A spherical soap bubble inside an air chamber at pressure $P_0 = 10^{5}\ \text{Pa}$ has a certain radius so that the excess pressure inside the bubble is $\Delta P = 144\ \text{Pa}$. Now, the chamber pressure is reduced to $8P_0/27$ 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 $\text{Pa}$ is ______.

Step 1: Identify the given parameters.

The initial chamber pressure is $P_0 = 10^{5}\text{ Pa}$, the initial excess pressure is $\Delta P_1 = 144\text{ Pa}$, and the final chamber pressure is $P_0' = \frac{8P_0}{27}$.

Step 2: Relate the excess pressure to the bubble radius.

For a spherical soap bubble with two surfaces, the excess pressure is determined by the surface tension $\sigma$ and radius $r$.

$$\Delta P = \frac{4\sigma}{r}$$

Step 3: Apply the ideal gas law at constant temperature.

Since the excess pressure is much smaller than the chamber pressure, the pressure inside the bubble is approximately equal to the chamber pressure.

$$P_1 V_1 = P_2 V_2$$ $$P_0 V_1 = \frac{8P_0}{27} V_2$$

Step 4: Determine the ratio of the radii.

The volume of a spherical bubble is proportional to the cube of its radius.

$$\frac{V_2}{V_1} = \frac{27}{8}$$ $$\frac{r_2}{r_1} = \left(\frac{27}{8}\right)^{1/3} = \frac{3}{2}$$

Step 5: Calculate the new excess pressure.

The excess pressure is inversely proportional to the radius of the bubble.

$$\frac{\Delta P_2}{\Delta P_1} = \frac{r_1}{r_2} = \frac{2}{3}$$ $$\Delta P_2 = 144 \times \frac{2}{3} = 96\text{ Pa}$$

Answer: 96