Question

The pressure in a bulb dropped from $2000$ to $1500mm$ 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 $4000mm$ of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of $74\min$.

Answer 1

Rate of effusion is expressed as
$-\frac{dp}{dt}=\frac{kp}{\sqrt{M}}$
$k=$ constant, $p=$ instantaneous pressure
$\Rightarrow -\frac{dp}{p}=\frac{kdt}{\sqrt{M}}$
Integration of above equation gives
$\ln \left(\frac{p_0}{p}\right)=\frac{kt}{\sqrt{M}}$
Using first information : $\ln \left(\frac{2000}{1500}\right)=\frac{k47}{\sqrt{32}}$
$\Rightarrow k=\frac{\sqrt{32}}{47}\ln \left(\frac{4}{3}\right)....$ (i)
Now in mixture, initially gases are taken in equal mole ratio, hence they have same initial partial pressure of $2000mm$ of $Hg$ each.
After $74\min$ :
For $O_2\ln \left(\frac{2000}{p_{O_2}}\right)=\frac{74k}{\sqrt{32}}$
Substituting $k$ from Eq. (i) gives
$\ln \left(\frac{2000}{p_{O_2}}\right)=\frac{74}{\sqrt{32}}\times \frac{\sqrt{32}}{47}\ln \left(\frac{4}{3}\right)$
$\ln \left(\frac{2000}{p_{O_2}}\right)=\frac{74}{47}\ln \left(\frac{4}{3}\right)$
Solving gives $p\left(O_2\right)$ at $74\min =1271.5mm$
For unknown gas: $\ln \left(\frac{2000}{p_g}\right)=\frac{74k}{\sqrt{79}}$
Substituting $k$ from (i) gives
$\ln \left(\frac{2000}{p_g}\right)=\frac{74}{\sqrt{79}}\times \frac{\sqrt{32}}{47}\ln \left(\frac{4}{3}\right)$
Solving gives: $p_g=1500mm$
$\Rightarrow$ After $74\min ,p\left(O_2\right):p(g)=1271.5:1500$
Also, in a mixture, partial pressure $\propto$ number of moles
$\Rightarrow n\left(O_2\right):n(g)=1:1.18$