Two soap bubbles coalesce, what is the surface tension?

[Quantum Mind]

Homework Statement

Two soap bubbles of equal radius 'r' coalesce isothermally to form a bigger bubble of radius 'R'. If the atmospheric pressure is P₀, find the surface tension of the soap solution.

Homework Equations

Surface energy = surface tension x increase in area (W = T.ΔA)
Excess pressure inside bubble = 4T/R i.e. T = p.R/4

The Attempt at a Solution

Don't know where to start. From the above equation, I thought I can proceed, but I neither know the surface energy nor the surface tension. I also do not know why the term 'p' has been given or how to relate it to the problem.

Probably I should take the area increase as the difference between the total areas of the smaller bubbles and the area of the larger bubble.

Area of smaller bubbles = 2.4∏r²
Area of bigger bubble = 4∏R²

Difference in area = 4∏(R² - 2r²)

W = T.4∏(R² - 2r²)

= p.R.4∏(R² - 2r²)/4

How do I proceed further?

 

[Quantum Mind]

Any hints ?

 

[Quantum Mind]

The answer is p₀(R³ - 2r³)/4(2r² - R²).

From this I gather that the volume of the bubbles has something to do with this problem and that the volume of the bigger bubble will be more than the combined volume of the two smaller ones. The area of the bigger bubble will however, be smaller than the areas of the two smaller bubbles.

How did this answer come about?

 

[Quantum Mind]

This is what I found on the Net, is there another way to do it without invoking the Young-Laplace equation?

Assume that the air inside the bubble is an ideal gas.

Because the temperature is constant, ideal gas law states that:
P∙V/n = R∙T = constant
(Pi pressure inside the bubble)

So state of air inside the single big bubble (1) and each of the two smaller bubbles can be compared as:
Pi₁∙V₁/n₁ = Pi₂∙V₂/n₂
<=>
Pi₁∙V₁ = (n₁/n₂)∙Pi₂∙V₂

The bigger bubble is formed from two smaller bubbles, so contains twice as much air, i.e.
(n₁/n₂) = 2
=>
Pi₁∙V₁ = 2∙Pi₂∙V₂

Assuming spherical shape the volumes for the bubbles are:
V₁ = (4/3)∙π∙R³
V₂ = (4/3)∙π∙r³
=>
Pi₁∙(4/3)∙π∙R³ = 2∙Pi₂∙(4/3)∙π∙r³
<=>
Pi₁∙R³ = 2∙Pi₂∙r³

The pressure inside the bubble can be found from Young-Laplace equation. From this equation you can derive the pressure difference between the inside and outside of a spherical bubble of radius r as:
ΔP = 2∙σ/R
But this is the relation for a simple bubble formed in a static fluid, e.g. an gas bubble in water. A soap bubble consist of two interlaced bubbles: there is soap fluid bubble in the atmospheric air and its filled with an air bubble. The formula above will only give you the pressure difference between the surrounding air and the soap film or between the soap film and the air inside the soap bubble. Assuming a thin film, i.e. outside radius and inside radius differ negligibly, you can add the two pressure differences, and get for the pressure difference between the air inside the soap bubble and the surrounding air:
ΔP = 4∙σ/R
=>
ΔP₁ = Pi₁ - P = 4∙σ/R
ΔP₂ = Pi₂ - P = 4∙σ/r
<=>
Pi₁ = P + (4∙σ/R)
Pi₂ = P + (4∙σ/r)

Hence,
(P + (4∙σ/R))∙R³ = 2∙(P + (4∙σ/r))∙r³
<=>
P∙R³ + 4∙σ∙R² = 2∙P∙r³ + 4∙σ∙2∙r²
<=>
P∙(R³ - 2∙r³) = 4∙σ∙(2∙r² - R²)
=>
σ = P∙(R³ - 2∙r³) / (4∙(2∙r² - R²))