Question

The pressure in bulb dropped from 2000 to 1500 mm of Hg in 48 minutes when the contained $O_2$ leaked through a small hole. The bulb was then completely evacuated. A mixture of $O_2$ and another gas of molar mass 79 in the molar ratio 1:1 at a total pressure of 4000 mm of Hg was introduced. Find the mole ratio of two gases (x:y) remaining in the bulb after a period of 74 minutes and calculate x+y (only integer part).

Answer 1

Pressure of $O_2(att=0)=2000mm$, say $n_1$ mole taken initially
Pressure of $O_2$ (at $t=47mi{n}_{=}1500mm$, say $n_2$ mole left after 47 minute
$\therefore$ For pure $O_2:\frac{P_1}{P_2}=\frac{n_1}{n_2}$
$\therefore \frac{n_1}{n_2}=\frac{2000}{1500}=\frac{4}{3}$
$\therefore n_2=\left(\frac{3}{4}\right)n_1$
$\therefore$ mole of $O_2$ diffused in 47 minute $=n_1-\left(3n_1/4\right)=\left(n_1/4\right)$
$\therefore$ mole of $O_2$ will diffuse in 74 minute
$=\frac{n_1}{4}\times \frac{74}{47}=\frac{74}{188}{n}_1$
$=0.3936$ (Assume $n_1=1$)
Now diffusion of $O_2$ in mixture also occurs at partial pressure of 2000 mm (because the ratio of gas and $O_2$ being 1:1)
When both $O_2$ and gas diffuses simultaneously at 2000 mm pressure, then for 74 minute,
$\frac{n_{O_2}}{74}\times \frac{74}{n_g}=\sqrt{\frac{79}{32}}$
$\therefore n_g=n_{O_2}\times \sqrt{\left(32/79\right)}=0.3936\times \sqrt{\left(32/79\right)}$
$=0.249$
$\therefore$ mole of $O_2$ left after 74 minute $=1-0.3936=0.6064$
also mole of gas left after 74 minute $=1-0.249=0.7510$
$\therefore n_{O_2}:n_g::0.6064:0.7510::1:1.236$
so value of $x+y=1+1.236=2.236$
integer part of this is 2.