An ideal gas expands in such a manner that its pressure and volume can be related by equation `PV^(2) = ` constant. During this process, the gas is

Options

A. heated
B. cooled
C. neither heated nor cooled
D. first heated and then cooled

Explanation

To solve the problem, we start with the given relationship between pressure (P) and volume (V) of an ideal gas during its expansion:

1. **Given Equation**:
\[ PV^2 = \text{constant} \]
Let's denote this constant as \( C \).

2. **Express Pressure in Terms of Volume**:
From the equation, we can express pressure as:
\[ P = \frac{C}{V^2} \]

3. **Using Ideal Gas Law**:
We know from the ideal gas law that:
\[ PV = nRT \]
Substituting the expression for \( P \) into the ideal gas law gives:
\[ \frac{C}{V^2} \cdot V = nRT \]
Simplifying this, we find:
\[ \frac{C}{V} = nRT \]

4. **Solving for Temperature (T)**:
Rearranging the equation to solve for temperature \( T \):
\[ T = \frac{C}{nR} \cdot \frac{1}{V} \]

5. **Analyzing the Relationship**:
From the equation \( T = \frac{C}{nR} \cdot \frac{1}{V} \), we can see that:
- \( C \), \( n \), and \( R \) are constants.
- Therefore, \( T \) is inversely proportional to \( V \).

6. **Understanding the Expansion Process**:
Since the gas is expanding, the volume \( V \) is increasing. According to the relationship we derived:
- As \( V \) increases, \( T \) decreases.

7. **Conclusion**:
If the temperature \( T \) is decreasing during the expansion process, this means that the gas is being cooled.

Thus, the correct answer is that during this process, the gas is cooled.