Question

Which of the following relation is correct? (\(v_{\mathrm{rms}}\): root mean square velocity, \(\bar{v}\): mean velocity and \(v_{\mathrm{mp}}\): most probable velocity)

• A. \( v_{\mathrm{rms}}>\bar{v}<v_{\mathrm{mp}} \)
• B. \( v_{\mathrm{rms}} v_{\mathrm{mp}} \)
• C. \( v_{\mathrm{rms}}>\bar{v}>v_{\mathrm{mp}} \)
• D. None of these

Hint:

Remember the order: RMS \(>\) Average \(>\) Most probable.

Answer 1

The Correct Option is C

To determine the correct relation among root mean square velocity (\( v_{\mathrm{rms}} \)), mean velocity (\( \bar{v} \)), and most probable velocity (\( v_{\mathrm{mp}} \)), it's essential to understand the context of kinetic theory of gases, which provides these expressions.

The velocities are related to the kinetic energy of gas molecules and are given by:

• \(v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}}\)
• \(\bar{v} = \sqrt{\frac{8kT}{\pi m}}\)
• \(v_{\mathrm{mp}} = \sqrt{\frac{2kT}{m}}\)

Here, \( k \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the mass of a molecule.

Analyzing these expressions, we see:

• The numerical factor for \( v_{\mathrm{rms}} \) (\(\sqrt{3}\)) is greater than that for \( \bar{v} \) (\(\sqrt{\frac{8}{\pi}}\)), which in turn is greater than the factor for \( v_{\mathrm{mp}} \) (\(\sqrt{2}\)).

Therefore, the order among these velocities is:

\( v_{\mathrm{rms}} > \bar{v} > v_{\mathrm{mp}} \)

This means the correct answer is:

\( v_{\mathrm{rms}} > \bar{v} > v_{\mathrm{mp}} \)

This answer can be logically confirmed by understanding that the root mean square velocity, which incorporates all square velocities, will always be the highest. The average or mean velocity represents an intermediate value, and the most probable velocity, being the peak of the Maxwell distribution, is the lowest among these three measurements.