pulmonary physics: balloons in closed chambers

Imagine we have an expanded balloon with radius $R_1$ and an internal pressure of $+20$ atm, in an room at $0$ atm of pressure. The system is at equilibrium. Now, I drop the pressure in the room to $-50$ atm. What happens then?

I don't think there is a unique solution to the problem as stated. First, at equilibrium, the net pressure across the balloon interface has to be $0$, so there has to be a $20$ atm elastic recoil force that wants to collapse the balloon. Once again after giving sufficient time to equilibrate, post $-50$ atm drop, you have a net pressure across the balloon interface of $0$ where $P_{\text{elastic recoil}}=P_\text{internal}-P_\text{external}$. If you only know the external pressure, you are left with two variables. Laplace's law does relate elastic recoil pressure to internal pressure if you know the radius, but you don't know the radius after the change. You can't use Boyle's law here either because you don't know the external pressure change and you don't have the radius to determine the volume change. Thoughts?

1 Answer

First thing, the units of 'atm' - atmosphere is an absolute measure - not a relative or gauge pressure. So while you can have fractions of an atmosphere (a vacuum), the lowest pressure one can achieve is zero atmospheres - a 'hard' vacuum.

Secondly when dealing with balloons, the pressure-size relation is nonlinear and elasticity actually decreases as they expand rather than increasing (recoil actually diminishes). Also elastic recoil is due to a property of the balloon's material -its 'stretchiness' , not the pressure differential across the wall. But pressure difference, and elasticity will ultimately determine radius.

So given these two facts I don't think your question makes any sense. But please consider recasting the question in light of my two comments.