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I'm teaching a home-schooled 13 year old some physics and chemistry, and last night's experiment was an attempt to measure the density of air.

Using an electronic balance that has proved pretty reliable, and weighs things up to several tens of grams to a precision of +/- 0.01 g, we weighed an empty balloon in a little cardboard platform (total about 7g).

Then we inflated it by blowing into it and weighed it again.

The weight difference was 0.3g.

We then captured the air as follows, using:

  • a large bowl of water
  • a 1l plastic kitchen measuring jug, calibrated in 50ml increments

The jug was filled with water by dipping under the surface in the bowl, and held inverted with the rim about 1cm below the surface of the water.

The air was released from the balloon under water, capturing it in the inverted jug, stopping the flow and refilling the jug with water when 800ml had been collected each time.

Total volume of air collected: 4000 ml, +/- 200 ml.

This gives a density of:

(1,000,000 / 4000) x 0.3 = 75g per cubic meter.

Whereas the accepted density of dry air is more like 1.3 kg per cubic meter.

So my result is a factor of 17 too small!!

My method was fairly crude, but there is clearly a colossal error here.

Certainly exhaled air is wet, and warmer than ambient, with more carbon dioxide and less oxygen, and hence of a different density - but not by a factor of 17!

Collecting the air under water could also have introduced errors - however, with each jugful of water I was careful to ensure that once I'd collected each lot of 800ml of air, the water level inside the jug was the same as the water level in the bowl, so that the air inside the jug would be pretty close to ambient atmospheric pressure. Again, errors here would be a few percent, not a huge factor!

I think the pressure in the balloon is irrelevant - the contained air weighs whatever it weighs - it's the collected volume that's important, not the volume in the balloon.

So, what am I doing wrong?


Edit to describe updated method:

Having recovered from my embarrassment at failing to do this properly the first time, some may be interested in a repeat of the experiment which has yielded a much better result:

I cut the valve out of a bicycle inner tube and glued it into the lid of a 2 litre pop bottle, having first drilled a suitable hole in the lid through which to poke the valve.

This enabled me to pump the bottle up to about 25 psi with a bicycle pump.

This time, unlike the balloon, the bottle stretches only very slightly under pressure, so the extra weight really is the extra weight of the air.

We collected the weighed extra air over water as before to measure its volume. 2.65 g air had a volume of 2.4 l.

The density came out as 1100 g/m3, which isn't too far from the accepted value of 1225.

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  • $\begingroup$ Was the weight difference negative? $\endgroup$ – Deep Oct 21 '16 at 11:03
  • $\begingroup$ No. The weight of the balloon plus cardboard platform plus air was 0.3g more than the weight of the empty balloon plus cardboard platform. The cardboard platform was just to stop the inflated balloon rolling off the balance. $\endgroup$ – ChrisA Oct 21 '16 at 11:08
  • $\begingroup$ If possible please include a figure illustrating your method of collecting air under water. $\endgroup$ – Deep Oct 21 '16 at 11:17
  • $\begingroup$ See physics.stackexchange.com/a/128484/26969 $\endgroup$ – Floris Oct 21 '16 at 11:56
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    $\begingroup$ you failed to account for Buoyancy $\endgroup$ – njzk2 Oct 21 '16 at 17:48
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Your method is the problem,

Imagine for an instant trying to measure the density of helium by the same method, your balance would measure a negative weight (the balloon rise), and hence a negative density. Which isn't the case.

Archimede's fault really

As @akhmeteli pointed out, pressure look like a far better explanation, if the pressure inside is 10% more ($1.1$ atm which seems realistic), the mass would also increase by 10%, and lead to the kind of mass difference you are expecting. My first explanation involved CO2, but this gas does not represent a sufficiently large portion of breathed air to explain anything.

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    $\begingroup$ Of course!!!!!!!!! And the fact that the pressure in the balloon is a greater too, so denser than ambient. I need to pump dry air into a vessel that doesn't inflate! Damn, much more difficult. $\endgroup$ – ChrisA Oct 21 '16 at 11:19
  • $\begingroup$ As far as I know, the concentration of CO2 in exhaled air is pretty low (about 5%), and the concentration of water vapor, which is lighter than air, is comparable to that. I would suspect the difference is due to the air pressure in the balloon being higher than the atmospheric pressure. $\endgroup$ – akhmeteli Oct 21 '16 at 11:22
  • $\begingroup$ The problem won't be to fill it with air : ), getting the air out is probably where trouble will start. But it should be doable via a pump or a syringe, depending on the size. $\endgroup$ – Jeannette Oct 21 '16 at 11:23
  • $\begingroup$ @Jeannette +1 for an excellent hypothesis. $\endgroup$ – Deep Oct 21 '16 at 11:23
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    $\begingroup$ @ChrisA "I'm now thinking of getting a lightweight box..." that's not a bad idea, but it might be better to get the inner tube from a bike tire, and pump it up to a higher pressure than your lungs can deliver. If you start with the tube filled with air at atmospheric pressure (not completely deflated), it won't expand much even without a box to restrain it. $\endgroup$ – alephzero Oct 21 '16 at 18:33
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Suppose the volume of your balloon is $V_0$ and lets call the volume of the air you collected $V_1$. Note that $V_1 \gt V_0$ because the air inside the ballon is under pressure because it's being compressed by the elastic walls of the balloon.

The extra weight of the ballon will be:

$$ \Delta m = \rho(V_1 - V_0) $$

So the density will be:

$$ \rho = \frac{\Delta m}{V_1 - V_0} $$

The problem with your experiment is that you only measured $V_1$ and you have no idea what the volume of the balloon is. You have calculated the density as:

$$ \rho = \frac{\Delta m}{V_1} $$

and because $V_1 \gg V_1 - V_0$ your density is coming out much too low.

Repeat the experiment, but this time before you deflate the balloon push it underwater and measure the volume of the water displaced. This will give you $V_0$. Then capture the air as before to give you $V_1$.

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  • $\begingroup$ A doubt: If pressure inside balloon tends to atmospheric pressure (i.e. tension in balloon surface tends to zero), $\rho$ in your formula grows without bound (because $V_1\to V_0$). @ChrisA is assuming $V_1\approx V_0$ to calculate density. If all the mass of air is transferred from balloon to water we must have $\Delta m=\rho_1V_1-\rho_0V_0=0$. $\endgroup$ – Deep Oct 21 '16 at 11:33
  • $\begingroup$ @Zero: no. If we reduce the tension so $V_1\rightarrow V_0$ then $\Delta m\rightarrow 0$. So the ratio $\Delta m/(V_1-V_0)$ remains the same. $\endgroup$ – John Rennie Oct 21 '16 at 12:20
  • $\begingroup$ How is $\Delta m$ related to tension in balloon? I thought it stood for mass of air inside balloon. $\endgroup$ – Deep Oct 22 '16 at 4:28
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The discrepancy may be caused by the following. When you blow air in the balloon and weigh the balloon, you actually measure the weight of the balloon in atmospheric air. To measure the actual weight of the balloon with air inside, you should take into account the buoyancy of the balloon. As the volume of the balloon with air in it is pretty high (about 4 l), this may be a significant factor, as the volume of an empty balloon should be negligible. Actually, the small density that you measured could well be the difference of the densities of the air in the balloon and the atmospheric air.

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Jeanettes answer is of course correct, but it begs the question: just how much higher than atmospheric is the pressure inside the balloon? Oddly enough, John Rennie's otherwise inadequate answer gives us the best clue for determining what the pressure is: immerse the balloon in water to get the compressed volume, and then release the balloon into a collection chamber to get the actual standard volume. The ratio of these volumes should give you the pressure difference as a ratio...if you divide your "measured" density by this pressure ratio, you should get close to the actual density.

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Your method is similar to Method 2 on the IoP Practical Physics website.

If immersing the balloon in water is necessary to find its volume $V_2$, there might be some improvement in accuracy in measuring the force required to submerge the balloon at the same time. This method replaces weighing the balloon in air, and compensates for any change in volume when it is immersed in water (although such a change is likely to be small).

You will need to suspend sufficient mass $M$ (including harness) from the balloon to submerge it just below the water level in a bucket or pedal bin. Then
$(\rho_w-\rho_2)V_2= M-\rho_w V_3$....(eqn 1)
where $\rho_w$ is the density of water, $\rho_2$ is the density of air in the submerged balloon, and $V_3$ is the volume of the weights and harness, which can be measured by the displacement method separately afterwards.

Additional measurements : (i) the volume $V_2$ by which the water level in the bucket or bin falls when the balloon is removed but the weights left in, and (ii) the volume $V_1$ of the air in the balloon at atmospheric pressure, which provides the relation
$\rho_2 V_2 = \rho_1 V_1$....(eqn 2)

Eqns 1 & 2 should be sufficient to enable you to find the density of air at 1 atmosphere, $\rho_1$.

Precautions : Ensure that the water is at room temperature, and that the air in the balloon has plenty of time to reach the same temperature (to avoid measuring the density of warm moist air). Fill the balloon using a hand pump. Suspend the weights from a net or harness draped it; it is likely to burst while being submerged if the weights are attached at the neck. Avoid the balloon touching the sides of the bin when the required mass $M$ is being determined.

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An object in air has a buoyancy (an upward force) equal to the weight of air displaced by the volume of the object. When you weigh the inflated balloon, you get the actual weight of the balloon plus the contents of the balloon minus the buoyancy of the inflated balloon. My advice is to retire the balloon and get a plastic two liter soda pop bottle. A little work is necessary in that an air valve must be fitted into a bottle top (I have used a valve from a blown out bicycle tire) and checked for a good seal. Weigh the bottle, bottle top and valve with the bottle filled with air at the ambient pressure and temperature of the room. The weight of the air in the bottle is approximately equal to the buoyancy of the bottle with the air in it thus cancelling out. There is the little matter of the volume of the plastic in the bottle. Use a tire pump to pump air into the bottle. About two atmospheres (30 lbs/sq in, Pardon the English units) seems to be safe. Reweigh the whole thing. The additional air in the bottle will add weight. Now do the calculation. Weight, or more properly mass, divided by volume should provide an answer much closer to the accepted value. Sources of systematic error are: 1. the slight expansion of the bottle under pressure; 2. the volume of the material of the bottle; 3. determining the actual volume the air under compression. 4. accuracy and precision of the measuring instruments. I have obtained good results demonstrating this in front of elementary school students in a classroom.

oops. The volume of the air must be corrected for the pressure. ie. Two atmospheres means that you added air equal two times the volume of the bottle.

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  • $\begingroup$ The balloon displaces the same amount of air whether it's inflated or not. Since when you inflate it, the air inside is after all just air. So there's no additional buoyancy - in fact due to the pressure required to inflate the balloon, the air inside will have negative buoyancy. A plastic bag filled with air will avoid this pressure difference. $\endgroup$ – Arunas Oct 21 '16 at 23:10
  • $\begingroup$ Yes, I agree. I had a go at stuffing a balloon into a box, but as I feared, it was very difficult. I have several inner tubes, so I'm going to try removing a valve from one and attempting your method! $\endgroup$ – ChrisA Oct 24 '16 at 10:27

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