When `2 g` of a gas `A` is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be `1 atm`. If `3 g` of another gas `B` is then heated in the same flask, the total pressure becomes `1.5 atm`. Assuming ideal gas behaviour, calculate the ratio of the molecular weights `M_(A)` and `M_(B)`.

Options

A. `1:3`
B. `1:1`
C. `2:1`
D. `3:1`

Answer

A

Explanation

From ideal gas equation, PV = nRT
`PV=(m/M)RT` or `M=m(RT)/(PV)`
Let the molecular wt. of A and B be `M_(A)` and `M_(B)` respectively.
Then `M_(A)=2(RT)/(1xxV),M_(B)(3xxRT)/(0.5xxV)`
`therefore(M_(A))/(M_(B))=(2RT)/Vxx(0.5V)/(3RT)=(2xx0.5)/3=1/3`
Therefore, the ratio `M_(A) : M_(B) = 1 : 3`