An air bubble is 50 cm below the surface of a lake. In that case, we assume that the bubble has only one surface to generate surface tension. The gauge pressure inside the bubble is 6.0 × 10³ Pa. What is radius of the bubble? (The density of the water is ñ_H2O = 1000 kg/m³, and the surface tension is ó = 0.073 N/m.) A)2.6 × 10^-4 m B)1.3 × 10^-4 m C)6.0 × 10^-7 m D)3.0 × 10^-7 m

Question

Quizplus · Accepted Answer

The gauge pressure inside the bubble can be expressed as the sum of the hydrostatic pressure due to the depth of the water and the pressure due to surface tension. The hydrostatic pressure is given by $P_h = \rho gh$, where $\rho = 1000 \, \text{kg/m}^3$ is the density of water, $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity, and $h = 0.5 \, \text{m}$ is the depth of the bubble. The pressure due to surface tension is given by $P_s = \frac{2\sigma}{r}$, where $\sigma = 0.073 \, \text{N/m}$ is the surface tension and $r$ is the radius of the bubble. The total gauge pressure is the sum of these two pressures: $P = P_h + P_s$.Given $P = 6.0 \times 10^3 \, \text{Pa}$, we can solve for $r$ by first calculating $P_h$ and then rearranging the equation for $P_s$ to solve for $r$.Calculating $P_h = \rho gh = 1000 \times 9.8 \times 0.5 = 4900 \, \text{Pa}$.Since $P = P_h + P_s$, we have $6000 = 4900 + \frac{2\sigma}{r}$.Rearranging for $r$, we get $r = \frac{2\sigma}{P - P_h}$.Substituting the given values, $r = \frac{2 \times 0.073}{6000 - 4900} = \frac{0.146}{1100} = 1.3 \times 10^{-4} \, \text{m}$.

An Air Bubble Is 50 Cm Below the Surface of a Lake