Question

The pressure in bulb dropped from $2000$ to $1500mmHg$ in $47$ mins, where the $O_2$ present in the bulb leaked through a small hole. The bulb was then completely evacuated. A mixture of $O_2$ and another gas of molecular weight of $79$ in the molar ratio $1:1$ at a total pressure of $400mmHg$ was introduced. Find the mole ratio of two gases remaining in the bulb after a period of $74$ mins.

• A. $1:1.236$
• B. $1:2.136$
• C. $1:3.336$
• D. $2:1.136$

Answer 1

Answer: (A)

$1:1.236$

Pressure of $O_2$ (at $t=0$) = 2000mm, $n_1$ mole taken initially

Pressure of $O_2$ (at $t=47$ min) = 1500, $n_2$mole left after 47min.

For pressure $O_2-\frac{P_1}{P_2}=\frac{n_1}{n_2}$

$\frac{n_1}{n_2}=\frac{2000}{1500}=\frac{4}{3}{n}_2=\frac{3}{4}{n}_1$

Mole of $O_2$defined in 47 minute$=n_1-\frac{{3n}_1}{4}=\frac{n_1}{4}$

Mole of $O_2$will diffused in 74 minutes

$=\frac{n_1}{4}\times \frac{74}{47}=\frac{74}{188}{n}_1$

$=0.3936(n_1=1)$

Now diffusion of $O_2$in mixture also occurs at partial pressure of 2000 nm

When both$O_2$& gas diffuses simultaneously at 2000nm.

Pressure then 74 minute

$\frac{n_0}{74}\times \frac{74}{n_g}=\frac{\sqrt{79}}{\sqrt{32}}$

$n_g={no}_2\times \frac{\sqrt{32}}{\sqrt{79}}=0.3936\times \frac{\sqrt{32}}{\sqrt{79}}=0.247$

Mole of $O_2$left after 74 minute=1-0.3936=0.6064

Also mole of a gas left after 74 minute=1-0.249=0.249=0.7510

${no}_2:n_g:0.6064:0.7510::1:1.236$

The correct option is $A$.