My friend puzzled me with this: Say I have a container that contains a vacuum and another container inside. And this inner container contains air. Also the outer container material is just barely strong enough to not implode due to the vacuum. If I remove the air from the inner container to outside of the outer container (for example through a tube that can be sealed), would the outer container implode?

My friend says it will implode because the total volume of the whole thing is decreased which leads to decreased inside air pressure. Mathematically, it sounds right. But I thought it was weird because there was really no interaction between the air in the inner container and the outer container in the first place. So my answer is that it will not implode.

But I'm not sure myself. What would actually happen and why?

EDIT: I actually think it will implode if the inner container was a balloon or something due to having less area for the remaining air molecules in the "vacuum" to bounce around. My question is for an inner container that doesn't change shape like a balloon.


If the volume between the two containers is a true vacuum (by definition no molecules present), taking away the air inside the inner container would have no effect.

The net pressure on the outer container would be atmospheric pressure, since the internal pressure is zero.

If the volume between the two containers is only a partial vacuum, and the inner container was ideally rigid there would still be no effect from removing the air from the innner container. But is the vacuum is only partial and the inner container elastic, removing the air from the inner container would cause the outer container to implode.

  • $\begingroup$ Thanks for the answer! So does the contents of the inner container not included in the calculations for the volume of the outer container? Is it because it's a closed system or something? $\endgroup$ – Cmile Feb 21 '14 at 16:40

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