$\text{A spherical soap bubble inside an air chamber at pressure}{\mathrm{ P}}_0={10}^5\text{Pa has a certain radius so that the}$

$\text{excess pressure inside the bubble is}\Delta P=144\text{Pa. }\text{Now, the chamber pressure is reduced to }\frac{8}{P_0}\text{so that}$

$\text{the bubble radius and its excess pressure change.}$ $\text{In this process, all the temperatures remain unchanged.}$

$\text{Assume air to be an ideal gas and the excess pressure}\Delta P\text{in both the cases to be much smaller than}$

$\text{the chamber pressure.}$ $\text{The new excess pressure}\mathrm{ \Delta }P\text{ in Pa is \_\_\_\_\_}$

The correct answer is 96 Pa.

Step 1: Understand the initial and final states of the bubble.
Let the initial state of the soap bubble and chamber be defined by:
- Initial chamber pressure, $P_0={10}^5\text{ Pa}$
- Initial excess pressure, $\Delta P_1=144\text{ Pa}$
- Initial radius of the bubble, $R_1$
- Surface tension of the soap bubble solution, $S$

For a spherical soap bubble with two surfaces, the excess pressure is given by:

$\Delta P=\frac{4S}{R}$

Thus, the initial excess pressure is:

$\Delta P_1=\frac{4S}{R_1}=144\text{ Pa}$

The total pressure inside the bubble initially is:

$P_{\text{in},1}=P_0+\Delta P_1=P_0+\frac{4S}{R_1}$

Step 2: Define the final state parameters.
When the chamber pressure is reduced to $P_0^{\text{'}}=\frac{8}{27}$, let the new radius be $R_2$ and the new excess pressure be $\Delta P_2$.
The new excess pressure is:

$\Delta P_2=\frac{4S}{R_2}$

The total pressure inside the bubble in the final state is:

$P_{\text{in},2}=P_0^{\text{'}}+\Delta P_2=\frac{8}{27}$

Step 3: Apply the Isothermal Ideal Gas Law.
Since the temperature remains constant throughout the process, the number of moles of air trapped inside the bubble is conserved. By Boyle's Law ($PV=\text{constant}$):

$P_{\text{in},1}{V}_1=P_{\text{in},2}{V}_2$

Substituting the volume of a sphere $V=\frac{4}{3}\pi R^3$:

$\left(P_0+\frac{4S}{R_1}\right)·\frac{4}{3}\pi R_1^3=\left(\frac{8}{27}{P}_0+\frac{4S}{R_2}\right)·\frac{4}{3}\pi R_2^3$

Step 4: Use the given approximation.
We are given that the excess pressure in both cases is much smaller than the chamber pressure ($\Delta P\ll P_0$).
Therefore, we can approximate:

$P_0+\frac{4S}{R_1}\approx P_0$

and

$\frac{8}{27}{P}_0+\frac{4S}{R_2}\approx \frac{8}{27}{P}_0$

Substituting these approximations back into the Boyle's law equation:

$P_0·R_1^3=\frac{8}{27}{P}_0·R_2^3$

Canceling $P_0$ from both sides:

$R_1^3=\frac{8}{27}{R}_2^3$

Taking the cube root on both sides:

$R_1=\frac{2}{3}{R}_2\Rightarrow R_2=\frac{3}{2}{R}_1$

Step 5: Calculate the new excess pressure.
Using the relation between excess pressure and radius:

$\Delta P_2=\frac{4S}{R_2}=\frac{4S}{\frac{3}{2}{R}_1}=\frac{2}{3}\left(\frac{4S}{R_1}\right)=\frac{2}{3}\Delta P_1$

Substitute $\Delta P_1=144\text{ Pa}$:

$\Delta P_2=\frac{2}{3}\times 144=96\text{ Pa}$

Thus, the new excess pressure inside the bubble is 96 Pa.