Question

At the same temperature the rms velocity of $H$ is \[2 \times {10^3}\,m{s^{ - 1}}\]. What will be the rms velocity of \[{O_2}\] molecule at the temperature:
A. \[10\] \[m{s^{ - 1}}\]
B. \[500\] \[m{s^{ - 1}}\]
C. \[0.5 \times {10^4}\] \[m{s^{ - 1}}\]
D. \[3 \times {10^3}\] \[m{s^{ - 1}}\]

Answer 1

Hint:Gas molecules are loosely packed and they keep on colliding with each other as well as with walls of container and their momenta change and pressure is exerted by gas on the walls of container. There is an assumption of kinetic theory of gases that no force of attraction is between their molecules.

Complete step by step answer:
To characterize the motion of molecules of gas three different velocities are defined. RMS velocity is the square root of the mean square of the speed of different molecules.
\[{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
where $M$ is molecular mass, $T$ is temperature, $R$ is gas constant
As given in question: Temperature for hydrogen and oxygen is same
Hydrogen \[{V_{rms}} = 2 \times {10^3}m{s^{ - 1}}\]
\[{V_{rms}} \propto \sqrt {\dfrac{1}{M}} \]
\[\Rightarrow \dfrac{{{V_{rms{\text{ of H}}}}}}{{{V_{rms{\text{ of }}{{\text{O}}_2}}}}} = \sqrt {\dfrac{{{M_{{O_2}}}}}{{{M_H}}}} \]
\[\Rightarrow \dfrac{{2 \times {{10}^3}}}{{{V_{rms{\text{ of }}{{\text{O}}_2}}}}} = \sqrt {\dfrac{{16}}{1}} \]
\[\Rightarrow {V_{rms{\text{ of }}{{\text{O}}_2}}} = \dfrac{{2 \times {{10}^3}}}{4}\]
\[\therefore {V_{rms{\text{ of }}{{\text{O}}_2}}}\]\[ = 500\,m{s^{ - 1}}\]

Hence, the correct answer is option B.

Note:The two velocities of gas molecules are most probable: the speed possessed by maximum of gas molecules and average speed which is the arithmetic mean of all speed of molecules \[{V_{rms}} > {V_{av}} > {V_{mp}}\].