Andrew Mason said:
I am not sure but here is my take on it: The pressure inside the bubble is a function of the surface tension of the bubble and the temperature. If temperature is constant, the higher the surface tension, the higher the pressure within the bubble. So if the two bubbles combine to form one bubble, the surface area of the combined bubble is smaller, meaning the membrane is thicker so the soap molecules are not stretched as much as before. Therefore, the surface tension is less than before and the pressure is less, so the volume increases.
AM
Andrew, I think that the pressure inside a soap bubble can be written as a function of the Surface Tension and the radius of the bubble.
A soap bubble has some definite (but very small) thickness and therefore has two layers, one towards the outside air and one towards the enclosed air, and between these layers is the soap solution.
So if the pressure of the air outside is P_{atm} and the pressure in the soap solution between these layers is P_1 and the pressure of the enclosed air is P_2
P_1 - P_{atm} = \frac{2S}{R}
where R is the radius of the soap bubble and S is the surface tension of the solution. I'm assuming the thickness of the soap bubble is negligible to the radius. So, again,
P_2- P_1 = \frac{2S}{R}
Adding,
P_2 - P_{atm} = \frac{4S}{R}.
And the pressure inside is greater than the pressure outside by an amount
\frac{4S}{R}
Also, for the OP's question, my assumption would be the mass of the air inside the two smaller soap bubbles will be conserved, and so the volume of the new, larger bubble will be the sum of the volumes of the two smaller bubbles. Since the radii of the smaller bubbles are known, the radius of the larger one can be found, and from this the pressure inside the new bubble can also be found.
What do you reckon?