I just found out that the famous Feynman Lectures on Physics are now online, so I'm going through them just for fun (I took Physics in College a long long time ago, but was too much content with too little time to actually understand it for real, so here I'm again learning it)

In chapter 1-2, Feyman is talking about simple intuitive properties of matter as a collection of atoms, and so he let us deduce how slowly lowering a piston on a confined gas increases the gas temperature. He says

A simple mental picture and that makes sense. But then he says that "a slow expansion will decrease the temperature"

I was trying to follow that using the same ideas the let us deduce the temperature will increase if we lower the piston, i.e. compress the gas, and I noticed that I can't quite follow it from the text alone. I need to introduce a new idea: "it is the direction of the moving piston what causes the atoms to speed up or slow down. If they collide in opposite directions, the atoms speed us. If they collide in the same direction, the atom slows down".

But the problem is I just made that up. Is it correct?

And if not, then what is the "equivalently simple" explanation? The way I see it, a simple model of the collision with the moving piston *seems* to explain the increase in temperature under slow compression (since the atoms speed up), but actually falls short, as it doesn't equally explain why it decreseaes under slow expansion.

It is a natural extension to the earlier picture - an approaching piston adds KE to the gas and a retreating one removes it ... due to collisions not being quite elastic. Say the gas particle sticks briefly to the walls before being bouncing off.

It's only an "intuitive" picture. It's actually not very helpful to think about thermodynamics that way: too easy to get tied in knots.

If the piston is moving inwards and a gas particle collides with it then it picks up speed. If the piston is moving away then a collision will decrease speed. This is easily calculable from momentum conservation. And since ##T \sim \langle v^2 \rangle ## we can see that ##\delta T \sim \langle \delta v \rangle## so if the piston is moving inwards ##\delta T > 0## and similarly for the case of the piston moving outwards, where ##\langle \rangle## represents an average quantity and ##T## is of course the temperature.

If you wish I can present a more detailed calculation that only uses basic mechanics (momentum and energy conservation).
It is based on problem 4.29 from Kleppner and Kolenkow.