Explaining Temp Increase When Compressing Gas: Kinetic Theory

[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 traveling 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]

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 traveling 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.

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 traveling freely between collisions, they are unlikely to radiate infrared radiation but can absorb infrared radiation. When the molecules collide, they are more likely to release infrared radiation.

When the gas is compressed and the molecules are closer together, their mean time between collisions is less, so there is more infrared 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 infrared radiation being absorbed than before the expansion.

 

[jtbell]

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 traveling freely between collisions, they are unlikely to radiate infrared radiation but can absorb infrared radiation. When the molecules collide, they are more likely to release infrared radiation.

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]

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!

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 infrared photon, but by what mechanism would the ideal gas atom emit an infrared 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 the speed of the compressing piston is negligible compared to the speed of the molecules in the gas?

"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]

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.

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 realize 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

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

"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.

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]

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

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]

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

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.]

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

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

 

[sophiecentaur]

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

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.

 

[sophiecentaur]

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.

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]

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

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]

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.

If this energy has not been extracted and now residing in the environment where in fact has it gone?

Thanks

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 won't 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]

In the meantime I had done some calculation with the increase in gas velocity being 2x the piston velocity per collision.


Thanks

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 won't 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.

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

 

[sgstudent]

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.

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?