Question

The pressure inside a soap bubble is given by $P_{in} = P_{out} + \frac{4S}{R}$, where $S$ is surface tension and $R$ is radius. Initially, the external pressure is $P_0$, so $P_{in,1} = P_0 + \frac{4S}{R}$. The initial volume is $V_1 = \frac{4}{3}\pi R^3$. Finally, the external pressure is $P_0/2$, and the radius is $R_2 = 5R/4$. The internal pressure is $P_{in,2} = \frac{P_0}{2} + \frac{4S}{5R/4} = \frac{P_0}{2} + \frac{16S}{5R}$. The final volume is $V_2 = \frac{4}{3}\pi (\frac{5R}{4})^3$. Since the air inside the bubble undergoes an isothermal process, $P_{in,1}V_1 = P_{in,2}V_2$. Substituting the expressions and simplifying leads to the relation $P_0 R = 96S$. Therefore, $\frac{RP_0}{12S} = \frac{96S}{12S} = 8$.

Answer 1

8

Answer 2

The pressure inside a soap bubble is given by $P_{in} = P_{out} + \frac{4S}{R}$.

Initially: $P_{in,1} = P_0 + \frac{4S}{R}$ $V_1 = \frac{4}{3}\pi R^3$

Finally: $P_{in,2} = \frac{P_0}{2} + \frac{4S}{5R/4} = \frac{P_0}{2} + \frac{16S}{5R}$ $V_2 = \frac{4}{3}\pi (\frac{5R}{4})^3$

For an isothermal process, $P_{in,1}V_1 = P_{in,2}V_2$. $(P_0 + \frac{4S}{R}) (\frac{4}{3}\pi R^3) = (\frac{P_0}{2} + \frac{16S}{5R}) (\frac{4}{3}\pi (\frac{5R}{4})^3)$

Simplifying this equation leads to $P_0 R = 96S$.

We need to find $\frac{RP_0}{12S}$. Substituting $P_0 R = 96S$: $\frac{RP_0}{12S} = \frac{96S}{12S} = 8$.