Question

The pressure in a bulb dropped from $2000$ to $1500$ mm of mercury in $47$ min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight $79$ in the molar ratio of $1:1$ at a total pressure of $4000$ mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of $74$ min.

Answer 1

The change of pressure of oxygen after $74$ min= $\frac{500}{47}\times 74=787.2mm$

The $1:1$ molar mixture of oxygen and another gas, each of them has an equal pressure of $2000mm$ because the total pressure is given to be $4000mm$ $Hg$.

The pressure of oxygen left after $74$ min

$=2000-787.2=1212.8mm$ $Hg$

Graham's law of diffusion:

$\frac{r_{gas}}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$

$\Rightarrow \frac{V_{gas}\left(diffused\right)}{V_{O_2}\left(diffused\right)}\times \frac{t_{O_2}}{t_{gas}}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$

$\Rightarrow \frac{P_{gas}\left(diffused\right)}{P_{O_2}\left(diffused\right)}\times \frac{t_{O_2}}{t_{gas}}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$

Both diffuse for the same time, so

$\Rightarrow \frac{P_{gas}\left(diffused\right)}{787.2}=\sqrt{\frac{32}{79}}$

$\Rightarrow P_{gas}\left(diffused\right)=500.8mm$ $Hg$

The pressure of gas left after $74$ min= $2000-500.8=1499.2mm$ $Hg$

Molar ratio of the gas and oxygen left= $\frac{1499.2}{1212.8}=1.236$ .