4.5.19 | revision #13568 | Savchenko Solutions

Statement

4.5.19. The outer radius of a soap bubble is $$R$$ , the thickness of its wall is $$h$$ . Find the air pressure inside the bubble. The air pressure outside the bubble is $$P_0$$ , the surface tension of water is $\sigma$ .

Solution

The pressure inside the bubble wall is:
$P_1 = P_0 + P_L = P_0 + \frac{2\sigma}{R}$

The pressure inside the bubble is:
$P = P_1 + \frac{2\sigma}{R-h}$

Combining the two equations, we get:
$P = P_0 + \frac{2\sigma}{R} + \frac{2\sigma}{R - h}$

$P = P_0 + 2\sigma \left( \frac{1}{R} + \frac{1}{R - h} \right)$

Answer

$P = P_0 + 2\sigma \left( \frac{1}{R} + \frac{1}{R - h} \right)$