Question

Two soap bubbles of radii \( r_1 \) and \( r_2 \) in vacuum coalesce isothermally to form a new bubble. The radius of the new soap bubble is

• A. \( \sqrt{r_1^2 + r_2^2} \)
• B. \( \sqrt{2 r_1 r_2} \)
• C. \( \sqrt{\frac{r_1 r_2}{2}} \)
• D. \( \frac{r_1 + r_2}{2} \)

Hint:

For soap bubbles coalescing isothermally, conserve the number of moles of air using the ideal gas law. The radius of the new bubble depends on the sum of the squares of the original radii.

Answer 1

The Correct Option is A

Since the process is isothermal and occurs in a vacuum, the pressure inside the soap bubbles is due to surface tension, and we can use the ideal gas law for the air inside the bubbles. The pressure inside a soap bubble is given by: \[ P = \frac{4\sigma}{r} \] where \( \sigma \) is the surface tension, and \( r \) is the radius. However, in a vacuum, the external pressure is zero, so the pressure inside is entirely due to surface tension. The number of moles of air inside each bubble can be found using the ideal gas law \( PV = nRT \). The volume of a bubble is \( V = \frac{4}{3} \pi r^3 \), so for the two bubbles: - Bubble 1: \( V_1 = \frac{4}{3} \pi r_1^3 \), pressure \( P_1 = \frac{4\sigma}{r_1} \), - Bubble 2: \( V_2 = \frac{4}{3} \pi r_2^3 \), pressure \( P_2 = \frac{4\sigma}{r_2} \). The number of moles in each bubble: \[ n_1 = \frac{P_1 V_1}{RT}, \quad n_2 = \frac{P_2 V_2}{RT} \] Substitute the pressures and volumes: \[ n_1 = \frac{\left( \frac{4\sigma}{r_1} \right) \left( \frac{4}{3} \pi r_1^3 \right)}{RT} = \frac{16\sigma \pi r_1^2}{3RT}, \quad n_2 = \frac{\left( \frac{4\sigma}{r_2} \right) \left( \frac{4}{3} \pi r_2^3 \right)}{RT} = \frac{16\sigma \pi r_2^2}{3RT} \] Total moles in the new bubble: \[ n = n_1 + n_2 = \frac{16\sigma \pi (r_1^2 + r_2^2)}{3RT} \] For the new bubble of radius \( r \), the pressure is \( P = \frac{4\sigma}{r} \), volume \( V = \frac{4}{3} \pi r^3 \), and moles: \[ n = \frac{\left( \frac{4\sigma}{r} \right) \left( \frac{4}{3} \pi r^3 \right)}{RT} = \frac{16\sigma \pi r^2}{3RT} \] Equate the total moles: \[ \frac{16\sigma \pi r^2}{3RT} = \frac{16\sigma \pi (r_1^2 + r_2^2)}{3RT} \] \[ r^2 = r_1^2 + r_2^2 \quad \Rightarrow \quad r = \sqrt{r_1^2 + r_2^2} \] So, the radius of the new soap bubble is \( \sqrt{r_1^2 + r_2^2} \).

Answer 2

Step 1: Understand the problem
- Two soap bubbles with radii \( r_1 \) and \( r_2 \) coalesce isothermally in vacuum.
- Find the radius \( R \) of the new soap bubble formed.

Step 2: Consider the surface energy
- Surface energy of a soap bubble is proportional to its surface area.
- Total surface energy before coalescence = surface energy of bubble 1 + surface energy of bubble 2.
- Since the process is isothermal, total surface energy is conserved.

Step 3: Write the surface area relation
\[ 4 \pi R^2 = 4 \pi r_1^2 + 4 \pi r_2^2 \]
\[ \Rightarrow R^2 = r_1^2 + r_2^2 \]

Final answer:
\[ R = \sqrt{r_1^2 + r_2^2} \]