Question

Let \( \bar{v}, v_{rms}, v_p \) denote the mean speed, root mean square speed and most probable speed of molecules of mass \( m \) in an ideal monoatomic gas at temperature \( T \). Which statements are correct?

• A. No molecules can have speed greater than \( \sqrt{2}v_{rms} \)
• B. No molecules can have speed less than \( \frac{v_p}{\sqrt{2}} \)
• C. \( v_p < \bar{v} < v_{rms} \)
• D. Average kinetic energy of a molecule is \( \frac{3}{4} m v_p^2 \)

Hint:

Gas speeds:

• \( v_p < \bar{v} < v_{rms} \).
• Average KE = \( \frac{3}{2} kT \).

Answer 1

The Correct Option is C

Concept: Maxwell distribution Relations: \[ v_p = \sqrt{\frac{2kT}{m}},\quad \bar{v} = \sqrt{\frac{8kT}{\pi m}},\quad v_{rms} = \sqrt{\frac{3kT}{m}} \] Step 1: Speed ordering. \[ v_p < \bar{v} < v_{rms} \] So (C) correct. Step 2: Statements A and B. Maxwell distribution allows wide range of speeds, so no strict limits ⇒ false. Step 3: Average KE. \[ \langle KE \rangle = \frac{3}{2}kT \] From \( v_p^2 = \frac{2kT}{m} \): \[ kT = \frac{mv_p^2}{2} \] \[ \langle KE \rangle = \frac{3}{2} \cdot \frac{mv_p^2}{2} = \frac{3}{4}mv_p^2 \] So (D) correct.