Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. Then

Options

A. No molecule can have a speed greater than
B. No molecule can have speed less than
C. `v_p lt vecv lt V_("m s")`
D. The average kinetic energy of a molecule is

Explanation

To solve the problem, we need to establish the relationships between the mean speed (\( \bar{v} \)), root mean square speed (\( v_{rms} \)), and most probable speed (\( v_p \)) of molecules in an ideal monoatomic gas at absolute temperature \( T \). We will also derive the kinetic energy of the molecules based on these speeds.

### Step-by-Step Solution:

1. **Define the Speeds**:
- The root mean square speed (\( v_{rms} \)) is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
- The most probable speed (\( v_p \)) is given by:
\[ v_p = \sqrt{\frac{2RT}{M}} \]
- The mean speed (\( \bar{v} \)) is given by:
\[ \bar{v} = \sqrt{\frac{8RT}{\pi M}} \]

2. **Establish Ratios**:
- To find the relationship between these speeds, we can take the ratios of \( v_{rms} \), \( \bar{v} \), and \( v_p \):
\[ \frac{v_{rms}}{v_p} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{3}{2}} \quad (1) \]
\[ \frac{v_{rms}}{\bar{v}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} = \sqrt{\frac{3\pi}{8}} \quad (2) \]
\[ \frac{\bar{v}}{v_p} = \frac{\sqrt{\frac{8RT}{\pi M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{4\pi}{1}} = 2\sqrt{\pi} \quad (3) \]

3. **Summarize Relationships**:
- From the ratios derived:
- \( v_{rms} > \bar{v} > v_p \)
- This indicates that the root mean square speed is the highest, followed by the mean speed, and the most probable speed is the lowest.

4. **Kinetic Energy Calculation**:
- The kinetic energy (\( KE \)) of a molecule is given by:
\[ KE = \frac{1}{2} m v_{rms}^2 \]
- Substituting \( v_{rms} \):
\[ KE = \frac{1}{2} m \left(\sqrt{\frac{3RT}{M}}\right)^2 = \frac{3}{2} \cdot \frac{RT}{M} \cdot m \]
- Since \( m = \frac{M}{N_A} \) (where \( N_A \) is Avogadro's number), we can express the kinetic energy in terms of \( v_p \):
\[ KE = \frac{3}{4} m v_p^2 \]

### Final Relationships:
- The relationships can be summarized as:
\[ v_{rms} : \bar{v} : v_p = \sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2} \]
- The kinetic energy of the molecule can be expressed as:
\[ KE = \frac{3}{4} m v_p^2 \]