Pressure of $1 \, g$ of an ideal gas $A$ at $27^{\circ} C$ is found to be $2 \, bar$. When $2 \, g$ of another ideal gas $B$ is introduced in the same flask at the same temperature,the pressure becomes $3 \, bar$. Find a relationship between their molecular masses.

(D) For ideal gas $A$,the ideal gas equation is $p_{A} V = n_{A} R T$ $(i)$.
For ideal gas $B$,the ideal gas equation is $p_{B} V = n_{B} R T$ $(ii)$.
Since $V$ and $T$ are constant,we have $\frac{p_{A} V}{n_{A}} = \frac{p_{B} V}{n_{B}} = R T$.
Substituting $n = \frac{m}{M}$,we get $\frac{p_{A} M_{A}}{m_{A}} = \frac{p_{B} M_{B}}{m_{B}}$.
Given $m_{A} = 1 \, g$,$p_{A} = 2 \, bar$,$m_{B} = 2 \, g$.
The partial pressure of gas $B$ is $p_{B} = p_{total} - p_{A} = 3 \, bar - 2 \, bar = 1 \, bar$.
Substituting the values: $\frac{2 \times M_{A}}{1} = \frac{1 \times M_{B}}{2}$.
Therefore,$4 M_{A} = M_{B}$.