My 7 year old daughter has been learning about states of matter in school and had to do an experiment to see if gasses have mass. The experiment involved balancing 2 inflated balloons on either end of a stick and then popping one of them. The idea being that, because all the air escapes from the popped balloon, the un-popped balloon will be heavier (because of the weight of air inside the balloon. The video of her experiment is linked here.
I posted this in my junior level fluid mechanics class discussion board and asked my students if this was a valid experiment to show that gasses have mass. My daughter was convinced (and excited to get to pop a balloon for class). I actually asked my students a more directed question which was:
Watch the video carefully and then give a reason why the experiment does not show what it claims to show. That is, what is the main reason that the un-popped balloon falls? Explain why, even if the experiment was run perfectly, the effect of the release of air [on the balance] would be tiny.
Next time I will ask a more open ended question. I got a few responses but, given that I did this near a test in the fall COVD semester, not as many as I had hoped. The two main answers were:
- The balloon fragments and so there is less latex on the popped side of the balance compared to the un-popped side so it falls toward the un-popped side
- There is an impulse generated by the balloon popping. This was not symmetric because of how the balloon was popped and ruptured. This generated an impulse and a temporary imbalance which was followed by the balance arm (the stick) sliding through the ribbon that was holding it up.
However, there is a third reason that this experiment would barely register in imbalance even if run perfectly. Below is a quick analysis of why.
Consider a perfectly balanced pair of balloons as shown in the diagram below. We assume that the balance mechanism is massless and symmetric as shown below
The balloons have the same mass and volume. A free body diagram of the green balloon before and after popping is shown in below.
The five forces acting on the balloon are the mass of the latex (balloon fabric), the mass of air in the balloon, the force due to the support, and the buoyancy forces. The buoyancy forces are due to the hydrostatic pressure acting on both the balloon fabric and the air in the balloon. Before the balloon is popped the force balance is given by
Therefore, the force due to the support is
Assuming that the balloon stays intact and that the change in volume of the latex as a result of deflation is negligible then the force balance after the deflation is
The support force is, therefore
As the changes in the weight and volume of the balloon fabric from the popping are negligible, we can calculate that the change in the magnitude of the force due to the support is
The weight of air in the balloon is
Where the subscript ‘air.B’ indicates the air in the balloon. The buoyancy force is
Where the subscript ‘air.a’ indicates the ambient air. Therefore, the change in the support force is
So there will be a change in the force not equal to the weight of air leaving the balloon being taken out of the force balance, but only due to the change in density in the air due to decompression. This is likely quite small.
To quantify this number we need to know the pressure inside the balloon. Fortunately there are research papers on this topic (https://www.researchgate.net/publication/239815213_Balloons_revisited). The answer to what is the pressure in the balloon is, obviously, that it depends on how full the balloon is. The paper linked above shows that there is an initial rapid increase in internal pressure followed as the balloon inflates by a slow decrease until some critical radius is reached beyond which it again increases rapidly with radius. We used 9” balloons (23 cm) and we did not overinflate them. We also didn’t test the elastic properties of the balloon fabric so the pressure we assume is only an estimate. But, if we inflated it to a radius of 10 cm then the internal pressure would be around 1.5 kPa. Assuming a room temperature of 22oC and atmospheric pressure (101.3 kPa) the density of dry air is 1.19565 kg/m3. At a gauge pressure of 1.5 kPa the density increases to 1.21336 kg/m3. I used https://www.omnicalculator.com/physics/air-density to get the densities. Plugging these densities and the volume of a 11 cm sphere into the last equation leads to
This is obviously very small and certainly undetectable by any domestic scale. It is equivalent to the weight of 0.06 ml of air.
So, ultimately, if you ran the perfect experiment with a perfect balance you would see an effect due to the balloon popping. But it is not as simple as removing the weight of air in the balloon from the force balance.
Other interesting balloon links
An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (firstname.lastname@example.org). I also welcome comments (through the comments section or via email) on improving the demonstrations.
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