When $2 g$ of a gaseous substance $A$ is introduced into an initially evacuated flask at $25^{\circ} C$ , the pressure is found to be $1$ atmosphere. $3 g$ of another gaseous substance $B$ is then added to it at the same temperature and pressure. The final pressure is found to be $1.5$ atmosphere. Calculate the ratio of molecular masses of $A$ and $B$ assuming ideal gas behaviour.

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Solution

Let the molecular masses of $A$ and $B$ be $M_{A}$ and $M_{B}$ respectively.
Pressure exerted by the gas $B = (1.5 - 1.0) = 0.5 a t m$
Volume and temperature are same in both the gases.
For gas $A : P = 1 a t m, w = 2 g, M = M_{A}$
We know that, $P V = \frac{w}{M} R T$
$1 \times V = \frac{2}{M_{A}} \cdot R T$ or $M_{A} = \frac{2 R T}{V} . . . . (i)$
For gas $B : P = 0.5 a t m, w = 3 g, M = M_{B}$
$0.5 \times V = \frac{3}{M_{B}} \cdot R T$ or $M_{B} = \frac{3 R T}{0.5 \times V} . . . . (i i)$
Dividing Eq. (i) by Eq. (ii),

$\frac{M_{A}}{M_{B}} = \frac{2 R T}{V} \times \frac{0.5 \times V}{3 R T}$
$= \frac{2 \times 0.5}{3} = \frac{1}{3}$
Thus, $M_{A} : M_{B} = 1 : 3$

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