Question

Match List - I with List - II. 

Choose the correct answer from the options given below : 
 

• A. A-III, B-II, C-IV, D-I
• B. A-III, B-I, C-II, D-IV
• C. A-II, B-I, C-IV, D-III
• D. A-IV, B-III, C-I, D-II

Hint:

A quick mnemonic for ordering the magnitudes of these gas speeds is R > A > M (Rams): Root mean square ($\sqrt{3} \approx 1.732$) $>$ Average ($\sqrt{8/\pi} \approx 1.595$) $>$ Most probable ($\sqrt{2} \approx 1.414$).

Answer 1

The Correct Option is A

Step 1: Concept:
This problem requires matching different statistical gas speeds derived from the Maxwell-Boltzmann distribution to their correct mathematical formulas.

Step 2: Key Formula or Approach:
For an ideal gas with molar mass $M$ at temperature $T$:
- Root mean square speed ($v_{rms}$) = $\sqrt{\frac{3RT}{M}}$
- Mean (average) speed ($v_{avg}$) = $\sqrt{\frac{8RT}{\pi M}}$
- Most probable speed ($v_{mp}$) = $\sqrt{\frac{2RT}{M}}$
- Mean relative speed ($v_{rel}$) of two colliding molecules = $\sqrt{2} \times v_{avg} = \sqrt{\frac{16RT}{\pi M}}$. Expressed using reduced mass ($\mu = m/2$) and the Boltzmann constant ($k$), this transforms to $\sqrt{\frac{8kT}{\pi\mu}}$.

Step 3: Step-by-step Explanation:

• A. Root mean square speed: By definition, $v_{rms} = (\frac{3RT}{M})^{1/2}$.
This matches III.

• B. Mean speed: The average speed of gas molecules is $v_{avg} = (\frac{8RT}{\pi M})^{1/2}$.
This matches II.

• C. Most probable speed: The speed at the peak of the Maxwell-Boltzmann curve is $v_{mp} = (\frac{2RT}{M})^{1/2}$.
This matches IV.

• D. Mean relative speed: The average relative speed between two colliding molecules is critical in collision theory. Using atomic mass $m$ and reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m}{2}$ (for identical molecules), the formula is $(\frac{8kT}{\pi\mu})^{1/2}$.
This matches I.
The final matching sequence is A-III, B-II, C-IV, D-I.

Step 4: Final Answer:
The matching pairs correspond perfectly to option (A).