Question

Let $\bar{v}, v_{ rms }$ and $v_{m p}$ 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
(1) $v_{m p} < \bar{v} < v_{ rms }$
(2) no molecule can have a speed greater than $\sqrt{2} v_{ rms }$
(3) the average kinetic energy of a molecule is $\frac{3}{4} m v_{m p}^{2}$
(4) no molecule can have a speed less than $v_{m p} / \sqrt{2}$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (D)

$\bar{v} =\sqrt{\frac{8 k_{B} T}{m \pi}} $
$v_{ rms } =\sqrt{\frac{3 k_{B} T}{m}},$
$ v_{m p}=\sqrt{\frac{2 k_{B} T}{m}}$
where $k_{B}$ is the Boltzmann's constant
$\therefore v_{ rms }>\bar{v}>v_{m p} $
or$ v_{m p}>\bar{v} < v_{ rms }$
Option (1) is correct.
Average kinetic energy of a molecule is
$E =\frac{1}{2} m v_{ rms }^{2} $
or $E =\frac{1}{2} m \frac{3 k_{B} T}{m}$
$E=\frac{3}{4} m \frac{2 k_{B} T}{m}$
or $\quad E=\frac{3}{4} m v_{p}^{2}$
Option (3) is correct.
Again, according to kinetic theory of gases, a molecule of a gas can have speed such that it lies between 0 and $\infty$. ie, $0 < v < \infty$.
Hence the options $(2)$ and (4) can never be correct: