JEE Advanced 2024 — Paper 2 — PHY-III solution

JEE Advanced2024Paper 2PHY-III

A spherical soap bubble inside an air chamber at pressure $P_{0} = 1 0^{5}$ Pa has a certain radius so that the excess pressure inside the bubble is $Δ P = 144$ Pa. Now, the chamber pressure is reduced to $8 P_{0} /27$ 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 $Δ P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $Δ P$ in Pa is ______.

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Solution

Let $R_{1}, R_{2}$ be the bubble radii in states 1 and 2, with internal pressures $P_{1} = P_{0} + Δ P_{1}$ and $P_{2} = \frac{8 P _{0}}{27} + Δ P_{2}$ . Surface tension gives $Δ P_{i} = \frac{4 T}{R _{i}}$ .

Since temperature is constant and the gas in the bubble is ideal, $P_{1} V_{1} = P_{2} V_{2}$ , i.e.

$(P_{0} + Δ P_{1}) R_{1}^{3} = (\frac{8 P _{0}}{27} + Δ P_{2}) R_{2}^{3} .$

Because $Δ P_{i} ≪ P_{0}$ , this simplifies to $P_{0} R_{1}^{3} \approx \frac{8 P _{0}}{27} R_{2}^{3}$ , giving $\frac{R _{2}}{R _{1}} = \frac{3}{2} .$

From $Δ P \propto 1/ R$ :

$Δ P_{2} = Δ P_{1} \cdot \frac{R _{1}}{R _{2}} = 144 \cdot \frac{2}{3} = 96 Pa .$

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