'Two spherical soap bubbles of radii r₁ and r₂ coalesce to form a bubble of radius R. If the surface tension of soap solution is T, atmospheric pressure is P, the change in volume of the air enclosed is V and the change in surface area is 'A'. Complete the expression using above statement, 3PV + ___ = 0

CONCEPT

The excess pressure inside a soap bubble of radius R and surface tension T is

\(P = \frac{4T}{R}\)

Also when the two spherical soap bubbles coalesce to form a new soap bubble temperature of the soap bubble remains constant.

By applying the following equation

\(P_1V_1 + P_2V_2 = P_3V_3\)

CALCULATION:
Absolute pressure inside the soap bubble of radius r₁

\(P_{abs1} = P_{atm} +\frac{4T}{r_1}\)

Absolute pressure inside the soap bubble of radius r₂

\(P_{abs2} = P_{atm} + \frac{4T}{r_2}\)

Absolute pressure inside the resulting soap bubble is r₃

\(P_{abs3} = P_{atm} + \frac{4T}{r_3}\)

Now applying the equation

\(P_1V_1 +P_2V_2 = P_3V_3\)

\(⇒ (P_{atm} +\frac{4T}{r_1})\frac{4\pi r_1^3}{3} + (P_{atm} +\frac{4T}{r_2})\frac{4\pi r_2^3}{3} = (P_{atm} + \frac{4T}{r_3})\frac{4\pi r_3^2}{3}\)

\(⇒ P_{atm}(\frac{4\pi r_1^2}{3} +\frac{4\pi r_2^2}{3} -\frac{4\pi r_3^2}{3}) + \frac{4T}{3}(4\pi r_1^2 +4\pi r_2^2 - 4\pi r_3^2) = 0 \)

\(⇒ P_{atm}(V_1 +V_2 -V_3) +\frac{4T}{3}(A_1 + A_2 - A_3) = 0 \)

\(⇒ P_{atm}V +\frac{4TS}{3} =0\)

⇒ 3PV + 4TA = 0 

Hence the completed expression is 

\(3PV + 4TA\)