Question
Derive an expression for the excess of pressure inside an air bubble.
$\delta\text{R}$ = Small increase in radius of bubble due to excess pressure
Work done, W = Force × Displacement. = (Excess pressure × Area) × Increase in radius$=\text{P}\times4\pi\text{R}^2\times\delta\text{R}$
Increase in surface area of bubble = Final surface area - Initial surface area$=4\pi(\text{R}+\delta\text{R})^2-4\pi\text{R}^2$
$=8\pi\text{R}(\delta\text{R})(\text{Neglecting }\delta\text{R}^2)$
$\therefore\text{P}\times4\pi\text{R}^2\times\delta\text{R}=8\pi\text{ R}(\delta\text{R})\times\sigma$
Increase in P.E. = increase in surface area × Surface tension$=8\pi\text{R}(\delta\text{R})\times\sigma$
Since the drop is in equilibrium.$\therefore\text{P}\times4\pi\text{R}^2\times\sigma\text{R}=8\pi\text{ R}(\delta\text{R})\times\sigma$
$\text{P}=\frac{2\sigma}{\text{R}}$
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