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 x 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (or relative density 1.20), what would be the pressure inside the bubble? (1 atmosphere pressure is 1.01 x 10⁵ Pa)

Answer 1

Here, surface tension,

T = 2.50 x 10^-2 Nm^-1

Radius, r = 5.00 mm = 5 x 10^-3 m

Excess pressure,

p = 4T/r (For a bubble) = {4 x 2.50 x 10^-2}/{5 x 10^-3} = 20 Pa

Now, p₁ - p₀ = p

⇒ p₁ = p + p₀ = 1.01 x 10⁵ + 20 = 1.0102 x 10⁵ Pa

If the bubble of the same radius as above is at depth of 40 cm inside the soap solution, then excess of pressure is given by,

p = 2T/r = {2 x 2.50 x 10^-2}/{5 x 10^-3} = 10 Pa

Also, outside pressure is,

p₀ = Pressure of water above the bubble + atmosphere pressure

= ρgh + 1.01 x 10⁵

Here, ρ = 1.2 x 10³ kg m^-3

g = 9.8 ms^-2

and h = 40 x 10^-2 m

p₀ = 1.2 x 10³ x 9.8 x 40 x 10^-2 + 1.0 x 10⁵

= 4.704 x 10³ + 1.01 x 10⁵

= 1.057 x 10⁵ Pa

If p₁ is the inside pressure, then

p₁ - p₀ = p

⇒ p₁ = p + p₀

= 1.057 x 10⁵ + 10 = 1.057 x 10⁵