Question

Short answer

The rms speed of an oxygen molecule is (a) \(461.2\;{\rm{m/s}}\) , and the molecule will move back and forth for (b) \(26.6\).

Step-by-step solution

Given data

The temperature is \(T = {\rm{0}}^\circ {\rm{C}}\).

The length of the room is \(d = 5\;{\rm{m}}\).

Understanding root mean square speed

In this problem, use the relation of rms speed to calculate the speed of the oxygen molecule. Consider that the particle has no preferred directions.

Evaluation of the rms speed of the molecule

The relation of rms speed can be written as:

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3kT}}{m}} \)

Here, kis the Boltzmann constant and mis the mass of the oxygen molecule.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}{v_{{\rm{rms}}}} &= \sqrt {\left( {\frac{{3\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {0^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{2\left( {16} \right)\left( {1.66 \times {{10}^{ - 27}}\;{\rm{kg}}} \right)}}} \right)} \\{v_{{\rm{rms}}}} &= 461.2\;{\rm{m/s}}\end{aligned}\)

Thus, \({v_{{\rm{rms}}}} = 461.2\;{\rm{m/s}}\) is the required rms speed.

Evaluation of the number of trips per second

Consider that the oxygen molecule has no preferred direction. So, the rms speed is calculated as:

\(\begin{aligned}{c}{v^2}_{{\rm{rms}}} &= {v_{\rm{x}}}^2 + {v_{\rm{y}}}^2 + {v_{\rm{z}}}^2\\{v^2}_{{\rm{rms}}} &= 3{v_{\rm{x}}}^2\\{v_{\rm{x}}} &= \frac{{{v_{{\rm{rms}}}}}}{{\sqrt 3 }}\end{aligned}\)

The total time required for crossing the room can be written as:

\(\begin{aligned}{c}t &= 2 \times \frac{d}{{{v_{\rm{x}}}}}\\t &= 2 \times \frac{d}{{\frac{{{v_{{\rm{rms}}}}}}{{\sqrt 3 }}}}\\t &= \frac{{2\sqrt 3 d}}{{{v_{{\rm{rms}}}}}}\end{aligned}\)

The number of round trips per second can be calculated as:

\(\begin{aligned}{l}N &= \frac{1}{t}\\N &= \frac{1}{{\left( {\frac{{2\sqrt 3 d}}{{{v_{{\rm{rms}}}}}}} \right)}}\\N &= \left( {\frac{1}{{\frac{{2\sqrt 3 \left( {5\;{\rm{m}}} \right)}}{{461.2\;{\rm{m/s}}}}}}} \right)\\N &= 26.6\end{aligned}\)

Thus, \(N = 26.6\) is the required number of trips per second.