Two spherical soap bubbles of radii a and b in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by

Options

A. ` (( a + b )) / ( 2 )`
B. ` ( ab ) /( a + b ) `
C. ` sqrt ( a ^ 2 + b ^ 2 ) `
D. ` a + b `

Explanation

To solve the problem of two spherical soap bubbles of radii \( a \) and \( b \) coalescing under isothermal conditions, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Excess Pressure in Soap Bubbles:**
- The excess pressure inside a soap bubble is given by the formula:
\[ P = \frac{4T}{R} \]
- For the first bubble with radius \( a \):
\[ P_1 = \frac{4T}{a} \]
- For the second bubble with radius \( b \):
\[ P_2 = \frac{4T}{b} \]

2. **Calculate the Volume of Each Bubble:**
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- For the first bubble:
\[ V_1 = \frac{4}{3} \pi a^3 \]
- For the second bubble:
\[ V_2 = \frac{4}{3} \pi b^3 \]

3. **Consider the Resulting Bubble:**
- Let the radius of the resulting bubble be \( r \).
- The excess pressure in the resulting bubble is:
\[ P_3 = \frac{4T}{r} \]
- The volume of the resulting bubble is:
\[ V_3 = \frac{4}{3} \pi r^3 \]

4. **Apply the Isothermal Condition:**
- Under isothermal conditions, the number of molecules (or moles) before and after coalescence remains constant. Therefore:
\[ n_1 + n_2 = n_3 \]
- Using the ideal gas law \( PV = nRT \), we can express the number of moles:
\[ n_1 = \frac{P_1 V_1}{RT}, \quad n_2 = \frac{P_2 V_2}{RT}, \quad n_3 = \frac{P_3 V_3}{RT} \]
- Substituting these into the equation gives:
\[ \frac{P_1 V_1}{RT} + \frac{P_2 V_2}{RT} = \frac{P_3 V_3}{RT} \]
- The \( RT \) cancels out, leading to:
\[ P_1 V_1 + P_2 V_2 = P_3 V_3 \]

5. **Substitute the Pressures and Volumes:**
- Substitute \( P_1, P_2, P_3, V_1, V_2, \) and \( V_3 \):
\[ \left(\frac{4T}{a}\right) \left(\frac{4}{3} \pi a^3\right) + \left(\frac{4T}{b}\right) \left(\frac{4}{3} \pi b^3\right) = \left(\frac{4T}{r}\right) \left(\frac{4}{3} \pi r^3\right) \]
- Simplifying this gives:
\[ \frac{16\pi T a^2}{3} + \frac{16\pi T b^2}{3} = \frac{16\pi T r^2}{3} \]

6. **Cancel Common Terms:**
- Cancel \( \frac{16\pi T}{3} \) from both sides:
\[ a^2 + b^2 = r^2 \]

7. **Final Result:**
- Taking the square root of both sides gives:
\[ r = \sqrt{a^2 + b^2} \]

### Conclusion:
The radius of the resulting bubble after coalescence is:
\[ r = \sqrt{a^2 + b^2} \]