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

Explanation

1. Excess of pressure inside air bubble in a liquid : consider an 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 force acting on the air bubble. For the hemispherical protion of the bubble, considering the forces acting on it, we get,
(i) The force due to surface tension acting towads right round the rim of length `2 pi R "is" F_T = 2 pi RT`
(ii) THe force due to outside pressure `P_1` is to the right acting across a cross sectional area of `pi R^2 "is" F_(P_1) = P_(1)piR^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)`
`P2pi R^2 = 2 pi RT + P_1 pi R^2 implies (P_2 - P_1) pi R^2 = 2 pi RT`
Excess pressure is `Delta P = P_2 - P_1 = (2T)/R`

2. Excess pressure inside a soap bubble : Consider a soap bubble of radius r and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble. Therefore, the force on the soap bubble due to surface tension is `2 xx 2 pi R T`. The various force acting on the soap bubble are,
(i) Force due to surface tension `F_t = 4 pi RT` towards right.
(ii) Force due to outside pressure , `F_(P_1) = P_1 pi R^2` towards right.
(iii) Force due to inside pressure, `F_(P_2) = F_(P_2) = P_2)piR^(2)` towards left
As the bubble is in equilibrium, `F_(P_2) = F_T + F_(P_1)`
`P_2 pi R^2 = 4 pi R T + P_1 pi R^2 implies (P_2 - P_1) pi R^2 = 4pi RT`
Excess pressure is `DeltaP = P_2 - P_1 = (4T)/R`.

3. Excess pressure inside the liquid drop : Consider a liquid drop of radius R and the surface tension of the lequid is T.
The various forces acting on the liquid drop are :
(i) Force due to surface tension `F_1 = 2 pi RT` towards right
(ii) Force due to outside pressrue `F_(P_1) = P_(1)piR^2` towards right.
(iii) Force due to inside pressure, `F_(P_2) = P_2)piR^(2)` towards left
As the bubble is in equilibrium, `F_(P_2) = F_T + F_(P_1)`
`P_2 pi R^2 = 2 pi R T + P_1 pi R^2 implies (P_2 - P_1) pi R^2 = 2pi RT`
Excess pressure is `DeltaP = P_2 - P_1 = (2T)/R`.