Solution Analysis
First, we compare the time scale of the expansion process with the mean collision time of the balls.
The piston moves a distance $h = 1.0 \text{ m}$ with velocity $u = 1.0 \text{ m/s}$. The time taken for expansion is:
$$ t_{exp} = \frac{h}{u} = \frac{1.0}{1.0} = 1.0 \text{ s} $$
The problem states the mean collision time of the balls is of the order of $10^3 \text{ s}$.
$$ \tau_{coll} \approx 10^3 \text{ s} $$
Since $t_{exp} \ll \tau_{coll}$, the gas particles do not collide with each other during the expansion. This means there is no energy redistribution between the $x, y,$ and $z$ components of velocity. The degrees of freedom are independent. The expansion only affects the vertical motion.
Consider the vertical component of velocity $v_y$. The rate of collision of a particle with the piston is given by $f = \frac{v_y}{2L}$, where $L$ is the instantaneous height.
For a piston moving away with speed $u$, the speed of the particle after reflection decreases by $2u$.
$$ \Delta v_y = -2u $$
The rate of change of speed is:
$$ \frac{dv_y}{dt} = f \cdot \Delta v_y = \left(\frac{v_y}{2L}\right)(-2u) = -\frac{v_y u}{L} $$
Since the piston velocity is $u = \frac{dL}{dt}$, we can write:
$$ \frac{dv_y}{dL} = \frac{dv_y / dt}{dL / dt} = \frac{-v_y u / L}{u} = -\frac{v_y}{L} $$
Separating variables and integrating from initial state ($1$) to final state ($2$):
$$ \int_{v_{y1}}^{v_{y2}} \frac{dv_y}{v_y} = -\int_{L_1}^{L_2} \frac{dL}{L} $$
$$ \ln\left(\frac{v_{y2}}{v_{y1}}\right) = -\ln\left(\frac{L_2}{L_1}\right) = \ln\left(\frac{L_1}{L_2}\right) $$
$$ v_{y2} = v_{y1} \left(\frac{L_1}{L_2}\right) $$
Given $L_1 = h$ and $L_2 = 2h$:
$$ v_{y2} = v_{y1} \left(\frac{h}{2h}\right) = \frac{v_{y1}}{2} $$
The total internal energy $U$ is the sum of kinetic energies associated with the $x, y,$ and $z$ components. Initially, the random motion implies the energy is equipartitioned: $$ U_{initial} = K_x + K_y + K_z $$ $$ K_x = K_y = K_z = \frac{U_{initial}}{3} $$ In the final state:
- Horizontal components ($v_x, v_z$) are unchanged because there are no collisions and vertical walls are stationary. Hence, $K’_x = K_x$ and $K’_z = K_z$.
- Vertical component ($v_y$) is halved ($v_{y2} = v_{y1}/2$). Kinetic energy is proportional to $v^2$, so: $$ K’_y \propto v_{y2}^2 = \left(\frac{v_{y1}}{2}\right)^2 = \frac{v_{y1}^2}{4} \implies K’_y = \frac{K_y}{4} $$