What is the excess pressure inside a bubble of soap solution of radius 500 mm, 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 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 `1.01xx10^(5)Pa`).

Explanation

(i) We are given that radius of the soap bubble, i.e., ` r = 5.0 mm = 5.0 xx 10 ^(-3) m `
Surface tension of the soap solution ` T = 2.5 xx 10 ^(-2 ) N//m`
Density of soap solution, ` rho = 1.2 xx 10 ^(3) kg //m ^ 3 `. Clearly, excess pressure inside soap bubble, i.e.,
` p = (4 T )/(r ) = ( 4 (2.5 xx 10 ^(-2) N//m ))/(5.0 xx 10 ^(-3) m ) = 20 Pa `
(ii) Express pressure inside the air bubble in soap solution, ` p = (2T ) /(r ) = (2(2.5 xx 10 ^(-2) N//m )) /(5.0 xx 10 ^(-3) m ) = 10 Pa `
Outside pressure on the air bubble formed at a depth of 40 cm (= 0.4 m) inside soap solution
= pressure due to 0.4 m column of soap solution + atmospheric pressure (1 atm )
` = (1.2 xx 10 ^(3) ) xx 9.8 xx 0.4 Pa + 1.013 xx 10 ^5 Pa = 0.047 xx 10 ^5 Pa + 1.01 xx 10 ^5 Pa = 1.06 xx 10 ^5 Pa `
(as density of soap ` 1.2 xx 10 ^(3) kg // m ^ 3 )` . Since the excess pressure (p = 10 Pa) is extremely small as compared to ` 1.06 xx 10 ^(5) Pa`, the total pressure inside the air bubble = ` 1.06 xx 10 ^(5) Pa`.