# Intuition on Kinetic Theory - Does temperature change during an adiabatic expansion?

I was reading the chapter "The Kinetic Theory of Gases " from the Feynman Lectures on Physics, where Feynman gives the intuition for why temperature is defined the way it is and what happens when you slowly compress a gas using a piston.

My understanding is, when we "push" on the piston, assuming that there is no loss of energy in the form of "heat" (I still don't know what heat is...in an intuitive sense), we transfer some energy to the atoms/molecules hitting the piston, thereby increasing the average kinetic energy of the atoms/molecules and hence increasing the temperature. (Temperature as I know is defined to be linearly proportional to the average kinetic energy of the atoms/molecules)

My question is, what would happen if we "pull" the piston....in this case..we certainly are not delivering any energy to the atoms/molecules...so it looks like the temperature should not change..

This also seems to be the case when we consider the formula $$P=nm\frac{}{3}$$. Here $$n$$ is the number of atoms per unit volume, $$m$$ is the mass of each atom and $$v$$ is the velocity of an atom which is averaged over all the molecules. So when we "pull" the piston, we are decreasing $$n$$ and hence decreasing $$P$$, but we are not doing anything to the average kinetic energy..so the temperature should not change..

But, I also know the formula that $$TV^{\gamma-1}=C$$ for an adiabatic process, where $$C$$ is a constant...now this formula says that $$T$$ changes as $$V$$ changes...

Where am I going wrong??

Thanks for any answers!!

EDIT: The setup is a can filled with some gas with a piston fitted at one of its ends

## 2 Answers

Compare with the static situation, without a piston, where all the collisons are elastic. By symmetry, for each collision on the left wall there is an equivalent collision on the right wall. So the total momentum of the molecules doesn't change.

As $$E_k = \frac{p^2}{2m}$$, the kinetic energy doesn't change.

But with a piston with velocity $$+v$$ to the right, the symmetry is broken. The ellastic collisions with the right wall bounce with the same speed in the frame of the moving wall. And that means less speed in the container's frame. The total momentum decreases and so the kinetic energy.

The opposite for a compressing piston.

"Pulling" the piston is the reverse of "pushing" it, in this case the gas does work into the environment, remember the $$dW=PdV$$. If the system does work it loses energy and the gas gets colder. Heat is a way to transfer energy in a microscopic way, as opposed to work, that is a macroscopic transfer. In the previous cases we assume that the piston moves slow enough so that the pressure P is well defined in the entire volume.

Finally, there is an irreversible process called free expansion. It is also adiabatic but in this case there is no piston and thus the gas expands freely and without making work. Here the temperature does stay constant.The formula you mention does not apply in this case, because it was derived assuming a slow process in which the gas makes work as it expands.

• But is there an intuitive picture as to why the average kinetic energy decreases?...Other than taking the analogy of the "reverse" case.. Jun 21, 2020 at 14:06
• Sure, the molecules hit the piston, and make a force in the direction of motion during some change in position. They make work and loose energy. Just imagine a collision between a ball and a wall that moves in the same direction
– user65081
Jun 21, 2020 at 14:08
• Oh...then what about the collision being elsatic? Jun 21, 2020 at 14:09
• Because someone is "pulling" the piston right? So it's not that the molecules are doing the pushing.. Jun 21, 2020 at 14:12
• it does not matter of the collision is elastic, but they are assumed to be unless there is some kind of irreversibly. You do not need to pull the piston, but you can, it does not matter as far as the process is infinitesimally slow, it is an idealization
– user65081
Jun 21, 2020 at 15:55