A soap bubble of radius 2.00 x 10^-4 m is blown from soapy water whose surface tension is 0.0370 N/m. The gauge pressure inside a spherical bubble is four times the surface tension divided by the radius. What is that pressure?
A185 Pa
B370 Pa
C1480 Pa
D740 Pa
Answer & Solution
Correct answer: D. 740 Pa
1. Gauge pressure inside a spherical bubble = 4 x surface tension / radius.
2. Numerator = 4 x 0.0370 N/m = 0.148 N/m.
3. Divide by the radius: (0.148 N/m) / (2.00 x 10^-4 m).
4. This gives 740 N/m^2, that is 740 Pa above the outside air.
5. Note that this is a gauge pressure, so the absolute pressure inside is 740 Pa above atmospheric.
6. The value 370 Pa uses a factor of two instead of four, and 185 Pa drops the factor entirely.
7. The value 1480 Pa doubles the correct result, which is what a factor of eight would give.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action_
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