Obtain an expression for the excess of pressure inside a (i) liquid drop (ii) liquid bubble (iii) air bubble.

Explanation

Excess of pressure inside air bubble in a liquid : consider air bubble of radius R inside a liquid having surface tension T. Let `P_1` and `P_2` be the pressures outside and inside the air bubble, respectively. Now, the excess pressure inside the air bubble is `Delta P = P_1 -P_2` . In order to find the excess pressure .inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get
(i) The force due to surface tension acting towards right around the rim of length `2pi R` is `F_T = 2pi RT`
(ii) The force due to outside pressure `P_1` is to the right acting across a cross sectional area of ` piR^2 ` is `F_(P_1) = P_1 pi R^2`
(iii) The force due to pressure `P_2` inside the bubble, acting to the left is `F_(P_2) = P_2 pi R^2`
As the air bubble is in equilibrium under the action of these forces, ` F_(P_2) = F_T + F_(P_1)`
Excess pressure is `DeltaP = P_2 - P_1 = (2T)/(R )`