What happens to a soap bubble as the outside pressure is increased? Imagine you blow a soap bubble at atmospheric pressure. The bubble equilibrates to some radius Ro with some internal pressure Po due to the surface tension To of the soap-water film. If you then place the bubble in a sealed chamber and gradually increase the pressure in the chamber from 1 atm to Po, what happens to the bubble? Logic would dictate that the bubble would shrinkas the outside pressure increases.

On the other hand, Laplace's law states that dP ~ T/R. As the outside pressure increases from 1atm to Po, the gauge pressure, dP, of the bubble decreases, which implies that the radius should increase (expand). So does the bubble shrink or expand? What is the source of the apparent contradiction?

Thanks!

What happens to a soap bubble as the outside pressure is increased? Imagine you blow a soap bubble at atmospheric pressure. The bubble equilibrates to some radius, $R_{0}$, with some internal pressure, $P_{0}$, due to the surface tension, $T_{0}$, of the soap-water film. If you then place the bubble in a sealed chamber and gradually increase the pressure in the chamber from 1 atm to $P_0$, what happens to the bubble? Logic would dictate that the bubble would shrink as the outside pressure increases.

On the other hand, Laplace's law states that $dP \propto \frac{T}{R}$. As the outside pressure increases from 1 atm to $P_0$, the gauge pressure, $dP$, of the bubble decreases, which implies that the radius should increase (expand). So does the bubble shrink or expand? What is the source of the apparent contradiction?

Thanks!