Given, Initially, p=2‌atm,T=300K , V=vL,mx=1g Finally, p=2+1=3‌atm,T=300K , my=2g,V=vL Let molar mass of gas X and Y be Mx and My respectively. Applying ideal gas equation to initial and final conditions. Initially, 2×V=
1
Mx
‌RT ...(i) Finally, 3×V=(
1
My
+
2
My
)‌RT ...(ii) Dividing (i) by (ii), we get My=4Mx.