What is the excess pressure inside a bubble of soap solution of radius 5.00mm, given that the surface tension of soap solution at the temperature `(20^(@)C)` is `2.50xx10^(-2)Nm^(-1)`? If an air bubble of the same dimension were formed at a 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? (1atm. is `1.01 xx 10^(5)Pa`).

Answer

`20 N//m^(2), 1.05714'10^(5)N//m^(2)`

Explanation

initially
clearly: `P_("air") - P_(0) = (2sigma)/(r)`
`:'` same level has pressure `P_(0)`.
and by mole conservation. `P_(0) xx l_(0) = P_(air) xx h …(2)`
`= (P_(0)l_(0))/(h) - P_(0) = (2sigma)/(r), h = (P_(0)l_(0))/((2sigma)/(r) + P_(0)) rArr h = 0.1 m`.
Hence the part inside the water is `0.11 - 0.1 = 0.01 m`.
If seal is broken, the atmospheric will be exerting the pressure which is lesser than `P_(air)`. Hence the liquid will rise up in the capillary.