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 $\frac{4 T}{r}$ and external pressure $P$.
$P_{i}=P+\frac{4 T}{r}$
So, $P_{a}=P+\frac{4 T}{a}, P_{b}=P+\frac{4 T}{b}$
and $ P_{c}=P+\frac{4 T}{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 $P V=\mu R T)$
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,
$\left(P+\frac{4 T}{a}\right)\left(\frac{4}{3} \pi a^{3}\right)+\left(P+\frac{4 T}{b}\right)\left(\frac{4}{3} \pi b^{3}\right)$
$=\left(P+\frac{4 T}{c}\right)\left(\frac{4}{3} \pi c^{3}\right)$
i.e., $4 T\left(a^{2}+b^{2}-c^{2}\right)=P\left(c^{3}-a^{3}-b^{3}\right) \,\,\,\ldots$ (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{T A}{\pi}=-\frac{3}{4 \pi} V P $
or $ T A+3 P V=0$