A sample of Br2(g) takes 30.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane?

Nov 17, 2016

Graham's law of effusion states that the rate of effusion ($R$) of a gas is inversely proportional to the square root of the molar mass ($M$) of the gas when temperature and pressure of gas remaining constant The formula can be written as:

$R \propto \frac{1}{\sqrt{M}}$

For two gases $B {r}_{2} \left(g\right) \mathmr{and} A r \left(g\right)$ the relation can be written as

 R_"Ar"/R_"Br" = sqrt(M_"Br"/M_"Ar")...(1)

If the times of effusion of same number of moles (n) of two gases (Ar and Br) or same volume (v) of the gases under same condition of temperature and pressure be ${t}_{\text{Ar" and t_"Br}}$ then

${R}_{\text{Ar"=n/t_"Ar" and R_"Br"=n/t_"Br}}$

So inserting these in (1) we get

 R_"Ar"/R_"Br" = sqrt(M_"Br"/M_"Ar")

 =>t_"Br"/t_"Ar" = sqrt(M_"Br"/M_"Ar").....(2)

Given

${t}_{\text{Br}} = 30 \min$

${M}_{\text{Ar"->"molar mass of Ar"=40" g/"mol}}$

${M}_{\text{Br"->"molar mass of" Br_2~~160" g/"mol}}$

So from (2) we get

$\implies \frac{30}{t} _ \text{Ar} = \sqrt{\frac{160}{40}} = 2$

$\implies {t}_{\text{Ar}} = \frac{30}{2} = 15 \min$