A spherical soap bubble inside an air chamber at pressure $$P_0 = 10^5$$ Pa has a certain radius so that the excess pressure inside the bubble is $$\Delta P = 144$$ 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 Pa is ________.

Let the outside (chamber) pressure initially be $$P_0 = 10^{5}\,\text{Pa}$$.
Inside the soap bubble the pressure is higher by the excess pressure $$\Delta P$$ produced by surface tension, so

$$P_{\text{in,1}} = P_0 + \Delta P.$$

For a soap bubble (two surfaces) the excess pressure is related to the surface tension $$T$$ and the bubble radius $$R$$ by

$$\Delta P = \dfrac{4T}{R} \qquad -(1)$$

The chamber pressure is now reduced to $$P_2 = \dfrac{8P_0}{27}$$ while temperature is kept constant. Because no air escapes from the bubble, the amount of gas inside remains the same, so the ideal-gas relation at constant temperature applies:

$$P_{\text{in,1}}V_1 = P_{\text{in,2}}V_2,$$

where $$V = \dfrac{4}{3}\pi R^{3}$$ is the bubble volume. Using $$P_{\text{in,1}} = P_0 + \Delta P$$ and $$P_{\text{in,2}} = P_2 + \Delta P_2$$, we get

$$(P_0 + \Delta P)R^{3} = (P_2 + \Delta P_2)R_2^{3}. \qquad -(2)$$

At the new equilibrium the excess pressure is

$$\Delta P_2 = \dfrac{4T}{R_2}. \qquad -(3)$$

Divide $$(1)$$ by $$(3)$$ to relate the two radii:

$$\dfrac{\Delta P}{\Delta P_2} = \dfrac{R_2}{R} \;\;\Longrightarrow\;\; R_2 = R\dfrac{\Delta P}{\Delta P_2}. \qquad -(4)$$

The statement “$$\Delta P$$ is much smaller than the chamber pressure” lets us neglect the terms $$\Delta P$$ and $$\Delta P_2$$ compared with $$P_0$$ and $$P_2$$. Thus $$P_{\text{in,1}}\approx P_0$$ and $$P_{\text{in,2}}\approx P_2$$, and equation $$(2)$$ simplifies to

$$P_0 R^{3} = P_2 R_2^{3}. \qquad -(5)$$

Substituting $$R_2$$ from $$(4)$$ into $$(5)$$:

$$P_0 R^{3} = P_2\!\left(R\dfrac{\Delta P}{\Delta P_2}\right)^{3} \;\;\Longrightarrow\;\; \dfrac{\Delta P}{\Delta P_2} = \left(\dfrac{P_2}{P_0}\right)^{1/3}. \qquad -(6)$$

Rearranging $$(6)$$ gives

$$\Delta P_2 = \Delta P\left(\dfrac{P_2}{P_0}\right)^{1/3}. \qquad -(7)$$

Insert the numerical values: $$\Delta P = 144\,\text{Pa}$$ and $$P_2/P_0 = \dfrac{8}{27}.$$ Because $$\left(\dfrac{8}{27}\right)^{1/3} = \dfrac{2}{3},$$

$$\Delta P_2 = 144 \times \dfrac{2}{3} = 96\,\text{Pa}.$$

Therefore, the new excess pressure in the bubble is 96 Pa.