FIGURE 20.13 shows a fast atom and a slow atom approaching the barrier from opposite sides. They undergo a perfectly elastic collision at the barrier. Although no net energy is lost in a perfectly elastic collision, in most such collisions the more-energetic atom loses energy while the less-energetic atom gains energy. In other words, there’s an energy transfer from the more-energetic atom’s side to the less-energetic atom’s side.

FIGURE 20.13 On average, collisions transfer energy from more-energetic atoms to less-energetic atoms.

The average translational kinetic energy per atom is directly proportional to the temperature: ${ϵ}_{\text{avg}}=\frac{3}{2}{k}_{\text{B}}T$. Because $T_{1\text{i}}>T_{2\text{i}}$, the atoms in system 1 are, on average, more energetic than the atoms in system 2. Thus on average the collisions transfer energy from system 1 to system 2. Not in every collision: sometimes a fast atom in system 2 collides with a slow atom in system 1, transferring energy from 2 to 1. But the net energy transfer, from all collisions, is from the warmer system 1 to the cooler system 2. In other words, heat is the energy transferred via collisions between the more-energetic (warmer) atoms on one side and the less-energetic (cooler) atoms on the other.

How do the systems “know” when they’ve reached thermal equilibrium? Energy transfer continues until the atoms on both sides of the barrier have the same average translational kinetic energy. Once the average translational kinetic energies are the same, there is no tendency for energy to flow in either direction. This is the state of thermal equilibrium, so the condition for thermal equilibrium is

where, as before, $ϵ$ is the translational kinetic energy of an atom.

Because the average energies are directly proportional to the final temperatures, ${ϵ}_{\text{avg}}=\frac{3}{2}{k}_{\text{B}}{T}_{\text{f}}$, thermal equilibrium is characterized by the macroscopic condition

In other words, two thermally interacting systems reach a common final temperature because they exchange energy via collisions until the atoms on each side have, on average, equal translational kinetic energies. This is a very important idea.

Equation 20.40 can be used to determine the equilibrium thermal energies. Because these are monatomic gases, $E_{\text{th}}=N{ϵ}_{\text{avg}}$. Thus the equilibrium condition ${\left({ϵ}_1\right)}_{\text{avg}}={\left({ϵ}_2\right)}_{\text{avg}}={\left({ϵ}_{\text{tot}}\right)}_{\text{avg}}$ implies

from which we can conclude

where in the last step we used moles rather than molecules.

Notice that $E_{1\text{f}}+E_{2\text{f}}=E_{\text{tot}}$, verifying that energy has been conserved even while being redistributed between the systems.

No work is done on either system because the barrier has no macroscopic displacement, so the first law of thermodynamics is

As a homework problem you can show that $Q_1=-Q_2$, as required by energy conservation. That is, the heat lost by one system is gained by the other. $\left|Q_1\right|$ is the quantity of heat that is transferred from the warmer gas to the cooler gas during the thermal interaction.

FIGURE 20.14 Equilibrium is reached when the atoms on each side have, on average, equal energies.

The main idea of this section is that two systems reach a common final temperature not by magic or by a prearranged agreement but simply from the energy exchange of vast numbers of molecular collisions. Real interacting systems, of course, are separated by walls rather than our unrealistic thin membrane. As the systems interact, the energy is first transferred via collisions from system 1 into the wall and subsequently, as the cooler molecules collide with a warm wall, into system 2. That is, the energy transfer is $E_1\to E_{\text{wall}}\to E_2$. This is still heat because the energy transfer is occurring via molecular collisions rather than mechanical motion.