Question

When $2 \,g$ of a gas $A$ is introduced into an evacuated flask kept at $25^{\circ} C$, the pressure is found to be one atmosphere. If $3 \,g$ of another gas $B$ is then added to the same flask, the total pressure becomes $1.5$ atm. Assuming ideal gas behaviour, calculate the ratio of the molecular weights $M_A : M_B.$

Answer 1

From the given information, it can be easily deduced that in the final mixture,
partial pressure of $A=1.0 \,atm$
partial pressure of $B=0.5\, atm$
Also $ n_{A}=\frac{p_{A} V}{R T}=\frac{V}{R T}$
$n_{B}=\frac{p_{B} V}{R T}=\frac{0.5 V}{R T}$
$\Rightarrow \frac{n_{B}}{n_{A}}=\frac{1}{2}=\frac{w_{B}}{M_{B}} \times \frac{M_{A}}{w_{A}}$
$=\frac{3}{2} \times\left(\frac{M_{A}}{M_{B}}\right)$
$\Rightarrow M_{A}: M_{B}=1: 3$