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Question
Derive an expression for excess pressure inside a drop of liquid.
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Solution
Consider a spherical drop as shown in the figure. Let pi be the pressure inside the drop and p0 be the pressure out side it. As the drop is spherical in shape, the pressure, pi, inside the drop is greater than p0, the pressure outside. Therefore, the excess pressure inside the drop is pi - p0.
Let the radius of the drop increase from r to r + ∆r, where ∆r is very small, so that the pressure inside the drop remains almost constant.
Let the initial surface area of the drop be A1 = 4πr2 , and the final surface area of the drop be A2 = 4π (r+∆r)2.
∴ A2 = 4π(r2 + 2r∆r + ∆r2)
∴ A2 = 4πr2 + 8πr∆r + 4π∆r2 (As ∆r is very small, ∆r2 can be neglected)
∴ A2 = 4πr2 + 8πr∆r
Thus, increase in the surface area of the drop is
`dA = A_2 – A_1 = 8πrDeltar`
Work done in increasing the surface area by `dA` is stored as excess surface energy.
`therefore dW = TdA= T (8πrDeltar)` ... (1)
This work done is also equal to the product of the force F which causes increase in the area of the drop and the displacement ∆r which is the increase in the radius of the bubble.
`therefore dW = F∆r` ... (2)
The excess force is given by,
(Excess pressure) ×(Surface area)
∴ F = (pi – p0) 4πr2 ... (3)
Equating Eq. (1), and Eq. (2), we get,
`T(8πrDeltar) = FDeltar`
`therefore T(8πrDeltar) = (p_i – p_0) 4πr^2Deltar` ... (using Eq. (3))
`therefore (p_i – p_0)=(2T)/r`
This equation gives the excess pressure inside a drop of liquid.
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