Question

Calculate the pressure inside a small air bubble of radius $0.01mm$ situated at a depth of $h=20m$ below the free surface of liquid of density ${\rho }_1={10}^{3}kg/m^3,{\rho }_2=800km/m^3$ and surface tension $T_2=7.5\times {10}^{-2}N/m$. The thickness of the first liquid is $h_1=15m$ and $h_2=25m$.

Answer 1

Given : $R=0.01mm=0.00001m$ $h=20$ m $h_1=15$ m ${\rho }_1={10}^{3}kg/m^3$

${\rho }_2=800kg/m^3$ $T_2=0.075N/m$

From the figure, $x=h-h_1=20-15=5m$

Pressure just outside the bubble $P=P_o+{\rho }_{1}g{h}_1+\rho gx$

$\therefore$ $P={10}^5+{10}^3\left(10\right)\left(15\right)+800\left(10\right)\left(5\right)=2.9\times {10}^{5}N/m^2$

Thus pressure inside the air bubble $P_i=P+\frac{2T_2}{R}$

$\therefore$ $P_i=2.9\times {10}^5+\frac{2\times 0.075}{0.00001}=3.05\times {10}^{5}N/m^2$