Question

Let $\overline{v}$, ${\overline{v}}_{rms}$ 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 the molecule is $m$. Then:

• A. no molecule can have a speed greater than $\left(\sqrt{2}{v}_{rms}\right)$
• B. no molecule can have a speed less than $\frac{v_p}{\left(\sqrt{2}\right)}$
• C. $\overline{v}<v_p<v_{rms}$
• D. the average kinetic energy of the molecules is $\frac{3}{4}\left(m{v}_p^2\right)$

Answer 1

The correct option is D the average kinetic energy of the molecules is $\frac{3}{4}\left(m{v}_p^2\right)$

Root mean square speed

$v_{rms}=\sqrt{\frac{3kT}{m}}$

Most probable speed

$v_{mp}=\sqrt{\frac{2kT}{m}}$

Average speed
$v_{av}=\sqrt{\frac{8kT}{\pi m}}$

$u_{mp}:u_{av}:u_{rms}=1:1.128:1.224$

From these expressions, we can see that

$v_p<\overline{v}<v_{rms}$
Again, $v_{rms}=v_p\sqrt{\frac{3}{2}}$

and average kinetic energy of a gas molecule $E_k=\frac{1}{2}m{v}_{rms}^2$
$E_k=\frac{1}{2}m{\left(\sqrt{\frac{3}{2}}{v}_p\right)}^2=\frac{1}{2}m\times \frac{3}{2}{v}_p^2=\frac{3}{4}m{v}_p^2$