In the process `pV^2=` constant, if temperature of gas is increased, then

Options

A. (a) change in internal energy of gas is positive
B. (b) work done by gas is positive
C. (c) heat is given to the gas
D. (d) heat is taken out from the gas

Explanation

To solve the problem, we need to analyze the given process where \( pV^2 = \text{constant} \) and determine the implications of an increase in the temperature of the gas.

### Step-by-Step Solution:

1. **Understanding the Process**:
The equation \( pV^2 = \text{constant} \) indicates a specific thermodynamic process. We can rewrite this in terms of the ideal gas law, \( pV = nRT \), where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature.

2. **Relating Pressure, Volume, and Temperature**:
From the equation \( pV^2 = \text{constant} \), we can express pressure \( p \) as:
\[ p = \frac{\text{constant}}{V^2} \]
Substituting this into the ideal gas law gives us:
\[ \frac{\text{constant}}{V^2} \cdot V = nRT \]
Simplifying this, we find:
\[ T = \frac{\text{constant}}{nR} \cdot \frac{1}{V} \]
This shows that temperature \( T \) is inversely proportional to volume \( V \).

3. **Effect of Increasing Temperature**:
If the temperature \( T \) of the gas is increased, according to the relationship derived, the volume \( V \) must decrease in order to maintain the constant value of \( pV^2 \).

4. **Change in Internal Energy**:
The internal energy \( U \) of an ideal gas is directly proportional to its temperature. Therefore, if the temperature increases, the internal energy will also increase, leading to a positive change in internal energy:
\[ \Delta U > 0 \]

5. **Work Done by the Gas**:
Work done \( W \) by the gas is given by:
\[ W = P \Delta V \]
Since the volume \( V \) is decreasing (as established earlier), \( \Delta V < 0 \). Therefore, the work done by the gas is negative:
\[ W < 0 \]

6. **Heat Transfer**:
For a polytropic process, the heat transfer \( Q \) can be related to the change in internal energy and work done:
\[ Q = \Delta U + W \]
Since \( \Delta U > 0 \) and \( W < 0 \), the overall heat transfer \( Q \) will be positive, indicating that heat is being added to the gas.

### Conclusion:
- **Change in Internal Energy**: Positive
- **Work Done by the Gas**: Negative
- **Heat Transfer**: Positive (heat is added to the gas)

### Final Answer:
The correct options are:
1. Change in the internal energy of the gas is positive.
2. Work done by the gas is negative.
3. Heat is given to the gas.