What happens to surface tension of bubble if pressure inside and outside outside become equal Imagine a soap bubble in water so that pressure outside the bubble is equal to pressure inside it .....
i.e excess pressure of soap bubble if it were in air, $P =\rho g h$
($\rho$, density of water)
 A: The situation of a soap-film bubble is that the curvature
of the surface IS the pressure inside (proportional to
the pressure) because that curvature and the surface
tension of the liquid are the only variables that matter.
So, a bubble wall that has the same pressure on both
sides, has zero curvature; this is the case for a twinned
soap bubble, two identical-diameter bubbles in contact,
the shared wall is... flat.
Soap Bubbles and the Forces Which Mould Them by C. V. Boys is recommended reading.
A: My guess is that according to Young-Laplace equation,
$$ \Delta p =-2\gamma H_{f} ,$$
where $H_{f}$ is mean curvature of a bubble surface, in case pressure difference $\Delta p = 0$, this gives per mean curvature definition,
$$ H_f = \frac 12 \left(\frac 1R_1 + \frac 1R_2 \right) ,$$
that when mean curvature $H_f$ is zero, radii of principal curvatures $R_1,R_2 \to \infty$. Infinite radius corresponds to flat surface. So bubble must transform into flat soap film.
But the problem is that you can't get a plane from a sphere by performing a homeomorphic topological transformation, unless you remove 1 point from a sphere. "Making a hole" in a bubble breaks bubble integrity, so the final answer is that because it's impossible for bubble to transform into flat surface,- likely it will blew up (disintegrate) in such process.
