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Derive an expression for the excess pressure inside a liquid drop |
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Answer» Here it goesExplanation: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.Initial Surface Area = 4πR²Final Surface Area = 4π(R±dr)² EXPANDING the following, we have-> 4πR²+4πdR²+8πRdRby neglecting DR², we have 4πR²+8πRdRIncrease in surface area =8πRdRWork done in enlarging the drop= Increase in surface energy= 8πRdR.SWork done= ρ × 4πRdR² ×dRSo we get P = 2S/R |
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