Why does an expanding gas cool?

In summary, refrigeration systems work by using heat to break the attraction between molecules of liquids at their boiling point, resulting in a decrease in kinetic energy and a lower temperature. This can be seen in the compression and expansion of gases, where compression increases temperature and expansion decreases it. In real refrigerators, other processes may also be used, such as the evaporation of freon which absorbs latent heat and the Joule-Thomson effect which allows a gas to cool during expansion even without work being done. In terms of pressurized gas as a form of refrigeration, it may not be as efficient as involving a phase change.
  • #1
aochider
13
0
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!
 
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  • #2
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
BOYLANATOR said:
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.

excellent visualization for that one. thanks!
 
  • #4
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
dauto said:
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)

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
aochider said:
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?
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
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
Chestermiller said:
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.

dauto said:
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.

thank you for the help!

might be more of an engineering question now, but couldn't 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
aochider said:
thank you for the help!

might be more of an engineering question now, but couldn't 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)?
Yes and yes.
 
  • #10
dauto said:
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.

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
Red_CCF said:
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
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 canceled by the viscous heating in the valve.

Chet
 
  • #12
Chestermiller said:
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 canceled by the viscous heating in the valve.

Chet

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
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
Red_CCF said:
I get that for an ideal gas enthalpy is pressure independent and temperature wouldn't change, but wouldn't a real gas always cool?

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
dauto said:
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.

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?

Also, what about for free expansion? I was reading this link: http://www.etomica.org/app/modules/sites/JouleThomson/Background2.html and they state that real gas always cool in free expansion?

Thanks
 
  • #16
Red_CCF said:
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?

Also, what about for free expansion? I was reading this link: http://www.etomica.org/app/modules/sites/JouleThomson/Background2.html and they state that real gas always cool in free expansion?

Thanks
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
Chestermiller said:
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

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
Red_CCF said:
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
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
Yeah, you have to consider enthalpy[1].

References:

[1] University of Arizona. The Joule Expansion. chem.arizona.edu[online]. 2014. Available from: http://www.chem.arizona.edu/~salzmanr/480a/480ants/jadjte/jadjte.html
 
  • #20
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.
 
  • #21
dauto said:
To complement Chestermiller's answer above, if you open the bottle of air in a vacuum than it wouldn't cool because there would be nothing for the air to do work on.
I disagree with this "no work" assumption of free expansion. Air has mass. If you open the bottle of pressurized air into a vacuum, the air mass gets accelerated, and that means work is being done. In fact if the bottle is floating in space, the bottle itself will accelerate in one direction due to the mass being thrust out the other direction - that's how rockets work in a vacuum. Puncture a compressed air container while orbiting in the vacuum of space and it will experience thrust, and the container will be moved. Work IS being done.
 
  • #22
I wonder if this problem could be solved as a uniform-state, uniform-flow process. A typical textbook problem, similar to this one, is:

Given an insulated, rigid container that is evacuated. Open a valve to allow ambient air at atmospheric pressure to enter. When the pressure equalizes, what is the temperature inside the container? The answer turns out to be γT, or 1.4 T, where T is the ambient air temperature in Kelvin. This is solved by starting out with an expression for the second law of thermodynamics; there's a good YouTube video here on how to solve it:


But what we have here is the opposite problem: Start out with an insulated, rigid container under pressure. Open a valve to allow the air to escape. When the pressure equalizes, what is the temperature inside the container?

That's actually harder than the textbook problem in that video.

I found an example where the problem was solved (Google for "Entropy Balance for the Depressurization of a Pressure Vessel" and see the isentropic solution on slide 27) but it relies on looking up values in a table.
 
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  • #23
Different versions of this problem have been solved many times on Physics Forums.
 
  • #24
Chestermiller said:
Different versions of this problem have been solved many times on Physics Forums.
Really? I have been browsing Physics Forums for a few days now (just registered today) and I haven't yet come across any version "solved" for an actual real-world example (in this case, final internal temperature after compressed air escapes from a bottle). There's been much discussion and various proposed approaches. The thread mentioned in the first post of this thread determined (also confirmed by experiment) that the formula for adiabatic expansion gives the wrong final temperature result, for example. It looks to me like uniform state uniform flow might work, but I haven't yet found a solution based on that. I think we also have a problem with enthalpy limiting the temperature reduction, assuming the air in the bottle isn't dry.

Unfortunately my knowledge of thermodynamics is rather limited (it's been 35 years since I took thermodynamics in college and haven't used it since then), so I find myself having to hunt around to see if others have solved it.

Post #6 (yours) in this thread gave me a different way to visualize the problem of temperature after air has escaped from a bottle, thanks. This is how I visualize it:
  • Imagine a perfectly rigid and insulating tube floating in the weightless vacuum of space, closed on one end.
  • There is a piston locked in the tube, holding in non-dry air.
  • We have one temperature sensor, at the closed end of the tube, measuring temperature T1, and one pressure sensor measuring P1.
  • We may know the initial volume of air, but it shouldn't matter.
  • The piston is released and the air expands, pushing the piston outward.
  • At some point, the piston encounters an obstruction and stops moving. At this time the pressure in the tube is P2.
  • Mass is irrelevant. Work is force × distance, and force is simply the pressure multiplied by the area of the piston, which travels a distance before stopping.
  • We don't know the final volume of the air, and we don't care, we just are concerned with the temperature sensor at the closed end of the tube.
  • This is a reversible process, not free expansion.
  • The question is, what is the temperature T2 of the air measured at the closed end of the tube?
The answer is apparently not the adiabatic expansion formula: ##T_2 = T_1 \left ( \frac{P_2}{P_1} \right ) ^ \frac{\gamma - 1}{\gamma}##

The other facts in the original problem, about letting the bottle come to ambient temperature before releasing the air, the volume of the bottle, the ambient pressure, and so forth, are distractions. The basic facts are: you have an initial P1 and T1, and you have a final P2, and need to know T2. Possibly the mass of air going through the neck of the bottle is relevant (analogous to the mass of air that moves past the initial position of the piston in my visualization), but I am not sure.
 
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  • #25
Anachronist said:
Really? I have been browsing Physics Forums for a few days now (just registered today) and I haven't yet come across any version "solved" for an actual real-world example (in this case, final internal temperature after compressed air escapes from a bottle). There's been much discussion and various proposed approaches. The thread mentioned in the first post of this thread determined (also confirmed by experiment) that the formula for adiabatic expansion gives the wrong final temperature result, for example. It looks to me like uniform state uniform flow might work, but I haven't yet found a solution based on that. I think we also have a problem with enthalpy limiting the temperature reduction, assuming the air in the bottle isn't dry.
First of all, welcome to Physics Forums. We all look forward to interacting and working with you.

The comment I made concerned the flow of outside air through a small hole in the wall into a chamber that initially contained vacuum (not the problem stated in this thread). Another version I have seen in Physics Forums involved a chamber split in half, with half filled with gas and the other half under perfect vacuum, separated by an insulated barrier; a perforation is made in the barrier.
Post #6 (yours) in this thread gave me a different way to visualize the problem of temperature after air has escaped from a bottle, thanks. This is how I visualize it:
  • Imagine a perfectly rigid and insulating tube floating in the weightless vacuum of space, closed on one end.
  • There is a piston locked in the tube, holding in non-dry air.
  • We have one temperature sensor, at the closed end of the tube, measuring temperature T1, and one pressure sensor measuring P1.
  • We may know the initial volume of air, but it shouldn't matter.
  • The piston is released and the air expands, pushing the piston outward.
  • At some point, the piston encounters an obstruction and stops moving. At this time the pressure in the tube is P2.
  • Mass is irrelevant. Work is force × distance, and force is simply the pressure multiplied by the area of the piston, which travels a distance before stopping.
  • We don't know the final volume of the air, and we don't care, we just are concerned with the temperature sensor at the closed end of the tube.
  • This is a reversible process, not free expansion.
  • The question is, what is the temperature T2 of the air measured at the closed end of the tube?
The answer is apparently not the adiabatic expansion formula: ##T_2 = T_1 \left ( \frac{P_2}{P_1} \right ) ^ \frac{\gamma - 1}{\gamma}##
I'm having trouble visualizing what you are describing. Can you please provide a diagram? I'm particularly wondering why you feel that this process is reversible.
The other facts in the original problem, about letting the bottle come to ambient temperature before releasing the air, the volume of the bottle, the ambient pressure, and so forth, are distractions. The basic facts are: you have an initial P1 and T1, and you have a final P2, and need to know T2. Possibly the mass of air going through the neck of the bottle is relevant (analogous to the mass of air that moves past the initial position of the piston in my visualization), but I am not sure.
Again, the original problem description was never clear to me, and still isn't.
 
  • #26
Chestermiller said:
I'm having trouble visualizing what you are describing. Can you please provide a diagram? I'm particularly wondering why you feel that this process is reversible.

Again, the original problem description was never clear to me, and still isn't.
Certainly. The original problem was the first post of this thread: https://www.physicsforums.com/threads/temperature-drop-in-an-expanding-gas.549795/

Original problem statement: A 2-liter soda bottle contains 5 bars of air pressure at 20°C. The surrounding environment is 1 bar at 20°C. What is the temperature inside the bottle after the pressure is released?

More general problem statement: Compressed air (a real gas) at pressure P1 and temperature T1 decompresses rapidly to pressure P2. What is the new temperature T2?

We can assume a perfectly rigid and insulating bottle, for simplicity. A complication might be that the air isn't dry. Someone asking questions about pressurizing plastic soda bottles is likely asking due to an interest in the hobby of water rocketry, because that's the primary reason why you'd want to pressurize a 2L bottle to 5 bars. So the air in the bottle will contain water vapor.

If we further assume that the bottle ejects its air into a perfectly insulating bag that provides no back pressure, then the process should be reversible: You can perform work to squeeze the same air back into the bottle, which should be the same work that pushed out the air in the first place. So in my previous message I proposed an alternative model that should be analogous to the original problem:

bottle_air_expansion_analogy.png


I thought at first that this is reversible. Work is done to move the piston out, and you can do the same work to move it back... but now I'm not so sure. My illustration above looks a lot like a Joule expansion, which isn't reversible.

Water rocketry became an interest of mine lately also, and as a former physics student 35 years ago, I'm trying to re-learn my weakest of all subjects, thermal physics. I haven't done any physics other than electromagnetism during my career, so effectively I'm re-learning thermodynamics from scratch. I'm trying to develop a numerical simulation of the thrust during the rocket flight, taking into account not only how internal pressure changes with volume as the water mass is ejected, but also how the pressure changes due to temperature cooling. When all the water is gone, there is still residual pressure in the bottle that provides a final burst of thrust as the air blows out... but at that time, the air has been cooled due to previous expansion to push out the water, and the air cools further as it expands from the bottle, which further reduces the pressure available for thrust.

My simulations using the ideal-gas assumption that temperature is constant results in over-optimistic thrust values not seen in actual practice. My simulations that account for pressure reductions due to cooling from adiabatic expansion may result in over-pessimistic thrust values because the adiabatic cooling calculation gives extremely low values of temperature for decompressed air; this fact is noted in the thread linked above, where a realistic final temperature for the original problem might be 10°C, not the -88°C given by the adiabatic expansion formula.
 
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  • #27
Anachronist said:
Certainly.
So in my previous message I proposed an alternative model that should be analogous to the original problem:

View attachment 198699

I thought at first that this is reversible. Work is done to move the piston out, and you can do the same work to move it back... but now I'm not so sure. My illustration above looks a lot like a Joule expansion, which isn't reversible.
Before getting into the soda bottle problem, I would like to first focus on this problem. Some questions:
1. Are you assuming that the piston is frictionless, or does it experience frictional forces from the wall of the cylinder?
2. Are you assuming that the piston has mass, or can its mass be assumed to be negligible?
 
  • #28
Chestermiller said:
Before getting into the soda bottle problem, I would like to first focus on this problem. Some questions:
1. Are you assuming that the piston is frictionless, or does it experience frictional forces from the wall of the cylinder?
2. Are you assuming that the piston has mass, or can its mass be assumed to be negligible?

I'm assuming the piston has negligible friction. This is analogous to air leaving the bottle without resistance, as if the bottle had no neck, just an opening at one end of a cylinder. The only resistance would be from displacing the air outside.

I assume that the piston also has mass, but I believe the mass should not matter, because the work performed on the piston (force times distance) is independent of mass. The more the mass, the slower the piston will accelerate, but its kinetic energy when it hits the stops will still be the same regardless of the mass. Furthering the analogy, the piston's mass would be analogous to the mass of air outside the bottle that gets displaced by the air escaping to the outside of the bottle. So a good assumption might be that the mass of the piston equals the mass of the compressed air.

One thing I neglected to mention: we know the mass of the compressed air. In the real-life bottle experiment, we would know the mass of compressed air in the bottle, if we already know its temperature, pressure, and density (assuming it's water-saturated - as it is in the real world, as evidenced by the water-relief valves found on workshop air compressors).

I was thinking of another way to go about solving this, using the work performed on the piston and subtract it from the energy of the compressed air, using a numerical technique with very small time steps. We know the initial force on the piston. That force will move the piston Δd in time step Δt, expending work ΔW. Then subtract ΔW from the energy in the air at each time step, which will have a corresponding effect of reducing both temperature and pressure (although I don't know the relationship yet), without needing to know initial or final volume. Then on the next time step, we have a lower pressure for performing work. Repeat until the piston hits the stop and you get the final temperature. I'm just brainstorming here - does that make any kind of sense? Using the bottle analogy, the work performed by the pressurized air might be the force on the air inside the neck of the bottle times the distance that air was moved through the neck.

Compressed air tanks are so common, and their internal heating or cooling are so commonly observed by everyone from auto mechanics to scuba divers, that you'd think the thermal physics of air escaping from a tank would be a textbook problem! But I cannot find anything online that shows anyone has actually solved it for a real-world example.
 
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  • #29
Anachronist said:
I'm assuming the piston has negligible friction. This is analogous to air leaving the bottle without resistance, as if the bottle had no neck, just an opening at one end of a cylinder. The only resistance would be from displacing the air outside.

I assume that the piston also has mass, but I believe the mass should not matter, because the work performed on the piston (force times distance) is independent of mass. The more the mass, the slower the piston will accelerate, but its kinetic energy when it hits the stops will still be the same regardless of the mass. Furthering the analogy, the piston's mass would be analogous to the mass of air outside the bottle that gets displaced by the air escaping to the outside of the bottle. So a good assumption might be that the mass of the piston equals the mass of the compressed air.
If the piston is massless, then the gas does no work whatsoever. From a force balance on the piston $$P_{gas}A=ma=0$$. So, in the limit of zero piston mass, $$P_{gas}\rightarrow 0$$. This is the gas pressure at the piston face. For an irreversible expansion like this, the pressure within the cylinder is non-uniform spatially, so even though the gas pressures at other locations are greater than zero, the value at the piston face approaches zero. So, negligible work is done by the gas on the piston, and no heat is transferred during this irreversible expansion. So, from the initial and final states of the system, $$\Delta U=Q-W=0$$

If the piston has mass, then the gas does work on the piston to accelerate it. But the kinetic energy gained by the piston converts back to internal energy when the piston is stopped by the detent. So, here again, $$\Delta U=0$$For an ideal gas, it its change in internal energy is zero, its change in temperature is zero. For a real gas, since internal energy also depends on specific volume, there is a small Joule-Thompson change in temperature (related to the non-ideal gas behavior).

I was thinking of another way to go about solving this, using the work performed on the piston and subtract it from the energy of the compressed air, using a numerical technique with very small time steps. We know the initial force on the piston. That force will move the piston Δd in time step Δt, expending work ΔW. I was thinking of using the kinetic theory of gases to subtract ΔW from the energy in the air at each time step, which will have a corresponding effect of reducing both temperature and pressure, without needing to know initial or final volume. Then on the next time step, we have a lower pressure for performing work. Repeat until the piston hits the stop and you get the final temperature. I'm just brainstorming here - does that make any kind of sense? Using the bottle analogy, the work performed by the pressurized air might be the force on the air inside the neck of the bottle times the distance that air was moved through the neck.
This solution method would not work because the gas within the cylinder is not uniform with respect to its spatial pressure distribution and temperature distribution. To get the details of what is happening temporally and spatially within the cylinder, one needs to solve the partial differential equations of gas dynamics, typically using computational fluid dynamics software. But the beauty of thermodynamics is that, for many problems like this one, one only needs to focus on the work and heat at the boundaries (if your interest is only in the final state of the system).
Compressed air tanks are so common, and their internal heating or cooling are so commonly observed by everyone from auto mechanics to scuba divers, that you'd think the thermal physics of air escaping from a tank would be a textbook problem! But I cannot find anything online that shows anyone has actually solved it for a real-world example.
Moran et al, in Fundamentals of Engineering Thermodynamics solve a problem for the air temperature in a tank as air leaks out (slowly). The problem can also be solved for rapid irreversible leakage.
 
  • #31
Chestermiller said:
For a real gas, since internal energy also depends on specific volume, there is a small Joule-Thompson change in temperature (related to the non-ideal gas behavior).
Well, that pretty much hits the nail on the head. We're dealing with a real gas, and the formula for adiabatic expansion gives an unrealistic result, so perhaps what I'm looking for is the temperature reduction (and corresponding effect on pressure and volume) due to that effect?

Chestermiller said:
Moran et al, in Fundamentals of Engineering Thermodynamics solve a problem for the air temperature in a tank as air leaks out (slowly). The problem can also be solved for rapid irreversible leakage.
I wondered why we couldn't just assume uniformity while the air expands, but I guess rapid irreversible leakage is a different problem.

Chestermiller said:
BOTTLE PROBLEM: See Example 2 in the following thread, imagining a membrane surrounding the air that was originally inside the bottle to separate it from the remainder of the air in the room:
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Aha! Yes, example 2 is a good description of the situation. In spite of being unable to think in terms of differential equations and integrals anymore, I did manage to follow that; it's quite clear. The fact that we get an always-positive ΔS (equation 9) means that the process is irreversible. Earlier in my search, I had seen statements to the effect that an adiabatic expansion is irreversible but up until now I didn't really understand why.
 
  • #32
I wanted to mention one more thing:

Chestermiller said:
For a real gas, since internal energy also depends on specific volume, there is a small Joule-Thompson change in temperature (related to the non-ideal gas behavior).

In the case of a 2-liter bottle with 5 bars of pressure at 20°C, rapidly released into a 1-bar environment at 20°C, the change in temperature isn't "small", it's rather significant; enough to cause fog condensation and a noticeable coldness to the bottle. But the temperature reduction is nowhere near as dramatic as the adiabatic expansion formula predicts.

I have another question, just for clarification. In the water rocket application, the bottle starts out with some water in it, say, half a liter in a 2-liter bottle. The pressure is typically pumped up to 7 bars (100 psi) above ambient pressure. When the rocket is launched, clearly the energy in the compressed air is doing work, forcing the mass of water out, and accelerating the rocket up. Would the expansion of the gas that forces out the water be considered an adiabatic process that can be well-modeled by the formula ##T_2 = T_1 \left ( \frac{p_2}{p_1} \right ) ^\frac{\gamma - 1}{\gamma}## ?

I suspect so... no gas is released during that water-thrust phase, it just expands enough to displace all the water. The problem in this thread has to do with what happens thermodynamically after the water is all gone and only air pressure remains in the bottle (in this example, the remaining pressure is about 5 bars, as stated in the original problem).
 
  • #33
Anachronist said:
I wanted to mention one more thing:

In the case of a 2-liter bottle with 5 bars of pressure at 20°C, rapidly released into a 1-bar environment at 20°C, the change in temperature isn't "small", it's rather significant; enough to cause fog condensation and a noticeable coldness to the bottle. But the temperature reduction is nowhere near as dramatic as the adiabatic expansion formula predicts.]
See what you get if you use the equation I gave for the final temperature in example 2.
I have another question, just for clarification. In the water rocket application, the bottle starts out with some water in it, say, half a liter in a 2-liter bottle. The pressure is typically pumped up to 7 bars (100 psi) above ambient pressure. When the rocket is launched, clearly the energy in the compressed air is doing work, forcing the mass of water out, and accelerating the rocket up. Would the expansion of the gas that forces out the water be considered an adiabatic process that can be well-modeled by the formula ##T_2 = T_1 \left ( \frac{p_2}{p_1} \right ) ^\frac{\gamma - 1}{\gamma}## ?
It depends on how much resistance the water is offering to the air expansion. The temperature will be somewhere between this equation, and the equation I gave in example 2. In my judgment, it will probably be closer to this equation. Do the calculation both ways and compare.
 
  • #34
Chestermiller said:
See what you get if you use the equation I gave for the final temperature in example 2.
OK, please pardon my dearth of thermodynamics knowledge here. Is this correct?

1. Calculate entropy change in terms of initial conditions and final pressure, using γ=1.4 (actually more like 1.398 for humid air according to one reference I found):
$$\Delta S=\frac{P_0V_0}{T_0}\frac{\gamma}{(\gamma-1)}\ln{\left[1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0}\right]}-\ln{\left[1-\frac{(P_0-P_F)}{P_0}\right]}$$

2. Then I'd need to calculate the final temperature from the entropy change by solving for TF here:
$$\Delta S=c_v \ln{\left ( \frac{T_F}{T_0} \right )}+R \ln{\left ( \frac{V_F}{V_0} \right )}$$
The problem is, I don't know VF, which would be the volume of all the air that escaped plus the volume in the bottle. I could use the ideal gas law ##P_0 V_0 = P_F V_F## but that assumes constant temperature, and the whole point of this exercise is that temperature isn't constant, so ##V_F## would be somewhat less than simply ##P_0 V_0 / P_F##, correct?
 
  • #35
Anachronist said:
OK, please pardon my dearth of thermodynamics knowledge here. Is this correct?

1. Calculate entropy change in terms of initial conditions and final pressure, using γ=1.4 (actually more like 1.398 for humid air according to one reference I found):
$$\Delta S=\frac{P_0V_0}{T_0}\frac{\gamma}{(\gamma-1)}\ln{\left[1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0}\right]}-\ln{\left[1-\frac{(P_0-P_F)}{P_0}\right]}$$

2. Then I'd need to calculate the final temperature from the entropy change by solving for TF here:
$$\Delta S=c_v \ln{\left ( \frac{T_F}{T_0} \right )}+R \ln{\left ( \frac{V_F}{V_0} \right )}$$
The problem is, I don't know VF, which would be the volume of all the air that escaped plus the volume in the bottle. I could use the ideal gas law ##P_0 V_0 = P_F V_F## but that assumes constant temperature, and the whole point of this exercise is that temperature isn't constant, so ##V_F## would be somewhat less than simply ##P_0 V_0 / P_F##, correct?
You use Eqn. 5 of Example 2.
 
<h2>1. Why does an expanding gas cool?</h2><p>When a gas expands, it does work on its surroundings by pushing against the walls of the container. This work requires energy, which is taken from the internal energy of the gas. As a result, the average kinetic energy of the gas particles decreases, leading to a decrease in temperature.</p><h2>2. How does the temperature of an expanding gas relate to its volume?</h2><p>According to the ideal gas law, the temperature of a gas is directly proportional to its volume. This means that as the gas expands and its volume increases, its temperature will decrease.</p><h2>3. Does the type of gas affect how much it cools when expanding?</h2><p>Yes, the type of gas does affect how much it cools when expanding. This is because different gases have different molecular weights and therefore different internal energies. Lighter gases will cool more when expanding compared to heavier gases.</p><h2>4. Can an expanding gas cool to absolute zero?</h2><p>No, an expanding gas cannot cool to absolute zero. This is because according to the third law of thermodynamics, it is impossible to reach absolute zero through a finite number of processes. The gas will continue to cool as it expands, but it will never reach absolute zero.</p><h2>5. How does the pressure of an expanding gas affect its cooling?</h2><p>The pressure of an expanding gas does not directly affect its cooling. However, it does affect the rate at which the gas expands, which in turn affects the rate at which it cools. Higher pressures will result in slower expansion and therefore slower cooling, while lower pressures will result in faster expansion and faster cooling.</p>

1. Why does an expanding gas cool?

When a gas expands, it does work on its surroundings by pushing against the walls of the container. This work requires energy, which is taken from the internal energy of the gas. As a result, the average kinetic energy of the gas particles decreases, leading to a decrease in temperature.

2. How does the temperature of an expanding gas relate to its volume?

According to the ideal gas law, the temperature of a gas is directly proportional to its volume. This means that as the gas expands and its volume increases, its temperature will decrease.

3. Does the type of gas affect how much it cools when expanding?

Yes, the type of gas does affect how much it cools when expanding. This is because different gases have different molecular weights and therefore different internal energies. Lighter gases will cool more when expanding compared to heavier gases.

4. Can an expanding gas cool to absolute zero?

No, an expanding gas cannot cool to absolute zero. This is because according to the third law of thermodynamics, it is impossible to reach absolute zero through a finite number of processes. The gas will continue to cool as it expands, but it will never reach absolute zero.

5. How does the pressure of an expanding gas affect its cooling?

The pressure of an expanding gas does not directly affect its cooling. However, it does affect the rate at which the gas expands, which in turn affects the rate at which it cools. Higher pressures will result in slower expansion and therefore slower cooling, while lower pressures will result in faster expansion and faster cooling.

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