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Ideal Gas Equation

Two grams of ...

Question

Two grams of gas $A$ are introduced in an evacuated flask at $25^{o} C$ . The pressure of the gas is $1 a t m$ . Now $3 g$ of another gas is introduced in the same flask, the total pressure becomes $1.5 a t m$ . The ratio of molecular mass $M_{A}$ and $M_{B}$ is $x : y$ . Find the value of $x \times y$

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Solution

For gas A : Its mass = $2 g$
$P_{A} = 1 a t m, T = 298 K$
For gas B : Its mass = $3 g$
$P_{B} = 0.5 a t m, T = 298 K$
According to Dalton's law of partial pressure
$P = P_{A} + P_{B}$
$1.5 = 1 + P_{B}$
$∴, P_{B} = 0.5 a t m$
Now,
$P V = n R T$
$P V = \frac{w}{m} R T$
For gas A
$P_{A} = 1 a t m, m = M_{A}, w = 2 g$
$1 \times V = \frac{2 \times R T}{M_{A}} . . . (i)$
For gas B
$P_{B} = 0.5 a t m, m = M_{B}, w = 3 g$
$0.5 \times V = \frac{3 \times R T}{M_{B}} . . . (i i)$
Dividing equation (i) and (ii)
$0.5 = \frac{3 \times M_{A}}{2 M_{B}}$
or
$\frac{M_{A}}{M_{B}} = \frac{1}{3}$
Hence, $1 \times 3 = 3$ .

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