Question

$X$ and $Y$ are two volatile liquids with molar weights $10{\text{g mol}}^{-1}$ and $40{\text{g mol}}^{-1}$ 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 $L=24\text{cm}$ as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of $300\text{K}$. Vapours of $X$ and $Y$ react to form a product which is first observed at a distance $d\text{cm}$ from the plug soaked in $X$. Consider $X$ and $Y$ to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.
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 compared to 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

Answer 1

The correct option is D Increased collision frequency of $X$ with the inert gas as compared to that of $Y$ with the inert gas

Mean free path is defined as the average distance travelled by the molecules between two consecutive collisions. The general formula of mean free path $\left(\lambda \right)$ is
$\lambda =\frac{RT}{\sqrt{2}\pi d^2{N}_{A}p}$
($d=$ diameter of molecule, $p=$ pressure inside the vessel)

$X$ is a lighter gas than $Y$. Hence, $X$ has a higher molecular speed. Due to the greater molar speed of $X$, it will have smaller mean free path and greater collision with the inert gas molecules. As a result, $X$ will take more time to travel a given distance along the straight line. Hence, $X$ and $Y$ will meet at a distance smaller than that calculated by Graham's law.