A spherical soap bubble inside an air chamber at pressure P0 = 105 Pa has a certain radius so that the excess pressure inside the bubble is P = 144 Pa. Now, the chamber pressure is reduced to 8P0/27 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 P in both the cases to be much smaller than the chamber pressure. The new excess pressure P in Pa is .
Answer
96
Solution
Since P P, pressure inside the bubble chamber pressure.
Isothermal (Boyle): P0r13 = 8P027r23 r2 = 32r1
Excess pressure P = 4Tr 1r: P2 = 144 23 = 96 Pa
Answer: 96