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?
 A: 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. 


*

*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!

*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 :)

A: It would expand until it popped, because the pressure from inside the balloon would be greater than the vacuum since there is no air pressure.
The only reason a balloon doesn't pop in a setting like Earth is that the atmospheric pressure is pushing on the balloon with the same force as the helium is pushing outward on the balloon.
