Physics Puzzle: Can You Reach 100 Pascals with 3 Bottles and a Pump?

[monzie]

Consider the following:

You are in a room containing an 3 empty metal bottles. Each bottle is equipped with a valve which when open connects the interior of the bottle to the room, and when shut seals the bottle.

You also have a pump that can generate an air pressure of 10 Pascals. It can be used to fill the bottles. The pump also has a valve at its inlet to which a bottle can be connected, thereby boosting its output pressre.

You also have any number of valves and pipes and the means to connect them together. These parts also fit the bottles.

Is there a way to fill one of the bottles to a pressure of 100 Pascals?

(Assume that all the components are ideal)

 

[NateTG]

Ugh, that's a tricky problem...

Assuming that you start at STP.

 

[Njorl]

Do you mean exactly 100 pascals? I can see that it would be easy to get more than that, and you only need 2 bottles, not three.

Njorl

Edited - I caught my mistake. Two bottles limits you to 20 pascals.

Hmmm...

 

[Adrian Baker]

Am I missing something here? 10 Pa is a very low pressure... Isn't atmospheric pressure around 100 000Pa?

Open the valve, let some air in?

 

[Njorl]

It is not possible to configure the bottles in such a way that the sum of the pressures is greater than 60 pascals.

Njorl

 

[Chi Meson]

So the pump can achieve a pressure at its output that is 10 Pa more than whatever the pressure is at its intake valve?

 

[Njorl]

One pascal is slightly more than one atmosphere (or is it slightly less?) I believe the ratio is 1.014 one way or the other.

Njorl

 

[Njorl]

Pascals, bars, what's the difference?

D'oh!

Njorl

 

[Chi Meson]

Originally posted by Njorl
One pascal is slightly more than one atmosphere (or is it slightly less?) I believe the ratio is 1.014 one way or the other.

Njorl

Uh, no. 1 pascal is "one Newton per square meter." ATM is 101,400 pascals. So that's 1.014 x 10^5 you're thinking.

 

[NateTG]

Easy answers:

1. Open the bottle -
one atmosphere is a bit more than 1*10⁵ pascals

Tricky answers -

2. Fill a bottle to 30 pascals, and then heat the room/bottle to about 700 C by running the pump or compressing gas. That will cause a pressure difference of 100 pascals between the inside of the bottle and the outside.

3. (Not sure if it's possible to get the bottle cold enough.) Use gas expansion to cool one of the bottles seal it, and let it return to room temperature.

4. (Against the rules I'm sure) Make a bunch of extra bottles out of pipes and valves, use fittings to increase alter the volume of the bottles, or use variable volume bottles.

5. (Probably also against the rules) build a pneumatic compressor out of the pipes and valves or squeeze the bottles or blow into one of the bottles.

So, let's assume that:
1. You're looking for a 100 pascal differential instead of 100 pascals of pressure.
2. The pump only generates a pressure differential between a single inlet and a single outlet. (Some pump mechanisms would allow 100 pascals with 1 bottle and fittings. Consider a loop from inlet to outlet with a T to a bottle.)
3. Temperature is constant everywhere. (So that temperature change cannot be used to generate pressure.)
4. You cannot introduce pressure from sources other than the pump.
5. The bottles have fixed volume.
6. The bottles (and the room) are the only containers available to you.

Now, it's clear that you can get one bottle to 30 pascal, one bottle to 20 pascal, and one bottle to 10 pascal. At that point it's impossible to add any more air to the system of bottles.

With appropriate bottle sizes, you can get arbitrarily close to having a bottle at 50 pascal.

 

[NateTG]

Originally posted by Njorl
It is not possible to configure the bottles in such a way that the sum of the pressures is greater than 60 pascals.

Njorl

You're assuming that the bottles are all the same size.

 

[Chi Meson]

Originally posted by Njorl
Pascals, bars, what's the difference?

A bar is based on "the weight of 1 kilogram" per square cm. This translates to 9.8 N per square cm, or 98,000 N per square meter. I think it has been readjusted so that now there are exactly 100,000 Pa in a bar.

Now THAT is a punchline waiting to happen!

 

[Njorl]

I know.

There's too many pressures associated with being a physicist. No, not job related stress, I mean units of pressure - Bars, pascals, atmospheres, torr, inches of mercury! G'ah! I can't take it!

Njorl

 

[monzie]

To clarify:

Yes the pump can achieve a pressure at its output that is 10 Pa more than whatever the pressure is at its intake valve.

I should have specified in the problem that all pressures were gauge pressures ie. relative to atmospheric pressure.

Also assume the atmosphere is composed of an ideal gas.

No heating or cooling allowed. (ie. assume you have to let the bottles and pump sit for a long period after any operation until its temperature is the same as ambient)

Consider this too:
Take a bottle, fill it to 10 Pa, then attach it to the inlet of the pump.
Take an empty bottle, attach it to the outlet of the pump.
Now turn the pump on.
The pump will push air from the inlet (Pin) to the outlet (Pout) until:
Pinlet_final + 10 = Poutlet_final

Now ideal gas law is PV = NRT. To make life simpler assume in this room that RT=1, and V (the volume of 1 bottle) is also 1. THen
P=N

Let
Ninlet_final=number of molecules in the inlet bottle after the pressure in both bottles stabilizes (no flow through the pump)

Ninlet_initial= initial number of molecules in the inlet bottle

Noutlet_final=molecules in outlet bottle when pressure stabilizes.

Noutlet_initial= molecules in outlet bottle before pumping.

Again
Pinlet_final + 10 = Poutlet_final
so Ninlet_final + 10 = Noutlet_final

But Ninlet_final = Ninlet_initial - Npumped

and Noutlet_final = Noutlet_initial + Npumped because the inlet of the pump is attached to the bottle, so the number of molecules removed from the inlet bottle must equal that added to the outlet bottle

So pumping from a bottle with 10 Pa into an empty bottle results in the inlet bottle having a final pressure of 0 and the outlet bottle having a final pressure of 10 pa.

If the outlet bottle is first pumped up to 10 pa, then a second bottle is pumped to 10 pa and attached to the inlet the final pressures are:

Poutlet_final = 15 Pa
PInlet_final = 5 Pa

If this process is repeated over and over ( transfer the outlet bottle to the inlet of the pump and pump into a bottle that is pumped to 10 Pa) the pressure of the outlet bottles tends towards 20 Pa. To see this, a steady state solution exists where the outlet bottle is 20 Pa. Moving it to the inlet and pumping into a bottle at 10 Pa will result in the inlet bottle dropping to 10 Pa and the outlet bottle climbing to 20 Pa and you haven't gained anything.

Now you say, OK, now use 2 bottles at 20 Pa. The final pressure will tend toward 30 Pa (Same argument as above). So with 3 bottles I think you are limited to 30 Pa, unless someone is clever enough to figure out how to get to 60 Pa without using more than 3 bottles.

 

[NateTG]

If the bottles are different sizes, then you can get a bit more pressure:

Let's say we have 1000 L, 1 L and 1 ml bottles.
Then we can get the 1000L bottle to 30 Pa using the technique you described.

Then hook the 1000L bottle into the inlet, and the 1L bottle (at 20 Pa) into the outled. You should get something close to 40 Pa in the 1L bottle.

Then take the 40 Pa 1L bottle, and put it on the inlet, and the 1ml bottle on the outlet. You should get just about 50 Pa.

 

[monzie]

Yes I hadn't thought of that - I guess 50 Pa with 3 bottles may be the limit. Is there a way to get arbitrarily large pressures with a limited number of bottles?

 

[NateTG]

Originally posted by monzie
Yes I hadn't thought of that - I guess 50 Pa with 3 bottles may be the limit. Is there a way to get arbitrarily large pressures with a limited number of bottles?

If you have bottles with variable(changable) volumes, then two bottles are sufficient for arbitrary pressure.

For n bottles of the same size, and a pump that produces a pressure difference of p the maximum pressure is np.

For bottles of varying sizes, the maximum is at least (2n-1)p.

 

[Loren Booda]

Shucks, I was going to wait for my quantum bottle-cavity containing a nitrogen molecule to probabilistically jump to 100 pascals.