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:

For a Maxwell distribution, \( v_p : \bar v : v_{rms} = \sqrt2 : \sqrt{8/\pi} : \sqrt3 \). Average kinetic energy is \( \tfrac32 kT \).

Answer 1

The Correct Option is C, D

Step 1: Understanding the Concept:
In an ideal gas the molecules move at many different speeds. Three speeds describe the spread: the most probable speed, the mean speed and the rms speed.

Step 2: Key Formulas:
\[ v_p = \sqrt{\frac{2kT}{m}}, \qquad \bar v = \sqrt{\frac{8kT}{\pi m}}, \qquad v_{rms} = \sqrt{\frac{3kT}{m}} \]
In units of \( \sqrt{kT/m} \), these are 1.414, 1.596 and 1.732.

Step 3: Check statement (C):
From the numbers above, \( 1.414 < 1.596 < 1.732 \). So \( v_p < \bar v < v_{rms} \). (C) is TRUE.

Step 4: Check statements (A) and (B):
The Maxwell distribution has a long tail with no upper cut-off. A few molecules always move faster than \( \sqrt2\, v_{rms} \approx 2.45\sqrt{kT/m} \). So (A) is FALSE.
The distribution also reaches down to zero speed. Some molecules move slower than \( v_p/\sqrt2 = \sqrt{kT/m} \). So (B) is FALSE.

Step 5: Check statement (D):
Average kinetic energy of a molecule is \( \tfrac32 kT \). From \( v_p^2 = 2kT/m \) we get \( kT = \tfrac12 m v_p^2 \).
\[ \langle KE \rangle = \frac32 \times \frac12 m v_p^2 = \frac34 m v_p^2 \]
So (D) is TRUE.

Final Answer:
Statements (C) and (D) are correct. \[ \boxed{\text{(C) and (D)}} \]