Pressure of the 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at Same temperature, the pressure become 3 bar. Relationship between their molecular masses is MB = xMA. Find the value of x.

pV = nRTFor A gas, pAV=nART -----(i) Similarly for B gas pBV=nBRT-------(ii)(V is same because the gases are taken in the same flask)nA=1MA, similarly nB=2MB (where, MA and MB are the molar masses of gases A and B respectively)pA=2 bar, p=pA+pB=3 bar  ∴ pB=3−2=1bar pApB=nBnB=1×MBMA×2⇒MBMA=2pApB=2×21 ⇒ MB=4MA Value of x =4