Question

Two soap bubbles combine under isothermal conditions to form a single soap bubble. If in the process, the change in volume is $V$ and change in area is $S$, then

($P$ is atmospheric pressure and $T$ is the ambient temperature).

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

Answer 1

The correct option is C $3PV+4TS=0$

Let $V_a{,V}_b,V_c,P_a,P_b,P_c$ as the volume and pressure of first, second and final liquid.

$P_a=P_0-\frac{4T}{r_a}{P}_b=P_0-\frac{4T}{r_b}{P}_c=P_0-\frac{4T}{r_c}$

As process is isothermal

$P_1{V}_1+P_2{V}_2=PV$

$(P-\frac{4T}{r_a})\frac{4}{3}\pi r_a^3+(P-\frac{4T}{r_b})\frac{4}{3}\pi r_b^3=(P-\frac{4T}{r_c})\frac{4}{3}\pi r_c^{3}P[r_c^3-(r_a^3+r_b^3)]-4T[r_a^2+r_b^2+r_c^2]=0$

Change in volume$=V$

$V=\frac{4}{3}\pi [r_c^3-(r_a^3+r_b^3)]$

Change in surface area

$S=4\pi [r_c^2+(r_a^2+r_b^2)]\frac{3PV}{4\pi }+\frac{ST}{\pi }=03PV+4ST=0$