JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Fluid Mechanics

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 ____.

Answer: 96

Solution

For a soap bubble, the excess pressure is

Δ P = \frac{4 T}{R}

, where

T

is surface tension and

R

is radius.

Case 1:

P = P_{0} + Δ P = P_{0} + \frac{4 T}{R}

Since

Δ P ≪ P_{0}

, we approximate

P \approx P_{0}

.

Case 2:
Chamber pressure becomes

P_{0}^{'} = \frac{8 P _{0}}{27}

, radius becomes

R_{1}

P_{1} = \frac{8 P _{0}}{27} + \frac{4 T}{R _{1}}

For isothermal process with ideal gas:

P V = P_{1} V_{1}

P_{0} \cdot \frac{4}{3} π R^{3} = \frac{8 P _{0}}{27} \cdot \frac{4}{3} π R_{1}^{3}

R^{3} = \frac{8 R _{1}^{3}}{27} \Rightarrow R_{1} = \frac{3 R}{2}

The new excess pressure:

Δ P_{1} = \frac{4 T}{R _{1}} = \frac{4 T}{3 R /2} = \frac{2}{3} \cdot \frac{4 T}{R} = \frac{2}{3} \times 144 = 96 Pa

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer

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JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Fluid Mechanics

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