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

Rate of effusion is expressed as
$-\frac{d p}{d t}=\frac{k p}{\sqrt{M}}$
$k=$ constant, $p=$ instantaneous pressure
$\Rightarrow -\frac{d p}{p}=\frac{k d t}{\sqrt{M}}$
Integration of above equation gives
$\ln \left(\frac{p_{0}}{p}\right)=\frac{k t}{\sqrt{M}}$
Using first information : $\ln \left(\frac{2000}{1500}\right)=\frac{k 47}{\sqrt{32}}$
$\Rightarrow k=\frac{\sqrt{32}}{47} \ln \left(\frac{4}{3}\right) ....$ (i)
Now in mixture, initially gases are taken in equal mole ratio, hence they have same initial partial pressure of $2000 \,mm$ of $Hg$ each.
After $74 \,\min$ :
For $O _{2} \ln \left(\frac{2000}{p_{ O _{2}}}\right)=\frac{74 k}{\sqrt{32}}$
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 $74 \min =1271.5 \,mm$
For unknown gas: $\ln \left(\frac{2000}{p_{g}}\right)=\frac{74 k}{\sqrt{79}}$
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}=1500 \,mm$
$\Rightarrow$ After $74\, \min , p\left( O _{2}\right): p(g)=1271.5: 1500$
Also, in a mixture, partial pressure $\propto$ number of moles
$\Rightarrow n\left( O _{2}\right): n(g)=1: 1.18$