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 (${20}^{o}C$) 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

Excess pressure inside the soap bubble is $20Pa$;
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 $=1.20$
∴ Density of the soap solution, $\rho =1.2\times {10}^{3}kg/m^3$
Air bubble formed at a depth, $h=40cm=0.4m$
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:
$P=\frac{4S}{r}$
$=\frac{4\times 2.5\times {10}^{-2}}{5\times {10}^{-3}}$

$=20Pa$
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
$P^{\prime }=\frac{2S}{r}$
$=\frac{2\times 2.5\times {10}^{-2}}{(5\times {10}^{-3})}$

$=10Pa$
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 $+h\rho g+P^{\prime }$
$=1.01\times {10}^5+0.4\times 1.2\times {10}^3\times 9.8+10$
$\approx 1.06\times {10}^{5}Pa$
Therefore, the pressure inside the air bubble is $1.06\times {10}^{5}Pa$