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

Number of moles of `A=(2)/(M_(A))`
Number of moles of `B=(3)/(M_(B))`
Initial pressure `P_(1) prop (2)/(M_(A)) prop 1 atm`
Final pressure `P_(2) prop (2)/(M_(A))+(3)/(M_(B)) prop 15 atm`
`(P_(1))/(P_(2))=(2//M_(A))/(2//M_(A)+3//M_(B))=(2//M_(A))/((2M_(B)+3M_(A))/(M_(A).M_(B)))`
`(2)/(3)=(2)/(M_(A))xx(M_(A).M_(B))/(2M_(B)+3M_(A))`
`(M_(A))/(3)=(M_(A).M_(B))/(2M_(B)+3M_(A))`
`(3)/(M_(A))=(3M_(A)+2M_(B))/(M_(A).M_(B))`
`3M_(B)=3M_(A)+2M_(B)`
`3M_(A)=M_(B)`
`M_(A) : M_(B) = 1:3`