States of Matter - Result Question 92
92. 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 $74 min$.
$(1981,3 M)$
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Solution:
Rate of effusion is expressed as $-\dfrac{d p}{d t}=\dfrac{k p}{\sqrt{M}}$
$k=$ constant,$p=$ instantaneous pressure
$\Rightarrow \quad-\dfrac{d p}{p}=\dfrac{k d t}{\sqrt{M}}$
Integration of above equation gives $\ln \left(\dfrac{p _0}{p}\right)=\dfrac{k t}{\sqrt{M}}$
Using first information : $\ln \left(\dfrac{2000}{1500}\right)=\dfrac{k 47}{\sqrt{32}}$
$ \Rightarrow \quad k=\dfrac{\sqrt{32}}{47} \ln \left(\dfrac{4}{3}\right) \hspace{15mm}…(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 $O _2 \quad \ln \left(\dfrac{2000}{p _{O _2}}\right)=\dfrac{74 k}{\sqrt{32}}$
Substituting $k$ from Eq. (i) gives
$ \ln \left(\dfrac{2000}{p _{O _2}}\right)=\dfrac{74}{\sqrt{32}} \times \dfrac{\sqrt{32}}{47} \ln \left(\dfrac{4}{3}\right) $
$ \ln \left(\dfrac{2000}{p _{O _2}}\right)=\dfrac{74}{47} \ln \left(\dfrac{4}{3}\right) $
Solving gives $p\left(O _2\right)$ at $74 min=1271.5 mm$
For unknown gas : $\ln \left(\dfrac{2000}{p _g}\right)=\dfrac{74 k}{\sqrt{79}}$
Substituting $k$ from (i) gives
$ \ln \left(\dfrac{2000}{p _g}\right)=\dfrac{74}{\sqrt{79}} \times \dfrac{\sqrt{32}}{47} \ln \left(\dfrac{4}{3}\right) $
Solving gives : $\quad p _g=1500 mm$
$\Rightarrow$ After 74 min, $p\left(O _2\right): p(g)=1271.5: 1500$
Also, in a mixture, partial pressure $\propto$ number of moles
$ \Rightarrow \quad n\left(O _2\right): n(g)=1: 1.18 $
BETA
