$U_{a v}, U_{r m s}$ and $U_{m p}$ are average, root mean square, and most probable speeds of the gas respectively at constant temperature T. The molar mass of the gas is M. Then

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no molecule can have a speed less than $U_{m p} / \sqrt{2}$

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

$U_{m p} < U_{a v} < U_{r m s}$

The Kinetic energy of a mole of gas is $\frac{3}{4} M U_{m p}^{2}$

answer is C, D.

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Detailed Solution

$U_{m p} < U_{a v} < U_{r m s}$

$U_{m p} = \sqrt{\frac{2 R T}{M}} U_{m p}^{2} = \frac{2 R T}{M}$

$\frac{3}{4} m {v^{2}}_{m p} = \frac{3}{2} R T$

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