Q5.5. Pressure of 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

becomes 3 bar. Find a relationship between their molecular masses.

let molecular masses be m1 and m2,

PV=nRT,P=(RT/V)(n1 + n2)

2 bar=(RT/V)*n1=(RT/V)*(1/m1)

3 bar=(RT/V)((1/m1)+(2/m2))

Approved

The Pressure of A + Pressure of B = 3 Bar (Given :Pressure when both gases are present in the flask under same conditions is 3bar)

Using Daltons law of partial Pressures, Pressure of B = 3-2=1bar

According to an ideal gas equation, For gas A

P₁V=n₁RT ==> P₁V=(m₁/M₁₎RT

Similarly for the gas B

P₂V=n₂RT ==> P₂V=(m₂/M₂₎RT

Diving both the Equ. We get P₁/P₂₌m₁M2/m₂M_{1 ==>}P₁m₂/P₂m₁₌M2/M₁

M2/M₁= P₁m₂/P₂m₁

So the ratio of their Molecular masses(M2/M_{1)=(2*2)/(1*1)=(4:1)---(Ans)}