# Child's experiment to weigh air

• I
• lloyd709
In summary, the book suggests that if you fill a balloon with air, the scale will tip demonstrating the weight of the air. However, this is not always accurate because the compressed air has more mass relative to the unfilled balloon.
lloyd709
TL;DR 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!)

Welcome, llyoyd709
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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Livio Arshavin Leiva
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.

Lnewqban
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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Motore and Lnewqban
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).

vanhees71
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.

vanhees71
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.

vanhees71 and Lnewqban
...
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.

davenn, Ibix, vanhees71 and 1 other person
I'll just say it's more fun to pop one of the baloons.

vanhees71 and Lnewqban
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?

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.

Lnewqban
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.

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)?

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.

vanhees71
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.

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.

Lnewqban
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?

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.

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.

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.

Vanadium 50 and Lnewqban
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?

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]

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.

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!

hutchphd
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.)

'

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.)

hutchphd
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) ...

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.

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

+

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.

vanhees71 and hutchphd
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.

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.

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.

One could use a jar, a roughing pump, vacuum gage and an analytical balance. Every well equipped kitchen should have these items.

sophiecentaur and hutchphd
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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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.

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