Pressure of 1 g ideal gas X at 300 K is 2 atm. When 2 g of another gas Y is introduced in the same vessel at same temperature, the pressure become 3 atm. The correct relationship between molar mass of X and Y is :

Explanation

From ideal gas equation,
We know that, `pv=nRT`
Now for gas A, `p_(A) V=n_(A)RT " " ...(i)`
For gas B, `p_(B)V=n_(B)RT" " ...(ii)`
Number of moles of A, `n_(A)=(1)/(M_(A))`
Number of moles of B, `n_(B)=(1)/(M_(B))`
where `M_(A)` and `M_(B)` are molar masses of gases A and B respectively
Also, `p_(A)=2"bar " p_("total")=p_(A)+p_(B)=3` bar
`:.p_(B)=3-2=1` bar
Now from equation (i) and (ii) `(p_(A))/(p_(B)) =(n_(A))/(n_(B)) =(1xx M_(A))/(M_(B) xx 2)`
`:. (M_(A))/(M_(B))=(2p_(A))/(p_(B)) =(2xx2)/(1)=4 rArr M_(B) =4M_(A)`