There is a soap bubble of radius $2.4 \times 10^{- 4} m$ inside a cylinder filled with air which is originally at the pressure of $P_{0} = 10^{5} N m^{- 2}$ The air in the cylinder is now compressed isothermally until the radius of the bubble is halved. Now the pressure of air in the cylinder is given by $8 P_{0} + α \times 10^{3} p a$ . The surface tension of the soap film is $0.08 N m^{- 1}$ Find the value of $α$

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Detailed Solution

Question Image $P_{1} = P_{0} + \frac{4 S}{r}$ $V_{1} = \frac{4}{3} π r^{3}$
$P_{1} V_{2} = P_{2} V_{2}, (P_{0} + \frac{4 s}{r}) \times \frac{4}{3} π r^{3} = (P_{0}^{'} + \frac{8 s}{r}) \times \frac{4}{3} π \frac{r^{3}}{8}$
$P_{0} + \frac{4 s}{r} = \frac{P_{0}^{'}}{8} + \frac{s}{r} \Rightarrow 8 P_{0} + \frac{24 s}{r} = 8 \times 10^{5} + 24 \times \frac{0.08}{2.4 \times 10^{- 4}} = 0.08 \times 10_{5} P a$