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)
Excess pressure inside the soap bubble is 20Pa; Pressure inside the air bubble is 1.06×105Pa Soap bubble is of radius, r=5.00mm=5×10−3m Surface tension of the soap solution, S=2.50×10−2Nm−1 Relative density of the soap solution =1.20 ∴ Density of the soap solution, ρ=1.2×103kg/m3 Air bubble formed at a depth, h=40cm=0.4m Radius of the air bubble, r=5mm=5×10−3m 1 atmospheric pressure =1.01×105Pa Acceleration due to gravity, g=9.8m/s2 Hence, the excess pressure inside the soap bubble is given by the relation: P=r4S =5×10−34×2.5×10−2 =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′=r2S =(5×10−3)2×2.5×10−2 =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 +hρg+P’ =1.01×105+0.4×1.2×103×9.8+10 ≈1.06×105Pa Therefore, the pressure inside the air bubble is 1.06×105Pa
