Question

The gas molecules have R.M.S. velocity of \[1000\,m{s^{ - 1}}\] . What is its average velocity?

Answer 1

Hint:According to the Kinetic Theory of Gases, the gas particles are in a continuous state of motion and show ideally elastic collisions in between. RMS stands for root mean square velocity and average velocity is the mean velocity. There is a plot between fractions of molecule versus speed which shows that the fraction of particles is less for root mean square velocity.

Complete step-by-step answer:The speed possessed by the gas molecules at a given temperature can be measured in various ways. We used these speeds or velocities to explain the movement of the gas particle. As you know we cannot take account of every single gas particle, and the velocity always keeps fluctuating so we interpret the velocity in terms of the average value.
The velocity possessed by the gas molecules can be expressed in three forms. These are;
ROOT MEAN SQUARE VELOCITY: It is given as the square root of the mean of squares of the velocities possessed by individual gas molecules. It is given by the expression:
${v_{rms}} = \,\sqrt {\dfrac{{3RT}}{M}} $
Where, $R = $Gas Constant, $T = $ temperature (in Kelvin) and $M = $ molar mass of the gas

AVERAGE VELOCITY: It is given as the mean or average value of the velocities possessed by the gas molecules. It is given by the expression:
${v_{avg}} = \,\sqrt {\dfrac{{8RT}}{{\pi M}}} $
Where, $R = $ gas constant, $T = $temperature (in Kelvin) and $M = $ molar mass of the gas
Now, we are given ${v_{rms}} = \,1000\,m{s^{ - 1}}$ and we have to find ${v_{avg}}$. We know that the ratio
Between Root mean square velocity and average velocity is:
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = \,\dfrac{{\sqrt {\dfrac{{3RT}}{M}} }}{{\sqrt {\dfrac{{8RT}}{{\pi M}}} }}$
If we solve it further it comes as $\sqrt {\dfrac{3}{{2.54}}} \, = \,\dfrac{{1.81}}{1} = \,1.81$
Now, putting the value of ${v_{rms}}$ and we get,
$\dfrac{{1000}}{{{v_{avg}}}} = \,\dfrac{{1.81}}{1}$
${v_{avg}}\, = \,\dfrac{{1000}}{{1.81}}\, = \,552.48$
Hence, the average velocity of the gas will be $552.48\,m{s^{ - 1}}$ .

Note: Other than Average velocity and root mean square velocity there is one more type of velocity possessed by the gas molecule, it is called most probable velocity. It is given as the velocity at which the most molecules in gas travel. It is given by the expression given below. According to the fraction of molecule distribution most molecules show this speed.
${v_{MPS}} = \,\sqrt {\dfrac{{2RT}}{M}} $