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)`.

Explanation

Let the molecular masses of A and B be `M_(A)` and `M_(B)` respectively.
Pressure exerted by the gas `B = (1.5 - 1.0) = 0.5` atm. Volume and temperature are same in both the gases.
For gas A: `P = 1 atm, w = 2 g, M = M_(A)`
We know that, `PV = (w)/(M) RT`
`1 xx V = (2)/(M_(A)). RT " or " M_(A) = (2RT)/(V)`....(i)
For gas B: `P = 0.5 atm, w = 3g, M = M_(B)`
`0.5 xx V = (3)/(M_(B)).RT " or " M_(B) = (3RT)/(0.5 xx V)`....(ii)
Dividing Eq. (i) by Eq. (ii).
`(M_(A))/(M_(B)) = (2RT)/(V) xx (0.5 xx V)/(3RT)`
`= (2 xx 0.5)/(3) = (1)/(3)`
Thus, `M_(A) : M_(B) = 1 : 3`