I Child's experiment to weigh air

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The discussion revolves around a child science experiment titled 'Weigh some air,' which suggests that filling a balloon with air will cause a balance scale to tip, indicating the weight of the air. Participants debate the physics behind this, with one arguing that the tipping occurs due to the increased density of the compressed air inside the balloon, while others emphasize that buoyancy and the weight of displaced air also play crucial roles. The conversation highlights the complexity of measuring air weight and the importance of experimental conditions, such as whether the air is compressed or not. Some suggest alternative experiments to demonstrate air weight more effectively, while others question the validity of the original experiment's conclusions. Ultimately, the discussion underscores the nuanced understanding of weight, buoyancy, and density in the context of air.
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I think a published experiment that demonstrates the weighing of air is incorrect and actually only demonstrates that compressed air weighs more than non compressed air.
I've just purchased a book of child scientific experiments and it has an experiment that is titled 'Weigh some air'. It shows that if you make a scales with a piece of wood and a pivot and then balance two empty balloons on a either end and then fill one of the balloons with air the scales will tip demonstrating the weight of the air.

Something didn't seem quite right to me and after giving it some thought I've come the conclusion (that my more highly physics qualified friend disagrees with) that the reason the scales tip is because the air inside the balloon is compressed slightly and therefore has higher density and less buoyancy within air.

Even though the balloon filled with air has more mass relative to the unfilled balloon it still actually weighs the same when weighed in air because 'weight' depends not only on gravity and mass but also on the fluid or gas that the weighing takes pace in (hence why objects 'weigh' less when weighed in water).

The other way I'm thinking about this is that if you somehow had a rigid container made of a hypothetical material that had the same weight/volume ratio as air (so that it 'weighed' nothing in air) and filled it with air (so the air was uncompressed) than it would still weigh nothing no mater how big you made the container.

I'm getting a lot of stick from my friend on this (he's basically calling me a numpty) so would be great to hear your opinions (a lot of beer depends on it!)
 
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Welcome, llyoyd709 :cool:
If you have more mass (number of molecules of air) within certain volume, you have greater force pulling down.

##Vertical~force=mg##

The elasticity of the walls of the balloon slightly increases the density (mass/volume) of the air that it contains, as soon as the internal pressure becomes greater than the atmospheric one.

You could compare the empty balloon to one that occupies the same volume of the inflated one, but with internal pressure equal to the atmospheric pressure (having rigid wall of same thickness, for example).
 
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lloyd709, welcome to PF.

lloyd709 said:
Something didn't seem quite right to me and after giving it some thought I've come the conclusion (that my more highly physics qualified friend disagrees with) that the reason the scales tip is because the air inside the balloon is compressed slightly and therefore has higher density and less buoyancy within air.

Yes, I think the density difference is the key. If the balloon air had the same density, then it's weight would be exactly balanced by the upward buoyant force on the balloon. Ask your friend to draw a free-body diagram of the air-filled balloon, and to be sure to include the buoyant force. It's probably more helpful if the weights of the balloon material and the enclosed air are represented as distinct forces.

To be honest, I am surprised that the weight difference is measurable with this method.
 
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lloyd709 said:
I think a published experiment that demonstrates the weighing of air is incorrect and actually only demonstrates that compressed air weighs more than non compressed air.

I think it depends on what you are trying to show. If you are trying to accurately weigh air - i.e. how many grams per liter - I agree, this is not a very accurate way to do it. If you, on the other hand, are trying to show that air has some weight - entirely appropriate for a child's science book - this is perfectly fine. Showing air has more weight per volume when it is compressed implies that it has some weight when uncompressed.
 
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lloyd709 said:
the scales will tip demonstrating the weight of the air.
That book is daft - if it really is saying that. Neglecting any pressure inside the balloon (and the plastic balloons, used for Helium, are not under much pressure at all), Archimedes' Principle says the upthrust on the full balloon will be equal to the weight of the atmospheric air that's been displaced. i.e. Only the plastic envelope will have an effective weight and that's the same in both cases.
There's a very convincing experiment / demonstration that needs School - type equipment to show that air 'has weight'.
You take a 1l round bottomed glass flask and weigh it. You then suck all (most of) the air out and weigh it again. (Obvs. you need a bung and a tap to maintain the emptiness). The difference is due to the weight of the air that occupied the flask. The density of air is about 1.2kg / m3 and there are 1000l in a m3. So the difference in mass will be about 1.2g. Not enough for your kitchen scales to register, perhaps but enough to show kids (if you can sell them the idea that 1.2g is significant).
 
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Vanadium 50 said:
Showing ait has more weight per volume when it is compressed implies that it has some weight when uncompressed.

Starting with only 1.2g/l, you have to try quite hard, either to get enough pressure or sensitive enough scales. The obvious 'proof' is the fact that a balloon full of Helium will lift a measurable load - but it does involve the concept of displacement, which may be a step too far for the OP's purposes.

Perhaps showing them a full dive bottle, weighing it and then weighing it again after letting the air out. (Explaining what 200 bar means and that the air inside would fill a wardrobe etc. etc.) That will give you a value in the region of a kg or two (depending on the bottle). It would be a nice, noisy (Hissssssssss!) demo which would be memorable for them. You just need to find yourself a scuba diver.
 
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Well, let's work it out. A balloon is "full" when the lungs can't provide any more overpressure. That's apparently about three-quarters of a pound per square inch. So air in the balloon is 5% denser than the air outside. Air is 1.3 mg/cm3. A 10cm radius balloon has about 4.4 g of air in it, so you're looking at a quarter of a gram difference.

Balance-beam scales can do this. A "piece of wood" is probably not able to. Furthermore, I doubt two "identical" balloons have the same weight to a quarter gram. My conclusion is that this can work, but the authors never tried it.

The correct way to do this is to have two filled balloons. Balance them. Let the air out of one, and see the other one is still heavier.
 
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Vanadium 50 said:
...
The correct way to do this is to have two filled balloons. Balance them. Let the air out of one, and see the other one is still heavier.

http://homeschoolden.com/2014/05/12/weather-unit-experiments-about-air/

AirHasWeight-675x308.jpg
 
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I'll just say it's more fun to pop one of the baloons. :wink:
 
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  • #10
Thanks for your comments but my friend on reading them thinks he is being vindicated overall. He simply comes back to me with the statement that 'a balloon is heavier due to weight of the air'. My thoughts are that weight depends on the environment that the weighing is being done in and if it is being done in air then the only reason a filled balloon will record a different weight is because the air is compressed and hence more dense. Of course, if the weighing was being done in a vacuum then agreed the balloon filled with air would record a higher weight.

I know I might be being a bit childish (we are grown men but as mentioned beer depends on it). Can you guys give your opinions on what would happen if there was absolutely no compression and the scales where perfect (but the experiment was still being undertaken in an air environment). Would the the scales tip?
 
  • #11
lloyd709 said:
f there was absolutely no compression

Then how does the balloon work?

A balloon is the size that it is when the inward (i.e. compression) force of the rubber matches the outward force of the overpressure.

No overpressure means it's empty. You're comparing an empty balloon with another empty balloon.
 
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  • #12
No compression hypothetically. I'm trying to confirm that the only reason the scales would tip in this experiment is due to the compression of the air - so if it were possible to contain the air in a container that was ridged but still weighed the same as the empty balloon (so no compression would take place) then a perfect set of scales would not tip - or would it? Clearly I can't do this experiment so that's why I'm asking you guys - but hope you will agree it's interesting.
 
  • #13
lloyd709 said:
No compression hypothetically. I'm trying to confirm that the only reason the scales would tip in this experiment is due to the compression of the air - so if it were possible to contain the air in a container that was ridged but still weighed the same as the empty balloon (so no compression would take place) then a perfect set of scales would not tip - or would it? Clearly I can't do this experiment so that's why I'm asking you guys - but hope you will agree it's interesting.
You could replace both balloons with two identical metal spheres of thin walls having valves to modify the internal pressures.
Internal pressure for one will be equal to atmospheric: it contains certain mass of air.
How could you double the mass of air contained within the second sphere (keeping same temperature)?

Please, see:
https://en.m.wikipedia.org/wiki/Ideal_gas_law

I believe your friend should start working for his beer as hard as you do and demonstrate his point.

:cool:
 
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  • #14
Are you confusing constant mass and constant volume?
  • If you pump more air into the same volume, it has higher mass and weighs more.
  • If you compress a given mass of air to use less volume, it has the same mass and same weight.

Blowing up a balloon changes both mass and volume at the same time, so it's more complex. But I think you can say that an inflated balloon weighs more than a deflated flaccid balloon, and that the air inside the inflated balloon is more dense than the surrounding ambient air. Therefore the weight of the inside air should be bigger than the weight of the ambient air displaced.

If that still sounds confusing, imagine holding the mass of the inflated balloon constant but then cooling it until the air inside becomes liquid nitrogen and liquid oxygen. The volume will decrease, the balloon will mostly deflate, but the mass stays constant. Certainly you could measure the mass of that liquid air on a scale.
 
  • #15
lloyd709 said:
but my friend on reading them thinks he is being vindicated overall

He's right. Look a the pictures @Lnewqban posted. If that doesn't convince you, nothing we write will.

Of course different experiments will produce different results. But you didn't bet on those.
 
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  • #16
Vanadium 50 said:
He's right. Look a the pictures @Lnewqban posted. If that doesn't convince you, nothing we write will.

Of course different experiments will produce different results. But you didn't bet on those.
It doesn't convince me - those pictures just re-state the experiment that I've already described. My thoughts are that the balloon filled with slightly compressed air (because it's under pressure) goes down (for want of a better term) because it is overall more dense than the air around it. I hypothesis that if you swapped one balloon for a sealed but ridged container that contained air under no pressure and the other balloon for any weight identical to the container (but not a container just a weight) then on a perfect set of scales there would be no tipping. Just so we are clear, do you think it would tip yes or no?
 
  • #17
lloyd709 said:
It doesn't convince me - those pictures just re-state the experiment that I've already described.
You are correct. You have two explanations for the same experimental result. The experiment, as is, cannot distinguish between them.
 
  • #18
anorlunda said:
Are you confusing constant mass and constant volume?
  • If you pump more air into the same volume, it has higher mass and weighs more.
  • If you compress a given mass of air to use less volume, it has the same mass and same weight.

Blowing up a balloon changes both mass and volume at the same time, so it's more complex. But I think you can say that an inflated balloon weighs more than a deflated flaccid balloon, and that the air inside the inflated balloon is more dense than the surrounding ambient air. Therefore the weight of the inside air should be bigger than the weight of the ambient air displaced.

If that still sounds confusing, imagine holding the mass of the inflated balloon constant but then cooling it until the air inside becomes liquid nitrogen and liquid oxygen. The volume will decrease, the balloon will mostly deflate, but the mass stays constant. Certainly you could measure the mass of that liquid air on a scale.
That doesn't confuse me. I'm not confusing mass or constant volume. The key point here is what 'weight' is. From my school physics I remember weight was mass multiplied by force which on Earth is it's gravity. But within this experiment you have to be more precise with measured weight because there are the forces of the air pressure all around the balloon. These will effect the 'weight' as defined as a net downward force that the filled balloon has. (Just like any object will weigh less under water than in air). Clearly the balloon filled with air has more mass than an unfilled balloon but it takes up more volume, but not quite proportionately more due to the compression of the air. This means overall the filled balloon has more 'weight'. My hypothesis is that if hypothetically there was no compression then the volume increase would be exactly in proportion to the mass increase so the net effect would be no change in measured weight over an unfilled balloon.
 
  • #19
lloyd709 said:
Summary:: I think a published experiment that demonstrates the weighing of air is incorrect and actually only demonstrates that compressed air weighs more than non compressed air.

I've just purchased a book of child scientific experiments and it has an experiment that is titled 'Weigh some air'. It shows that if you make a scales with a piece of wood and a pivot and then balance two empty balloons on a either end and then fill one of the balloons with air the scales will tip demonstrating the weight of the air.

Something didn't seem quite right to me and after giving it some thought I've come the conclusion (that my more highly physics qualified friend disagrees with) that the reason the scales tip is because the air inside the balloon is compressed slightly and therefore has higher density and less buoyancy within air.

Even though the balloon filled with air has more mass relative to the unfilled balloon it still actually weighs the same when weighed in air because 'weight' depends not only on gravity and mass but also on the fluid or gas that the weighing takes pace in (hence why objects 'weigh' less when weighed in water).

The other way I'm thinking about this is that if you somehow had a rigid container made of a hypothetical material that had the same weight/volume ratio as air (so that it 'weighed' nothing in air) and filled it with air (so the air was uncompressed) than it would still weigh nothing no mater how big you made the container.

I'm getting a lot of stick from my friend on this (he's basically calling me a numpty) so would be great to hear your opinions (a lot of beer depends on it!)

Note that the book only wanted to weigh "some air". It didn't specify that this is the weight of air at STP. You can make the same argument with temperature variation as well, not just pressure.

So here, if you don't read more than what it intended, all it wants to show is that a volume of air, ANY air, might have detectable weight. It doesn't even want to know what the weight is, so accuracy of the measurement is not even necessary. It only wants to show that a volume of air has a "weight", nothing more, nothing less.

Zz.
 
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  • #20
ZapperZ said:
Note that the book only wanted to weigh "some air". It didn't specify that this is the weight of air at STP. You can make the same argument with temperature variation as well, not just pressure.

So here, if you don't read more than what it intended, all it wants to show is that a volume of air, ANY air, might have detectable weight. It doesn't even want to know what the weight is, so accuracy of the measurement is not even necessary. It only wants to show that a volume of air has a "weight", nothing more, nothing less.

Zz.
OK, I agree the experiment shows that air has mass. It does this by compressing it and showing that in this compressed state within a ballon, surrounded by uncompressed air, the ballon's weight will increase. So yes the kid will sort of get the idea but it could confuse (if what I think is happening is really happening).

Or maybe I'm wrong, in which case can you confirm that if you had a rigid container filled with air under no pressure on one side of a set of perfect scales and just a plain weight but exactly equal to the weight of the rigid container on the other side of the scales the scales would tip towards the side of the ridged container filled with air?
 
  • #21
lloyd709 said:
Or maybe I'm wrong, in which case can you confirm that if you had a rigid container filled with air under no pressure on one side of a set of perfect scales and just a plain weight but exactly equal to the weight of the rigid container on the other side of the scales the scales would tip towards the side of the ridged container filled with air?
You are correct that such an experiment would show the two to be in balance. The extra weight of the filled container would be exactly compensated by the extra buoyant force on it.

An incompletely inflated mylar balloon would be a good choice for such a test. I think that has been pointed out already. [Though I cannot find such a posting now]
 
  • #22
jbriggs444 said:
You are correct that such an experiment would show the two to be in balance. The extra weight of the filled container would be exactly compensated by the extra buoyant force on it.

An incompletely inflated mylar balloon would be a good choice for such a test. I think that has been pointed out already.
Thank you that's what I thought.
 
  • #23
lloyd709 said:
Thank you that's what I thought.
I tried to show pupils the Conservation of Mass by using a flask containing an acid and a carbonate, standing on a chemical balance, with the CO2 blowing up a balloon. Unfortunately for me, when the balloon blows up it occupies more volume and its bouyancy in air is increased, so the total mass seems to decrease!
 
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  • #24
There is one more fly in the ointment here...air from your lungs has `100% relative humidity so on a dry day this could significantly lower the density of the balloon interior air. Enough to destroy the effect.

tech99 said:
when the balloon blows up it occupies more volume and its bouyancy in air is increased, so the total mass seems to decrease!

I would think the CO2 would work (so long as the gas is not warm). It should be significantly heavier than O2 (and I don't understand the reference to size above...denser is denser.)

'
 
  • #25
It's amazing what a range of opinions and experiments there are in this thread; some good ingenuity in there. Problem is that the numbers in most of them are 'anyone's guess' because the conditions that the air under test are so 'mild'. The demos rely on some subtle explanations which, although PF members will clearly get them. I'm not sure the kids would.
appreciate 'just a bit more air in a blown up balloon'.

That's where the more 'obvious' result of my suggested Vacuum Experiment in which the mass change is 1.2g for 1l or so (of actually removed air) and, even better, the scuba bottle, which contains 200 times its own volume of air and weighs much more than 1kg extra when full of compressed air.

I think the scuba bottle gets it because it's loud and sexy. (Typical of loud, sexy people.)
 
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  • #26
Can you do the "rocket powered chair" demo as an adjunct ? My all time favorite demo (although easier to do with the usual CO2 bottle) ...
 
  • #27
hutchphd said:
There is one more fly in the ointment here...air from your lungs has `100% relative humidity so on a dry day this could significantly lower the density of the balloon interior air. Enough to destroy the effect.
I would think the CO2 would work (so long as the gas is not warm). It should be significantly heavier than O2 (and I don't understand the reference to size above...denser is denser.)

'
The balloon displaces about 1g of air so it experiences lift.
 
  • #28
tech99 said:
The balloon displaces about 1g of air so it experiences lift.
Not if it is filled with 1.2g of CO2

+
 
  • #29
tech99 said:
The balloon displaces about 1g of air so it experiences lift.
The lift force is the weight of Air Displaced - less than the weight of air (or whatever else) in the pressurised balloon.
 
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  • #30
Here's my reasoning. Take 1000 cc of air at STP. Blow it into a rubber balloon and tie off the balloon. Because of the tension of the balloon, the air inside will have a pressure slightly higher than 1 atm, so the volume of that air will now be less than 1000 cc, so it will not displace 1000 cc of air, and it will be heavier; certainly not neutral buoyancy.

I'm ignoring changes in temperature as we fill the balloon.

Also, as I said before, you could chill the air in the balloon to liquefy it without changing its mass. Now that mass occupies much less than 1000 cc.
 
  • #31
sophiecentaur said:
The lift force is the weight of Air Displaced - less than the weight of air (or whatever else) in the pressurised balloon.
Agree. I am probably wrong here but I expected a 1 litre balloon filled with CO2 to weigh very slightly more (due to the excess pressure) than Molecular Mass / Gay Lussac Constant = 44/22.4 = 1.96g. But I forgot to subtract the bouyancy, which is the displacement of 1 litre of air, which is 1.23g. This bouyancy means that we see a reduction in total weight as the balloon fills.
 
  • #32
lloyd709 said:
Thanks for your comments but my friend on reading them thinks he is being vindicated overall. He simply comes back to me with the statement that 'a balloon is heavier due to weight of the air'. My thoughts are that weight depends on the environment that the weighing is being done in and if it is being done in air then the only reason a filled balloon will record a different weight is because the air is compressed and hence more dense.
If filling it increases the weight, then it must have some weight. You have lost the bet.
 
  • #33
One could use a jar, a roughing pump, vacuum gage and an analytical balance. Every well equipped kitchen should have these items.
 
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  • #34
Mister T said:
If filling it increases the weight, then it must have some weight. You have lost the bet.
Clearly air has weight, I'm not saying it doesn't. What I'm saying is that it's not the weight of this air in the blown up balloon that is causing the scales to tip. This is because the experiment is being conducted in an air environment and as such if hypothetically there was no compression the scales would not tip. The reason the scales tip is because the air compresses. I'm probably not explaining this quite how I should so please can someone that gets it translate to physics language so others will.

If I really have got things wrong please explain but it's not enough to convince me by saying air has weight so I'm wrong - that's what my mate keeps saying and I agree with him on that bit!
 
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  • #35
lloyd709 said:
If I really have got things wrong please explain but it's not enough to convince me by saying air has weight so I'm wrong - that's what my mate keeps saying and I agree with him on that bit!
The issue is that when you put air molecules into the balloon the scale tips, showing that the air has weight. It makes no difference if you do it in a vacuum, you get the same result for the same reason.
 
  • #36
lloyd709 said:
This is because the experiment is being conducted in an air environment and as such if hypothetically there was no compression the scales would not tip. The reason the scales tip is because the air compresses.
I understand what you're asking, so let's do a couple of thought experiments:
1) You have two equal rigid sealed containers on a balance in a vacuum. You fill one of them with air at STP and the other one remains empty (vacuum inside). Clearly the one filled with air will tip down demonstrating that air has weight.
2) You have two equal rigid sealed containers on a balance in STP air. You fill one of them with air at STP and the other one remains empty (vacuum inside). Clearly the one filled with vacuum will tip up demonstrating that vacuum has less weight than air (or more precisely that air has weight).

Of course both of this scenarios are analog to the one where both containers are in STP air and one has more air in it (air is compressed) and thus demonstarting that air has weight. So I think you lost the bet.
 
  • #37
Mister T said:
The issue is that when you put air molecules into the balloon the scale tips, showing that the air has weight. It makes no difference if you do it in a vacuum, you get the same result for the same reason.
If you filled the balloon with Hydrogen, that would also have weight. The Displacement of Air has to count in both cases. So your argument is not comprehensive enough.
 
  • #38
lloyd709 said:
Summary:: I think a published experiment that demonstrates the weighing of air is incorrect and actually only demonstrates that compressed air weighs more than non compressed air.

Something didn't seem quite right to me and after giving it some thought I've come the conclusion (that my more highly physics qualified friend disagrees with) that the reason the scales tip is because the air inside the balloon is compressed slightly and therefore has higher density and less buoyancy within air.

I don't know the the exact wording of your bet, but the statement in blue is exactly correct, assuming issues of temperature and humidity do not enter.
Do not back down, this is a matter of honor, and beer is involved.
 
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  • #39
lloyd709 said:
If I really have got things wrong please explain but it's not enough to convince me by saying air has weight so I'm wrong - that's what my mate keeps saying and I agree with him on that bit!
The balloons originally balanced so they each weigh the same.

You then blew into one balloon which pushed more air into it.

When you put it back on the balance you see it is now heavier than the other balloon.

Hence the air you blew into it must have had weight to cause the balloon to be heavier than it was before.

Therefore air has weight.

Note we don't know how much the air weighs - only that it must weigh something.

In fact, air at sea level has a density of about 1.2 kg/cubic metre at 25C, which is about 1/1000th of the density of water.

A small passenger carrying hot air balloon called a "77" has a volume of 77,000 cubic feet or about 2,200 cubic metres.

Now, 2,200 cubic metres of air weighs about two and a half tons on its own so, measured in a vacuum, it would tip the scales at two and a half tons.

However, if measured in the open air, it displaces 77,000 cubic feet of air around it so it is effectively "floating in air" and the air in it not measure anything on the scales.
 
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  • #40
Frodo said:
The balloons originally balanced so they each weigh the same.

We don't actually know that. The balance was constructed specifically to balance those two empty balloons whether they weighed the same or not.
Frodo said:
You then blew into one balloon which pushed more air into it.

When you put it back on the balance you see it is now heavier than the other balloon.

Hence the air you blew into it must have had weight to cause the balloon to be heavier than it was before.

You do two things when you inflate a balloon, add air and displace air. They both affect the scale. I'm just repeating what others have already said. The point is, unless you already understand buoyancy then the conclusion you came to is accidental. For all you know increasing the volume increases the weight. That's not how buoyancy works but if you aren't certain air has weight you don't know that.

To come to the conclusion that you did presumes that you have already decided that air has weight.
 
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  • #41
Do the experiment under water or with a balloon full of Hydrogen then you will get different results. In all cases, there will be substance in the balloon that has weight but sometimes the scale will move up and sometimes down. There is no correlation between what you do and direction of movement. The idea, as it stands is flawed.
You have to introduce something else - upthrust due to displacement of a fluid. The thread has clocked over fifty posts and many of them seem to ignore this. Boats and Helium balloons have to use Archimedes in order to work. Let's do some Physics chaps.
 
  • #42
To be clear all I'm trying to confirm is that the scales tip because the air is compressed and only because it is compressed. If it wasn't compressed then the scales would not tip. Doing a bit of other research it seems that it's all to do with buoyancy, the difference between 'actual' weight and 'apparent' weight and Archimedes principle that states that buoyant force provided by a fluid or gas is equal to the weight of that fluid or gas displaced. So according to this principle, IN THE ABSENCE OF COMPRESSION (hypothetically speaking), when the balloon is being blown up a buoyant force is being created exactly equal to the air displaced and hence in this situation the scales would still be in equilibrium and the experiment would not demonstrate that air has mass. Now bring in a bit of compression and the buoyant force created is now less than the displacement of the air and so the scales tip. So the concussion is that the cause of the scales tipping in this experiment is due to the compression of the air in the balloon and hence it's increase in density. Clearly the air has to have mass for it to be able to compress so, OK, looking at it this way you can say that it shows that air has mass but that is not the explanation given in two examples of the experiment description that I have read.

I've also just discussed it with my wife's brother who is a lecturer in physics (he was tied up so was hard to get hold of until now) and he says I'm completely right and it's a very misleading experiment that is poorly understood.
 
  • #43
lloyd709 said:
IN THE ABSENCE OF COMPRESSION (hypothetically speaking)
Why hypothetically? Just replace the rubber balloons with some bags that are not stretched, just filled. This won't tip the scale in air, but would in vacuum.

lloyd709 said:
...he says I'm completely right and it's a very misleading experiment that is poorly understood.
I agree that it is misleading when done in air to weigh air.
 
  • #44
lloyd709 said:
So according to this principle, IN THE ABSENCE OF COMPRESSION (hypothetically speaking), when the balloon is being blown up a buoyant force is being created exactly equal to the air displaced
Absolutely. A balloon without an elastic envelope would not show the effect. (A vast proportion of party balloons are like this, these days) Moreover, the phone experiment would actually indicate that air has no weight if the conclusion were based on ignoring any consideration of buoyancy.
 
  • #45
A simple demonstration can be done with a $15 centigram pocket scale, a small, narrow-neck glass bottle, and one of those inexpensive wine vacuum pumps. That will show you that the removed air makes the bottle+wine stopper weigh less. If you then invert the bottle in a basin of water and remove the stopper it will suck up some water. In this way it's possible to determine how much air was removed and you can crudely measure the density of air, for about $25.
 
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  • #46
Two additional considerations:

(1) The inner surface of a burst balloon is rather wet, due to condensed water from exhaled breath. Couldn't this condensed water contribute more to the increased apparent weight than the compressed air? I examined it using a precision scale. The excess weight of a balloon blown with dry air from a bicyle pump was 0.40 gram. After deflation, the excess weight was 0.00 g (gram). The excess weight of the same balloon blown with exhaled breath was initally 0.40 g, and it increased in a few minutes to 0,61 g. After deflation, the excess weight was 0.14 g. When cut open, the inner surface was wet. After removing the water with tissue, the excess weight was 0.0. In conclusion, the condensed water does not contribute more than the compressed air.

What excess weight was to be expected? Air in the balloon had a temperature of 28°C, which is a mixture of air from the alveoli and air from the lung dead space, presumably unnmodified ambient air of 22°C. So the balloon contained 40% alveolar air and 60% ambient air. The blown balloon had a volume of 5 L, so it contained 2 L of alveolar air.

Alveolar air has a humidity of 100%, which at 37°C implies an absolute humidity of 0,044 g/L. So those 2 L contained 0.09 g. That is a bummer, more water was created in the balloon than was present in the humid air. Maybe droplets from my throat were added to the air while blowing.

(2) The air in the balloon (28°C) is hotter than the ambient air (22°C), this will reduce the apparent weight with 0,13 gram
 
  • #47
How much of the moisture you exhale would already be condensed before entering the balloon? Water vapor is less dense than N2 and O2 so humidity actually reduces the weight as compared to dry air for a given volume/temperature/pressure. The increase in weight with time is most likely due to the cooling of the air in the balloon. This will cause the balloon to shrink and hence reduce the upward buoyant force.
 
  • #48
JT Smith said:
How much of the moisture you exhale would already be condensed before entering the balloon? Water vapor is less dense than N2 and O2 so humidity actually reduces the weight as compared to dry air for a given volume/temperature/pressure. The increase in weight with time is most likely due to the cooling of the air in the balloon. This will cause the balloon to shrink and hence reduce the upward buoyant force.
Within the body, the temperature of the exhaled breath should still be essentially the same as body temperature. It will not have condensed. However, once the breath contacts something at lower than body temperature (such as a mirror held close to the lips), condensation can and will occur.

Some decrease in density over time is certainly due to the cooling of the air in the balloon. Some decrease will occur due to the condensation of water vapor. A quick trip to Google says about 3% decrease in volume due to condensation (going from about 0.06 atm vapor pressure down to 0.03 atm vapor pressure. A quick trip to the calculator says about 4% decrease in volume due to temperature (going from 310 K to 298K).

The decrease in density due to vapor content is a bit more troublesome to assess. Let us assume 50% relative humidity of the ambient air versus 100% relative humidity in exhaled breath that has been allowed to cool. We are talking about 18 molecular weight vapor versus approximately 29 average molecular weight dry air (60% as dense). And we are talking about half (50% relative humidity) of the 3% saturated vapor pressure at 25 degrees Celsius. So the discrepancy is 40% of 50% of 3% = 0.6 percent at 25 degrees. And 40% of 50% of 6% = 1.2 percent at 37 degrees.

If we put these numbers together then, in the absence of a pressure difference one would expect:

5.2% (4% thermal plus 1.2% vapor) lower density compared to ambient atmosphere immediately after the balloon is filled. The air-filled balloon should weigh less than expected due to the extra buoyancy.

decreasing to 2.4 % (-3% condensation plus 0.6% vapor) greater density compared to ambient atmosphere after temperature equilibriates and the breath has had time to condense. The air-filled balloon should weigh more than expected due to the condensation.

[These figures are quoted more precisely than the back of the envelope calculation above actually justifies]
 
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  • #49
@jbriggs444 You make me extremely lazy because I can always rely on you to put some figures to this sort of problem. (keep it up!) You put it all in the proper proportion.

Problem is, that, rightly or wrongly, I read the OP and this bit probably weighed too heavily in how I have approached the thread:
lloyd709 said:
I've just purchased a book of child scientific experiments
Imo, the experiment is a bad one because it misdirects any child with the comparison 'full balloon vs empty balloon' and that, in itself is flawed. PF has had to do its usual thing and dig deeper and deeper into the phenomenon until the original idea has been trampled and lies bleeding on the sidewalk. Any deeper (sometimes valid) analysis is beyond a 'kids' analysis and, unless you (one) can give them much more exaggerated versions of the experiment, you'd best not even to have started.
 
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  • #50
sophiecentaur said:
original idea has been trampled and lies bleeding on the sidewalk.
Love this mental picture.
 
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