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

Here surface tension of soap solution at room temperature

T = 2.50 x 10^-2 Nm^-1, radius of soap bubble, r = 5.00 mm = 5.00 x 10^-3 m.

:. Excess pressure inside soap bubble,  `p = p_i - p_0 = (4T)/r`

`= (4xx2.50xx10^(-2))/(5.00xx10^(-3)) = 20.0 Pa`

When an air bubble of radius `r  = 5.00 xx 10^(-3)` m is formed at a depth h = 40.0 cm  = 0.4 m inside a container containing a soap solution of relative density 1.20 or density `rho = 1.20 xx 10^3 kg m^(-3)`, then excess pressure

`P = P_i - P_0 = (2T)/r`

`:. P_i = P_0 + (2T)/r = (P_a + hrhog) + (2T)/r`

`=[1.01 xx 10^5 + 0.4 xx 1.2 xx 10^3 xx 9.8 + (2xx2.50xx10^(-2))/(5.00xx10^(-3))] Pa`

`= (1.01 xx 10^5 + 4.7 xx 10^3 + 10.0)Pa`

`= 1.06 xx 10^5` Pa

Answer 2

Excess pressure inside the soap bubble is 20 Pa;

Pressure inside the air bubble is `1.06 xx 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 = 4S/r`

`= (4xx2.5xx10^(-2))/(5xx10^(-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 = (2S)/r`

`= (2xx2.5xx10^(-2))/(5xx10^(-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 xx 10^5 +0.4 xx 1.2 xx 10^3 xx 9.8 + 10`

`= 1.057 xx 10^5 Pa`

`= 1.06 xx 10^(5) Pa`

Therefore, the pressure inside the air bubble is `1.06 xx 10^5 Pa`