Question

Two spherical soap bubbles coalesce. If V is the consequent change in volume of the contained air and S is the change in the total surface area and T is the surface tension of the soap solution, then $(P_0$ is atmospheric pressure)

• A. $3P_{0}V+4ST=0$
• B. $4P_{0}V+3ST=0$
• C. $P_{0}V+4TS=0$
• D. $4P_{0}V+ST=0$

Answer 1

Answer: (A)

Let $a$ and $b$ be the radii of two soap bubbles before coalescing and $c$ be the radius of bigger bubble.
$\therefore P_a=P_0+\frac{4T}{a}$,
$V_a=\frac{4}{3}\pi a^3$
$P_b=P_0+\frac{4T}{b}$,
$V_b=\frac{4}{3}\pi b^3$
$P_c=P_0+\frac{4T}{c}$,
$V_c=\frac{4}{3}\pi c^3$
where $P_0$ is the atmospheric pressure.
Now, as mass is conserved,
$\therefore n_a+n_b=n_c$ or $\frac{P_a{V}_a}{R{T}_a}+\frac{P_b{V}_b}{R{T}_b}=\frac{P_c{V}_c}{R{T}_c}$
At constant temperature,
$P_a{V}_a+P_b{V}_b=P_c{V}_c$
$\therefore \left(P_0+\frac{4T}{a}\right)\left(\frac{4}{3}\pi a^3\right)+\left(P_0+\frac{4T}{b}\right)\left(\frac{4}{3}\pi b^3\right)$
$=\left(P_0+\frac{4T}{c}\right)\left(\frac{4}{3}\pi c^3\right)$
or $4T\left(a^2+b^2-c^2\right)=P_0\left(c^3-a^3-b^3\right)$
or $\frac{4}{3}T4\pi \left(a^2+b^2-c^2\right)=P_0\frac{4}{3}\pi \left(c^3-a^3-b^3\right)$
or $\frac{4T}{3}S=-P_{0}V$ or $3P_{0}V+4ST=0$