When 2g of a gas A is introduced into an evacuated flask kept at `25^@C`, the pressure is found to be one atmosphere. If 3 g of another gas B are then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of molecular weights `M_A : M_B`.

Explanation

The ideal gas equation is
`PV=nRT=(w)/(M)RT` `PV=nRT=(w)/(M)RT` ["`w=`weight of substance", "`M=`molecular weight"]
`:. M=w(RT)/(PV)`
Let the molecular weights of `A` and `B` be `M_(A)` and `M_(B)`, respectively, Therefore,
`M_(A)=(2xxRT)/(1xxV)`
and `M_(B)=(3xxRT)/(0.5xxV)`
Since pressure due to gas `B` is `1.5-1.0=0.5 atm`)
Hence,
`(M_(A))/(M_(B))=(2RT)/(V)xx(0.5V)/(3RT)=(2xx0.5)/(3)=(1)/(3)`
`:. M_(A) : M_(B)= 1: 3`