Example 11.11 from College Physics, 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

Surface Tension: Pressure Inside a Bubble

Calculate the gauge pressure inside a soap bubble $2 . 00 \times 10^{- 4} m$ in radius using the surface tension for soapy water in Table 11.3. Convert this pressure to mm Hg.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The radius is given and the surface tension can be found in Table 11.3, and so $P$ can be found directly from the equation $P = \frac{4γ}{r}$ .
Solution
Substituting $r$ and $γ$ into the equation $P = \frac{4γ}{r}$ , we obtain
$P = \frac{4γ}{r} = \frac{4 (0.037 N/m)}{2 . 00 \times 10^{- 4} m} = 740 {N/m}^{2} = 740 Pa .$
11.49

We use a conversion factor to get this into units of mm Hg:

$P = (740 {N/m}^{2}) \frac{1.00 mm Hg}{133 {N/m}^{2}} = 5.56 mm Hg .$
11.50
Discussion
Note that if a hole were to be made in the bubble, the air would be forced out, the bubble would decrease in radius, and the gauge pressure would reduce to zero, and the absolute pressure inside would decrease to atmospheric pressure (760 mm Hg).

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Solution

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Strategy

Solution

Discussion