Question

Five particles have velocities $1\;ms^{-1},\;2\;ms^{-1},\; 3\;ms^{-1},\; 4\;ms^{-1}$ and $5\;ms^{-1}$. The mean square velocity of the particles will be:
$5\;(ms^{-1})^2$
$11\;(ms^{-1})^2$
$7\;(ms^{-1})^2$
$9\;(ms^{-1})^2$

Answer 1

We know that the kinetic energy of a particle constituting a gas is proportional to the square of its velocity. If the average kinetic energy of a gas is the mean of individual kinetic energies possessed by its constituents, determine the relation between the average kinetic energy and particle velocities. This relation will result in a collective expression for the velocity of the gas, which is nothing but the mean squared velocity. To this end, plug in the individual velocities to arrive at an appropriate solution.

Formula Used:
Mean square velocity $v_{ms} = \dfrac{v_1^2+v_2^2+…..+v_n^2}{n}$

Complete step-by-step solution:
Let us begin by first understanding the assumptions posed by the kinetic theory of gases following which we shall understand why it is important to analyse the collective behaviour and velocity of molecules constituting a gas.
The simplest kinetic theory model is built on the assumptions that a gas is made up of a large number of identical particles or molecules constantly moving in random directions, having negligible intermolecular forces and volume, undergo perfectly elastic collisions and have an average kinetic energy that is conserved by virtue of their random movement.
Now, the kinetic energy of a particle constituting a gas is proportional to the square of the velocity of the particle.
$KE = \dfrac{1}{2}mv^2 \Rightarrow KE \propto v^2$
Average$\;KE$ of a gas containing$\;n$ particles will be:
$KE_{avg} = \dfrac{KE_1 +KE_2 + ….. + KE_n}{n}$
This means that the average kinetic energy of a gas will be proportional to the average of the velocity squared of each particle.
$KE_{avg} = \dfrac{KE_1 +KE_2 + ….. + KE_n}{n} = \dfrac{\left(\dfrac{1}{2}mv_1^2 + \dfrac{1}{2}mv_2^2 + …. + \dfrac{1}{2}mv_n^2\right)}{n}$
$\Rightarrow KE_{avg} \propto \dfrac{v_1^2 + v_2^2 + … + v_n^2}{n}$
Thus, the average kinetic energy of a gas is proportional to an expression containing the individual velocity of particles constituting the gas, which is nothing but the mean square velocity of the gas.
Thus, the mean square velocity of a system of particles can be defined as the mean of the squares of the velocity of individual particles, i.e.,
$v_{ms} = \dfrac{v_1^2+v_2^2+…..+v_n^2}{n}$
Given that $v_1 = 1\;ms^{-1}$, $v_2 = 2\;ms^{-1}$, $v_3= 3\;ms^{-1}$, $v_4 = 4\;ms^{-1}$ and $v_5 = 5\;ms^{-1}$, so $n=5$.
The mean square velocity of this system of particles will be:
$v_{ms} = \dfrac{1^2+2^2+3^2+4^2+5^2}{5} = \dfrac{1+4+9+16+25}{5} = \dfrac{55}{5} = 11\;(ms^{-1})^2$
Therefore, the correct choice would be B. $11\;(ms^{-1})^2$

Note: Do not get confused between the mean square velocity, the root mean square velocity and average velocity of a gas. The mean square (ms) velocity is the mean of the squares of the velocity of individual particles, i.e.,
$v_{ms} = \dfrac{v_1^2+v_2^2+…..+v_n^2}{n}$
The root mean square (rms) velocity is the square root of the average of the square of n individual velocities, i.e.,
$v_{rms} = \sqrt{\dfrac{(v_1^2 + v_2^2 + ….. + v_n^2)}{n}}$
Whereas, $v_{average} = \dfrac{(v_1 + v_2 +….. +v_n)}{n}$
The reason we use ms or rms velocities instead of the average velocity in most cases of kinetic treatment of gases is due to the fact that for a typical gas sample, the net velocity is zero since the constituent particles are moving in all directions. Therefore, the average velocity would yield no quantifiable result. This is why sometimes the ms or rms velocities are referred to as ms or rms speeds since they do not indicate the resultant velocity direction. In fact, the ms or rms velocities are independent of the direction of any of its individual velocities and consider only the magnitude.