# 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 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.

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.

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.

sophiecentaur
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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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.

sophiecentaur and nasu
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.

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.

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.

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.

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.

sophiecentaur

(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

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.

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]

hutchphd
@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.

jbriggs444
sophiecentaur said:
original idea has been trampled and lies bleeding on the sidewalk.
Love this mental picture.

sophiecentaur
Fail while repeating the experiment, by balancing two identical folded garbage bags.
Almost fill one bag with air by holding open the neck while moving it through the air, (so it does not gain extra moisture from lungs, and you do not hyperventilate).
Tie off the opening, without compressing the contents.
Since buoyancy in atmosphere is 1:1 with the contained air, the weights will still balance.

We know that it is the compressed air in the balloon that is more dense than the atmosphere.

So now the fun begins. Balance two elastic balloons, Inflate one just slightly, the other to maybe twice the diameter. Which will weigh more on the balance?
To answer that question you must realize that the pressure (and so the density) of the gas in a balloon is reduced as it is inflated. Notice how hard it is to start inflating an elastic balloon, but it gets easier as the radius of curvature increases and the pressure falls. Is there some ratio where the large and small balloons balance?

Two values for a given pressure but not a single ratio of radii.

That's a cool experiment, the two balloons connected with a tube. I just fooled my wife with it. I wonder, can you pop a balloon that way or will the pressure start rising again as you near the popping radius?

When you blow up a balloon in the usual manner, you're using your exhaled breath, which is depleted in O2, and has an equivalent amount of CO2 in its place, which is denser and, indeed, weighs more than air. If you used a compressor and actually filled the balloon with "air", it would not appear to gain weight when filled, and the "experiment" would fail.
The pressure in a regular "party" balloon is only slightly above one atmosphere, and the added density due to the higher pressure is very slight.

James Demers said:
When you blow up a balloon in the usual manner, you're using your exhaled breath, which is depleted in O2, and has an equivalent amount of CO2 in its place, which is denser and, indeed, weighs more than air.
The CO2 and O2 change is extremely small. The air you breath out is close to saturated with H2O at body temperature, some of which will condense inside the balloon. "Hot" and/or "Wet" air weighs less than dry air.

There is no question that the internal pressure will increase the density of the air in the balloon. A 1 psi internal pressure will increase the density by about ( 1 + 14.7 ) / 14.7 = 1.07 = 7%

The whole experiment is totally misleading for 'kids'.
We really shouldn't be taking it further without a caveat at the start of every 'PF worthy' post.

sophiecentaur said:
The whole experiment is totally misleading for 'kids'.
We really shouldn't be taking it further without a caveat at the start of every 'PF worthy' post.
It is not quite as bad as the Crookes radiometer which "proves" that light has negative mass.

sophiecentaur
Does anyone make a 'real' Crooke's Radiometer or is it just an experiment we can do in space?

sophiecentaur said:
Does anyone make a 'real' Crookes Radiometer or is it just an experiment we can do in space?

Note that there is no apostrophe in either designation. The guy's names are "Crookes" and "Nichols".

And they met in St James's park, no doubt. Ignorance is no excuse - mea culpa.

jbriggs444

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