Two containers A and B have the same volume. Container A contains 5 moles of `O_(2)` gas. Container B contains 3 moles of He and 2 moles of `N_(2)`. Both the containers are separately kept in vacuum at the same temperature. Both the containers have very small orifices of the same area through which the gases leak out. Compare the rate of effusion of `O_(2)` with that of He gas mixture.

Explanation

Since both the containers are in the same conditions of P, V and T,
`r_(O_(2))/(r_("mix"))=sqrt((M_("mix"))/(M_(O_(2)))`
As the mixture contains three moles of He and 2 moles of `N_(2)`, the effective molecular weight of the mixture would be.
`(3)/(5)xx4+(2)/(5)xx28=13.6`
`therefore (r_(O_(2)))/(r_("mix"))=sqrt((13.6)/(32))=0.652`
Though this solution looks OK, there is one big flaw in it.
The error is that we have assumed that `He and N_(2)` from vessel B would effuse out with the same rate. This assumption was made in because we have taken the composition of the gas mixture coming out of the vessel to be same as that of the mixture that was inside the vessel. It should be duly noted that the two mixtures (inside and the are that effused out) have different compositions. Therefore first we must find the composition of the gas mixture coming out of the vessel B.
`(r_(N_(2)))/(r_("He"))=(2)/(3) sqrt((4)/(28))=(2)/(3) sqrt((1)/(7))=0.252`
This means that initially the ratio of moles of `N_(2)` to the moles of He coming out of the vessel are in the molar ratio of 0.252 and not `(2)/(3)`
Let moles of the He coming out to be x
`therefore` Moles of `N_(2)` coming out is 0.252 x
`therefore (n_(N_(2)))/(n_("total"))=(0.252 x)/(1.252x)=0.2`
`(n_("He"))/(n_("total"))=0.8 rArr M_("mix") =0.2xx28+0.8xx4=8.8`
`therefore (r_(O_(2)))/(r_("mix"))=sqrt((8.8)/(32))=0.52`