Two soap bubbles combine under isothermal conditions to form a single soap bubble. If in this process, the change in volume is V and change in area is S, then

Let the volume and pressure of the first, second and final liquid be,

${\mathrm{V}}_{\mathrm{a}}, {\mathrm{V}}_{\mathrm{b}}, {\mathrm{V}}_{\mathrm{c}}, {\mathrm{P}}_{\mathrm{a}}, {\mathrm{P}}_{\mathrm{b}}, {\mathrm{P}}_{\mathrm{c}}$

We know that,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{a}}={\mathrm{P}}_0-\frac{4\mathrm{T}}{{\mathrm{r}}_{\mathrm{a}}} \\ {\mathrm{P}}_{\mathrm{b}}={\mathrm{P}}_0-\frac{4\mathrm{T}}{{\mathrm{r}}_{\mathrm{b}}} \\ {\mathrm{P}}_{\mathrm{c}}={\mathrm{P}}_0-\frac{4\mathrm{T}}{{\mathrm{r}}_{\mathrm{c}}} \end{gathered}$$

The process is isothermal,

${\mathrm{P}}_1{\mathrm{V}}_1+{\mathrm{P}}_2{\mathrm{V}}_2=\mathrm{PV}$

$\left(\mathrm{P}-\frac{4\mathrm{T}}{{\mathrm{r}}_{\mathrm{a}}}\right)\frac{4}{3}{{\mathrm{\pi r}}_{\mathrm{a}}}^3+\left(\mathrm{P}-\frac{4\mathrm{T}}{{\mathrm{r}}_{\mathrm{b}}}\right)\frac{4}{3}{{\mathrm{\pi r}}_{\mathrm{b}}}^3=\left(\mathrm{P}-\frac{4\mathrm{T}}{{\mathrm{r}}_{\mathrm{c}}}\right)\frac{4}{3}{{\mathrm{\pi r}}_{\mathrm{c}}}^3$

$\mathrm{P}[{{\mathrm{r}}_{\mathrm{c}}}^3-({{\mathrm{r}}_{\mathrm{a}}}^3+{{\mathrm{r}}_{\mathrm{b}}}^3\left)\right]-4\mathrm{T}[{{\mathrm{r}}_{\mathrm{a}}}^2+{{\mathrm{r}}_{\mathrm{b}}}^2-{{\mathrm{r}}_{\mathrm{c}}}^2]=0$

Change in volume V

$\mathrm{V}=\frac{4}{3}\mathrm{\pi }{[{\mathrm{r}}_{\mathrm{c}}}^3-({{\mathrm{r}}_{\mathrm{a}}}^3+{{\mathrm{r}}_{\mathrm{b}}}^3)]$

surface area

$S=4\mathrm{\pi }\left[\left[{{\mathrm{r}}_{\mathrm{c}}}^3-({{\mathrm{r}}_{\mathrm{a}}}^3+{{\mathrm{r}}_{\mathrm{b}}}^3)\right]\right]$

$\therefore \frac{3\mathrm{PV}}{4\mathrm{\pi }}+\frac{\mathrm{ST}}{\mathrm{\pi }}=0\Rightarrow 3\mathrm{PV}+4\mathrm{ST}=0$

Hence the correct answer is $3\mathrm{PV}+4\mathrm{ST}=0.$