A spherical soap bubble inside an air chamber at pressure $P_0=10^5 \mathrm{~Pa}$ has a certain radius so…

A spherical soap bubble inside an air chamber at pressure $P_0=10^5 \mathrm{~Pa}$ has a certain radius so that the excess pressure inside the bubble is $\Delta P=144 \mathrm{~Pa}$. Now, the chamber pressure is reduced to $8 P_0 / 27$ so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P$ in $\mathrm{Pa}$ is ________ .

Solution

Case-1 $\begin{aligned} & \mathrm{P}-\mathrm{P}_0=\Delta \mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \\ & \mathrm{P}=\left(\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{R}}\right)\end{aligned}$ Case-2
$\begin{aligned} & \mathrm{P}_1-\frac{8 \mathrm{P}_0}{27}=\Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1} \\ & \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\end{aligned}$ Constant temperature process $\begin{aligned} & \mathrm{PV}=\mathrm{P}_1 \mathrm{~V}_1 \\ & \left(\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4}{3} \pi \mathrm{R}^3=\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\right) \frac{4}{3} \pi \mathrm{R}_1^3 ;\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right),\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}\right) \rightarrow \text { (Neglected) } \\ & \mathrm{R}=\frac{2}{3} \mathrm{R}_1 \Rightarrow \mathrm{R}_1=\frac{3}{2} \mathrm{R} \\ & \Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}=\frac{4 \mathrm{~T}}{3 \mathrm{R}} \times 2=\frac{2}{3} \times(144)=96 \mathrm{~Pa}\end{aligned}$ `

Asked in: JEE Advanced 2024 (Paper 2)

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