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 . What would be the ratio of their molecular masses ?

Explanation

Pressure of `1g` of an ideal gas `A` at `27^(@)C (300 K)=2 "bar"` pressure of introducing `2 g` of another gas `B` in the same flask `=3 "bar"`
Pressure of ideal gas `B` in the same flask `=1 "bar"`
`P=W/(MV)=RT`
`2=1/(M_(A)xxV)RT` …(i)
`1=2/(M_(B)xxV)RT` ...(ii)
From equations (i) and (ii), we get
`2=M_(B)/(2M_(A))` or `M_(B)=4M_(A)`