Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S change in total surface area, T is the surface tension and P atmospheric pressure, then which of the following relation is correct?

To solve this problem, we need to find the correct relation between the change in volume (V), change in surface area (S), surface tension (T), and atmospheric pressure (P) when two soap bubbles coalesce into one. We start by recalling the physics of soap bubbles.

A soap bubble has an excess pressure inside due to surface tension. Since a soap bubble has two surfaces (inner and outer), the excess pressure is given by $$\Delta P = \frac{4T}{r}$$, where $$r$$ is the radius. Therefore, the pressure inside a bubble is $$P_{\text{inside}} = P + \frac{4T}{r}$$, with $$P$$ being atmospheric pressure.

Let the two initial bubbles have radii $$r_1$$ and $$r_2$$. After coalescing, they form a single bubble of radius $$r$$. The air inside is conserved, and assuming constant temperature, we use the ideal gas law. The number of moles of air is conserved.

For the first bubble, the number of moles is:

$$n_1 = \frac{\left(P + \frac{4T}{r_1}\right) \cdot \frac{4}{3}\pi r_1^3}{RT}$$

For the second bubble:

$$n_2 = \frac{\left(P + \frac{4T}{r_2}\right) \cdot \frac{4}{3}\pi r_2^3}{RT}$$

The total initial moles are:

$$n_1 + n_2 = \frac{\frac{4}{3}\pi}{RT} \left[ \left(P + \frac{4T}{r_1}\right) r_1^3 + \left(P + \frac{4T}{r_2}\right) r_2^3 \right]$$

Simplifying the terms inside the brackets:

$$\left(P + \frac{4T}{r_1}\right) r_1^3 = P r_1^3 + 4T r_1^2$$

$$\left(P + \frac{4T}{r_2}\right) r_2^3 = P r_2^3 + 4T r_2^2$$

So:

$$n_1 + n_2 = \frac{\frac{4}{3}\pi}{RT} \left[ P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 \right]$$

For the final single bubble:

$$n = \frac{\left(P + \frac{4T}{r}\right) \cdot \frac{4}{3}\pi r^3}{RT} = \frac{\frac{4}{3}\pi}{RT} \left[ P r^3 + 4T r^2 \right]$$

Since moles are conserved, $$n_1 + n_2 = n$$:

$$P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 = P r^3 + 4T r^2$$

Rearranging:

$$P r_1^3 + P r_2^3 - P r^3 + 4T r_1^2 + 4T r_2^2 - 4T r^2 = 0$$

$$P (r_1^3 + r_2^3 - r^3) + 4T (r_1^2 + r_2^2 - r^2) = 0 \quad ...(1)$$

Now, define the changes. The change in volume of contained air, $$V$$, is the final volume minus the initial total volume. The volume of a bubble is $$\frac{4}{3}\pi r^3$$, so:

$$V = \frac{4}{3}\pi r^3 - \left( \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 \right) = \frac{4}{3}\pi (r^3 - r_1^3 - r_2^3)$$

Solving for $$r^3 - r_1^3 - r_2^3$$:

$$r^3 - r_1^3 - r_2^3 = \frac{3V}{4\pi} \quad ...(2)$$

The change in total surface area, $$S$$, is defined as the final surface area minus the initial total surface area. For a soap bubble, the surface area is often considered as the area of one surface (e.g., the outer surface), which is $$4\pi r^2$$. Thus:

Initial total surface area = $$4\pi r_1^2 + 4\pi r_2^2 = 4\pi (r_1^2 + r_2^2)$$

Final surface area = $$4\pi r^2$$

So:

$$S = 4\pi r^2 - 4\pi (r_1^2 + r_2^2) = 4\pi (r^2 - r_1^2 - r_2^2)$$

Solving for $$r^2 - r_1^2 - r_2^2$$:

$$r^2 - r_1^2 - r_2^2 = \frac{S}{4\pi} \quad ...(3)$$

Note that equation (1) contains $$r_1^3 + r_2^3 - r^3$$ and $$r_1^2 + r_2^2 - r^2$$. Using equations (2) and (3):

$$r_1^3 + r_2^3 - r^3 = - (r^3 - r_1^3 - r_2^3) = - \frac{3V}{4\pi}$$

$$r_1^2 + r_2^2 - r^2 = - (r^2 - r_1^2 - r_2^2) = - \frac{S}{4\pi}$$

Substitute these into equation (1):

$$P \left( - \frac{3V}{4\pi} \right) + 4T \left( - \frac{S}{4\pi} \right) = 0$$

$$- \frac{3PV}{4\pi} - \frac{4T S}{4\pi} = 0$$

Simplify the second term:

$$- \frac{3PV}{4\pi} - \frac{T S}{\pi} = 0$$

Multiply both sides by $$-1$$:

$$\frac{3PV}{4\pi} + \frac{T S}{\pi} = 0$$

To eliminate denominators, multiply both sides by $$4\pi$$:

$$4\pi \cdot \frac{3PV}{4\pi} + 4\pi \cdot \frac{T S}{\pi} = 0$$

$$3PV + 4 T S = 0$$

Since $$T S = ST$$, we write:

$$3PV + 4ST = 0$$

This matches option B.

Hence, the correct answer is Option B.