Find the excess pressure inside a soap bubble of radius 1.5 cm in air (S = 0.030 N/m); an air…

Step 1. (a) $\dfrac{4 \times 0.030\ \text{N/m}}{0.015\ \text{m}} = 8.0\ \text{Pa}$.

Step 2. (b) $\dfrac{2 \times 0.072\ \text{N/m}}{2.0 \times 10^{-4}\ \text{m}} = 720\ \text{Pa}$.

Step 3. (c) $1.013 \times 10^{5}\ \text{Pa} + 1000\ \text{kg/m}^3 \times 9.8\ \text{m/s}^2 \times 0.50\ \text{m} + 720\ \text{Pa} = 106\,920\ \text{Pa}$.