Question

At \[27^\circ {\rm{C}}\] , the ratio of rms velocities of ozone to oxygen is
A. \[\sqrt {\dfrac{3}{5}} \]
B. \[\sqrt {\dfrac{4}{3}} \]
C. \[\sqrt {\dfrac{2}{3}} \]
D. 0.25

Answer 1

Hint: The molecules of gas move in different directions with different speeds and they undergo collisions with each other and with the walls of the container. Therefore, to find the average speed of these gaseous molecules, the root mean square (rms) speed is calculated.

Complete Step by Step Solution:
We know, rms velocity can be find out by the following formula,
 \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]

As R is gas constant and given that both the oxygen and ozone gases are at same temperature of \[27^\circ {\rm{C}}\], molar masses of each gas decide its rms speed.

So, \[{v_{rms}} = \sqrt {\dfrac{1}{M}} \], that means, rms velocity is indirectly proportional to molar mass of gases. So, the ratio of the rms velocities of ozone and oxygen gas is also ration of molar masses.
Therefore,
\[\dfrac{{{v_{rms({{\rm{O}}_{\rm{3}}})}}}}{{{v_{rms({{\rm{O}}_{\rm{2}}})}}}} = \sqrt {\dfrac{{{M_{{O_2}}}}}{{{M_{{{\rm{O}}_{\rm{3}}}}}}}} \]

Now, we have to calculate the molar masses of ozone and oxygen.
Ozone is composed of three oxygen atoms. Its formula is \[{{\rm{O}}_{\rm{3}}}\] . So, its molar mass is 48 u. And we know that oxygen gas consists of two atoms of oxygen. Therefore, oxygen possesses a molar mass of 32 u.

Now, we have to put the values of molar masses of both the gases in the above equation. Therefore, the above equation becomes,
  \[\dfrac{{{v_{rms({{\rm{O}}_{\rm{3}}})}}}}{{{v_{rms({{\rm{O}}_{\rm{2}}})}}}} = \sqrt {\dfrac{{32}}{{48}}} = \sqrt {\dfrac{2}{3}} \]
Therefore, the ratio is \[\sqrt {\dfrac{2}{3}} \].
Hence, option C is right.

Note: It is to be noted that RMS speed gives only average speed not velocity because velocity is a vector quantity and possesses both direction and speed. But speed is a scalar quantity and has only magnitude.