A spherical soap bubble has internal pressure | vedclass

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Question

(C) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{r}$.Initially, the pressure difference is $P_0 - P_1 = \frac{4S}{r_0}$.

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$3$

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(C) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{r}$.Initially, the pressure difference is $P_0 - P_1 = \frac{4S}{r_0}$.Given $P_1 = \frac{8P_0}{9}$, we have $P_0 - \frac{8P_0}{9} = \frac{4S}{r_0}$, which simplifies to $\frac{P_0}{9} = \frac{4S}{r_0}$, so $P_0 = \frac{36S}{r_0}$.The initial internal pressure is $P_{in, initial} = P_0 = \frac{36S}{r_0}$.Since the temperature is constant, the number of moles of air inside the bubble remains constant. Using the ideal gas law $PV = nRT$, we have $P_{in, initial} V_{initial} = P_{in, final} V_{final}$.Finally, the enclosure is evacuated, so $P_{outside} = 0$. The final internal pressure is $P_{in, final} = \frac{4S}{r}$, where $r$ is the final radius.Substituting the values: $(\frac{36S}{r_0})(\frac{4}{3}\pi r_0^3) = (\frac{4S}{r})(\frac{4}{3}\pi r^3)$.This simplifies to $36S r_0^2 = 4S r^2$, which gives $9r_0^2 = r^2$.Therefore, $r = 3r_0$, and the ratio $\frac{r}{r_0} = 3$.

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