Tale Of 2 Balloons

Classical Mechanics Level 3

Two identical balloons are inflated such that one balloon is bigger than the other before placing both of them at the opposite ends of a pipe with a closed valve.

What will happen if you open this valve?

Details and Assumptions

Both balloons remain in the elastic regime.
Both balloons are inflated by the same air source.
Both balloons contain enough air to sustain the intrinsic shape of the balloon, i.e. they're not floppy.

All the air comes into the smaller balloon The bigger one expands; the smaller shrinks They become equal in size Nothing will happen

1 solution

Worranat Pakornrat
Oct 2, 2016

The bigger balloon will become even bigger and the smaller one will become even smaller because the air pressure inside the big one is, in fact, less than that in the small one, causing the air to rush in from the higher pressure to the lower one.

Despite more air and larger radius, the big balloon actually has thinner membrane and more surface area compared to the small one. The force acting on its surface, on the other hand, varies directly with radius $R$ ; however, the pressure is force per the balloon's area, which varies with $R^2$ for the bigger radius $R$ . Overall, the air pressure inside will inversely vary with $R$ .

As a result, the bigger the radius, the less pressure it contains. Practically, we can feel that it seems difficult to blow the balloon at first but gets easier as it gets bigger and so requires less lung force to push for.

More factors are at play than those described here. While there is a range of inflations which produce the behavior in the solution, it is equally true that there are ranges which don't. The pressure inside the balloon is typically only a small fraction of 1% above ambient. This is not sufficient to explain the difficulty in the initial stage of blowing up the balloon. The dominant factor is the elasticity of the material. Assuming the balloon doesn't burst, there is a point where the material reaches its elastic limit and pressure starts to increase again.

Tom Capizzi - 4 years, 8 months ago

I've added in the assumption that both balloons remain in the elastic regime throughout the exchange. I've also given them the same inflation process to avoid hysteresis.

Josh Silverman Staff - 4 years, 8 months ago

It's a lot more complicated than this solution. If initial states of the two balloons are just right, it is possible for equilibrium to be reached with both balloons the same size. See: here

Bob Kadylo - 4 years, 8 months ago

As written above, more factors are at play and I disagree with the solution. Pressure in more inflated balloon would be usually higher than pressure in less inflated balloon because higher elastic tension of thin rubber/plastic envelope. As one notice when blowing a balloon, it's much easier at beginning (less pressure) than at end (more pressure needed). What will usually happen in my opinion when valve is open is larger balloon will become smaller but still stay larger than smaller balloon as rubber/plastic is not fully elastic and expanded (plasticity) in larger balloon.

michael de seguin - 4 years, 8 months ago

As one notice when blowing a balloon, it's much easier at beginning (less pressure) than at end (more pressure needed).

This is not true.

Josh Silverman Staff - 4 years, 8 months ago

With the new assumption that both balloons remain in elastic regime, I agree with you (my assumption, may be wrong, was larger ballon reaching limit of elasticity & pressure increasing). An experiment would not be too hard to do (setup with 3 valves so can inflate each balloon separately, both being attached to the setup and then close blow thru 3rd valve before opening 2 others). However this experiment may not be conform to the elastic regime assumption.

michael de seguin - 4 years, 7 months ago

Tale Of 2 Balloons

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