# If you put a latex balloon in a vacuum, how much would it expand?

If you put a latex balloon in a vacuum, how much would it expand? And would it pop? Assume it doesn't leak.

EDIT:

Some numbers: Ambient pressure is 100 KPa, balloon is perfectly spherical with a diameter of 300 mm, deflated it has a diameter of 25mm, temperature is always at equilibrium.

Question components stated more formally:

• What pressure does the balloon need to have been inflated to to reach the 300 mm radius?
• What is the relationship between the (gauge) pressure and volume? I think this simply comes down to PV=nRT
• If there is some sort of spring constant involved, what is this value for a typical latex balloon?
• What is the relationship between the pressure (or volume) and the tension on the material the balloon is made from?
• What is the limit the tension can reach before popping?

Semi-related: how much would a balloon expand if you sealed it deflated and put it in a vacuum?

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Is there anything in the balloon? –  Greg P Apr 17 '11 at 2:02
A normal balloon filled with air or helium. –  Random832 Apr 17 '11 at 2:10
How much air or helium? If the amount is small enough, I don't think it would pop. –  Greg P Apr 17 '11 at 2:23
and is it closed or not? Even if closed, I'd expect diffusion of gasses through the latex material to equalise pressure during the evacuation process, at least to an extent. –  jwenting Apr 17 '11 at 14:39

It would depend on two things - how much air is in the balloon and the tensile strength of latex. To see why, I hope you'll find the following useful.

In a balloon on earth near the surface, the pressure inside depends on how much air you blow into it. But in order for the balloon to be in static equilibirium, that internal pressure has to be matched by the (constant) atmospheric pressure that opposes outward expansion PLUS the elastic surface tension of the surface, which also tends to oppose outward expansion (wanting to minimize the surface area). $$P_i = P_0 + S$$ If you blow more air into it, $P_i$ increases, and since $P_0$ is constant at the same height, the surface tension has to increase for LHS and RHS to match. This will continue until the surface tension exceeds the tensile strength of the balloon, at which point the balloon pops.

Now let's say you have blown only enough air into the balloon to make it taut -- if you let this balloon go so that it floats up, the atmospheric pressure $P_0$ starts dropping with height, and so again, the surface tension has to keep rising so that the RHS matches the constant internal pressure $P_i$.

Two things could happen now.

1. Even before $P_0$ has dropped to zero (in the vacuum of outer space), the surface tension has had to become so high that it exceeds the tensile strength and makes the balloon pop. This is what happens to most helium balloons let go - they eventually pop even before reaching vacuum - because whatever material the surface is made of has a tensile strength too low to be able to sustain the minimum amount of internal pressure required to make the balloon taut once the external pressure has dropped enough. Note that this minimum amount of internal pressure could be made smaller by reducing the size of the balloon but you can't do this beyond the limit at which the balloon's mass overcompensates and makes it impossible for it to float in the first place!
2. The other possibility is that the tensile strength is high enough that even after $P_0$ is zero, the surface tension required to match the internal pressure doesn't exceed the tensile strength and you'll have a nice happy balloon floating in the vacuum. This is what happens with space stations, for instance :)
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""If you blow more air into it, Pi increases,"" for the first "blow" into a balloon it takes more pressure than for later blows. I think that this is due to the small radius at first. Pressure is inversly related to balloon radius, but at the same time "surface tension" of the balloon increases. The pressure/radius relation for soap bubbles is well known, the same thing for rubber balloons I do not know. –  Georg Apr 17 '11 at 9:42
"if you let this balloon go so that it floats up... the surface tension has to keep rising so that the RHS matches the constant internal pressure Pi." I think you have a small error here. As the atmospheric pressure drops, the balloon will expand, the tension term will increase, but Pi will also drop. I'm not sure of the exact relationship describing this, which probably has a lot to do with the spring constant of the balloon, but I'm pretty sure that the internal pressure will not be exactly constant. –  Colin K Apr 17 '11 at 20:11
If I can manage to turn this into a concrete answer with the numbers I've arbitrarily selected in the edit, I'll accept this answer once I've done the math. –  Random832 Apr 20 '11 at 21:44