Question

A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is

• A. 2√2:√3
• B. 2√2:1
• C. 1:2
• D. 2:1

Answer 1

The Correct Option is D

Step 1: Use the Ideal Gas Law to Determine the Molar Masses of Gases X and Y 

The ideal gas law relates pressure and the number of moles:

\[ P \propto n \quad \text{or} \quad P \propto \frac{\text{mass}}{\text{molar mass}}. \]

For Gas X:

• \( P_X = 2 \) atm
• Mass of X = 10 g

Let the molar mass of gas X be \( M_X \). Then:

\[ \frac{\text{mass of X}}{M_X} \propto P_X \Rightarrow M_X \propto \frac{\text{mass of X}}{P_X} = \frac{10}{2} = 5. \]

For Gas Y:

• \( P_Y = P_{\text{total}} - P_X = 6 - 2 = 4 \) atm
• Mass of Y = 80 g

Let the molar mass of gas Y be \( M_Y \). Then:

\[ M_Y \propto \frac{\text{mass of Y}}{P_Y} = \frac{80}{4} = 20. \]

Step 2: Calculate the Ratio of Root Mean Square Velocities

The root mean square (RMS) velocity of a gas is given by:

\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \]

where \( M \) is the molar mass. The ratio of \( v_{\text{rms}} \) for X and Y is:

\[ \frac{v_{\text{rms, X}}}{v_{\text{rms, Y}}} = \sqrt{\frac{M_Y}{M_X}}. \]

Substituting \( M_X = 5 \) and \( M_Y = 20 \):

\[ \frac{v_{\text{rms, X}}}{v_{\text{rms, Y}}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2. \]

Thus, the ratio is:

\[ v_{\text{rms, X}} : v_{\text{rms, Y}} = 2 : 1. \]