Two soap bubbles combine to form a single bubble. In this process, the change in volume and surface area are…
Two soap bubbles combine to form a single bubble. In this process, the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure, and $T$ is the surface tension of the soap solution, the following relation is true :
$4 P V+3 T A=0$
$3 P V-4 T A=0$
$4 P V-3 T A=0$
$3 P V+4 T A=0$
Solution
Let radii of two soap bubbles are $a$ and $b$ respectively and radius of single larger bubble is $c$.
As excess pressure for a soap bubble is $\frac{4 T}{r}$ and external pressure $P$.
$P_i=P+\frac{4 T}{r}$
So, $\quad P_a=P+\frac{4 T}{a}, P_b=P+\frac{4 T}{b}$
and $\quad P_c=P+\frac{4 T}{c}$ ...(i)
and $\quad V_a=\frac{4}{3} \pi a^3, \quad V_b=\frac{4}{3} \pi b^3$
and $\quad V_c=\frac{4}{3} \pi c^3$ ...(ii)
Now as mass is conserved.
$\mu_a+\mu_b=\mu_c$
i.e., $\quad \frac{P_a V_a}{R T_a}+\frac{P_b V_b}{R T_b}=\frac{P_c V_c}{R T_c} \quad($ as $P V=\mu R T)$
As temperature is constant, i.e., $T_a=T_b=T_c$
So, $\quad P_a V_a+P_b V_b=P_c V_c$
which in the light of Eqs. (i) and (ii) becomes,
$\left(P+\frac{4 T}{a}\right)\left(\frac{4}{3} \pi a^3\right)+\left(P+\frac{4 T}{b}\right)\left(\frac{4}{3} \pi b^3\right)$ $=\left(P+\frac{4 T}{c}\right)\left(\frac{4}{3} \pi c^3\right)$
i.e., $4 T\left(a^2+b^2-c^2\right)=P\left(c^3-a^3-b^3\right)$ ...(iii)
Now, $\quad V=\frac{4}{3} \pi\left(a^3+b^3-c^3\right)$
and $\quad A=4 \pi\left(a^2+b^2-c^2\right)$
$\therefore \quad \frac{T A}{\pi}=-\frac{3}{4 \pi} V P$
or $\quad 4 T A+3 P V=0$