When I push a piston in a cylinder downwards, I'll give more kinetic energy to the particles. That's easy to understand. But, what if I transfer heat to my system in order the gas to expand, why does particles lose kinetic energy? I know that the heat goes to the system, and this energy is then converted in work on the piston. But, why particles lose kinetic energy when pushing the piston?

  • $\begingroup$ If you throw a ball to a wall it bounces back with nearly the same velocity. If you throw a ball towards a moving to you bus (don't do it, please) would it come back faster as in the case with the wall? The same happens with the gas molecules and the piston. $\endgroup$ Commented Apr 23, 2017 at 17:10
  • $\begingroup$ @HolgerFiedler of course, in the case of the moving bus the ball will come back at me with higher speed. In the case of the wall, it comes back at me with same speed. But that's not what I'm asking. I'm asking on a wall that is free to translate on some rails (per example). $\endgroup$ Commented Apr 23, 2017 at 21:18

1 Answer 1


You push the piston down, you give KE to the gas. The gas pushes the piston back up, the gas gives KE to you. It is just the reverse of the same process.

The gas particles gain KE because they are bouncing off an object that is moving towards them. They lose KE when they are bouncing off an object that is moving away from them.

These are elastic collisions (total KE is conserved) so the law of restitution applies : relative speed of approach equals relative speed of separation. When the piston moves down at speed $u$, and the gas particles approach it with perpendicular speed $v$, the relative speed of approach is $v+u$, so the gas particles rebound with speed $v+2u$. Their component of speed in the direction of the piston motion increases by $2u$ from each collision with it.

At room temperature the speed of air molecules is already around $\sqrt3 v=460 m/s$. The factor of $\sqrt3$ arises because the molecular speed has equal components $v$ in 3 perpendicular directions. So $v=266m/s$ in one direction. If the vessel is $0.5m$ across the particles collide with the piston 266 times per second. After one second (266 collisions) their speed increases by $2u+4u+6u+...+532u=2u(1+2+3+...+266)=71,000u$ approx. If u is only $1mm/s=0.001m/s$ this is an increase of 71m/s in the direction of the piston.

The gas "thermalises," sharing this increase with the other 2 component directions; after such "thermalising" the increase in each direction is $\frac{71}{\sqrt3}=41m/s$. Absolute temperature is proportional to $v^2$ so temperature increases to $300K\times (\frac{266+41}{266})^2=400K$ during this compression.

If the gas is expanding, the component of speed of the molecules decreases by the same amount of $41 m/s$, so the temperature falls to $300K\times (\frac{266-41}{266})^2=215K$ during this expansion.

My calculation illustrates how a piston moving at 1mm/s can have a significant effect on the speed of molecules which are already moving at about 460m/s. It does not attempt to be rigorous. For example, I have assumed that thermalisation occurs only at the end of the expansion/compression, but it may happen during, which reduces the change in temperature.

  • $\begingroup$ yes, I know, but I asked "why", I think that's because of the linear momentum. Why would kinetic energy decrease when I pull the piston in a process that no heat is transferred to the system? I simply pull it out. No heat is transferred, no collisions happen, I just pull it up. Why KE decreases? $\endgroup$ Commented Apr 23, 2017 at 14:20
  • $\begingroup$ The temperature of the gas only changes when there are collisions. If there are no collisions with the piston (eg the piston moves away much faster than the speed of molecules) then there is no change in the temperature of the gas. This is a "free expansion". In theory the same effect could be achieved for compression, if the molecules are all on one side of the vessel when the piston moves, so there are no collisions. But with around $10^{23}$ molecules moving at random, such a situation would happen more rarely than the age of the universe. $\endgroup$ Commented Apr 23, 2017 at 15:06
  • $\begingroup$ ahah very good. In your case, considering that I pull the piston with velocity u, and particles have velocity "v", their relative speed to the piston is "v-u", so when it bounces back it comes "v-2u" $\endgroup$ Commented Apr 25, 2017 at 10:42

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