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(A) Let the initial pressure of the air inside the soap bubble be $P_{in,1}$.The excess pressure inside a soap bubble is given by $\Delta P = \frac{4\

Vedclass · Accepted Answer

$8p_1 + \frac{24\sigma}{r}$

Vedclass · Answer

(A) Let the initial pressure of the air inside the soap bubble be $P_{in,1}$.The excess pressure inside a soap bubble is given by $\Delta P = \frac{4\sigma}{r}$.Thus, $P_{in,1} - p_1 = \frac{4\sigma}{r} \implies P_{in,1} = p_1 + \frac{4\sigma}{r} \quad ...(i)$Since the temperature $T$ is constant, Boyle's Law $(PV = 	ext{constant})$ applies to the air inside the bubble.When the radius becomes $r' = r/2$, the volume $V$ becomes $V' = \frac{4}{3}\pi(r/2)^3 = \frac{1}{8}V$.Therefore, the new pressure inside the bubble $P_{in,2} = 8P_{in,1} = 8(p_1 + \frac{4\sigma}{r}) = 8p_1 + \frac{32\sigma}{r}$.The new pressure outside the bubble is $p_2$. The excess pressure equation for the new radius $r/2$ is:$P_{in,2} - p_2 = \frac{4\sigma}{r/2} = \frac{8\sigma}{r}$.Substituting $P_{in,2}$:$(8p_1 + \frac{32\sigma}{r}) - p_2 = \frac{8\sigma}{r}$.$p_2 = 8p_1 + \frac{32\sigma}{r} - \frac{8\sigma}{r} = 8p_1 + \frac{24\sigma}{r}$.

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