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.