Temperature drop in an expanding gas?

  • #1
Hi, I'm prototyping a non electrical refrigeration system and need some help with a calculation.

I have a 2 litre bottle.
I pump in air until the internal pressure is 5 bar.
I allow the heat increase to dissipate back to the ambient 20 C.
I release the air.

What will the temperature of the air now be?



Answers and Replies

  • #2
The air remaining in the bottle, or the air released or what?
  • #3
Well, they should be the same, no?
It'll start to warm up on contact with the outside air, but what will the temperature of the air leaving the bottle initially be when it has come back down to 1 bar?
  • #4
If you release the air quickly, it will behave like an adiabatic expansion.

That means you have the relation P1V1γ = P2V2γ, where γ=7/5 for air.

Together with the ideal gas formula PV=nRT, you can deduce the final temperature.
  • #5
Something's not right.

P1.V1.γ = P2.V2.γ
5 bar . 2 L . 1.4 = 1 bar . V2 . 1.4
gives V2 as 10 litres.

to calculate molar mass:

5 bar . 2 L = n . R . 293.15 K
gives n as 0.41571 mol

final temp:

1 bar . 19 L = .41571 mol . R . T
= 293.15 K

or 20 C, which is the room temperature.

  • #6
The formula is to-the-power-of-gamma:

[tex]P_1 \cdot (V_1)^γ = P_2 \cdot (V_2)^γ[/tex]
[tex]5 \textrm{ bar} \cdot (2 \textrm{ L})^{1.4} = 1 \textrm{ bar} \cdot (V_2)^{1.4}[/tex]

As you did it, it worked out as an isothermal process (constant temperature).
  • #7
Ok, now I get the volume as 6.31385 L

1 bar . 6.31385 L = .41571 mol . R . T
= 185.091 K (kelvins) / -88.059 °C

This can't be right..
  • #8
I believe your calculation is correct now.
Did you try it out in practice?
  • #9
It can't possibly be -88 C, that's the boiling point of CO2...

I did a quick test on this, at 5 bar, a 1.2 litre bottle, and 19 C room temp, the air came out at 10.5 C.

At 7 bar and 16 C ambient it was 5.5 C.

It wasn't exactly an optimised rig, but close ish.
  • #10
Hmm, I can't explain that.
It does make me wonder how "adiabatic" it is...

Let me check with the specialists on PF.
I'll get back to you.
  • #11
Maybe the molar mass is off for some reason?
  • #12
Maybe the molar mass is off for some reason?

The number of moles "n" you found is correct (this is not the molar mass).

Note that the number of moles "n", and also the volume of your bottle (2 L), are irrelevant to the problem.
They cancel in the calculation.

The only relevant data is the initial pressure (5 bar), the final pressure (1 bar), the initial temperature (293 K), and the gamma of the gas (7/5).
The assumptions we make, is that the expansion is quick enough to be adiabatic, and that air behaves sufficiently like a perfect diatomic gas (which it does).
  • #14
So it's not reversible and all bets are off?
  • #15
So it's not reversible and all bets are off?
The process is not reversible, but all bets are not off.

What work is the escaping gas doing? Use the answer to this question in conjunction with the first law of thermodynamics.
  • #16
As it is the air is only expanding freely, so no significant work.

But a few people now have suggested I'll get a greater drop in temperature if I have it working on something; spinning a turbine or similar. There's not really any work needs to be done in the system, so I'd likely have the air acting against gravity/inertia/friction in some way just for the point of dissipating energy.
What specifically this would be I'll have to figure out.

So can this be effectively calculated? Or should I just build the thing and see what happens.
  • #17
According to the First Law, ΔU=Q+W(external)

The external work is that of the ambient air, which is at 1 bar pressure. So W=-Pa ΔV.

But we do not know ΔV. And the expansion is not a quasi-static process, when the gas is in equilibrium with itself and with its surroundings. The overall pressure of the gas is just undefined. The question is, when does the gas stop to expand. It should be when the pressure inside is equal to the ambient pressure Pa=1 bar. But the gas is in turbulent motion when expanding, when does it stop?
If we accept 1 bar as the final pressure, the ideal gas law is one equation for the final volume and temperature. If the process goes on without heat transfer, the change of energy is equal to the external work. So the second equation is ΔU = 5/2nRΔT=-PaΔV. This can be solved, but still results in low temperature. (226 K)

I guess, assuming an adiabatic process is not realistic.
When and with what thermometer was the temperature of the gas measured?
I am also concerned about the initial pressure. Was it measured when the gas in the vessel was at room temperature really?

  • #18
Hmm, here's a thought experiment.
Mind you, I'm still learning here myself.

Suppose we put a divider in the bottle halfway and then we open the bottle.
Half of the content would make a free expansion to the outside.

The other half would expand to the full content, and with some care it would expand as a quasi-static process which is also adiabatic.
In this case, work is done by pushing the other air outside.
Would this work as a reversible adiabatic process?
(The temperature should drop in this case from 20 °C to -61 °C, which still seems like more than we can hope for.)
  • #19
> assuming an adiabatic process is not realistic.

I'm not so much wanting to know what temperature it will ultimately end up as the amount of coldth in it, so I can see what kind of minimum temp is possible in an insulated container.

> When and with what thermometer was the temperature of the gas measured?

Glass and alcohol, proper chemist's one. Takes a couple seconds to adjust, but I was measuring over several minutes.

> Was it measured when the gas in the vessel was at room temperature really?

It may have been slightly warmer. I was cooling the bottle with room temp water, but it probably didn't come all the way down before I released it. It didn't feel warm to the touch by then tho.
I did an earlier run with a lower pressure and the thermometer in the bottle. It came up by about 10 degrees, then cooled back to room temp over a couple minutes.
Actually, come to think of it, if I was to measure the maximum temp then it should be possible to work out how much energy has left the air, no?

I_like_Serena, not really equipped to interact with that question...
  • #20
Excuse some of the formula ( as they look awkward)
STEP 5 is not the way to calculate the final temperarure.

Step 1. A volume of air at 1 bar is compressed into a container of 1 litre to a pressure of 5 bar. What is the initial volume of air?

P1V1γ = Constant, where γ=1.41 for air

P1V1γ = P2V2γ,
V1γ = P2/P1 * V2γ
= (5 bar / 1 bar ) * 1 lire
V1 = 3,13 litre

Step 2. How much work is done compressing the gas?
K = P1V1^ γ
= 1x10^5 Pa x 3.13x10(exp-3) m^2
= 3.13 x 10(exp2) =100

W = K(V2^(1-γ) - V1^(1-γ) )/ (1-γ)
= 313 ( 1(exp-3)^(-0.41) - 3.13(exp-3)^(-0.41) ) / ( -0.41)
= 313( 17 - 10.6 ) / (-0.41)
= 313(-15.6)
= -4886 J

Step 3. What is its temperature
P1 V1/T1 = P2 V2 / T2
T2 = (P2/P1)(V2/V1)(T1)
= (5/1)(1/3.13)(273K) = 436K ie 163 C

Step 4. The gas is allowed to return to temperature of 273 K. What is its pressure.
P2 V2 / T2 = P3 V3 / T3
P3 = P2 (V2/V3) (T3/T2)
= 5bar ( 1/1) ( 273K/436K )
= 3.13 bar

Step 5. The gas now expands adiabatically against a piston. What is the temperarature.
Incorrect by the way
P1V1γ = P2V2γ, where γ=7/5 for air
T V^(γ-1) = constant
P^(1-γ)T^γ = constant

P4^(1-γ)T4^γ = P3^(1-γ)T3^γ

T4^γ = (P3/P4)^(1-γ) * T3^γ
= (3.13 bar/1bar)^(1-γ) * (273K)^γ
= (0,626) ( 2722 )
= 1704
T4 = 195K or -77C
Incorrect answer

Step 6.W = P*V = work on exterior air
P = 1x10^5 Pa = (Nt/m^2)
1 liter = .1x.1x.1 m^3 = 1*10^ -3m^3
W = 100 J

Compare this work (100J) done by the gas ( on the atmosphere ),
to the work (-4886) done (by you) on the gas in step 2,

There is lost work due to two reasons:
1. the gas was allowed to cool after the compression
3. the gas was allowed a free expansion into the atmosphere.
Since there is lost work the temperature of the air during the expansion will drop very little - if it dropped as much as in step 5 that would be a violation of conservatin of energy.

if you have a free expansion of a gas into a vacuum, the temperature of the gas remains constant since there is no work done and no heat change, so the change in internal energy = 0. ( internal energy is a function of temperature only.)

In this case of an expansion into the atmosphere some work is done by the gas.
so the process is neither isentropic (constant entropy) nor isenthalpic ( constant enthalpy)

To determine the change in temperature, we would need to know the enthapy of the gas before and after the expansion.
A change in this case of about 10C sounds about right.

EDIT: PS I was originally going to respond as the first post much earlier, using the same as I but I noticed I ran into the same delemma as I Luv serenna so had to think about this one quite a while.
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  • #21
Wow, nice reply, thanks.

> 436K ie 163 C

Seems a bit high...
At 5 bar and 2.5 litres I measured the temp increase inside the bottle at something around 10 C.

> 3.13 bar

I'm using a bike pump with a pressure gauge to pressurise the bottle. Bringing it up to 5 bar and cooling the air, the pressure doesn't really seem to decrease much if anything, tho this might just be how the gauge works...

> There is lost work due to two reasons: 1. the gas was allowed to cool after the
> compression

So not letting the air cool would lead to a decrease in the expanded air temp? This seems counter intuitive, and the times that I vented the air without leaving it as long, the temp was higher.

> the gas was allowed a free expansion into the atmosphere.

This is starting to look like the crux of the matter. I can further decrease the temp by having the gas do work as it expands, so what would be examples of work? Anything where it is exerting force? Like spinning a turbine or kicking up sand/water?

  • #22
My case ideally assumes a cylinder of air being compressed from 1 bar to 5 bar with no heat loss during compression, with a single piston doing the compression on all the air at once. After 5 bar the air is only then allowed to cool. Your situation with a bicycle pump is thus modeled from the ideal, since there is no way to tell how much cooling is going on while doing the actual action of pumping.

> 436K ie 163 C
That temperature would be for a single quick compression with no heat loss.
With your bicycle pump, you have heat loss from the parts of the pump and the bottle. Feel the bottom of the pump and after many strokes and it is quite hot.

> 3.13 bar
Your pressure will be higher for the above reason. The air has already lost some heat while you are pumping up to 5 bar. Therefor , your air in the bottle at 5 bar will be at a lower temperature.

V1 = 3,13 litre
Similarily your amount of air pumped in is also larger somewhat.

> There is lost work due to two reasons:
I put that in there to explain that the work you put into reach the pressure of 5 bar, is lost in two separate processes. One by by letting it cool down. Second by the expansion to the atmosphere.
If you pump up to 5 bar and expand immediately, then the lost work would be only from the expansion to the atmosphere.
In which case the work against the atmosphere would be the same.

The change in temperature drop would be also somewhat, not much, greater, as the enthapy of the compressed air is not the same as in my case. You would have a bit more air expanding into the atmosphere from a slightly higher pressure.

There is an error in my calculation for the lost work.
It should be Wlost = Patm(Vf-Vi) = 1 bar x ( 1 litre - 3,13 litre ) = 213 Joules

> the gas was allowed a free expansion into the atmosphere.
For a reversible adiabatic compression and expansion there is no heat flow into or out of the system. The work you put in can be (theoretically) recovered by having the gas expand against a resistance. the gas will return to its original pressure, temperature, and volume.
You can try that with your bicycle pump. If the nozzle end is held closed somehow so no air escapes, and you press down and immediately let the handle back up, the handle will return close to its original position. If you push down and hold it there so the heat can flow out, and then let the handle back up, the handle will return to some lower position (ie some lost work). If you push down, hold it there for the heat to flow out, and then reelease the nozzle for the air to flow out, the handlw does not return up at all ( and you have lost all the work you put in ).

I hope that helps out some more.

Edit: It is surprising that this is not part of a textbook example. Everyone has a bicycle pump and wondered about this very same question that is so close to home, so to speak. No one has knocked the analysis yet, so hopefully it passes scrutiny.
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  • #23
Ok, cool, that all makes sense.
So I guess the whole thing is a bit too real-worldy to accurately calculate. I can accept that.

So now the question is, and this may be slightly more engineering than physics, what would be the best way forward for me to optimise the cooling effect? What's the best kind of work, best way to realize that, and other principles that can be utilised.
  • #24
In a more real world approach, the work done could be a fan that increases the heat removal after compression. The work done would decrease the cold temp, and the fan could speed up the high temp dispersal. Also one way you might measure how much heat is absorbed by the gas expansion, would be to release the gas in a tub of water.
measuring the temperature drop of the water would tell you how much heat was removed.
  • #25
Thought about that, but the cooling of the air is done by the water in the tank, which is being circulated for free by the air bubbling through it.
There's not really anything needs doing...

Measuring coldth with a volume of water is a good idea, might do that.

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