QJEEAdv24Chf36d: JEE Advanced 2024 Chemistry - Class 11 JEEAdv24Chf36d Solution: A closed vessel contains of an ideal gas at , which exerts…

To solve this problem, we'll start by using the ideal gas law and the formula for the root mean square (rms) velocity of a gas.

The ideal gas law is given by:

$PV = nRT$

where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature. The rms velocity ( $v_{rms}$ ) of a gas is given by:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

where $M$ is the molar mass of the gas. We need the rms velocities of gases $X$ and $Y$ .

First, let's find the number of moles of each gas. For gas $X$ , let’s denote the molar mass as $M_X$ . The number of moles is:

$n_X = \frac{10 \, \text{g}}{M_X}$

For the initial gas $X$ at $300 \, \text{K}$ and $2 \, \text{atm}$ , we can write:

$P_X V = n_X RT$

When gas $Y$ is added, the total pressure becomes $6 \, \text{atm}$ . Let the molar mass of $Y$ be $M_Y$ . The number of moles of gas $Y$ added is:

$n_Y = \frac{80 \, \text{g}}{M_Y}$

Now, the total number of moles in the vessel is $n_X + n_Y$ , and this total exerts a pressure of $6 \, \text{atm}$ at the same volume and temperature:

$P_{total} V = (n_X + n_Y) RT$

Given the total pressure is $6 \, \text{atm}$ , solving for the volume $V$ and equating, we get two equations:

$2V = \frac{10}{M_X} RT \quad ......\text{(i)}$

$6V = \left( \frac{10}{M_X} + \frac{80}{M_Y} \right) RT \quad ......\text{(ii)}$

Dividing (ii) by (i) and solving for $\frac{80}{M_Y}$ , we get:

$\frac{80}{M_Y} = 2 \frac{10}{M_X}$

$\frac{80}{M_Y} = \frac{20}{M_X}$

$M_Y = 4 M_X$

Now, applying the formula for the rms velocity, the ratio of rms velocities of $X$ and $Y$ is:

$\frac{v_{rms, X}}{v_{rms, Y}} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{4 M_X}{M_X}} = 2$

So, the ratio of root mean square velocities of $X$ and $Y$ is $2:1$ .

Thus, the correct answer is Option D: $2:1$ .