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  • 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 ?

Q 10.20 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 × 105Pa).

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Excess Pressure inside a bubble is given by

\Delta P=\frac{4T}{r}            (It's double the usual value because of the presence of 2 layers in case of soap bubble)

where T is surface tension and r is the radius of the bubble

\\\Delta P=\frac{4\times 2.5\times 10^{-2}}{5\times 10^{-3}}\\ \Delta P=20\ Pa

Atmospheric Pressure is P_{a}=1.01\times 10^{-5}\ Pa

The density of soap solution is \rho _{s}=1.2\times 10^{3}\ kg\ m^{-3}

The pressure at a depth of 40 cm (h) in the soap solution is 

\\P_{s}=P_{a}+\rho _{s}gh\\ P_{s}=1.01\times 10^{5}+1.2\times 10^{3}\times 9.8\times 40\times 10^{-2}\\ P_{s}=1.01\times 10^{5}+4704\\ P_{s}=1.057\times 10^{5}\ Pa

Total Pressure inside an air bubble at that depth

\\P_{t}=P_{s}+\frac{2T}{r}\\ P_{t}=1.057\times 10^{5}+\frac{2\times 2.5\times 10^{-2}}{5\times 10^{-3}}\\ P_{t}=1.057\times 10^{5}+10\\ P_{t}=1.0571\times 10^{5}\ Pa

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Sayak

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