Soap Bubbles and Laplace's Law - contradicts intuition

Question

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!

Answer 1

The source of the contradiction is the assumption that the pressure inside the bubble remains constant as the pressure outside the bubble increases. This is not the case. If the volume of the bubble changes then the pressure on the inside of the bubble will also change.

To rectify this, let us assume that the interior of the bubble consists of an ideal gas at constant temperature. Thus inside the bubble we have, for some constant k, $$ P_iR^3=k $$ where $P_i$ is the internal pressure and $R$ the radius of the bubble.

The equation for the surface tension is $$ P_i - P_o = \frac{\gamma} {R} $$ where $P_o$ is the external pressure.

Combining these we get that $$ \frac{k} {R^3} - P_o= \frac{\gamma} {R} $$ This is a difficult equation to solve for $R$, but we can implicitly differentiate wrt $P_o$ $$ -\frac{dR} {dP_o} \frac{3k} {R^4} - 1= - \frac{dR} {dP_o} \frac{\gamma} {R^2} $$ $$ \frac{dR} {dP_o} \left( \frac{3k} {R^4} - \frac{\gamma} {R^2} \right) = - 1 $$ $$ \frac{dR} {dP_o} \frac{3k - \gamma R^2 } {R^4} = - 1 $$ $$ \frac{dR} {dP_o} = \frac{-R^4}{3k - \gamma R^2} $$

From our earlier eqiation $$ \frac{k} {R^3} - P_o= \frac{\gamma} {R} $$ $$ \gamma R^2=k- P_o R^3 $$

Therefore $$ \frac{dR} {dP_o} = \frac{-R^4}{2k + P_o R^3} $$ $$ \frac{dR} {dP_o} < 0 $$ and as the external pressure increases, the radius of the bubble decreases.

Edit

To get the above in terms of the actual measured surface tension of the fluid we must consider the geometry of the fluid in the film. For a thin film bubble, the bubble material exists for radii $R_1<r<R_2$. Denoting the pressure of the fluid in the film $P_f$ $$ P_f - P_o = \frac{2\sigma} {R_2} $$ $$ P_f - P_i = \frac{-2\sigma} {R_1} $$ for the measured surface tension $\sigma$. Thus $$ P_i - P_o = \frac{2\sigma} {R_2} +\frac{2\sigma} {R_1} $$ Next we define $$ R^{-1} = \frac{R_1^{-1}+R_2^{-1}} {2} $$ the harmonic mean on the radii, then $$ P_i - P_o = \frac{4\sigma} {R} $$ thus, from above $$ \gamma = 4\sigma $$

To be able to use the ideal gas law as stated we require $R_1 \approx R$ such that the volume of gas inside the bubble is proportional to $R^3$ to a good accuracy.

Note that this means that the effective surface tension of the bubble is double what the surface tension of a single surface would be, the two sides of the bubble compound.