Two soap bubble are combined isothermally to form a big bubble of radius R. If `DeltaV` is change in volume, `DeltaS` is change in surface area and `P_(0)` is atmospheric pressure then show that `3P_(0)(DeltaV)+4T(DeltaS)=0`

Options

A. `4pV+3SA=0`
B. `3pV-4SA=0`
C. `4pV-3SA=0`
D. `3pV+4SA=0`

Answer

To solve the problem, we need to analyze the situation of two soap bubbles combining into a larger bubble and derive the relationship between the change in volume (ΔV) and the change in surface area (ΔS) under isothermal conditions.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the radii of the two soap bubbles be \( r_A \) and \( r_B \).
- The pressure inside a soap bubble is given by \( P = P_0 + \frac{4T}{r} \), where \( P_0 \) is the atmospheric pressure and \( T \) is the surface tension.

2. **Calculate the Initial Pressures**:
- For bubble A:
\[ P_A = P_0 + \frac{4T}{r_A} \]
- For bubble B:
\[ P_B = P_0 + \frac{4T}{r_B} \]

3. **Calculate the Pressure of the Combined Bubble**:
- After combining, let the radius of the new bubble be \( R \):
\[ P = P_0 + \frac{4T}{R} \]

4. **Apply the Isothermal Condition**:
- The isothermal condition implies that the total pressure times the total volume before combining equals the total pressure times the total volume after combining:
\[ P_A V_A + P_B V_B = P V \]
- The volumes of the bubbles are given by:
\[ V_A = \frac{4}{3} \pi r_A^3, \quad V_B = \frac{4}{3} \pi r_B^3, \quad V = \frac{4}{3} \pi R^3 \]

5. **Substituting the Volumes**:
- Substitute the expressions for pressure and volume into the equation:
\[ \left(P_0 + \frac{4T}{r_A}\right) \left(\frac{4}{3} \pi r_A^3\right) + \left(P_0 + \frac{4T}{r_B}\right) \left(\frac{4}{3} \pi r_B^3\right) = \left(P_0 + \frac{4T}{R}\right) \left(\frac{4}{3} \pi R^3\right) \]

6. **Simplifying the Equation**:
- Cancel \( \frac{4}{3} \pi \) from both sides:
\[ \left(P_0 + \frac{4T}{r_A}\right) r_A^3 + \left(P_0 + \frac{4T}{r_B}\right) r_B^3 = \left(P_0 + \frac{4T}{R}\right) R^3 \]
- Rearranging gives:
\[ P_0 (r_A^3 + r_B^3 - R^3) + 4T \left(\frac{r_A^3}{r_A} + \frac{r_B^3}{r_B} - \frac{R^3}{R}\right) = 0 \]

7. **Expressing Change in Volume and Surface Area**:
- The change in volume \( \Delta V \) is:
\[ \Delta V = \frac{4}{3} \pi (r_A^3 + r_B^3 - R^3) \]
- The change in surface area \( \Delta S \) is:
\[ \Delta S = 4\pi R^2 - 4\pi r_A^2 - 4\pi r_B^2 \]

8. **Substituting ΔV and ΔS**:
- Substitute \( \Delta V \) and \( \Delta S \) into the rearranged equation:
\[ P_0 \Delta V + 4T \left(-\frac{\Delta S}{4\pi}\right) = 0 \]
- This leads to:
\[ 3P_0 \Delta V + 4T \Delta S = 0 \]

### Final Result:
Thus, we have shown that:
\[ 3P_0 \Delta V + 4T \Delta S = 0 \]