Question

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 :

• A. 4 PV + 3 TA = 0
• B. 3 PV - 4 TA = 0
• C. 4 PV - 3 TA = 0
• D. 3 PV + 4 TA = 0

Answer 1

Answer: (D)

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 $\frac{4T}{r}$ and external pressure $P$.
$P_i=P+\frac{4T}{r}$
So, $P_a=P+\frac{4T}{a},P_b=P+\frac{4T}{b}$
and $P_c=P+\frac{4T}{c}...(i)$
and $V_a=\frac{4}{3}\pi a^3,$
$V_b=\frac{4}{3}\pi b^3$
and $V_c=\frac{4}{3}\pi c^3...(ii)$
Now as mass is conserved.
${\mu }_a+{\mu }_b={\mu }_c$
i.e., $\frac{P_a{V}_a}{R{T}_a}+\frac{P_b{V}_b}{R{T}_b}=\frac{P_c{V}_c}{R{T}_c}($ as $PV=\mu RT)$
As temperature is constant,
i.e., $T_a=T_b=T_c$
So, $P_a{V}_a+P_b{V}_b=P_c{V}_c$
which in the light of Eqs. (i) and (ii) becomes,
$\left(P+\frac{4T}{a}\right)\left(\frac{4}{3}\pi a^3\right)+\left(P+\frac{4T}{b}\right)\left(\frac{4}{3}\pi b^3\right)$
$=\left(P+\frac{4T}{c}\right)\left(\frac{4}{3}\pi c^3\right)$
i.e., $4T\left(a^2+b^2-c^2\right)=P\left(c^3-a^3-b^3\right)\dots$ (iii)
Now, $V=\frac{4}{3}\pi \left(a^3+b^3-c^3\right)$
and $A=4\pi \left(a^2+b^2-c^2\right)$
$\therefore \frac{TA}{\pi }=-\frac{3}{4\pi }VP$
or $TA+3PV=0$