Question

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?

Short answer

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

Step-by-step solution

Given data

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

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

Understanding the average molecular kinetic energy

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.

Determination of the speed of the paper clip

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 \(3.5 \times {10^{ - 9}}\;{\rm{m/s}}\).