Question

Two soaps bubbles (surface tension $T$) coalesce to form a big bubble under isothermal condition. If in this process the change in volume be $V$ and the surface area be $S$, then the correct relation is ($P$ is atmospheric pressure):

• A. $PV+TS=0$
• B. $3PV+4TS=0$
• C. $3PV+TS=0$
• D. $4PV+3TS=0$

Answer 1

Answer: (B)

$3PV+4TS=0$

Since the condition is isothermal, we have
$P{V}_f=P_1{V}_1+P_2{V}_2$, where $P$ and $V_f$ are pressure and volume of combined bubble.

$(P+\frac{4T}{R})\left(\frac{4}{3}\pi R^3\right)=(P+\frac{4T}{R_1})\left(\frac{4}{3}\pi R_1^3\right)+(P+\frac{4T}{R_2})\left(\frac{4}{3}\pi R_2^3\right)$

$\Rightarrow P[\frac{4}{3}\pi R^3-\frac{4}{3}\pi R_1^3-\frac{4}{3}\pi R_2^3]=\frac{4T}{3}(4\pi {R_1}^2+4\pi {R_2}^2-4\pi R^2)$

$\Rightarrow PV=-\frac{4T}{3}S$ $(\because$ Change in volume $=V$ and Change in surface area $=S$)
$\Rightarrow 3PV+4TS=0$