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

eddywalrus

Say you compress a closed container full of gas (you reduce its volume). You are not holding the pressure constant or anything.

As you do this, the temperature increases, but why? My best guesses are:
- some of the energy you use for pressing down on the container is transferred to heat energy and thus increasing the temperature (?)
- the force exerted on pushing the container down is passed onto colliding particles, thus increasing their kinetic energy

But these two explanations would only increase the average kinetic energy by a small amount, if at all.

Is there an explanation for this using kinetic theory?

sophiecentaur

If you compress a gas (say with a piston) the piston will be moving inwards. The molecules that bounce against it will rebound with a higher velocity (however slowly you do it, you will still get the same answer). This means they will transfer this increased kinetic energy to all the other molecules. They will then be striking the walls of the container faster. There is more change of momentum at each collision - so more force - so more pressure.
Now - the molecules are travelling faster (average speed) so this implies a rise in temperature. If you cool the container down to its original temperature there is STILL a different situation, despite the fact that the molecules have the same average speed as before. There is less distance for the molecules to travel between collisions with the walls so there will still be more collisions per second. This means that there will STILL be an increase in pressure.

The "small amount" in your post is Just Enough to produce the right answer.

robinpike

Surely the speed of the compressing piston is negligible compared to the speed of the molecules in the gas?

I would have thought that the increase in temperature is more likely due to the increased rate of collisions between the molecules...

When the molecules are travelling freely between collisions, they are unlikely to radiate infra-red radiation but can absorb infra-red radiation. When the molecules collide, they are more likely to release infra-red radiation.

When the gas is compressed and the molecules are closer together, their mean time between collisions is less, so there is more infra-red radiation being released than before the compression.

The opposite happens when a gas is expanded - the molecules are further apart and their mean time between collisions increases, so there is more infra-red radiation being absorbed than before the expansion.

jtbell

You get an increase in temperature even for an ideal gas, in which you assume either that the molecules don't interact at all, or that they collide completely elastically without losing kinetic energy. Remember, PV = nRT is for an ideal gas!

robinpike

How can that be true - that ideal gases collide completely elastically without losing kinetic energy?

For in that case, how would an ideal gas ever lose heat? Since an ideal gas atom (as they are typically mono-atomic) can always absorb an infra-red photon, but by what mechanism would the ideal gas atom emit an infra-red photon?

mikeph

An ideal gas does not interact with radiation.

The additional energy comes from elastic collisions with the compressing wall. Even if the wall moves very slowly, it is far more massive than an atom and the elastic collision will increase the atom's speed quite a lot (momentum conservation). Since the temperature of a gas is proportional to the square of the rms speed of the molecules, this compression leads to a very noticeable increase in temperature.

eg. for an adiabatic compression, (T'/T) = (V/V')^(gamma - 1)

So if you halve the volume (V/V' = 2) of a diatomic gas (gamma = 7/5), the temperature will increase by a factor of 2^0.4 = 1.32. In terms of Kelvin that's the equivalent of going from room temperature to about 125 C!

sophiecentaur

"Surely"????
It may seem counter-intuitive but you won't be able to show that the statement is wrong. Remember that every molecule that strikes the piston will have its speed increased by the massive piston. This can happen quickly or slowly, depending on the speed of the piston and, however quickly or slowly the change is made, the KE is redistributed amongst all the molecules to produce a different velocity distribution.
The argument to justify Boyle's Law (constant temperature situation) is fairly easy to grasp because it just depends on the reduced amount of space and the resulting frequency of collisions. To derive the full Gas Law is a bit more long winded and I recommend that you hunt around on the web for a website that presents it at a level you can cope with. Just think of the actual work done on a gas to compress it to half its volume. All that energy has to go somewhere and the only way it can be transferred for an ideal gas is by speeding up the molecules by contact with the piston.

dipole

It has more to do with quantum mechanics, if you really want to talk about it in a way that gets at the fundamental reasons behind it all.

Temperature is a measure of how "spread out" the particles are over all possible energy levels. At low temperatures, the particles are concentrated mostly in the lowest available energy levels, where as at higher temperatures they're more spread out.

If when you compress a volume, you decrease the spacing between energy levels - this is is a result from quantum mechanics.

If you decrease the spacing between energy levels, but don't change the total energy, then the particles have to spread out and occupy higher energy levels in order to keep everything conserved, and by definition this is a higher temperature.

I think this is really the best way to think about it.

sophiecentaur

You are right, of course but it may not be necessary to understand this in depth as long as you can accept that there will be some sort of energy distribution amongst the molecules. That is a reasonable intermediate step in getting a handle on this, I feel. I realise that QM is necessary to deal with the 'ultraviolet catastrophe' but it's a huge amount extra to take on board all in one go.

Austin0

Quote by robinpike

Hi I may be off base but I see some questions here.
I understood that ideally slow compression or expansion was specifically to remove the momentum of the piston from consideration. Equivalent to ideally slow clock transport to remove time dilation as a factor.
SO for expansion; no work is done by the gas to remove kinetic energy through transference of momentum to the piston. Or vise versa with compression
Temperature is purely a function of internal kinetic energy so internal collisions are zero sum events and their increase in frequency through decreasing spatial relationships should have no effect on total energy, although it does have an affect on pressure which is dependent on number of collisions with the container walls.
As I remember it the increase in internal temperature is related to the Van der Walls force.
As such is a result of spatial reduction and the decrease in the mean distance between nuclei .
I think for some monatomic gases, expansion for some pressure ranges actually results in temperature increase. Or reciprocally compression results in temperature decrease which seems to contradict any concept of piston imparted increased velocities.
This force imparts an acceleration to proximate gas molecules dependent on distance and so does result in an overall increase in the mean probable velocity.
If I am incorrect in any of these assumptions I will be happy to learn of it..

sophiecentaur

I think you may be thinking in terms of isothermal changes or non-ideal gases. It has to be true that, if you insulate a mass of an ideal gas and do work in compressing it - and there is work involved in moving a piston against pressure (force times distance), however slowly you do it - then conservation of energy must apply (no?). The only place the energy can go must be into the internal energy of the gas. In an ideal gas, this must result in an increase in average K.E. - that is an increase in temperature.
When Van der Waall's forces are involved (non-ideal gases), you can get some of the work done transferred to potential energy and that will modify the temperature change - in either direction, I seem to remember.

sophiecentaur

Look at the Wiki article on Joule Kelvin Effect. They make the distinction nicely between volume changes with and without work being done.

Austin0

Thanks for the reference . In the meantime I had done some calculation with the increase in gas velocity being 2x the piston velocity per collision.
Doing even a rough quesstimate of the number of collisions based on initial gas velocity and the average length of the container during compression, it was quite clear that no matter how negligible the piston velocity the net increase must be a significant percentage of the initial average velocity, even without considering the exponential increase in collision frequency with increasing velocity.
SO I was badly mistaken in my idea of the effect of slow transport and the relative effects of the Van der Wall force.
It makes me wonder about the conceptual basis of slow clock transport???
Can you fool mother nature??
Like a photon that can pop into existence out of nothing because it happens to quick for the conservation cops to take action ;-)

Thanks

sophiecentaur

Glad your sums have convinced you about the reasonableness of that. Of course, the actual distribution of velocities after the change in volume needs the application of Quantum Theory to avoid the high energy problem.
I think the slow clock transport thing with light speed is a different matter - possibly to do with the fact that one involves Energy transfer(?).

A.T.

The faster the molecules are, the more bounces with the moving wall will occur per time.

sophiecentaur

That's why both volume and temperature account for the pressure - according to the gas laws.

the_emi_guy

A way to reconcile intuition with sophiecentaur's good explanation is to imagine a container of gas at absolute zero, i.e. molecules absolutely at rest. Inserting the piston to compress the gas leaves particles with motion (increased temp) due to their encounter with the inward moving piston. Expansion cannot lower temp below absolute zero because the piston will not interact with the stationary gas molecules as it moves away.

Interesting, it is possible to "cheat" Charle's Law in the expansion case if the piston moves very fast, fast enough that no gas molecules collide with it while it is in motion.
In this (impractical) case, the expansion would be simultaneously isothermal and adiabatic.