# Temperature drop in an expanding gas?

1. Nov 12, 2011

### sugarandfat

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?

Thanks,

Daniel.

2. Nov 12, 2011

### Staff: Mentor

The air remaining in the bottle, or the air released or what?

3. Nov 12, 2011

### sugarandfat

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. Nov 13, 2011

### I like Serena

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. Nov 14, 2011

### sugarandfat

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:

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

final temp:

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

or 20 C, which is the room temperature.

..?

6. Nov 14, 2011

### I like Serena

The formula is to-the-power-of-gamma:

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

As you did it, it worked out as an isothermal process (constant temperature).

7. Nov 14, 2011

### sugarandfat

Ok, now I get the volume as 6.31385 L

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

This can't be right..

8. Nov 14, 2011

### I like Serena

I believe your calculation is correct now.
Did you try it out in practice?

9. Nov 14, 2011

### sugarandfat

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. Nov 14, 2011

### I like Serena

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. Nov 14, 2011

### sugarandfat

Maybe the molar mass is off for some reason?

12. Nov 14, 2011

### I like Serena

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).

13. Nov 14, 2011

### D H

Staff Emeritus
Correct. It's a free expansion.

14. Nov 14, 2011

### sugarandfat

So it's not reversible and all bets are off?

15. Nov 15, 2011

### D H

Staff Emeritus
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. Nov 15, 2011

### sugarandfat

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. Nov 15, 2011

### ehild

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?

ehild

18. Nov 15, 2011

### I like Serena

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. Nov 15, 2011

### sugarandfat

> 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. Nov 15, 2011

### 256bits

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
or
T V^(γ-1) = constant
or
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

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.