Question

Two soaps bubbles (surface tension T) coalesce to form a big bubble under isothermal condition. If in the process the change in volume be V and the surface area be S, then the correct relation is (P is the atmospheric pressure):
(A) $PV + TS = 0$
(B) $3PV + 4TS = 0$
(C) $3PV + TS = 0$
(D) $4PV + 3TS = 0$

Answer 1

Hint : As the two bubbles coalesce together to form a big bubble the formation will follow Boyle's law, i.e. the sum of the product of pressure and volume of the soaps bubbles will be equal to the product of pressure and volume of the bubble formed.
Mathematically Boyle’s law is written as,
${P_1}{V_1} = {P_2}{V_2}$
Where, $m_1$ and $m_2$ are pole strengths and r is the distance in between them.

Complete Step By Step Answer:
Let the pressure inside the two bubbles which coalesce together be Pa and Pb and their radius be a and b respectively.
Then the pressure inside the bubble 1 will be the sum of the atmospheric pressure and excess pressure and will be given by,
${P_a} = P + \dfrac{{4T}}{a}$
Where, P is the atmospheric pressure and T is the surface tension
Similarly the pressure inside the bubble 2 will be given by,
${P_b} = P + \dfrac{{4T}}{b}$
The Volume of the bubbles will be given by,
${V_a} = \dfrac{4}{3}\pi {a^3}$ And ${V_b} = \dfrac{4}{3}\pi {b^3}$
Let the radius of the bubble formed by the two above bubbles be c, then the Volume and pressure inside it will be given by,
${V_c} = \dfrac{4}{3}\pi {c^3}$ and ${P_c} = P + \dfrac{{4T}}{c}$
Now, according to the Boyle’s law,
${P_1}{V_1} = {P_2}{V_2}$
${P_a}{V_a} + {P_b}{V_b} = {P_c}{V_c}$
$(P + \dfrac{{4T}}{a})(\dfrac{4}{3}\pi {a^3}) + (P + \dfrac{{4T}}{b})(\dfrac{4}{3}\pi {b^3}) = (P + \dfrac{{4T}}{c})(\dfrac{4}{3}\pi {c^3})$
Rearranging the above equation will give us,
$(P + \dfrac{{4T}}{c})(\dfrac{4}{3}\pi {c^3}) - [(P + \dfrac{{4T}}{a})(\dfrac{4}{3}\pi {a^3}) + (P + \dfrac{{4T}}{b})(\dfrac{4}{3}\pi {b^3})] = 0$
$P(\dfrac{4}{3}\pi {c^3} - \dfrac{4}{3}\pi {a^3} - \dfrac{4}{3}\pi {b^3}) + \dfrac{{4T}}{3}(4\pi {c^3} - 4\pi {a^3} - 4\pi {b^3}) = 0$
As we are given the change in the volume be V and change in the surface area be S, so putting these in above equations, we will get,
$PV + \dfrac{{4T}}{3}S = 0$
Solving this we will get,
$3PV + 4TS = 0$
Hence, B is the correct option.

Note :
According to Boyle's law the absolute pressure exerted by an ideal gas is inversely proportional to the volume it occupies if the temperature remains constant and here we were given the isothermal condition hence we used Boyle's law.