Let v¯,v¯rms and vp 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 the molecules is m. Then:

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Let $\bar{v}, {\bar{v}}_{r m s}$ 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 the molecules is m. Then:

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no molecule can have a speed greater than $\sqrt{2} v_{r m s}$

no molecule can have a speed less than $\frac{v_{p}}{(\sqrt{2})}$ .

$\bar{v} < v_{p} < v_{rms}$

the average kinetic energy of the molecules is $\frac{3}{4} ({mv}_{p}^{2}) .$

answer is D.

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Detailed Solution

Average kinetic energy:

$\begin{array}{r} = \frac{1}{2} {mv}_{rms}^{2} = \frac{1}{2} m {(\sqrt{\frac{3 KT}{m}})}^{2} \\ = \frac{3 KT}{2} = \frac{3}{4} m {(\sqrt{\frac{2 KT}{m}})}^{2} \\ = \frac{3}{4} {mv}_{p}^{2} \end{array}$