Question

Let \( \bar{V} \), \( V_{\text{rms}} \), \( V_p \) denote the mean speed, root mean square speed, and most probable speed of the molecules of mass \( m \) in an ideal monoatomic gas at absolute temperature \( T \) Kelvin. Which statement(s) is/are correct?

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

Hint:

In an ideal monoatomic gas, the most probable speed, the mean speed, and the root mean square speed are related by the formulas \( V_p<\bar{V}<V_{\text{rms}} \). Always remember that the distribution of molecular speeds is not uniform, and different speeds correspond to different probabilities.

Answer 1

The Correct Option is C

We are given that \( \bar{V} \), \( V_{\text{rms}} \), and \( V_p \) are the mean speed, root mean square speed, and most probable speed of molecules in an ideal monoatomic gas. These quantities are related to the temperature \( T \) and mass \( m \) of the molecules. The expressions for these speeds are as follows: 1. The most probable speed \( V_p \) is given by: \[ V_p = \sqrt{\frac{2kT}{m}} \] 2. The root mean square speed \( V_{\text{rms}} \) is given by: \[ V_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] 3. The mean speed \( \bar{V} \) is given by: \[ \bar{V} = \sqrt{\frac{8kT}{\pi m}} \] where: - \( k \) is the Boltzmann constant, - \( T \) is the absolute temperature, - \( m \) is the mass of a molecule. Step 1: Analyzing Option (A) - Speed Greater than \( \sqrt{2} V_{\text{rms}} \) From the Maxwell-Boltzmann distribution, the speeds of molecules follow a probability distribution. Although some molecules can have speeds much higher than the most probable speed \( V_p \), the probability density rapidly decreases as speed increases. Therefore, no molecule can have a speed greater than \( \sqrt{2} V_{\text{rms}} \), making this statement true. Thus, Option (A) is correct. Step 2: Analyzing Option (B) - Speed Less than \( \frac{V_p}{\sqrt{2}} \) The speed \( V_p \) corresponds to the most probable speed, which is the speed at which the maximum number of molecules are moving. The distribution of speeds has a wide spread, and some molecules will have speeds less than \( \frac{V_p}{\sqrt{2}} \), so this statement is not correct. Thus, Option (B) is incorrect. Step 3: Analyzing Option (C) - \( V_p<\bar{V}<V_{\text{rms}} \) From the equations for the speeds: \[ V_p = \sqrt{\frac{2kT}{m}}, \quad \bar{V} = \sqrt{\frac{8kT}{\pi m}}, \quad V_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] we know that \( V_p<\bar{V}<V_{\text{rms}} \), as: \[ V_p = \sqrt{\frac{2}{3}} V_{\text{rms}} \quad \text{and} \quad \bar{V} = \sqrt{\frac{8}{3\pi}} V_{\text{rms}} \] Thus, the relationship \( V_p<\bar{V}<V_{\text{rms}} \) is correct. Hence, Option (C) is correct. Step 4: Analyzing Option (D) - Kinetic Energy Formula The average kinetic energy per molecule is given by: \[ E_{\text{kinetic}} = \frac{3}{2} kT \] The statement in Option (D) suggests the kinetic energy is proportional to \( \frac{3}{4} m V_p^2 \), which is not correct. The correct relationship for the kinetic energy is based on the temperature of the gas and not directly on the speed. Therefore, this statement is incorrect. Thus, Option (D) is incorrect. Conclusion The correct statement is: \[ \boxed{(C) \, V_p<\bar{V}<V_{\text{rms}}} \]