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 $\left(P_{0}\right.$ is atmospheric pressure)

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

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.
$\therefore P_{a}=P_{0}+\frac{4 T}{a}$,
$V_{a}=\frac{4}{3} \pi a^{3} $
$P_{b}=P_{0}+\frac{4 T}{b}$,
$V_{b}=\frac{4}{3} \pi b^{3} $
$ P_{c}=P_{0}+\frac{4 T}{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{4 T}{a}\right)\left(\frac{4}{3} \pi a^{3}\right)+\left(P_{0}+\frac{4 T}{b}\right) \left(\frac{4}{3} \pi b^{3}\right) $
$=\left(P_{0}+\frac{4 T}{c}\right)\left(\frac{4}{3} \pi c^{3}\right)$
or $4 T\left(a^{2}+b^{2}-c^{2}\right)=P_{0}\left(c^{3}-a^{3}-b^{3}\right)$
or $ \frac{4}{3} T 4 \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{4 T}{3} S=-P_{0} V$ or $3 P_{0} V+4 S T=0$