Question

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

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

Hint:

Apply Boyle law to the total gas, using the excess pressure 4S/r inside a soap bubble.

Answer 1

The Correct Option is A

Step 1: Gas in each bubble
Gas pressure $P_i = P+\frac{4S}{r_i}$ and volume $V_i = \frac43\pi r_i^3$.

Step 2: Isothermal mixing
$\sum P_iV_i$ is conserved since the temperature is fixed and the gas amounts add.

Step 3: Collect S terms
Terms in $P$ give $P(r_1^3+r_2^3-R^3)$ and terms in $S$ give $4S(r_1^2+r_2^2-R^2)$. Setting the sum to zero gives option (A).

Final Answer:
Option A. \[ \boxed{\text{(A)}\ S=\frac{P(R^3-r_1^3-r_2^3)}{4(r_1^2+r_2^2-R^2)}} \]