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

Explanation

(i) Excess presure inside the liquid drop : Consider a liquid drop of radius R and the surface tension of the liquid is T as shown in Figure.

The various forces acting on the liquid drop are,
(i) Force due to surface tension `F_(T) = 2pi 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)) = P_(2)pi R^(2)` toward left
As the drop is in equilibrium,
`F_(P_(2)) = F_(T) + F_(P_(1))`
`P_(2) pi R^(2) = 2pi RT + P_(1) pi R^(2)`
`rArr (P_(2) - P_(1)) pi R^(2) = 2pi RT`
Excess pressure is `Delta P = P_(2) - P_(1) = (2T)/(R )`
(ii) Excess pressure inside a soap bubble
Consider a soap bubble of radius R and the surface tension of the soap bubble be T as shown. A soap bubble has two liquid 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 2pi RT`. The various forces acting on the soap bubble are

(i) Force due to surface tension `F_(T) = 4pi RT` towards right (ii) Force dur to inside pressure,
`F_(P_(1)) = P_(1) pi R^(2)` towards right
(iii) Force due to inside pressure,
Force dur to inside pressure,
`F_(P_(2)) = P_(2) pi R^(2)` towards left
As the bubble is in equilibrium,
`F_(P_(2)) = F_(T) + F_(P_(1))`
`P_(2) pi R^(2) = 4pi RT + P_(1)pi R^(2)`
`rArr (P_(2) - P_(1)) pi R^(2) = 4pi RT`
Excess pressure is `Delta P = P_(2) - P_(1) = (4T)/(R )`.
(iii) Excess of pressure inside air bubble in a liquid.
Consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. 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 `DeltaP = P_(1) - P_(2)`
To find the axcess pressure inside the air bubble, let us consider the forces acting on the air bubble are
(i) The force due to surface tension acting toward right arounf the rim of length `2pi R` is `F_(T) = 2pi RT`
(ii) The force due to outside pressure acting `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 `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))`
`P_(2)pi R^(2) = 2pi RT + P_(1)pi R^(2)`
`rArr (P_(2) - P_(1))piR^(2) = 2pi RT`
Excess pressure is `Delta P = P_(2) -P_(1) = (2T)/(R )`.