Question

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 \_\_\_\_.

Hint:

For isothermal processes, use the ideal gas law to relate pressure and volume, and consider surface tension for excess pressure in bubbles.

Answer 1

The process is isothermal. For an isothermal process, \(P_1 V_1 = P_2 V_2\). Initial pressure and volume: \[ P_1 = P_0, \quad V_1 = \frac{4}{3} \pi r_1^3. \] Final pressure and volume: \[ P_2 = \frac{8P_0}{27}, \quad V_2 = \frac{4}{3} \pi r_2^3. \] From isothermal conditions: \[ P_1 V_1 = P_2 V_2 \quad \Rightarrow \quad P_0 \cdot r_1^3 = \frac{8P_0}{27} \cdot r_2^3. \] Simplify: \[ r_2 = \frac{2r_1}{3}. \] The excess pressure inside a bubble is: \[ \Delta P = \frac{4T}{r}. \] Using \(\Delta P \propto \frac{1}{r}\): \[ \frac{\Delta P_2}{\Delta P_1} = \frac{r_1}{r_2}. \] Substitute \(r_2 = \frac{2r_1}{3}\): \[ \frac{\Delta P_2}{144} = \frac{3}{2} \quad \Rightarrow \quad \Delta P_2 = \frac{2}{3} \cdot 144 = 96 \, \text{Pa}. \]