Solution of Problem 3 (2025) by JK

The assumption $\lambda \gg H^2/vt$ implies that the process time is much longer than the thermal diffusion time within the gas ( $t \gg H^2/v\lambda$ ). This allows us to assume the gas remains in thermal equilibrium at all times, described by a single RMS speed $v$ . Consequently, instead of tracking the entire gas, we can find the average energy transfer rate by analyzing the interaction of a single representative molecule with the piston and then averaging over the Maxwell distribution.
From this viewpoint, the key physical quantities—the frequency of collisions with the piston and the velocity distribution of colliding particles—are determined solely by the gas’s thermal state. Therefore, our analysis correctly describes the thermalized gas ( $H \gg \lambda$ ) even though we model the molecule’s round-trip journey ballistically.
Let us denote with $p_+$ and $p_-$ the probabilities for a molecule with vertical speed $w$ to collide with the piston during its downward (approaching) and upward (receding) stroke, respectively. Since the collision frequency is proportional to the relative speed $w \pm u$ , we find:

\frac{p_+}{p_-} = \frac{w + u}{w-u}, \quad \text{which implies} \quad p_{\pm} = \frac{1}{2}\left(1 \pm \frac{u}{w}\right).

The molecular velocity after collision is $-w \pm 2u$ . The mass-normalized expected change in kinetic energy, averaged over one probabilistic interaction cycle, is therefore:

\langle \Delta(\tfrac{1}{2}w^2) \rangle = \frac{1}{2}\left[ p_+(-w+2u)^2 + p_-(-w-2u)^2 \right] - \frac{1}{2}w^2 = 4u^2.

It is remarkable that the expected energy gain per molecule per interaction, $4mu^2$ , is independent of the molecule’s own speed.
The mass-normalized rate of energy gain for a single molecule is this quantity multiplied by its collision frequency, $f = |w|/2H$ . To find the heating rate for the entire gas, we average this over the Maxwell distribution for all molecules:

\frac{\mathrm{d}}{\mathrm{d}t}\langle \tfrac{1}{2}v^2 \rangle = \langle 4u^2 f \rangle = \frac{2u^2}{H}\langle |w| \rangle.

Using the kinetic theory result $\langle|w|\rangle = v\sqrt{2/3\pi}$ and the identity $\mathrm{d}\langle\frac{1}{2}v^2\rangle = v,\mathrm{d}v$ , we obtain the differential equation:

v,\mathrm{d}v = \frac{2u^2}{H} v\sqrt{\frac{2}{3\pi}},\mathrm{d}t \quad \implies \quad \mathrm{d}v = \frac{u^2}{H}\sqrt{\frac{8}{3\pi}},\mathrm{d}t.

Integrating this from the initial speed $v$ to the final speed $2v$ yields:

v = \frac{u^2}{H}\sqrt{\frac{8}{3\pi}} t \quad \implies \quad t = \frac{Hv}{u^2}\sqrt{\frac{3\pi}{8}}.