Question

What is the excess pressure (in atm) inside a soap bubble with a radius of 1.5 cm and surface tension of 3 × 10^-2 N/m? [1 atm = 101.3 kPa]

Answer 1

Data : R = 1.5 × 10^-2 m, T = 3 × 10^-2 N/m, 1 atm = 1.013 × 10⁵ Pa

The excess pressure inside a soap bubble is

p – p₀ = \(\frac{4T}R\)

= \(\frac{4\times3\times10^{-2}}{1.5\times10^{-2}}\) = 8pa

= \(\frac{8}{1.013\times10^5}\) atm = 7.897 x 10^-5 atm