Question

Two spherical bubbles coalesce isothermally. $V$ is the consequent change in volume and $S$ is the total change in surface area, then

Let $P$ be atmospheric pressure outside the bubbles and $T$ be surface tension.

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

Answer 1

The correct option is A $3PV+4ST=0$

Let A, B coalesce to form C.
Radius of $A,B,C$ be $r_1,r_2,r$ respectively.
Pressure inside bubbles $A,B,C$ be $P_1,P_2,P_3$ respectively.
Their volumes be $V_1,V_2,V_3$
For isothermal situation,
$P_1{V}_1+P_2{V}_2=P_3{V}_3$
$(P+\frac{4T}{r_1})\left(\frac{4}{3}\pi r_1^3\right)+(P+\frac{4T}{r_2})\left(\frac{4}{3}\pi r_2^3\right)$
$=(P+\frac{4T}{r})\left(\frac{4}{3}\pi r^3\right)$
$4T(r_1^2+r_2^2-r^2)=-P_0(r_1^3+r_2^3-r^3)$
Given that $S$ is change in surface area and $V$ is change in volume
Hence,
$S=4\pi (r^2-r_1^2-r_2^2)$
$V=\frac{4}{3}\pi (r^3-r_1^3-r_2^3)$
$4T\frac{-S}{4\pi }-\frac{3}{4\pi }PV=0$
$4TS+3PV=0$