Question

Two soap bubbles coalesce to form a single bubble. If $V$ is the subsequent change in volume of contained air and $S$ the change in total surface area, $T$ is the surface tension and $P$ atmospheric pressure, which of the following relation is correct?

• A. $4PV+3ST=0$
• B. $3PV+4ST=0$
• C. $2PV+3ST=0$
• D. $3PV+2ST=0$

Answer 1

Answer: (B)

$3PV+4ST=0$

Let moles of air in firs t bubble of radius $R_1$ be $n_1$ and that in second bubble of radius $R_2$ be $n_2$.
From ideal gas equation, number of moles of gas $n=\frac{PV}{Rt}$
where $t$ is the temperature of gas.
Since, number of moles of gas is conserved.
So, $n_1+n_2=n^{\prime }$
where $n_3$ is the number of moles of gas in the final single bubble of radius $R^{\prime }$.
Or $\frac{P_1{V}_1}{Rt}+\frac{P_2{V}_2}{Rt}=\frac{P^{\prime }{V}^{\prime }}{Rt}$
We get $P_1{V}_1+P_2{V}_2=P^{\prime }{V}^{\prime }$
Pressure inside the soap bubble of radius $R^{\prime }$ is given by $P^{\prime }=P+\frac{4T}{R^{\prime }}$
Volume of the bubble $V^{\prime }=\frac{4\pi }{3}{R}^{\prime 3}$
$\therefore$ $(P+\frac{4T}{R_1})\left(\frac{4\pi }{3}{R}_1^3\right)$ $+(P+\frac{4T}{R_2})\left(\frac{4\pi }{3}{R}_2^3\right)$ $=(P+\frac{4T}{R^{\prime }})\left(\frac{4\pi }{3}{R}^{\prime 3}\right)$
Or $P(\frac{4\pi }{3}{R}_1^3+\frac{4\pi }{3}{R}_2^3-\frac{4\pi }{3}{R}^{\prime 3})+\frac{4T}{3}(4\pi R_1^2+4\pi R_2^2-4\pi R^{\prime 2})=0$
Or $PV+\frac{4T}{3}S=0$
$⟹3PV+4ST=0$