JEE Advanced 2024 Paper 2 · Q13 · Surface Tension

Question

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 _____.

SolveKar AI · Accepted Answer

Excess pressure inside soap bubble: $\Delta P = 4T/r$, so $r = 4T/\Delta P$. Initial radius $r_1 = 4T/144$ m. Inside-bubble pressure $P_1 = P_0 + 144 \approx P_0$ (since 144 ≪ $10^5$).

Moles of trapped air constant. Isothermal ideal gas: $P_1 V_1 = P_2 V_2$, with $V \propto r^3$.

Approx (since $\Delta P \ll$ chamber pressure): $P_1 \approx P_0$, $P_2 \approx 8P_0/27$. So

$\dfrac{r_2^3}{r_1^3} = \dfrac{P_1}{P_2} = \dfrac{P_0}{8P_0/27} = \dfrac{27}{8}$ ⇒ $\dfrac{r_2}{r_1} = \dfrac{3}{2}$.

New excess pressure: $\Delta P_2 = 4T/r_2 = (4T/r_1)\cdot(r_1/r_2) = 144 \cdot \dfrac{2}{3} = 96$ Pa.

Units: $T$ in N/m; $r$ in m; $\Delta P$ in Pa = N/m$^2$. Consistent. Answer: $96$.

JEE Advanced 2024 Paper 2 · Q13 · Surface Tension

Solution