Question

The RMS velocity of dihydrogen is \( \sqrt{7} \) times more than that of dinitrogen. If \( T_{\text{H}_2} \) and \( T_{\text{N}_2} \) are the temperatures of dihydrogen and dinitrogen, then the correct relationship between them is

• A. \( T_{\text{H}_2} = T_{\text{N}_2} \)
• B. \( T_{\text{H}_2} \textgreater T_{\text{N}_2} \)
• C. \( T_{\text{H}_2} = \sqrt{7} T_{\text{N}_2} \)
• D. \( T_{\text{H}_2} = \frac{T_{\text{N}_2}}{2} \)

Hint:

Always square the RMS ratio relation to remove the square roots from the formulas for simpler algebraic manipulation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
RMS velocity formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \). Relation given: \( v_{H_2} = \sqrt{7} \times v_{N_2} \). Note: "times more than" usually means \( v + \sqrt{7}v \), but in physics/chem MCQ context, it often means "times of". The options suggest a direct multiplicative relationship. Let's check calculations for "times".
Step 3: Detailed Explanation:
\( M_{H_2} = 2 \) g/mol. \( M_{N_2} = 28 \) g/mol. Given: \( v_{H_2} = \sqrt{7} v_{N_2} \). Squaring both sides: \( v_{H_2}^2 = 7 v_{N_2}^2 \) Substitute formula: \( \frac{3RT_{H_2}}{M_{H_2}} = 7 \left( \frac{3RT_{N_2}}{M_{N_2}} \right) \) \( \frac{T_{H_2}}{2} = 7 \frac{T_{N_2}}{28} \) \( \frac{T_{H_2}}{2} = \frac{7 T_{N_2}}{28} = \frac{T_{N_2}}{4} \) \( T_{H_2} = \frac{2}{4} T_{N_2} = \frac{T_{N_2}}{2} \)
Step 4: Final Answer:
The relationship is \( T_{\text{H}_2} = \frac{T_{\text{N}_2}}{2} \).