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 $\left({20}^{\circ }C\right)$ is $2.50\times {10}^{–2}N{m}^{–1}$? 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 $1.01\times {10}^{5}Pa)$.

Answer 1

Given,

Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is $1.06\times {10}^{5}Pa$
Soap bubble is of radius, $r=5.00mm=5\times {10}^{–3}m$
Surface tension of the soap solution, $S=2.50\times {10}^{-2}N{m}^{-1}$
Relative density of the soap solution = 1.20
$\therefore$ Density of the soap solution, $\rho =1.2\times {10}^{3}kg/m^3$
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, $r=5mm=5\times {10}^{–3}m$
1 atmospheric pressure = $1.01\times {10}^{5}Pa$
Acceleration due to gravity, $g=9.8m/s^2$
Hence, the excess pressure inside the soap bubble is given by the relation:
$P=\frac{4S}{r}$
$P=\frac{4\times 2.5\times {10}^{-2}}{5\times {10}^{-3}}$
$P=20Pa$
Therefore, the excess pressure inside the soap bubble is $20Pa$.
The excess pressure inside the air bubble is given by the relation:
$P^{\prime }=\frac{2S}{r}$

($\because$ When the air bubble is inside the liquid, there is one surface - the water to air surface.)
$\Rightarrow P^{\prime }=\frac{2\times 2.5\times {10}^{-2}}{5\times {10}^{-3}}$
$P^{\prime }=10Pa$
Therefore, the excess pressure inside the air bubble is $10Pa$.
At a depth of 0.4 m, the total pressure inside the air bubble = Atmospheric pressure + hρg + P’
$=1.01\times {10}^5+0.4\times 1.2\times {10}^3\times 9.8+10$
$=1.057\times {10}^{5}Pa$
$=1.06\times {10}^{5}Pa$
Therefore, the pressure inside the air bubble is $1.06\times {10}^{5}Pa.$