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

Total pressure = 1.5 atm.
Moles of A `= (2)/(M_(A))`
Moles of B `= (3)/(M_(B))`
Total moles `= (2)/(M_(A)) + (3)/(M_(B))`
Partial pressure of A `= (2//M_(A))/(2//M_(A) + 3//M_(B)) xx 1.5`
and partial pressure of B `= (3//M_(B))/(2//M_(A) + 3//M_(B)) xx 1.5`
As given, partial pressure of A = 1 atm.
Partial pressure of B = 1.5 - 1 = 0.5 atm.
Thus, `((2//M_(A))/(2//M_(A)+3//M_(B)) xx 1.5)/((3//M_(B))/(2//M_(A) + 3//M_(B)) xx 1.5) = (1)/(0.5) = 2`
or `(2)/(3) xx (M_(B))/(M_(A)) = 2` or `(M_(A))/(M_(B)) = (1)/(3)`