The process is isothermal. For an isothermal process, \( P_1 V_1 = P_2 V_2 \).
\[ P_1 = P_0, \quad V_1 = \frac{4}{3} \pi r_1^3 \]
\[ P_2 = \frac{8P_0}{27}, \quad V_2 = \frac{4}{3} \pi r_2^3 \]
\[ P_1 V_1 = P_2 V_2 \quad \Rightarrow \quad P_0 \cdot r_1^3 = \frac{8P_0}{27} \cdot r_2^3 \]
Simplifying:
The excess pressure inside a bubble is:
\[ \Delta P \propto \frac{1}{r} \]
Substitute \( r_2 = 2r_1 \) into the equation:
\[ r_2 = 2r_1^3 \] \[ \Delta P = 4 \cdot T \cdot \frac{1}{r} \]
The ratio of the pressures is:
\[ \frac{\Delta P_2}{\Delta P_1} = \frac{r_1}{r_2} \] Substituting \( r_2 = 3r_1 \): \[ \frac{\Delta P_2}{144} = \frac{3}{2} \quad \Rightarrow \quad \Delta P_2 = \frac{2}{3} \cdot 144 = 96 \, \text{Pa} \]
The excess pressure inside the bubble is \( \Delta P_2 = 96 \, \text{Pa} \).