Question

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10^–2 N m^–1 ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 10⁵ Pa).

Answer 1

Given Data

• Radius of bubble: \( r = 5.00 \,\text{mm} = 5.00 \times 10^{-3} \,\text{m} \)
• Surface tension of soap solution: \( T = 2.50 \times 10^{-2} \,\text{N m}^{-1} \)
• Depth of air bubble in solution: \( h = 40.0 \,\text{cm} = 0.40 \,\text{m} \)
• Relative density of soap solution: \( 1.20 \Rightarrow \rho = 1.20 \times 1000 = 1200 \,\text{kg m}^{-3} \)
• Acceleration due to gravity: \( g \approx 9.8 \,\text{m s}^{-2} \)
• Atmospheric pressure: \( P_{\text{atm}} = 1.01 \times 10^{5} \,\text{Pa} \)

1. Excess pressure inside the soap bubble

A soap bubble in air has two liquid surfaces (inner and outer), so the excess pressure inside it is \( \Delta P_{\text{soap}} = \dfrac{4T}{r} \).

\( \Delta P_{\text{soap}} = \dfrac{4 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} \) \( = \dfrac{10.0 \times 10^{-2}}{5.00 \times 10^{-3}} = 20 \,\text{Pa} \)

Excess pressure inside the soap bubble = \( 20 \,\text{Pa} \).

2. Pressure inside an air bubble at depth 40 cm

An air bubble inside a liquid has only one liquid surface, so its excess pressure is \( \Delta P_{\text{air}} = \dfrac{2T}{r} \).

\( \Delta P_{\text{air}} = \dfrac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} \) \( = \dfrac{5.0 \times 10^{-2}}{5.00 \times 10^{-3}} = 10 \,\text{Pa} \)

The hydrostatic pressure at depth \( h \) in the soap solution is \( P_{\text{hydro}} = \rho g h = 1200 \times 9.8 \times 0.40 \,\text{Pa} \).

\( P_{\text{hydro}} = 1200 \times 9.8 \times 0.40 \approx 4704 \,\text{Pa} \)

Total pressure just outside the air bubble at depth \( h \) is \( P_{\text{out}} = P_{\text{atm}} + P_{\text{hydro}} \).

\( P_{\text{out}} = 1.01 \times 10^{5} + 4704 \approx 1.057 \times 10^{5} \,\text{Pa} \)

Pressure inside the bubble will be this outside pressure plus excess pressure \( \Delta P_{\text{air}} \).

\( P_{\text{in}} = P_{\text{out}} + \Delta P_{\text{air}} \approx 1.057 \times 10^{5} + 10 \approx 1.0571 \times 10^{5} \,\text{Pa} \)

Pressure inside the air bubble at 40 cm depth \( \approx 1.06 \times 10^{5} \,\text{Pa} \).

Physical understanding

• Soap bubble: two surfaces (inside and outside), so the curvature effect is stronger and gives \( \Delta P = 4T/r \).
• Air bubble in liquid: only one liquid–air surface, so \( \Delta P = 2T/r \); the bubble also feels extra pressure due to the depth of liquid above it.