AP EAPCET 23-Aug-2021 Shift 2 Paper

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.