Solution 11: Atmospheric Pressure from Coalescing Bubbles
Objective: Find the atmospheric pressure $P_0$ given that two bubbles of diameters $a$ and $b$ coalesce into a single bubble of diameter $c$ under isothermal conditions.
1. Principle of ConservationSince the process is isothermal (temperature remains constant), the total number of moles of gas is conserved. Applying Boyle’s Law ($PV = \text{constant}$): $$ P_a V_a + P_b V_b = P_c V_c $$
2. Pressure EquationsThe pressure inside a soap bubble is the sum of the atmospheric pressure $P_0$ and the excess pressure due to surface tension. Note that a soap bubble has two surfaces (inner and outer), so the excess pressure is $\frac{4\sigma}{r}$ or $\frac{8\sigma}{diameter}$.
- For bubble $a$: $P_a = P_0 + \frac{8\sigma}{a}$
- For bubble $b$: $P_b = P_0 + \frac{8\sigma}{b}$
- For bubble $c$: $P_c = P_0 + \frac{8\sigma}{c}$
Using the volume of a sphere $V = \frac{4}{3}\pi (\frac{d}{2})^3 = \frac{\pi}{6}d^3$: $$ \left(P_0 + \frac{8\sigma}{a}\right) \frac{\pi}{6}a^3 + \left(P_0 + \frac{8\sigma}{b}\right) \frac{\pi}{6}b^3 = \left(P_0 + \frac{8\sigma}{c}\right) \frac{\pi}{6}c^3 $$
Canceling common terms ($\frac{\pi}{6}$): $$ (P_0 a^3 + 8\sigma a^2) + (P_0 b^3 + 8\sigma b^2) = (P_0 c^3 + 8\sigma c^2) $$
4. Solving for Atmospheric Pressure ($P_0$)Group the $P_0$ terms on one side and $\sigma$ terms on the other: $$ P_0 (a^3 + b^3) – P_0 c^3 = 8\sigma c^2 – 8\sigma (a^2 + b^2) $$ $$ P_0 (a^3 + b^3 – c^3) = 8\sigma (c^2 – a^2 – b^2) $$
$$ P_0 = \frac{8\sigma (c^2 – a^2 – b^2)}{a^3 + b^3 – c^3} $$
Multiplying numerator and denominator by $-1$ to match the standard form: