# Cooling of an expanding gas

1. Mar 21, 2014

### aochider

I've been a bit confused about when/why an expanding gas cools.

Here is what I've heard so far:

1. Refrigeration systems work based on using heat that is used merely to break attraction of molecules of liquids at their boiling point (eg, freon released through an expansion valve gets cooler because particles break the attraction between the freon molecules, so some energy is lost to that "breaking" process, resulting in less KE in the particles involved and thus a lower temp). A common explanation of this, however, is that an expanding gas cools, not the boiling point bit.
2. In Feynman's lectures he talks about how if you compress a gas, you are imparting energy onto the gas's particles and the KE and temperature rises (ie, a piston/compressor is "hitting" the particles faster). Then he says that expansion lowers the temperature. "So, under slow compression, a gas will increase in temperature, and under slow expansion it will decrease in temperature."
3. The OP in this thread talks about how the air is measurably cooler after compressing, cooling, and then releasing it. https://www.physicsforums.com/showthread.php?t=549795
4. The ideal gas law seems to *not* imply a temperature decrease in the above scenarios because pressure decreases as volume increases, so T should stay the same. But maybe I'm wrong about this because maybe p and V change at different rates which causes T to noticeably fluctuate.

So, I'm mostly struggling to visualize 2 and 3 above. Why does Feynman say that expansion will decrease temperature? Also, why was the uncompressed air cooler? I'm trying to visualize those on a molecular level, but it's hard to see why things would get cooler in those cases.

Thanks!

2. Mar 21, 2014

### BOYLANATOR

To visualize 2: Imagine you are bouncing a tennis ball up and down on a racket. You move your racket upwards as it hits the ball and the ball bounces higher - it's kinetic energy increases. If you move your racket down you can "cushion" the impact and the ball loses kinetic energy.

Similarly a gas particle hitting the wall of an expanding container loses energy. Particles losing kinetic energy is a temperature decrease.

3. Mar 21, 2014

### aochider

excellent visualization for that one. thanks!

4. Mar 21, 2014

### dauto

3. Compress the air doing work on it with a piston and it gets hotter. Let if cool back to room temperature. Expand the air back to its original volume letting the air to perform work on the piston and it gets cooler.

4. Yes p and V change in different ways and T drops in a expansion (As described by Feynman).

The process described above works, but in real refrigerators other processes may be used. For instance

1. When freon evaporates it absorbs latent heat from the environment cooling down.

It's also possible to cool a non-ideal gas through the Joule–Thomson effect which allows a gas to cool during expansion even if no work is being done. (Some gasses will warm up instead of cool down)

5. Mar 21, 2014

### aochider

thanks!

for 3, does this apply to a pressurized container that was opened in the free atmosphere? so you pressurize a bottle with a pump, detach the pump, let the container cool, and then open the lid. air rushes out and at least momentarily is cooler than the surrounding air. there is no piston for the escaping air to do work on. i believe that was the OPs experiment in the cited thread. why would the air flowing out be cooler?

6. Mar 21, 2014

### Staff: Mentor

It is doing work on the air that it is displacing (to push the air back), just as if the air it is displacing were a piston.

7. Mar 21, 2014

### dauto

To complement Chestermiller's answer above, if you open the bottle of air in avacuum than it wouldn't cool because there would be nothing for the air to do work on.

8. Mar 21, 2014

### aochider

thank you for the help!

might be more of an engineering question now, but couldnt you use pressurized gas as a form of refrigeration technology? or is the cooling effect not great enough and that is why we need to involve the phase change part (like in current refrigeration technology)?

9. Mar 21, 2014

### Staff: Mentor

Yes and yes.

10. Mar 21, 2014

### Red_CCF

Why wouldn't the air (taking it as a real gas) cool? I always thought in a Joule-Thomson process or free expansion to a vacuum, temperature always decrease as molecular KE is converted to PE due to a lower pressure at the exit/end state?

Thanks

11. Mar 21, 2014

### Staff: Mentor

In the Joule-Thompson process applied to an ideal gas, there is no change in temperature for the gas. Mechanistically, the expansion cooling occurring within the valve is exactly cancelled by the viscous heating in the valve.

Chet

12. Mar 21, 2014

### Red_CCF

I get that for an ideal gas enthalpy is pressure independent and temperature wouldn't change, but wouldn't a real gas always cool? Was the mechanism I am visualizing by which temperature decreases correct (particle KE converted to PE as P decreases at constant specific energy)?

Thanks

13. Mar 21, 2014

### dauto

Well, we're talking about ideal gas. Also note that real gasses due to Joule-Thomson effect, somewhat paradoxically, sometimes warm up instead of cool down when they expand.

14. Mar 21, 2014

### Staff: Mentor

Real gases often closely approach ideal gas behavior in actual practical situations. That's why the ideal gas model is of such widespread application and widespread value.

Chet

15. Mar 21, 2014

### Red_CCF

How does that work? I looked at a couple of refrigerant P-h graph and temperature trends downwards as pressure goes down at constant specific enthalpy?

Thanks

16. Mar 21, 2014

### Staff: Mentor

The reference you provided looks very solid. As far as your molecular interpretation is concerned, I must admit I'm not too adept at doping out that type of thing. I'm more of a continuum mechanics guy.

Your reference indicates that, just beyond the ideal gas region, at constant enthalpy, the temperature decreases with pressure and, at constant internal energy, the temperature decreases with volume. That seems good enough for me. However, I don't rule out regions further beyond the ideal gas region where, for some gases, the reverse behavior might come into play. Have you seen any p-h diagrams where this is shown?

Chet

17. Mar 21, 2014

### Red_CCF

Hi

From some searching I found this wiki article talking about inversion temperature (http://en.wikipedia.org/wiki/Inversion_temperature) found as 27/4T_c, which is above the vapor dome which I didn't pay any attention to and many diagrams cut off. I found this P-h diagram http://www.4shared.com/file/yU4FnV_U that extends further and it shows that indeed temperature does increase at constant h with decreasing P far above the vapor dome. It looks like I learned something today!

The wiki link talks about attractive/repulsive intermolecular forces; what specific forces would these include? Also what happens when a supercritical fluid with has T > T_inv goes through a JT process?

Thanks very much

18. Mar 22, 2014

### Staff: Mentor

To find the answer to your JT question, just examine the p-H diagram that you were referring to. I guess that 6.75 Tc refers to the prediction of some EOS model (probably VanDer Waals).

Chet

19. May 23, 2014

### CBHarding

20. Jan 23, 2017

### Dirk Gerwin

Here would be a simple experiment to test this. Use a cylinder and piston with a sensor that can read temperature in the cylinder. fill the cylinder with air at atmospheric pressure. Since you know the volume of the cylinder you can calculate the mass of the air inside the cylinder. Now apply a measured amount of force (energy) to the piston compressing the volume of air to a smaller volume (I would suggest a ratio of compression similar to what is used in a diesel engine) and measure the change in temperature in the cylinder (Do this during the compression because energy will quickly be taken from air in the form of heat transferred from the air to the cylinder).
These are the results I think you would find. The rise in temperature would be greater than what the energy applied to the piston could account for. Since no other energy has been added to the air in the piston something else must account for the rise in temperature. I believe that this is due to the higher density of the energy in the compressed gas. When the gas expands it cools because the same amount of energy is spread out over a greater volume. Think of the science experiment that almost every young budding physicist has done, setting paper on fire with a magnifying glass. The magnifying glass focuses (compresses) the light going through it, increasing its intensity (temperature), but does not add any energy to the light. The same amount of energy focused in a smaller area (volume) makes it strong enough to ignite the paper (makes the air hot enough to ignite the diesel fuel being sprayed into a diesel cylinder). I hope this makes sense to everyone.