Why Does an Expanding Gas Cool Down?

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.
  • #36
Chestermiller said:
You use Eqn. 5 of Example 2.
How embarrassing. I was focusing on the end result in Example 2 and trying to back out temperature from entropy, and somehow forgot about equation 5.

Using γ=1.4, P0=5 bars, PF=1 bar, T0=20°C=293K:

Adiabatic expansion: ##T_F= T_0 \left ( \frac{P_F}{P_0} \right ) ^\frac{\gamma - 1}{\gamma} = 185\text{K} = -88°\text{C}##

Eq 5 from example 2: ##T_F=T_0 \left[1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0}\right] = 226\text{K} = -47°\text{C}##

The equation 5 result looks more reasonable but still seems colder than what would be observed in the real world. I would guess that's likely due to two things not accounted for here: (1) the air is being forced through a constriction (does that matter?), and (2) water vapor in the air effectively chokes down the reduction in temperature, so it doesn't go as low.
 
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  • #37
Anachronist said:
How embarrassing. I was focusing on the end result in Example 2 and trying to back out temperature from entropy, and somehow forgot about equation 5.

Using γ=1.4, P0=5 bars, PF=1 bar, T0=20°C=293K:

Adiabatic expansion: ##T_F= T_0 \left ( \frac{P_F}{P_0} \right ) ^\frac{\gamma - 1}{\gamma} = 185\text{K} = -88°\text{C}##

Eq 5 from example 2: ##T_F=T_0 \left[1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0}\right] = 226\text{K} = -47°\text{C}##

The equation 5 result looks more reasonable but still seems colder than what would be observed in the real world. I would guess that's likely due to two things not accounted for here: (1) the air is being forced through a constriction (does that matter?),
The constriction would make it closer to the -88 within the chamber.
and (2) water vapor in the air effectively chokes down the reduction in temperature, so it doesn't go as low.
Condensation of water vapor couldn't contribute much.

I would think maybe heat from the container wall might modulate the temperature drop of the gas, but that would have to be looked at.
 
  • #38
Chestermiller said:
Condensation of water vapor couldn't contribute much.
I mentioned that because someone in another forum told me "Any moisture in the air will make it difficult (require extreme expansion) to cool it below 0°C because of water's significant Enthalpy of fusion." He may have a point there, in that the temperature will likely get stuck at 0°C until the water has precipitated out.
I would think maybe heat from the container wall might modulate the temperature drop of the gas, but that would have to be looked at.
That's a good point. Even though the wall is thin and made of a somewhat non-thermally-conductive polymer material, it still has a much larger mass than the air inside (50g versus a tiny fraction of a gram of air).

Thank you so much for your help on this. Too bad the original poster of this thread hasn't had any activity lately. He or she would be pleased to see the conclusion.

I think I have a path forward now to refine my simulation of the rocket's flight. It's an interesting problem with several variables that change rapidly over time: mass, mass flow rate, pressure, and temperature, which in turn affect thrust, acceleration, velocity, air resistance, and altitude. I'm finding that certain combinations of initial water volume, ballast, and nozzle diameter achieve more altitude than others, and with further refinements based on thermodynamics, I suspect that the world record altitude of 85.7 meters achieved by an unreinforced 2-liter bottle rocket is probably right near the upper theoretical limit of what is possible given the constraints imposed (e.g. no more than 100psi initial pressure, 2-liter soda bottle, and the ballast weight required to hold recording instrumentation). So if that record is broken, it won't be by much.
 
  • #39
Anachronist said:
I mentioned that because someone in another forum told me "Any moisture in the air will make it difficult (require extreme expansion) to cool it below 0°C because of water's significant Enthalpy of fusion." He may have a point there, in that the temperature will likely get stuck at 0°C until the water has precipitated out.
People on Physics Forums often like to wave their hands a lot without actually making any quantitative calculations. Why don't you assume that the air starts at 20 C with a relative humidity of 80% and calculate how much heat is required to both condense the water vapor to liquid and to freeze the resulting liquid? That will settle the issue once and for all.
That's a good point. Even though the wall is thin and made of a somewhat non-thermally-conductive polymer material, it still has a much larger mass than the air inside (50g versus a tiny fraction of a gram of air).
Do the calculation and see what you get. Are you sure that an empty 2 liter bottle weights that much?
 
  • #40
Chestermiller said:
People on Physics Forums often like to wave their hands a lot without actually making any quantitative calculations. Why don't you assume that the air starts at 20 C with a relative humidity of 80% and calculate how much heat is required to both condense the water vapor to liquid and to freeze the resulting liquid? That will settle the issue once and for all.
Yes, that's one of my objectives. As I said earlier, I'm having to re-learn my weakest subject in physics from 35 years ago from scratch. It's amazing how having a child interested in a new hobby can motivate a father (yeah, I'm pretty old to be a dad, but that's how it is; his schoolmates think I'm his grandpa). Eventually I want him to appreciate the science that can be done with this hobby. At the moment I have no idea how to account for the temperature (and pressure) modulation effects of moisture, although I suspect it has something to do with enthalpy -- and I still need to re-learn that. I have a physics background, but in the area of thermodynamics I'm basically a layman.
Do the calculation and see what you get. Are you sure that an empty 2 liter bottle weights that much?
Yes, a 2-L round-shouldered polyethylene terephthalate soda bottle weighs 50g, other styles weigh more or less, plus or minus a few grams.
 
  • #41
Anachronist said:
Yes, that's one of my objectives. As I said earlier, I'm having to re-learn my weakest subject in physics from 35 years ago from scratch. It's amazing how having a child interested in a new hobby can motivate a father (yeah, I'm pretty old to be a dad, but that's how it is; his schoolmates think I'm his grandpa). Eventually I want him to appreciate the science that can be done with this hobby. At the moment I have no idea how to account for the temperature (and pressure) modulation effects of moisture, although I suspect it has something to do with enthalpy -- and I still need to re-learn that. I have a physics background, but in the area of thermodynamics I'm basically a layman.
Well, if you want to make the calculation, I can help you.
 

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