# How is work done on a body account for increase in internal energy?

Considering that I have a container with a compressible gas inside it and a pump at the top. As i push down on it, I'll be doing work on the system.

As I am applying pressure on the system the particles would come close to each other, their kinetic energies will decrease.

I don't understand how internal energy of the system increases when work is done on it by an external agency ( like a pump).

And according to the definition the sum of kinetic energies of the particles of a system is it's internal energy. Or is it that when the gases come close to each other their Kinetic energies decrease and potential energies increase?

• Yes the internal temperature will increase as you are doing work on the system. I recall solving this in college but I forget the specifics. I think we just did an energy balance. It is complex as the compressed gas is also potential energy. The reverse is true when you throttle the gas out of the tank - gas will cool. Commented Jan 11, 2018 at 14:22
• Why do you say the kinetic energy decreases? The kinetic energy and the density are two different things. Commented Jan 11, 2018 at 14:26
• Your assertion that the kinetic energies decrease is incorrect. Commented Jan 11, 2018 at 14:43
• The particles come close so their motion will be kinda restricted, and kinetic energy is the energy possessed by an object by the virtue of it's motion Commented Jan 11, 2018 at 15:24

For an ideal gas there are no interactions between molecules, so the internal energy is the total kinetic energy of all of its molecules. Compressing the gas increases the kinetic energy (and therefore also internal energy and temperature) of the gas, whereas allowing the gas to expand decreases kinetic energy.

When the walls of the container are fixed then the molecules which rebound from the walls have the same KE after rebounding as they had before. The velocity is reversed but the kinetic energy remains unchanged. The temperature of the gas neither increases nor decreases.

When you push down on the piston to compress the gas, the molecules rebound from the piston with a higher speed than when they approach it. Collisions with a piston which is moving towards the gas molecules increase the average speed of molecules and therefore also the temperature. When the piston is moving away from the molecules this reduces the temperature of the gas.

The fact that the piston usually moves very slowly (eg 1 mm/s) compared with the speed of gas molecules (about 400 m/s) might suggest that the motion of the piston can only have a miniscule effect on the speed of molecules. However, the high speed of the molecules means that there is a large number of collisions, each of which increases the speed of a molecule by a tiny amount, so the total effect can be significant.

eg In a container which has a height of 25 cm each molecule will collide with the piston around 800 times per second. If the piston moves for 5 cm at 1 mm/s that is 40,000 collisions per molecule, increasing its speed by a total of about 40 m/s, around 10%. Temperature is proportional to $v^2$, which increases by a factor of 1%. The temperature increases from 300 K by about 3 K.

The work done by the piston on the gas is the work done on its molecules. This is equal to the increase in the KE of the molecules, which is the increase in internal energy.

In a free expansion the volume of the gas is increased suddenly, without the molecules having time to collide with the piston. This does not result in any reduction in temperature. This shows that it is not the change in volume which changes the temperature of the gas but the work done by the piston, either imparting to or taking away kinetic energy from its molecules.

• The mechanism by which the internal kinetic energy and temperature of a gas increases under adiabatic compression has been a subject of interest to me for some time now. I had yet to find a satisfying explanation but I felt it had to do with the gas molecules impacting the moving piston, with a higher rebound velocity (kinetic energy) when the piston moved toward them (gas compression) and lower rebound velocity (kinetic energy) when the piston moved away from them Commented Apr 14, 2020 at 14:57
• Then, as a result of continued research I happened on your answer above. It confirmed my thoughts, so of course I agree with it. Here’s a further thought. Consider that the increase in kinetic energy of a gas as a result of adiabatic compression work is an example of the work energy theorem: The net work done on and object equals its change in kinetic energy. In this case, the work done by the adiabatic compression equals the increase in the average kinetic energy of the gas. Commented Apr 14, 2020 at 14:57
• Yes that is how I understand it also. Commented Apr 14, 2020 at 20:45
• Beautiful answer! If I may ask, how did you develop this intuition? @sammygerbil Commented Feb 19, 2021 at 8:40

If the volume of the container in which the gas resides is reduced, on rebound with a wall the speed of molecules of the gas will be larger than their speed before they hit the wall.

So now the molecules have a bit more kinetic energy after hitting a wall than then they had before hitting the wall and that extra kinetic energy is then distributed throughout all the molecules by inter-molecular collisions.

The net effect is that the average kinetic energy of the molecules has increased ie the internal energy has increase.
That extra kinetic energy has come from the work done by the external force moving the walls of the container inwards.

• Thank you :) Just one more question , Does that mean when work is done on the surrounding by the system internal energy must decrease? Commented Jan 11, 2018 at 15:32
• @susanJ Yes, if there is no heat transfer. You could have the system doing work on the surroundings and heat input to the system from the surroundings and then the internal energy would not decrease as much or the internal energy could increase if the heat input was large enough. Commented Jan 11, 2018 at 16:34
• How does reducing the wall distance make them rebound faster after impact? That doesn't sound possible unless the wall were moving at the time.
– JMac
Commented Jan 11, 2018 at 16:49
• @JMac Work is only done when the volume changes which means that the walls must move. When the volume is constant no work is done. Commented Jan 11, 2018 at 16:51
• @Farcher That wording just seems really odd unless you specify that it's due to the movement of the wall. Like changing it to "if the volume is reducing...". If the volume is already reduced, there shouldn't be any change on impact; they would already be at the increased velocity.
– JMac
Commented Jan 11, 2018 at 17:56