Let $v_{A V G}$ , $v_{R M S}$ and $v_{M P}$ denote mean speed, rms speed and most probable speed, respectively, of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then,

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the average kinetic energy of a molecule is $\frac{3}{4} m v_{M P}^{2}$

no molecule can have a speed less than $\frac{v_{M P}}{\sqrt{2}}$

$v_{M P} > v_{A V G} > v_{R M S}$

no molecule can have a speed greater than $\sqrt{2} v_{R M S}$

answer is D.

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Detailed Solution

According to the kinetic theory of gases, a molecule of a gas can have any speed between 0 and ¥. Thus, (1) and (2) are obviously wrong.

Also, $v_{R M S} = \sqrt{\frac{3 R T}{M}}; v_{A V G} = \sqrt{\frac{8 R T}{π M}}; v_{M P} = \sqrt{\frac{2 R T}{M}}$ . So, $v_{R M S} > v_{A V G} > v_{M P}$
$E_{A V G} = \frac{1}{2} m v_{R M S}^{2} = \frac{1}{2} m (\frac{3 k T}{m}) = \frac{3}{2} k T$ (1)
But $v_{M P} = \sqrt{\frac{2 k T}{m}}$
$\Rightarrow k T = \frac{1}{2} m v_{M P}^{2}$ . (2)
Using (2) in (1),
$\Rightarrow E_{A V G} = \frac{3}{4} m v_{M P}^{2}$ .