Calculate Gas Density at Vacuum & Water Content

[Yuri B.]

What is the formula to calculate gas (air) density at negative pressures, as well as water content ? For instance, a vessel with air with volume of 1 m3, density 1 kg/m3, and water content 10g /m3 get vacuumed from 1BarA to 0.1BarA isothermally.

 

[sophiecentaur]

I don't think you mean "negative pressures". You must mean pressures below atmospheric.

The density will be, as always, Mass/Volume.

You can calculate the volume at the new pressure using Boyle's Law. The Mass will be the same - so you can work out the density.

What you may need to think about is how the original density is affected by the presence of 10g of Water / m³. (You are assuming, presumably, that the water is there as vapour, initially.) Is the water vapour there in addition or does it replace some of the air?

 

[Yuri B.]

Hello!
(The water as it is present in the air around us. )

What will be the mass of the air - the above 1 kg - reduced to when we vacuumed the container down to 0.1 bars absolute ?

 

[sophiecentaur]

The total mass sill be the same ( you just put some of it elsewhere) - you could imagine just increasing the chamber size to ten times the volume.
There is a small problem and that is where did you get your numbers from in the original scenario? Did you make them up, choosing 'round figures' or find them from somewhere? These figures would only apply for a particular temperature (around 90Celcius)

I have a feeling that the answer to the question is trivial, as you have been given so much info. The gas mixture should just follow Boyle's Law so density will just be proportional to pressure as long as the temperature doesn't change.

 

[Yuri B.]

..where did you get your numbers from in the original scenario?...density will just be proportional to pressure as long as the temperature doesn't change.

(the scenario is - vacuumating a hermetic container)

So, is it right:
1 kg/m3 at 1 BarA,
0.1kg/m3 at 0.1 BarA,
...
0.0001 kg/m3 at 0.0001 BarA
...?

 

[sophiecentaur]

(the scenario is - vacuumating a hermetic container)

So, is it right:
1 kg/m3 at 1 BarA,
0.1kg/m3 at 0.1 BarA,
...
0.0001 kg/m3 at 0.0001 BarA
...?

I asked where the numbers come from in your question. The scenario is pretty clear.
But, yes, Boyle's law will work fine as long as the water doesn't condense (vapour pressure needs to be high enough)

 

[Yuri B.]

A citation: "Boyle's Law:
Doubling the pressure on a gas halves its volume, as long as the temperature of the gas and the amount of gas aren't changed."

My qwestion:
Halving the pressure on a gas , as long as the temperature of the gas and the voulume of gas aren't changed, does halve its amount ?

 

[nasu]

The answer to the question does not follow from Boyle's law but from the conditions of the experiment. If the container is closed (not necessarily fixed volume) then the quantity of gas does not change. Otherwise it may change. If it does you cannot use Boyle's law to relate the states of the two different quantities of gas.
Boyle's law applies to a fixed quantity of gas (number of particles does not change).
If the number of particles changes during the process you may use the equation of state for the gas (PV=nRT) but not Boyle's law.

In this problem the number of particles in the container definitely changes. By "vacuumed" I suppose he means that some of the gas is removed.
The volume and temperature are constant and the pressure and number of particles will change. The equation of state for the two cases will read:
State1: P1V=N1 R T
State 2: P2V=N2 R T
So N1/N2=P1/P2 (which you could write right away knowing that the pressure is proportional to the particle density).

Here N1 and N2 include all particles, air, water vapor, whatever is in the container. The ratio between components after the process depends on the "vacuuming" process. You may assume that it stays the same as in the beginning but is not necessarily so.

 

[sophiecentaur]

A citation: "Boyle's Law:
Doubling the pressure on a gas halves its volume, as long as the temperature of the gas and the amount of gas aren't changed."

My qwestion:
Halving the pressure on a gas , as long as the temperature of the gas and the voulume of gas aren't changed, does halve its amount ?

Use the word Mass

It doesn't halve the TOTAL amount / mass. I already made that point, earlier on. One litre of the gas has no idea whether it's in a 1l container or is part of the contents of a 1,000l container. The laws still apply

But, if you are OK with the practical implications of the first para of your post then where's the problem with the second para?

 

[sophiecentaur]

The answer to the question does not follow from Boyle's law but from the conditions of the experiment. If the container is closed (not necessarily fixed volume) then the quantity of gas does not change. Otherwise it may change. If it does you cannot use Boyle's law to relate the states of the two different quantities of gas.
Boyle's law applies to a fixed quantity of gas (number of particles does not change).
If the number of particles changes during the process you may use the equation of state for the gas (PV=nRT) but not Boyle's law.

In this problem the number of particles in the container definitely changes. By "vacuumed" I suppose he means that some of the gas is removed.
The volume and temperature are constant and the pressure and number of particles will change. The equation of state for the two cases will read:
State1: P1V=N1 R T
State 2: P2V=N2 R T
So N1/N2=P1/P2 (which you could write right away knowing that the pressure is proportional to the particle density).

Here N1 and N2 include all particles, air, water vapor, whatever is in the container. The ratio between components after the process depends on the "vacuuming" process. You may assume that it stays the same as in the beginning but is not necessarily so.

You are being a bit too literal about this. Boyle's Law says exactly what will happen to the Density. We're talking about intrinsic quantities and not extrinsic quantities - volume is not relevant.

 

[Yuri B.]

Thank you Nasu and Sophiecentaur !

 

[nasu]

You are being a bit too literal about this. Boyle's Law says exactly what will happen to the Density. We're talking about intrinsic quantities and not extrinsic quantities - volume is not relevant.

Well, I meant something like this when I added the parenthesis. My point was that you don't need Boyle's law (or equation of state) if you know that the pressure is proportional to density. It seems that you may call this relationship Boyle's law (or maybe a consequence of it).

 

[Yuri B.]

...The ratio between components after the process depends on the "vacuuming" process. You may assume that it stays the same as in the beginning but is not necessarily so.

Hello Nasu
Water may turn to ice and stay in the container - but we can add heat and not let the water to turn to ice. What else may prevent different particles be removed proportionally - both to each other and to the pressure decreasing ?

 

[nasu]

I was thinking more at the effect of the device (pump) you use to vacuum. Some may introduce water vapor (water aspirator) others may introduce oil vapor. Also various molecules may stick to the walls in different ways. But I guess this is not so important at the pressure you are talking about.

 

[sophiecentaur]

I was thinking more at the effect of the device (pump) you use to vacuum. Some may introduce water vapor (water aspirator) others may introduce oil vapor. Also various molecules may stick to the walls in different ways. But I guess this is not so important at the pressure you are talking about.

I should say that the above counts as 'second year work'. The question they've been given, with all those round figures, was approximately appropriate for one specific temperature. Fine details like how a vacuum pump works are just going to confuse students, I should have thought.

 

[nasu]

I should say that the above counts as 'second year work'. The question they've been given, with all those round figures, was approximately appropriate for one specific temperature. Fine details like how a vacuum pump works are just going to confuse students, I should have thought.

I thought this is about some personal (real or imagined) project rather than a school homework for a student. Sorry If I had the wrong impression.

 

[sophiecentaur]

I looked at the original figures and they looked, to me, as if they'd been made up as an example. They represent an unusual situation.

 

[nasu]

I had the same impression. Only that my guess was that the poster made it up rather than it was part of some assignment. I think that the use of the term "negative pressure" to describe this situation made it unlikely to be part of a physics assignment.
Maybe he can shade some light on the actual purpose or origin of the problem.

 

[Yuri B.]

It seems the gas density is related to the gas residue in the vacuum % found from a Vacuum Conversion Table.
PS. Preconcern was vacuumation of systems and knowledge of how much actually there is left in a system - especially water - at different vacuum degrees.

 

[sophiecentaur]

It seems the gas density is related to the gas residue in the vacuum % found from a Vacuum Conversion Table.
PS. Preconcern was vacuumation of systems and knowledge of how much actually there is left in a system - especially water - at different vacuum degrees.

If the water is there as vapour then why should the proportion change any more than the proportion of Oxygen and Nitrogen? It's just a mixture of gases.

Going back to the OP, this 10g of Water: is it liquid water in the container? If it is then the situation is very different.