Why does temperature increase when volume decreases? (Kinetic theory explanations)

sophiecentaur

That's the equivalent of opening the side into a vacuum. No work is done then so no temperature drop for an ideal gas, I think.

Austin0

Actually I first encountered this material in the past while researching the basis of compressive cooling. All the literature seemed to agree that such systems worked because gases cooled through expansion. This seemed to me to be a misconception equivalent to water as ice cooling through melting. In this case the ice has not only less KE than water at the same temp 0 deg C but greatly less internal energy.
So a system designed to cool the gas itself would be totally adiabatic except for the extraction interface in the compressed phase. It seemed to me that if the energy extracted was only equal to the energy added by compression that the system would not work.
Clearly in the case of gases where compression brought about a state transition this made available for extraction a significant amount of internal energy which could then leave the system as KE or radiation. SO the gas would have a much lower energy content even before the expansion phase.In other cases I understood that extra energy was available for extraction through actualization of Van der Waals potential.
SO if we start with a volume of gas, say nitrogen, at environmental temp and bring it down to liquid temperature there is a huge net loss of energy.
If this energy has not been extracted and now residing in the environment where in fact has it gone???
The ideas that it was somehow "disappeared" through internal throttling seems like energy down the rabbit hole.
Or if there are two equal volumes with equivalent PT that are expanded into twice the volume, one through throttling and one through free expansion , in the first case there is a reduction of temp but not in the second case.(ignoring Van der Waals effect).
In the first case it is stated that there is work done even though there is no exchange with the outside. But if there is no work done on the valve, no transference of momentum, and the final condition is identical to free expansion as far as displacement of the gas itself where is the result of this work. I.e., where did the energy go??
Thanks

sophiecentaur

I didn't get the details of all that but when you add all the Kinetic Energies and all the Potential Energies and the Work in or out, the sum will be zero. It's just a matter of modelling the particular substance right. The devil must be in the detail of the changes of Potential Energy as the volume changes and the work done. I don't think you can ignore Van der Vaal because that blanket term accounts for the departure from idea l behaviour.

the_emi_guy

Austin,
I think you are missing an important step in your analysis.

First , you are correct that when the gas is compressed adiabatically then cooled back to its original temperature, the energy leaving the system in the form of heat is equal to the the energy that was just added in the form of work pushing the piston down. The thermal energy within system is the same before and after since the gas temperatures are the same.

However, in the expansion phase the system is doing work on the environment. Think of it this way: the piston has gas pressure pushing outward while the piston is moving outward. This is like a spring expanding, it is releasing energy to the external environment (or better yet, think about the gas molecules losing some of their kinetic energy to the outward moving piston at each collision). This is not as obvious since this rarely represents *useful* work in a cooling system. It is this lost energy that leads to a lower temperature (and pressure) when we expand back to the original volume.

The PV diagram trajectories for this would be:

1 - compression phase: adiabat up and to the left.
2 - cooling back to original temp: vertical down.
3 - expansion to original volume: adiabat down and to the right.

Your final position will be *below* the original position, i.e. same volume, lower pressure.

chingel

In the reference frame of a wall, when there is an elastic collision between the wall and a small ball then the component of the speed perpendicular to the wall is reversed, assuming it wont start rotating. So if you have a ball hitting the wall straight on 5 m/s, it will come back (roughly) at 5 m/s. Now if the wall is moving towards the ball at 2 m/s, the wall sees the ball hitting it at 7 m/s and bouncing back at 7 m/s, but in the stationary frame you would see the ball hitting the wall at 5 m/s and bouncing back at 9 m/s, so that's where it gets its extra speed.

Austin0

Thanks for the input but this was already covered. ;-)

sgstudent

Hi why would that be Charles's Law? In a normal case when i decrease the pressure by moving my piston up won't the pressure of the gas decrease?

So shouldn't this be a Boyle's Law rather than a Charles's Law case?