# Is this kids experiment a legitimate way to show that air has mass?

Consider the experiment in this link.

The experiment includes using a ruler as a lever, with an inflated balloon on one side and a balloon which is not inflated on the other.

The aim of the experiment is to show that air has mass.

I have seen many kids performing similar experiments.

But, if the air pressure inside the balloon is equal to that outside, then the buoyant force will cancel out the weight of the air inside the balloon, won't it?

• But why do you think that the air pressure inside the balloon is equal to the outside? Aren't you doing work when you blow up a balloon? Commented Jul 28, 2014 at 17:49
• Wayback Machine link in case that's ever needed. Nevertheless, it would be good if you can expand somewhat more on the experiment: (a) exactly what you're supposed to do, (b) what the observations are, and (c) the inferences from them. Commented Jul 28, 2014 at 18:26

I can think of at least four things going on in this experiment that need pointing out:

1. When you inflate a balloon by mouth, the air is warm: this makes the air inside the inflated balloon slightly lighter than the air it displaced
2. The air inside the balloon has 100% relative humidity at 37C, and condensation will quickly form on the inside of the balloon as the air inside cools down.
3. The air inside the balloon contains carbon dioxide, which has higher density than room air (molecular mass of 12+16+16 = 44 amu, vs oxygen at 32 amu and nitrogen at 28 amu - ignoring small isotopic effects, and ignoring Argon).
4. The pressure inside the balloon is larger than outside - this increases the density

So how large are each of these effects?

1. Warm air: 37C vs 20C results in drop in density of 0.945x (293 / 310) or -5.5%
2. Moisture: partial pressure of water at 37C is 47.1 mm Hg source which is about 0.061 atmospheres. Assuming that pressure is constant, this water (mass 18 amu) displaces air (mean mass 29 amu), so the density of the air decreases by 0.061 * (29 - 18) / 29 = 2.3%. If we allow the air outside the balloon to have 60% relative humidity (with saturated vapor pressure of 10.5 mm Hg), it would be slightly less dense than dry air (10.5*0.6/760*(29-18)/29 = 0.3%) making the net difference -2.0%. Note that much of this moisture will condense when the balloon cools down - little droplets will form on the inside of the balloon. With the air inside still saturated, its density will be 0.1% lower than on the outside; the net result amounts to 2.9% of the mass of the air in the balloon.
3. Carbon dioxide: the exhaled air has 4 - 5 % carbon dioxide source: wikipedia, with an equivalent drop in oxygen. The density of exhaled air is therefore higher than that of inhaled air by 0.045 * (44 - 32) / 29 = +1.9%
4. Pressure in the balloon: from this youtube video - time point 3:43 I estimate the pressure increase in the balloon at 23 mm Hg, resulting in an increase in density of 2.9%

Summarizing in a table:

factor      effect   at room T
temperature -5.5%       0.0%
moisture    -2.0%       2.9%
CO2          1.9%       1.9%
pressure     2.9%       2.9%
net         -2.7%       7.7%


A freshly inflated balloon will thus have only a slightly lower density than the air it displaced, because the temperature + moisture effect is greater than the other two. After you wait a little while, the temperature will equalize and the density of the air inside the balloon will be greater - by 7.7%, with more than half of that not caused by the pressure in the balloon...

In summary: the experiment described in your link measures the difference in density between air in a balloon, and ambient air. Since the density of the air inside the balloon is higher than the density outside the balloon, one may conclude that the air inside the balloon has finite density. One may NOT conclude that the medium outside the balloon (which we believe to be "dry air") has any density at all - since nothing in this measurement tells us about the air outside the balloon.

If you did the experiment carefully with a balloon initially filled with warm air, and you allowed the air to cool down, you might be able to tell that the balance shifts - in other words, that there must be a change in the buoyancy experienced by the balloon as it cools down. THAT would be an experiment to demonstrate "air has mass" (volume of balloon decreases, and it experiences less buoyancy). From the experiment as described (popping the balloon), we learn that "exhaled air has mass". That is not the same thing.

If you used an air pump (balloon pump) to inflate the balloon, the first three components would go away and you are left with the difference due to the pressure only - 2.9% of the mass of the air in the balloon.

That's a perfectly good way to show that the density of air increases with pressure, and therefore that air must have a mass.

When the scales tilt down on the side of the unburst balloon it shows the volume contained within the balloon has a higher mass than the same volume of air at atmospheric pressure. This means the density must be greater. we don't know the pressure inside the balloon, though we know it must be higher than atmospheric pressure because the elastic skin of the balloon is exerting a force on the air inside it.

John Rennie's answer correctly points out that the correct chain of reasoning here is a bit complex:

• On one side of the balance we have a column of air going up to space, a balloon, and some compressed air, and have displaced an amount of the air column equal to the volume of the compressed air.

• On the other side we have a similar column of air and a baloon.

• The first side dips, and so has more mass.

• Therefore air is compressible into a form that has greater density than the mass of air displaced.

• Therefore air has mass.

This complex chain of reasoning is likely to be lost on all but the brightest students. And a student could easily point out that a balloon containing a mixture of air and helium could be produced so as to cause the other side of the balance to dip, but from this we would not conclude that air-plus-helium is massless.

A better experiment would be to have two identical rigid bottles, evacutate the air from one, and show that the evacuated bottle has less mass, either by balancing them against each other or by measuring their inertia. That of course has the down side of requiring specialized equipment.

An experiment that I like that shows that air has mass is as follows:

• Get a helium balloon, a bowling ball and a station wagon.
• Put the balloon in the front seat and the bowling ball in the back.
• Get up to speed and keep the objects stationary with respect to the car.
• Hit the brakes.

The bowling ball rolls forwards, indicating that it has mass. It has inertia, therefore it has mass. The helium balloon floats backwards. What causes the balloon to go backwards? The air pressure must have increased at the front of the car and decreased at the back when the brakes were applied. Some of the air from the back of the car kept going forwards when the brakes were applied, therefore the air, like the bowling ball, has inertia, and therefore mass.

While the experiment you describe is legitimate, it has a complexity that could make it difficult to understand ("oh, my breath has mass!").

The simpler way to do this is to put a container (I've seen a basketball used, but balloons work as well) on a scale or a balance with regular weights on the other side. Use a bicycle pump to add air to your container, and see the mass of the container increase.
It's difficult to argue that the pump is importing anything other than the same air that surrounds your container, and the effect is clear. This worked well for me in my youth.

Sometimes it's helpful to explain the rationale of experimental design to a student as testing "competing alternative hypotheses."

A positive result by this experiment would be strong evidence against the alternative hypothesis that the air blown into the balloon has no mass.

A negative result however would not prove that the air has no mass. Further experimentation would be necessary to understand why the unexpected result was obtained.

Aside, perhaps the best way to do this is to follow the experimental approach of Lavoisier - and demonstrate the conservation of mass. Weigh a solid compound, heat the solid to drive off a gas (capture the gas) and then measure the weight lost by the solid and (less easily) the weight of the gas. Increasingly better experimental technique should show quantitatively better agreement with conservation of mass.

The answer to your question: "But, if the air pressure inside the balloon is equal to that outside, then the buoyant force will cancel out the weight of the air inside the balloon, won't it?" is YES! Unfortunately this is not the case for an inflated elastic balloon. The pressure inside the balloon is increased by the elastic pressure of the balloon reducing the volume and therefore the buoyancy force.

The experiment in the link given is flawed in that 1) the substance inside the balloon isn't the same as outside and due to the use of an elastic balloon, it has a higher density. 2) The statement in the experiment that "Air actually weighs 14.7 pounds per square inch at sea level" is in fact the pressure not weight.

The differing results of the experiment are due to experimental errors.

The mass of air might be more accurately determined with an experiment using a mass dependant property such as inertia (m=F/a).

I offer the following in hopes to clarify the pressure, temperature, substance, container issue.

The "mass" of a non-rigid container and the substance with which it is filled is greater than the container alone, by the "mass" of the substance contained in it.

Since measurable “weight” is dependent on things such as gravitational force and buoyancy, it is difficult to experimentally measure the weight differences.

If the non-rigid container is filled with the same substance as outside the container, and is at the same temperature and pressure, the “weight” of the empty and filled container will be the same. The weight of the substance inside the container will be cancelled by the buoyancy from the substance outside the container. This is valid for both liquid and gaseous substances, as long as the container alone has negative buoyancy.

If you feel led to try to experimentally “measure” the weight difference, I would suggest using a digital kitchen scale that can measure at least 0.1 ounces and a lever arm arrangement to increase the sensitivity. If you can get a lever arm that is light enough, you might be able to a measure sensitivity of 0.01 ounces. You will be limited by the range of the scale and the weight of the lever mechanism and container. At sea level and at 15 °C, air has a density of approximately 1.225 kg/m3 (0.001225 g/cm3, 0.0023769 slug/ft3, 0.0765 lbm/ft3) according to ISA (International Standard Atmosphere). This means a cubic foot of air weighs more than 8 ounces, it should be easy to measure if the captured air is adding to the weight of the container.

Put a hook at the end of the lever arm and hang an unopened garbage bag on it. If you use the tare weigh option it will allow you to directly measure any change in weight. After you have nulled out the weight of the bag and tying mechanism, remove it from the hook and scoop air into it. Tie the opening shut slowly to make sure no pressure has been applied to the air inside other than the pressure from the weight of the hanging bag. You should not see any weight on the scale indicating there was no increase in weight by adding the air. Keep in mind that if you are able to make the scale sensitive enough, you may see fluctuations due to seismic or acoustic vibrations.

Another interesting experiment showing the affect of temperature and pressure on the balloon would be to use your lever arm scale mechanism to measure a breath-filled large standard “elastic” party balloon outside on a very cold Minnesota day. If your scale is sensitive enough, you should be able to see the weight increase as the hot carbon monoxide/air mixture in the balloon cools. This would be due to the decrease in buoyancy as the balloon shrinks.