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

Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is 1.06 × 10 5 Pa
Soap bubble is of radius, r = 5.00 mm = 5 × 10 - 3 m
Surface tension of the soap solution, S = 2.50 × 10 - 2 Nm -1
Relative density of the soap solution = 1.20
∴Density of the soap solution, ρ = 1.2 × 103 kg/m3
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 × 10–3 m
1 atmospheric pressure = 1.01 × 105 Pa
Acceleration due to gravity, g = 9.8 m/s2
Hence, the excess pressure inside the soap bubble is given by the relation :

$P =\frac{ 4S}{ r}$

$= \frac{4 × 2.5 × 10 ^{- 2}}{ 5 × 10 ^{- 3}}$

= 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.

The excess pressure inside the air bubble is given by the relation :

$P = \frac{2S }{ r}$

$=\frac{ 2 × 2.5 × 10 ^{- 2} }{5 × 10 ^{- 3}}$

= 10 Pa
Therefore, the excess pressure inside the air bubble is 10 Pa.

At a depth of 0.4 m, the total pressure inside the air bubble

= Atmospheric pressure + hρg + P'

= 1.01 × 105 + 0.4 × 1.2 × 10 3 × 9.8 + 10

= 1.057 × 10 5 Pa

= 1.06 × 10 5 Pa

Therefore, the pressure inside the air bubble is 1.06 × 10 5 Pa.