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 Δ𝑃 = 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 Δ𝑃 in both the cases to be much smaller than the chamber pressure. The new excess pressure Δ𝑃 in Pa is ______.

The correct answer is 96.

The excess pressure inside a spherical soap bubble of radius $R$ and surface tension $T$ is given by the formula:
$\mathrm{\Delta P}=\frac{4T}{R}$

Initially, the chamber pressure is $P_0={10}^5\text{Pa}$ and the initial excess pressure is ${\mathrm{\Delta P}}_1=144\text{Pa}$.
The total pressure inside the bubble is the sum of the chamber pressure and the excess pressure:
$P_1=P_0+{\mathrm{\Delta P}}_1$

Since we are given that the excess pressure in both cases is much smaller than the chamber pressure ($\mathrm{\Delta P}\ll P_{\text{chamber}}$), we can approximate the total pressure inside the bubble as the chamber pressure:
$P_1\approx P_0$

Similarly, in the final state, the chamber pressure is reduced to $P_0^{\prime }=\frac{8P_0}{27}$. The final total pressure inside the bubble is:
$P_2=P_0^{\prime }+{\mathrm{\Delta P}}_2\approx P_0^{\prime }=\frac{8P_0}{27}$

Assuming air to be an ideal gas and that the temperature remains constant during the process, we apply Boyle's Law ($PV=\text{constant}$) to the air trapped inside the bubble:
$P_1{V}_1=P_2{V}_2$

Since the volume of a sphere of radius $R$ is $V=\frac{4}{3}\pi R^3$, this relation simplifies to:
$P_1{R}_1^3=P_2{R}_2^3$

Substituting the approximated pressures:
$P_0{R}_1^3\approx \left(\frac{8P_0}{27}\right)R_2^3$
Dividing both sides by $P_0$:
$R_1^3\approx \frac{8}{27}{R}_2^3$

Taking the cube root of both sides gives the relation between the initial and final radii of the bubble:
$R_1\approx \frac{2}{3}{R}_2\Rightarrow R_2\approx \frac{3}{2}$

Now, we can express the new excess pressure ${\mathrm{\Delta P}}_2$ in terms of the initial excess pressure ${\mathrm{\Delta P}}_1$:
${\mathrm{\Delta P}}_2=\frac{4T}{R_2}=\frac{4T}{\frac{3}{2}}=\frac{2}{3}\left(\frac{4T}{R_1}\right)=\frac{2}{3}{\mathrm{\Delta P}}_1$

Given that ${\mathrm{\Delta P}}_1=144\text{Pa}$, we calculate:
${\mathrm{\Delta P}}_2=\frac{2}{3}\times 144=96\text{Pa}$