A spherical balloon contains air at temperature `T_(0)` and pressure `P_(0)`. The balloon material is such that the instantaneous pressure inside is proportional to the square of the diameter. When the volume of the balloon doubles as a result of heat transfer, the expansion follows the law

Options

A. `PV =` constant
B. `PV^(2//5) =` constant
C. `PV^(-1) =` constant
D. `PV^(-2//3) =` constant

Explanation

To solve the problem, we need to analyze the relationship between the pressure, volume, and diameter of the balloon as it expands. Here's a step-by-step solution:

### Step 1: Understand the relationship between pressure and diameter
We are given that the pressure inside the balloon (P) is proportional to the square of the diameter (d). This can be expressed mathematically as:
\[ P \propto d^2 \]
or
\[ P = k \cdot d^2 \]
where \( k \) is a constant of proportionality.

### Step 2: Relate volume and diameter
For a spherical balloon, the volume (V) is given by the formula:
\[ V = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 = \frac{\pi d^3}{6} \]
Thus, we can express the diameter in terms of volume:
\[ d = \left( \frac{6V}{\pi} \right)^{1/3} \]

### Step 3: Analyze the change in volume
We know that the volume of the balloon doubles as a result of heat transfer:
\[ V_2 = 2V_1 \]
This means:
\[ \frac{V_2}{V_1} = 2 \]

### Step 4: Relate the diameters before and after expansion
Using the relationship between volume and diameter, we can write:
\[ \frac{d_2}{d_1} = \left( \frac{V_2}{V_1} \right)^{1/3} = 2^{1/3} \]

### Step 5: Relate the pressures before and after expansion
Using the relationship between pressure and diameter, we can express the pressures before and after the expansion:
\[ \frac{P_2}{P_1} = \left( \frac{d_2}{d_1} \right)^2 = \left( 2^{1/3} \right)^2 = 2^{2/3} \]

### Step 6: Combine the relationships
From the relationships derived, we can express the pressures and volumes in a combined form:
\[ P_2 \cdot V_2^{-\frac{2}{3}} = P_1 \cdot V_1^{-\frac{2}{3}} \]
This indicates that the product of pressure and volume raised to the power of \(-\frac{2}{3}\) remains constant during the expansion.

### Conclusion
Thus, we conclude that during this expansion process, the relationship follows:
\[ PV^{-\frac{2}{3}} = \text{constant} \]