Why Does an Expanding Gas Cool Down?

[Anachronist]

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?

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.

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.

 

[Anachronist]

I wanted to mention one more thing:

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).

 

[Chestermiller]

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.

 

[Anachronist]

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 T_F 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 V_F, 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?

 

[Chestermiller]

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 T_F 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 V_F, 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.

 

[Anachronist]

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, P_F=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.

 

[Chestermiller]

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, P_F=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.

 

[Anachronist]

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.

 

[Chestermiller]

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?

 

[Anachronist]

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.

 

[Chestermiller]

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.

 

[Anachronist]

Well, if you want to make the calculation, I can help you.

Thanks - I will start another thread on that. For the original poster of this thread, the problem is solved.

Update: The new thread is here: https://www.physicsforums.com/threads/pressure-and-temperature-of-expanding-humid-air.914421/

 

[Anachronist]

New discussion started on related problem with gas expanding quickly in a 2-liter bottle, involving the effects of humidity: https://www.physicsforums.com/threads/pressure-and-temperature-of-expanding-humid-air.914421/