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PhyHardAdv 2024 • Paper 2

New Excess Pressure in Soap Bubble After Isothermal Chamber Pressure Reduction

A spherical soap bubble inside an air chamber at pressure P0=105 PaP_0 = 10^5\text{ Pa} has a certain radius so that the excess pressure inside the bubble is ΔP=144 Pa\Delta P = 144\text{ Pa}. Now, the chamber pressure is reduced to 8P0/278P_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 ΔP\Delta P in both the cases to be much smaller than the chamber pressure. The new excess pressure ΔP\Delta P in Pa is ________.

Official Numerical Answer96

Topics & Concepts

Mechanical Properties of FluidsSurface TensionKinetic Theory of Gases

Step-by-Step Solution

To find the new excess pressure inside the soap bubble, we analyze the isothermal expansion of the trapped air inside it.

1. Initial State:

  • Chamber pressure: P1=P0=105 PaP_1 = P_0 = 10^5 \text{ Pa}
  • Initial excess pressure: ΔP1=144 Pa\Delta P_1 = 144 \text{ Pa}
  • Initial radius of the bubble: R1R_1

The excess pressure for a soap bubble with surface tension TT is given by: ΔP1=4TR1\Delta P_1 = \frac{4T}{R_1}

The absolute pressure inside the bubble initially is: Pin,1=P1+ΔP1=P0+ΔP1P_{\text{in}, 1} = P_1 + \Delta P_1 = P_0 + \Delta P_1

Given that the excess pressure is much smaller than the chamber pressure (ΔP1≪P0\Delta P_1 \ll P_0): Pin,1≈P0P_{\text{in}, 1} \approx P_0

2. Final State:

  • Chamber pressure: P2=827P0P_2 = \frac{8}{27} P_0
  • New excess pressure: ΔP2\Delta P_2
  • New radius of the bubble: R2R_2

The new excess pressure is: ΔP2=4TR2\Delta P_2 = \frac{4T}{R_2}

The absolute pressure inside the bubble finally is: Pin,2=P2+ΔP2=827P0+ΔP2P_{\text{in}, 2} = P_2 + \Delta P_2 = \frac{8}{27} P_0 + \Delta P_2

Since ΔP2≪P2\Delta P_2 \ll P_2, we approximate: Pin,2≈827P0P_{\text{in}, 2} \approx \frac{8}{27} P_0

3. Isothermal Process for the Trapped Air:

Since the temperature remains constant, the trapped air inside the bubble undergoes an isothermal expansion. Applying Boyle's Law (PinV=constantP_{\text{in}} V = \text{constant}): Pin,1V1=Pin,2V2P_{\text{in}, 1} V_1 = P_{\text{in}, 2} V_2

Substituting V=43πR3V = \frac{4}{3} \pi R^3 and the approximated pressures: P0(43πR13)=(827P0)(43πR23)P_0 \left( \frac{4}{3} \pi R_1^3 \right) = \left( \frac{8}{27} P_0 \right) \left( \frac{4}{3} \pi R_2^3 \right)

Simplifying the equation gives: R13=827R23R_1^3 = \frac{8}{27} R_2^3

R23R13=278\frac{R_2^3}{R_1^3} = \frac{27}{8}

Taking the cube root on both sides: R2R1=32\frac{R_2}{R_1} = \frac{3}{2}

4. Calculating New Excess Pressure:

Using the relation between excess pressure and radius: ΔP2ΔP1=R1R2=23\frac{\Delta P_2}{\Delta P_1} = \frac{R_1}{R_2} = \frac{2}{3}

Therefore, the new excess pressure is: ΔP2=23ΔP1=23×144 Pa=96 Pa\Delta P_2 = \frac{2}{3} \Delta P_1 = \frac{2}{3} \times 144 \text{ Pa} = 96 \text{ Pa}

Final Answer: The new excess pressure ΔP\Delta P is 96.

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New Excess Pressure in Soap Bubble After Isothermal Chamber Pressure Reduction | Physics PYQ Solution - JEE Challenger