Question

derive an expression for the excess pressure inside a liquid drop​

Answer 1

Answer:

Here it goes

Explanation:

Consider a spherical liquid drop of radius R. Let S be the surface tension of the liquid due to its spherical shape, there is excess pressure p inside the drop over that on outside. This excess prssure acts normally ouwards. Let the radius of the drop increase from R to R+-dr under the pressure P.

1. Initial Surface Area = 4πR²
2. Final Surface Area = 4π(R±dr)²
3. Expanding the following, we have-> 4πR²+4πdR²+8πRdR

by neglecting DR², we have 4πR²+8πRdR

Increase in surface area =8πRdR

Work done in enlarging the drop= Increase in surface energy= 8πRdR.S

Work done= ρ × 4πRdR² ×dR

So we get P = 2S/R