Step-by-step Solution:

1. Understand the initial conditions for gas X.

• A closed vessel of volume $V$ contains gas $\mathbf{X}$.
• Mass of gas $\mathbf{X}$, $m_X = 10 \mathrm{~g}$.
• Temperature, $T = 300 \mathrm{~K}$.
• Pressure exerted by gas $\mathbf{X}$, $P_X = 2 \mathrm{~atm}$.
• Let the molar mass of gas $\mathbf{X}$ be $M_X$.

Using the ideal gas equation, $PV=nRT$, where $n = \frac{\text{mass}}{\text{molar mass}}$. For gas $\mathbf{X}$: $P_X V = n_X R T$ $2V = \frac{10}{M_X} RT \quad \cdots (1)$

2. Understand the final conditions after adding gas Y.

• Gas $\mathbf{Y}$ is added to the same vessel.
• Mass of gas $\mathbf{Y}$, $m_Y = 80 \mathrm{~g}$.
• The temperature remains constant at $T = 300 \mathrm{~K}$.
• The total pressure becomes $P_{total} = 6 \mathrm{~atm}$.
• Let the molar mass of gas $\mathbf{Y}$ be $M_Y$.

According to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures of the individual gases. $P_{total} = P_X + P_Y$ Since gas $\mathbf{X}$ is still in the container at the same temperature and volume, its partial pressure remains $P_X = 2 \mathrm{~atm}$. $6 \mathrm{~atm} = 2 \mathrm{~atm} + P_Y$ $P_Y = 6 - 2 = 4 \mathrm{~atm}$

Now, apply the ideal gas equation for gas $\mathbf{Y}$: $P_Y V = n_Y R T$ $4V = \frac{80}{M_Y} RT \quad \cdots (2)$

3. Determine the ratio of molar masses ($M_X/M_Y$).

• We have two equations:
  1. $2V = \frac{10}{M_X} RT$
  2. $4V = \frac{80}{M_Y} RT$
• Divide equation (2) by equation (1) to eliminate $V$, $R$, and $T$: $\frac{4V}{2V} = \frac{\frac{80}{M_Y} RT}{\frac{10}{M_X} RT}$ $2 = \frac{80/M_Y}{10/M_X}$ $2 = \frac{80}{M_Y} \times \frac{M_X}{10}$ $2 = 8 \times \frac{M_X}{M_Y}$ $\frac{M_X}{M_Y} = \frac{2}{8} = \frac{1}{4}$ This implies $M_Y = 4M_X$.

4. Calculate the ratio of root mean square (rms) velocities.

• The formula for the root mean square velocity of a gas is given by: $v_{rms} = \sqrt{\frac{3RT}{M}}$

• At a constant temperature $T$, the rms velocity is inversely proportional to the square root of the molar mass, $v_{rms} \propto \frac{1}{\sqrt{M}}$.

• The ratio of the rms velocities of gas $\mathbf{X}$ and gas $\mathbf{Y}$ is: $\frac{v_{rms, X}}{v_{rms, Y}} = \frac{\sqrt{\frac{3RT}{M_X}}}{\sqrt{\frac{3RT}{M_Y}}} = \sqrt{\frac{M_Y}{M_X}}$

• Substitute the ratio of molar masses we found in step 3: $\frac{v_{rms, X}}{v_{rms, Y}} = \sqrt{\frac{4}{1}} = 2$

• Therefore, the ratio of the root mean square velocities is $v_{rms, X} : v_{rms, Y} = 2 : 1$.

5. Conclusion The ratio of root mean square velocities of $\mathbf{X}$ and $\mathbf{Y}$ at $300 \mathrm{~K}$ is $2:1$. This matches option D.