91Ó°ÊÓ

Kinetic energy can be defined as the energy of the object that is owned by the object due to its movement from one location to another.

As the object's velocity increases, the value of the object's kinetic energy also improves.

Given data:

The mass of the oxygen molecule at room temperature is\(m = 5.31 \times {10^{ - 26}}\;{\rm{kg}}\).

The kinetic energy is \(KE = 6.21 \times {10^{ - 21}}\;{\rm{J}}\).

The expression for the kinetic energy is given by:

\(KE = \frac{1}{2}m{v^2}\)

Rewrite the above equation to find the value of the velocity.

\(v = \sqrt {\frac{{2KE}}{m}} \)

Substitute the values in the above equation.

\(\begin{aligned}{l}v = \sqrt {\frac{{2\left( {6.21 \times {{10}^{ - 21}}\;{\rm{J}}} \right)}}{{\left( {5.31 \times {{10}^{ - 26}}\;{\rm{kg}}} \right)}}} \\v = 483.6{\rm{ m/s}}\\v \approx 484\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\end{aligned}\)

Thus, the speed of the oxygen molecule is \(484\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\).