Question: (I)What speed would a 1.0-g paper clip have if it had the same kinetic energy as a molecule at 22°C?

The temperature of the molecule is \(T = 22^\circ {\rm{C}} = 295\;{\rm{K}}\).

The mass of the paper clip is \(m = 1\;{\rm{g}}\).

The average kinetic energy of the molecule is proportional to the absolute temperature of the gas.

The relation of average kinetic energy is given by the following:

\(KE = \frac{3}{2}kT\)

Here,\(k\)is the Boltzmann constant, and Tis the absolute temperature of the gas.

Since the paper clip has the same kinetic energy as the molecule,

\(\begin{aligned}{c}\frac{1}{2}m{v^2} &= \frac{3}{2}kT\\v &= \sqrt {\frac{{3kT}}{m}} \end{aligned}\)

Here, vis the speed of the paper clip, and m is the mass of the paper clip.

Substitute the values in the above equation.

\(\begin{aligned}{l}v &= \sqrt {\left( {\frac{{3\left( {1.38 \times {{10}^{ - 23}}\,{\rm{J/K}}} \right) \times 295\;{\rm{K}}}}{{\left( {1\;{\rm{g}} \times \frac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right)}}} \right)} \\v &= 3.5 \times {10^{ - 9}}\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the paper clip is \(3.5 \times {10^{ - 9}}\;{\rm{m/s}}\).