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A single argon atom (molar mass 0.040 kg/mol) moves to and fro along the axis of a cylinder…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A single argon atom (molar mass $0.040\ \text{kg/mol}$ ) moves to and fro along the axis of a cylinder, between its closed end and a piston, as shown in the figure. At first the gap is $0.30\ \text{m}$ and the atom moves at $450\ \text{m/s}$ . The piston is pushed in at a steady $0.20\ \text{m/s}$ , much slower than the atom, until the gap is $0.10\ \text{m}$ . The atom bounces elastically off both ends. Take $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ .

A horizontal cylinder with its closed end on the left and a piston on the right. One atom sits on the axis between them, with a double arrow showing that it moves to and fro at 450 m/s. The gap between the closed end and the piston is 0.30 m at the start. An arrow on the piston shows it being pushed in, to the left, at 0.20 m/s.

By how much does the speed of the atom grow at each bounce off the piston?
Show that the product of the atom's speed and the gap stays very nearly the same. What is the speed of the atom when the gap is $0.10\ \text{m}$ ?
About how many times does the atom hit the piston on the way?
How much work does the piston do on the atom?
The cylinder now holds a real monatomic gas, whose atoms move in all directions and collide with each other. It is squeezed in the same slow way, with insulating walls, from a gap of $0.30\ \text{m}$ to $0.10\ \text{m}$ . By what factor does its temperature rise?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Treat one bounce exactly, then add up many bounces as a smooth change. Finally compare with a gas whose atoms move in three dimensions.

Step 1. One bounce (a). In the piston's frame the atom arrives at $v + u$ and leaves at $v + u$ . Back in the cylinder's frame it leaves at $v + 2u$ . So each hit adds $2u = 0.40\ \text{m/s}$ .

Step 2. Speed and gap (b). One round trip takes $\dfrac{2L}{v}$ . During it the gap shrinks by $dL = -\dfrac{2Lu}{v}$ , and the speed grows by $dv = 2u$ . So $\dfrac{dv}{v} = -\dfrac{dL}{L}$ , which means $vL$ stays fixed. Then

\displaystyle v_1 = v_0\,\frac{L_0}{L_1} = 450\ \text{m/s} \times \frac{0.30\ \text{m}}{0.10\ \text{m}} = 1350\ \text{m/s}

Step 3. Number of hits (c). The speed rises by $900\ \text{m/s}$ in steps of $0.40\ \text{m/s}$ : about $2250$ hits, over $1.0\ \text{s}$ .

Step 4. Work (d). The atom's mass is $m = 6.642 \times 10^{-26}\ \text{kg}$ . The work done on it is its gain in kinetic energy:

\displaystyle W = \tfrac12 m\left(v_1^2 - v_0^2\right) = 5.38 \times 10^{-20}\ \text{J}

Step 5. A real gas (e). For a monatomic gas squeezed slowly with no heat flow, $TV^{2/3}$ is fixed. The volume falls by a factor of $3$ , so the temperature rises by $3^{2/3} = 2.08$ .

Answer: (a) $0.40\ \text{m/s}$ . (b) $1350\ \text{m/s}$ . (c) about $2250$ . (d) $5.38 \times 10^{-20}\ \text{J}$ . (e) $2.08$ .

Check. The single atom's kinetic energy rises by $3^2 = 9$ times. That is the law $TV^{\gamma - 1}$ with $\gamma = 3$ , the value for a gas with one degree of freedom, since $\gamma = 1 + 2/f$ . In a real gas, collisions share the energy the piston gives among all three directions, so only one third of it shows up along the axis, and the temperature rises less.

Watch out. The atom does not keep its kinetic energy: a moving wall does work on whatever bounces off it. That work is the microscopic picture of the work $-P\,dV$ done on a gas.

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