- #1

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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.

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- #1

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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

Drakkith

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

- #3

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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

I like Serena

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That means you have the relation P

Together with the ideal gas formula PV=nRT, you can deduce the final temperature.

- #5

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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

I like Serena

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[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

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PV=nRT

1 bar . 6.31385 L = .41571 mol . R . T

= 185.091 K (kelvins) / -88.059 °C

This can't be right..

- #8

I like Serena

Homework Helper

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I believe your calculation is correct now.

Did you try it out in practice?

Did you try it out in practice?

- #9

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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

I like Serena

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It does make me wonder how "adiabatic" it is...

Let me check with the specialists on PF.

I'll get back to you.

- #11

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Maybe the molar mass is off for some reason?

- #12

I like Serena

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The number of moles "n" you found is correct (this is not the molar mass).Maybe the molar mass is off for some reason?

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

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Correct. It's a free expansion.Something's not right.

- #14

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

- #15

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The process is not reversible, but all bets are not off.So it's not reversible and all bets are off?

What work is the escaping gas doing? Use the answer to this question in conjunction with the first law of thermodynamics.

- #16

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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

ehild

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

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 P

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=-P

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

I like Serena

Homework Helper

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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

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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

256bits

Gold Member

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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**

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.

P1V1γ = Constant, where γ=1.41 for air

P1V1γ = P2V2γ,

V1γ = P2/P1 * V2γ

= (5 bar / 1 bar ) * 1 lire

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)

P1 V1/T1 = P2 V2 / T2

T2 = (P2/P1)(V2/V1)(T1)

= (5/1)(1/3.13)(273K) =

P2 V2 / T2 = P3 V3 / T3

P3 = P2 (V2/V3) (T3/T2)

= 5bar ( 1/1) ( 273K/436K )

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

Incorrect answer

P = 1x10^5 Pa = (Nt/m^2)

1 liter = .1x.1x.1 m^3 = 1*10^ -3m^3

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

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

To determine the change in temperature,

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

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> 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?

Cheers.

- #22

256bits

Gold Member

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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.

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.

**V1 = 3,13 litre **

Similarily your amount of air pumped in is also larger somewhat.

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**

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.

That temperature would be for a single quick compression with no heat loss.> 436K ie 163 C

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.

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.> 3.13 bar

Similarily your amount of air pumped in is also larger somewhat.

I put that in there to explain that the work you put in to reach the pressure of 5 bar, is lost in two seperate processes. One by by letting it cool down. Second by the expansion to the atmosphere.> There is lost work due to two reasons:

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

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

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

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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 realise that, and other principles that can be utilised.

- #24

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measuring the temperature drop of the water would tell you how much heat was removed.

- #25

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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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