A spherical soap bubble inside an air chamber | vedclass

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(C) For a spherical soap bubble,the excess pressure is given by $\Delta P = \frac{4T}{R}$.In the first case,the total pressure inside the bubble is $P

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(C) For a spherical soap bubble,the excess pressure is given by $\Delta P = \frac{4T}{R}$.In the first case,the total pressure inside the bubble is $P = P_0 + \Delta P = P_0 + \frac{4T}{R}$.Since $\Delta P \ll P_0$,we can approximate the internal pressure as $P \approx P_0$.In the second case,the chamber pressure is $P_0' = \frac{8P_0}{27}$. The new radius is $R_1$ and the new excess pressure is $\Delta P_1 = \frac{4T}{R_1}$.The total pressure inside the bubble is $P_1 = P_0' + \Delta P_1 = \frac{8P_0}{27} + \frac{4T}{R_1} \approx \frac{8P_0}{27}$.Since the temperature remains constant,we apply Boyle's Law: $PV = P_1 V_1$.Substituting the volumes $V = \frac{4}{3}\pi R^3$ and $V_1 = \frac{4}{3}\pi R_1^3$:$P_0 \cdot R^3 = P_0' \cdot R_1^3$$P_0 \cdot R^3 = \left(\frac{8P_0}{27}ight) R_1^3$$R^3 = \frac{8}{27} R_1^3 \implies R = \frac{2}{3} R_1 \implies R_1 = \frac{3}{2} R$.Now,the new excess pressure is $\Delta P_1 = \frac{4T}{R_1} = \frac{4T}{(3/2)R} = \frac{2}{3} \left(\frac{4T}{R}ight)$.Given $\Delta P = \frac{4T}{R} = 144 \ Pa$,we get $\Delta P_1 = \frac{2}{3} 	imes 144 = 96 \ Pa$.

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