States of Matter - Result Question 50
Passage
$X$ and $Y$ are two volatile liquids with molar weights of $10 \mathrm{~g} \mathrm{~mol}^{-1}$ and 40 g $\mathrm{mol}^{-1}$ respectively. Two cotton plugs, one soaked in $X$ and the other soaked in $Y$, are simultaneously placed at the ends of a tube of length $L=24 \mathrm{~cm}$, as shown in the figure. The tube is filled with an inert gas at 1 atm pressure and a temperature of $300 K $. Vapours of $X$ and $Y$ react to form a product which is first observed at a distance $d \mathrm{~cm}$ from the plug soaked in $X$. Take $X$ and $Y$ to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.
(2014 Adv.)
50. The experimental value of $d$ is found to be smaller than the estimate obtained using Graham’s law. This is due to
(a) larger mean free path for $X$ as a compared of that of $Y$
(b) larger mean free path for $Y$ as compared to that of $X$
(c) increased collision frequency of $Y$ with the inert gas as compared to that of $X$ with the inert gas
(d) increased collision frequency of $X$ with the inert gas as compared to that of $Y$ with the inert gas
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Answer:
Correct Answer: 50. (d)
Solution:
X is a lighter gas than Y , hence X has greater molecular speed. Due to greater molecular speed of X , it will have smaller mean free path and greater collision frequency with the incrt gas molecules. As a result X will take more time to travel a given distance along a straight line. Hence X and Y will meet at a distance smaller than one calculated from Graham’s law.
Hence, (d) is the correct choice.
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