Question

Let $\overline v ,{v_{rms}}$ and ${v_p}$respectively, denote the mean speed, root mean square speed and most probable speed of the molecule in an ideal monatomic gas at absolute temperature T. the mass of a molecule is m. then
This question has multiple correct options
A. No molecule can have a speed greater than $\sqrt 2 {v_{rms}}$
B. No molecule can have a speed less than $\dfrac{{{v_p}}}{{\sqrt 2 }}$
C. ${v_p} < \overline v < {v_{rms}}$
D. The average kinetic energy of a molecule is $\dfrac{3}{4}m{v_p}^2$

Answer 1

we know that the average kinetic energy of a gas molecule is directly proportional to RMS speed. RMS speed is directly proportional to the temperature. The particles of a gas have a different range of speeds. The speed possessed by the maximum number of molecules in a gas at constant temperature is called the most probable speed.
Formula used:
${v_{rms}} = \sqrt {\dfrac{{{v_1}^2 + {v_2}^2 + {v_3}^2 + {v_4}^2.......}}{N}} = \sqrt {{{\left( {\overline v } \right)}^2}}$

Complete step-by-step answer:
We are interested to find the correct statements about the molecule. Let us first start with the root mean square speed,
We know that the root mean square speed is stated as follows,
Root mean square speed: Root means square speed is defined as the square root of the mean of squares of the speed of different molecules. That is,
${v_{rms}} = \sqrt {\dfrac{{{v_1}^2 + {v_2}^2 + {v_3}^2 + {v_4}^2.......}}{N}} = \sqrt {{{\left( {\overline v } \right)}^2}}$
We also know that,${v_{rms}} = \sqrt {\dfrac{{3kT}}{m}}$………….. (1)
Where m is the mass of the gas, k is Boltzmann constant, T is the absolute temperature.

The most probable speed of molecule m is stated as the particles of a gas have a different range of speeds. The speed possessed by the maximum number of molecules in a gas at constant temperature is called the most probable speed. For example: if speed of 10 molecules of gas are 1,2,2,3,3,3,4,5,6,6 km/s then the most probable speed is 3km/s. that is a speed which is in maximum number.
Most probable speed,${v_p} = \sqrt {\dfrac{2}{3}} {v_{rms}} = 0.816{v_{rms}}$…………… (2)
We are also well familiar with Average speed which is the arithmetic mean of the speed of the molecules in a gas at a given temperature.
${v_{av}} = \sqrt {\dfrac{8}{{3\pi }}} {v_{rms}} = 0.92{v_{rms}}$ ………. (3)

From equations (1), (2) and (3), we can say,
${v_{rms}} > {v_{av}} > {v_p}$
That is,
${v_p} < \overline v < {v_{rms}}$ .
We can conclude that Most probable speed is very less among these, and RMS speed is more among all.
Statement (c) is a valid and true statement.
Now let consider the equation (2),
From equation (2), ${v_p} = \sqrt {\dfrac{2}{3}} {v_{rms}} = 0.816{v_{rms}}$
We write it as ${v_{rms}} = \sqrt {\dfrac{3}{2}} {v_p}$ …………… (4)
We know that the average kinetic energy of the gas molecule is given by the equation,
$K = \dfrac{1}{2}m{v_{rms}}$
Now substitute the value of RMS speed from equation (4),
We get, $K = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{3}{2}} {v_p}} \right)^2}$
On Simplifying,
$K = \dfrac{1}{2}m\left( {\dfrac{3}{2}{v_p}^2} \right)$
Therefore, the average kinetic energy of a gas molecule,$K = \dfrac{3}{4}m{v_p}^2$.
Thus statement (d) is also a valid statement.
Hence, correct options are (C) and (D).
Additional information:
There is no atmosphere on the moon because the escape speed of gases on the moon is less than that of the root mean square speed due to the low gravity of the moon. Thus the moon does not have any atmosphere on it.
There is no hydrogen in the atmosphere of the earth. The root mean square speed of hydrogen is less than that of nitrogen hence it easily escapes from the atmosphere of the earth.
Note:
The average distance a molecule can travel without collision is called the mean free path.
With the rise in temperature, RMS speed increases.
Moon has no atmosphere because; RMS speed of gas molecule is more than the escape velocity.
At the T=0K RMS speed of gas, molecules will be zero.