Name: Angle of Contact
Uploaded: 2019-07-19
Duration: 6 min 57 s
Description: What is the contact angle? Which factor does the contact angle depend on? Why does the contact angle for the same liquid on different surfaces differ? Let's...

Question

Deduce expressions for the excess pressure inside a liquid drop

Accepted Answer

.Excess pressure inside a liquid drop

consider a spherical liquid drop of radius $R$ . Let $σ$ with the surface tension of the liquid. Due to its spherical shape, there is an excess pressure p inside the drop over that on outside. This excess pressure acts normally outwards.

Let the radius of the drop increase from $R$ to $R + d R$ under the excess pressure $p$ .

Initial surface area $= 4 {πR}^{2}$ ...(I)

Final surface area $= 4 π {(R + dR)}^{2}$

Final surface area $= 4 π (R^{2} + 2 R d R + d R^{2})$

$= 4 π R^{2} + 8 π R d R$ ...(II)

[ $d R^{2} =$ very small so neglected]

Increase in surface area $(△ A) = = 4 {πR}^{2} + 8 π R d R - 4 {πR}^{2}$

$= 8 πRdR$ ....(III)

Work done in enlarging the drop $=$ Increase in surface energy ...(IV)

[Increase in surface energy $= △ A \times σ$ ]

$σ =$ Surface Tension

But work done $= Force \times displacement$

$[Force = Pressure \times area]$

Work done $= Pressure \times area \times displacement$

Work done $= p \times 4 {πR}^{2} \times dR$

From equation (III) and (IV)

$P \times 4 π R^{2} d R = 8 π R d R σ$

$p = \frac{2 σ}{R}$ Excess pressure inside water drop.

Deduce expressions for the excess pressure inside a liquid drop

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