Question

Under isothermal conditions, two soap bubbles of radii \(r_1\) and \(r_2\) combine to form a single soap bubble of radius \(R\). The surface tension of soap solution is (\(P = \text{outside pressure}\))

• A. \(\frac{P(R^3 + r_1^3 + r_2^3)}{4(r_1^2 - r_2^2 + R^2)}\)
• B. \(\frac{P^2 + r_1^2 + r_2^2}{4(r_1^2 + r_2^2 + R^2)}\)
• C. \(\frac{P(R^3 - r_1^3 - r_2^3)}{4(r_1^2 + r_2^2 - R^2)}\)
• D. \(\frac{P(R^2 - r_1^2 - r_2^2)}{4(r_1^3 + r_2^3 - R^3)}\)

Hint:

For soap bubble, excess pressure = \(\frac{4T}{r}\) (two surfaces). For isothermal combination, use \(P_1V_1 + P_2V_2 = P_3V_3\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Two soap bubbles combine isothermally to form one bubble. We need surface tension \(T\) in terms of outside pressure \(P\) and radii.

Step 2: Key Formula or Approach:
For a soap bubble, excess pressure inside = \(\frac{4T}{r}\). Pressure inside = \(P + \frac{4T}{r}\). For isothermal process, \(P_1V_1 + P_2V_2 = P_3V_3\) (since number of moles is conserved: \(n_1+n_2=n_3\) and \(PV = nRT\), so \(P_1V_1 + P_2V_2 = P_3V_3\)).

Step 3: Detailed Explanation:
Volume of bubble = \(\frac{4}{3}\pi r^3\).
Inside pressure for bubble of radius \(r\): \(P_{\text{in}} = P + \frac{4T}{r}\).
Using isothermal condition: \( (P + \frac{4T}{r_1}) \cdot \frac{4}{3}\pi r_1^3 + (P + \frac{4T}{r_2}) \cdot \frac{4}{3}\pi r_2^3 = (P + \frac{4T}{R}) \cdot \frac{4}{3}\pi R^3\).
Cancel \(\frac{4}{3}\pi\): \[ (P r_1^3 + 4T r_1^2) + (P r_2^3 + 4T r_2^2) = P R^3 + 4T R^2. \] \[ P(r_1^3 + r_2^3) + 4T(r_1^2 + r_2^2) = P R^3 + 4T R^2. \] \[ 4T(r_1^2 + r_2^2 - R^2) = P(R^3 - r_1^3 - r_2^3). \] \[ T = \frac{P(R^3 - r_1^3 - r_2^3)}{4(r_1^2 + r_2^2 - R^2)}. \]

Step 4: Final Answer:
Option (C) is correct.