Question

Two soap bubbles of radii $a$ and $b$ combine to form a single bubble of radius $c.$ If $P$ is the external pressure, then the surface tension of the soap solution is

• A. $\frac{P(c^3+a^3+b^3)}{4(a^2+b^2-c^2)}$
• B. $\frac{P(c^3-a^3-b^3)}{4(a^2+b^2-c^2)}$
• C. $\frac{P(c^3-4a^3-4b^3)}{4(a^2+b^2-c^2)}$
• D. $\frac{P(c^3-2a^3-2b^3)}{4(a^2+b^2-c^2)}$

Answer 1

The correct option is B $\frac{P(c^3-a^3-b^3)}{4(a^2+b^2-c^2)}$

Assuming isothermal conditions,

$\left(P+\frac{4\sigma }{a}\right)\left(\frac{4}{3}\pi a^3\right)+\left(P+\frac{4\sigma }{b}\right)\left(\frac{4}{3}\pi b^3\right)$

$=(P+\frac{4\sigma }{c})\left(\frac{4}{3}\pi c^3\right)$
or $P[a^3+b^3-c^3]=4\sigma [c^2-a^2-b^2]$

$\sigma =\frac{P(c^3-a^3-b^3)}{4(a^2+b^2-c^2)}$