Question

two spehrical soap bubble r1 and r2 in vaccum colaces isothermally find the radius of resetting bubble

Answer 1

that was not in the option so thank u for trying. this was a question in BE entrance examination

Answer 2

Tricky. Your first step is to realize that the pressure inside the bubbles is not the same. From thermodynamics it is possible to derive the general formula:
\[p_i = p_o + \frac{2 \gamma}{r}\]
where p_i is the pressure inside a bubble of radius r, gamma is the surface tension of the liquid that makes up the bubble, and p_o is the pressure outside, here zero because the bubbles are a in vacuum.

Now a key fact we'll need to calculate the size of the coalesced bubble is the amount of gas in the bubble. Let's assume the ideal gas equation of state, so we can calculate the moles n from the pressure inside each bubble p. Then we have:
\[n_1 = \frac{p_1 V_1}{R T} = \frac{2 \gamma}{r_1} \frac{\frac{4}{3}\pi r_{1}^{3}}{RT}\]
and similarly for n_2. Clearly the moles in the combined bubble n = n_1 + n_2, so:
\[n = \frac{2 \gamma}{R T}\frac{4}{3} \pi \left(r_{1}^{2} + r_{2}^{2} \right)\]
Now we can use the capillarity formula (top) to calculate the radius of the combined bubble from its pressure, and we'll get the pressure from the ideal gas e.o.s.:
\[p = \frac{2 \gamma}{r} = \frac{n R T}{V} = \frac{n R T}{\frac{4}{3} \pi r^3}\]
Solving for r gives:
\[r^2 = \frac{R T}{2 \gamma \frac{4}{3} \pi} n\]
Plugging in our results for n above:
\[r^2 = \frac{R T}{2 \gamma \frac{4}{3} \pi} \frac{2 \gamma}{R T} \frac{4}{3} \pi \left( r_{1}^2 + r_{2}^2 \right)\]
All the constants cancel, conveniently, leaving us with:
\[r^2 = r_{1}^{2} + r_{2}^{2}\]
A very nice problem!