JEE Advanced · Physics · 11. Properties of Fluids
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 ________ .
- A 92
- B 94
- C 98
- D 96
Answer & Solution
Correct Answer
(D) 96
Step-by-step Solution
Detailed explanation
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
\(\mathrm{PV}=\mathrm{P}_1 \mathrm{~V}_1 \)
\( \left(\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4}{3} \pi \mathrm{R}^3=\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\right) \frac{4}{3} \pi \mathrm{R}_1^3 ;\) \(\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right),\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}\right) \rightarrow \text { (Neglected) } \)
\( \mathrm{R}=\frac{2}{3} \mathrm{R}_1 \Rightarrow \mathrm{R}_1=\frac{3}{2} \mathrm{R} \)
\( \Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}=\frac{4 \mathrm{~T}}{3 \mathrm{R}} \times 2=\frac{2}{3} \times(144)=96 \mathrm{~Pa}\)

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