# About a balloon in a balloon in a balloon in a balloon in a balloon

Suppose we manage to put $x$ identical (blown up) balloons into one another. The balloons are identical and snap if the inside air pressure is $l$ times as big as the atmospheric pressure (if the outside of the balloon is the atmosphere).

The first balloon is in a state just below snapping, so we can say that the inside air pressure is almost $l$ times as big as the atmospheric pressure. The second balloon is in a state so that is will just snap if the first balloon snaps. The third balloon inside this (second) balloon is in such a state so that it will just snap if the first and second balloon are snapped. The fourth balloon inside the third balloon is again in such a state that it just snaps after the first, second and third one are snapped. And so on until the $x$-th balloon.

Now we saw that the pressure inside the first balloon is almost $l$ times as big as the atmospheric pressure. For the second balloon the inside pressure is $l^2$ as big as the atmospheric pressure, the inside pressure of the third balloon is $l^3$ as big as the atmospheric pressure, and so on until the $x$-th balloon, in which the air pressure is $l^x$ times as big as the atmospheric pressure (although these pressures are bigger than $l$ times the atmospheric pressure they won't snap, because this happens only for balloons without blown up balloons around them).

Is my reasoning right that if we let the first balloon snap by a needle, all the others will snap, one after another, or am I overlooking something? For example the fact that each balloon expands after the surrounding balloon snaps (which would imply that the enclosed balloons would have to be under greater pressure)? How would the ditribution of pressures in the balloons have to be to let them snap one aftyer another if you let the first one snap?

• I would say that the successive pressures should be 2l, 3l, etc, not exponentials, because what you do is keep difference of pressure of l between successive balloons. And, I agree, they will expand and break. – user126422 May 1 '17 at 12:20
• This seems like what you would expect to happen. As you start breaking balloons the total energy contained by pressure shouldn't decrease, but the amount of material to contain it will. If they were on the verge of breaking already this should cascade it. This isn't even including the shockwave created when the balloon breaks, which should send an impulse which would slightly amplify the initial pressure. – JMac May 1 '17 at 14:06
• Like @WillyBillyWilliams says, the progression is arithmetic. Not geometric. Because pressures add. As a side note, there is at least one instance of this device being used in science fiction and that book (Seveneves) attributes it to the Soviet space program. – dmckee --- ex-moderator kitten May 1 '17 at 16:43
• I can see that the progression is of course arithmetic. But does $pV$ (the energy in each balloon) stays the same if, after the first balloon snaps, the second exerts work on the balloon thereby reducing its energy? The same holds for the balloons that follow. So shouldn't the pressures be bigger in every subsequent balloon than the arithmetic sequence describe? – descheleschilder May 1 '17 at 18:08