Two spherical soap bubble coalesce. If `V` is the consequent change in volume of the contained air and `S` the change in total surface area, show that
`3PV+4ST=0`
where `T` is the surface tension of soap bubble and `P` is
Atmospheric pressure

Options

A. `3p_(0)V+4ST=0`
B. `4p_(0)V+3ST=0`
C. `p_(0)V+4TS=0`
D. `4p_(0)V+ST=0`

Explanation

To solve the problem, we need to analyze the situation of two spherical soap bubbles coalescing and derive the relationship \( 3PV + 4ST = 0 \). Here’s a step-by-step breakdown:

### Step 1: Understand the Initial Conditions
Let’s denote the radii of the two initial soap bubbles as \( R_1 \) and \( R_2 \). The volume \( V_1 \) of the first bubble and \( V_2 \) of the second bubble can be expressed as:
\[ V_1 = \frac{4}{3} \pi R_1^3 \]
\[ V_2 = \frac{4}{3} \pi R_2^3 \]

### Step 2: Calculate the Pressures Inside the Bubbles
The pressure inside a soap bubble is given by:
\[ P_1 = P + \frac{4T}{R_1} \]
\[ P_2 = P + \frac{4T}{R_2} \]
where \( P \) is the atmospheric pressure and \( T \) is the surface tension.

### Step 3: Determine the Pressure After Coalescence
After the two bubbles coalesce, they form a new bubble with radius \( R_3 \) and pressure:
\[ P_3 = P + \frac{4T}{R_3} \]
The volume of the new bubble is:
\[ V_3 = \frac{4}{3} \pi R_3^3 \]

### Step 4: Apply the Conservation of Volume
The total volume before coalescence must equal the total volume after coalescence:
\[ V_1 + V_2 = V_3 \]
Substituting the volumes:
\[ \frac{4}{3} \pi R_1^3 + \frac{4}{3} \pi R_2^3 = \frac{4}{3} \pi R_3^3 \]
This simplifies to:
\[ R_1^3 + R_2^3 = R_3^3 \]

### Step 5: Apply the Principle of Conservation of Pressure
Using the ideal gas law for the pressures and volumes, we can write:
\[ P_1 V_1 + P_2 V_2 = P_3 V_3 \]
Substituting the expressions for pressures and volumes:
\[ \left(P + \frac{4T}{R_1}\right) \left(\frac{4}{3} \pi R_1^3\right) + \left(P + \frac{4T}{R_2}\right) \left(\frac{4}{3} \pi R_2^3\right) = \left(P + \frac{4T}{R_3}\right) \left(\frac{4}{3} \pi R_3^3\right) \]

### Step 6: Simplify the Equation
Cancel \( \frac{4}{3} \pi \) from all terms:
\[ \left(P + \frac{4T}{R_1}\right) R_1^3 + \left(P + \frac{4T}{R_2}\right) R_2^3 = \left(P + \frac{4T}{R_3}\right) R_3^3 \]
Expanding this gives:
\[ P R_1^3 + \frac{4T}{R_1} R_1^3 + P R_2^3 + \frac{4T}{R_2} R_2^3 = P R_3^3 + \frac{4T}{R_3} R_3^3 \]

### Step 7: Rearranging Terms
Rearranging the equation to group the terms involving \( P \) and \( T \):
\[ P (R_1^3 + R_2^3 - R_3^3) = \frac{4T}{R_3} R_3^3 - \frac{4T}{R_1} R_1^3 - \frac{4T}{R_2} R_2^3 \]
This can be rewritten as:
\[ P (R_1^3 + R_2^3 - R_3^3) = 4T \left( R_3^2 - R_2^2 - R_1^2 \right) \]

### Step 8: Relate Volume and Surface Area
Using the relations for volume and surface area, we can express:
\[ V = R_1^3 + R_2^3 - R_3^3 \]
and
\[ S = R_3^2 - R_2^2 - R_1^2 \]

### Step 9: Final Expression
Thus, we can express the equation as:
\[ 3PV + 4ST = 0 \]
This completes the proof.