Question

Let $\overrightarrow{v}$, v$_{rms}$ 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 a molecule is m. Then

• A. no molecule can have a speed greater than $\sqrt 2v_{rms}$
• B. no molecule can have speed less than $v_p\sqrt 2$
• C. $v_p < \overrightarrow{v} < v_{rms}$
• D. the average kinetic energy of a molecule is $\frac{3}{4}mv_p^2$

Answer 1

Answer: (C), (D)

$v_{rms}=\sqrt{\frac{3RT}{M}}, \overrightarrow{v}=\sqrt{\frac{8}{\pi}.\frac{RT}{M}}=\sqrt{\frac{2.5RT}{M}}$
and $ \, \, \, \, \, \, \, \, \, \, \, \, \, \, v_p=\sqrt{\frac{2RT}{M}}$
From these expressions we can see that
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, v_p < \overrightarrow{V} < v_{rms}$
Secondly $ \, \, \, \, \, \, \, \, \, \, \, v_{rms}=\sqrt{\frac{3}{2}}v_p$
and average kinetic energy of a gas molecule
$\, \, \, \, \, \, \, \, \, =\frac{1}{2}mv^2_{rms}=\frac{1}{2}m\bigg(\sqrt{\frac{3}{2}v_p}\bigg)^2=\frac{3}{4}mv^2_p$