Ratio of Root Mean Square Velocities of Ideal Gases

To find the ratio of the root mean square ($u_{\text{rms}}$) velocities of ideal gases X and Y, we use the ideal gas equation and the expression for $u_{\text{rms}}$.

Step 1: Determine the partial pressure of each gas

For a closed vessel of fixed volume $V$ at constant temperature $T = 300\text{ K}$:

• The partial pressure exerted by gas X is: $P_X = 2\text{ atm}$
• When gas Y is added, the total pressure becomes $P_{\text{total}} = 6\text{ atm}$. According to Dalton's Law of Partial Pressures: $P_{\text{total}} = P_X + P_Y$ $P_Y = P_{\text{total}} - P_X = 6\text{ atm} - 2\text{ atm} = 4\text{ atm}$

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Step 2: Relate partial pressures to the number of moles

Using the ideal gas equation $PV = nRT$, at constant $V$ and $T$, pressure is directly proportional to the number of moles ($P \propto n$): $\frac{P_X}{P_Y} = \frac{n_X}{n_Y}$

Substituting the values of partial pressures: $\frac{n_X}{n_Y} = \frac{2}{4} = \frac{1}{2}$

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Step 3: Calculate the ratio of molar masses

The number of moles $n$ of a gas is given by $n = \frac{w}{M}$, where $w$ is the mass and $M$ is the molar mass.

• For gas X: $n_X = \frac{10}{M_X}$
• For gas Y: $n_Y = \frac{80}{M_Y}$

Substituting these expressions into the mole ratio: $\frac{n_X}{n_Y} = \frac{\frac{10}{M_X}}{\frac{80}{M_Y}} = \frac{10}{80} \times \frac{M_Y}{M_X} = \frac{1}{8} \cdot \frac{M_Y}{M_X}$

Equating this to $\frac{1}{2}$: $\frac{1}{8} \cdot \frac{M_Y}{M_X} = \frac{1}{2}$ $\frac{M_Y}{M_X} = \frac{8}{2} = 4$

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Step 4: Find the ratio of root mean square velocities

The root mean square velocity of an ideal gas is given by: $u_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

Since both gases are at the same temperature $T = 300\text{ K}$: $\frac{u_{\text{rms}, X}}{u_{\text{rms}, Y}} = \sqrt{\frac{M_Y}{M_X}}$

Substituting $\frac{M_Y}{M_X} = 4$: $\frac{u_{\text{rms}, X}}{u_{\text{rms}, Y}} = \sqrt{4} = 2$

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

Correct Answer: (D) $2 : 1$