Question

X and Y are two volatile liquids with molar weights of 10 g mol^–1 and 40 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 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere 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 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.

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Linked Question 1

The value of d in cm (shown in the figure), as estimated from Graham's law, is

$\frac{r_{x}}{r_{y}} = \sqrt{\frac{40}{10}} = \sqrt{\frac{M_{y}}{M_{x}}}$

$\frac{x}{24 - x} = 2$

x = 48 – 2x

x = 16

Linked Question 2

The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to

increases collision frequency of X with the inert gas as compared to that of Y with the inert gas.

larger mean free path for Y as compared to that of X.

increased collision frequency of Y with the inert gas as compared to that of X with the inert gas.

larger mean free path for X as compared to that of Y

Mean free math $(λ) = \frac{1}{\sqrt{2} {πσ}^{2} N^{*}}$

Here σ is same but no information is given about N*, so we are considering it same for both, therefore mean free path is same for both gases and collision frequency of gas X will be higher than gas Y & it will travel less distance than expected. Out of all given option D is best answer.