Question

When $2g$ of a gas $A$ is introduced into an evacuated flask kept at $25^{\circ }C$, the pressure is found to be one atmosphere. If $3g$ 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 $A=1.0atm$
partial pressure of $B=0.5atm$
Also $n_A=\frac{p_{A}V}{RT}=\frac{V}{RT}$
$n_B=\frac{p_{B}V}{RT}=\frac{0.5V}{RT}$
$\Rightarrow \frac{n_B}{n_A}=\frac{1}{2}=\frac{w_B}{M_B}\times \frac{M_A}{w_A}$
$=\frac{3}{2}\times \left(\frac{M_A}{M_B}\right)$
$\Rightarrow M_A:M_B=1:3$