Question

Statement-1: Two gases \( \mathrm{H_2} \) and \( \mathrm{O_2} \) are having the same average kinetic energy, then they have the same temperature.
Statement-2: \( \mathrm{H_2} \) and \( \mathrm{O_2} \) will have the same \( v_{\mathrm{rms}} \) at the same temperature.

• A. Statement-1 & Statement-2 both are correct and Statement-2 is correct explanation of Statement-1.
• B. Both Statement-1 & Statement-2 correct but Statement-2 is not correct explanation of Statement-1.
• C. Statement-1 true and Statement-2 is false.
• D. Both are false.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The average kinetic energy of a gas molecule depends solely on its absolute temperature. However, the root mean square velocity (\( v_{rms} \)) depends on both the temperature and the molar mass of the gas.

Step 2: Key Formula or Approach:
1. Average Kinetic Energy (\( KE_{avg} \)) = \( \frac{3}{2} kT \).
2. Root Mean Square Velocity (\( v_{rms} \)) = \( \sqrt{\frac{3RT}{M}} \).

Step 3: Detailed Explanation:
- Statement-1: Since \( KE_{avg} = \frac{3}{2} kT \), if two gases have the same average kinetic energy, they must be at the same temperature \( T \). This statement is true.
- Statement-2: At the same temperature, \( v_{rms} \propto \frac{1}{\sqrt{M}} \). Since the molar mass of \( \ce{H2} \) (2 g/mol) is much smaller than that of \( \ce{O2} \) (32 g/mol), \( \ce{H2} \) will have a much higher \( v_{rms} \) than \( \ce{O2} \). This statement is false.

Step 4: Final Answer:
Statement-1 is true and Statement-2 is false.