Pressure of 1 g of an ideal gas A at `27^(@)C` is found to be 2 bar, when 2 g of another gas B is introduced in the same flask at same temperature. The pressure becomes 3 bar. Find a relationship between their molecular masses.

Explanation

Suppose molecular masses of A and B `M_(A)` and `M_(B)` respectively. Then their number of moles will be
`m_(A)=(1)/(M_(A))," "n_(B)=(2)/(M_(B))`
`P_(A)=2"bar", P_(A)+P_(B)=3"bar",i.e.,P_(B)=1"bar"`
Applying the relation, PV=nRT
`P_(A)V=n_(A)RT,P_(B)V=n_(B)RT`
`:." " (P_(A))/(P_(B))=(n_(A))/(n_(B))=(1//M_(A))/(2//M_(B))=(M_(B))/(2M_(A))`
`"or " (M_(B))/(M_(A))=2xx(P_(A))/(P_(B))=2xx(2)/(1)=4" " or " " M_(B)=4M_(A)`.