An ideal monatomic gas undergoes process `PV^(1.25)` = constant .Then

Options

A. upon increase in pressure temperature decreases
B. upon increase in pressure heat is absorbed by gas
C. if heat is given to gas its internal energy increases
D. heat is absorbed by gas if volume increases

Answer

To solve the problem regarding the ideal monatomic gas undergoing the process \( PV^{1.25} = \text{constant} \), we will analyze the implications of this equation step by step.

### Step-by-Step Solution:

1. **Understanding the Process**:
The equation \( PV^{1.25} = \text{constant} \) suggests that the product of pressure \( P \) and volume \( V \) raised to the power of 1.25 remains constant throughout the process. This indicates a specific relationship between pressure and volume during the transformation.

**Hint**: Recognize that the relationship given can be used to derive other thermodynamic properties.

2. **Using the Ideal Gas Law**:
The ideal gas law states that \( PV = nRT \), where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature. Since \( PV = nRT \), we can express pressure in terms of volume and temperature:
\[ P = \frac{nRT}{V} \]

**Hint**: Substitute \( P \) in the process equation to see how it relates to temperature.

3. **Substituting into the Process Equation**:
Substitute \( P \) from the ideal gas law into the process equation:
\[ \left(\frac{nRT}{V}\right)V^{1.25} = \text{constant} \]
This simplifies to:
\[ nRT V^{0.25} = \text{constant} \]

**Hint**: This shows that as volume changes, the temperature must adjust to keep the product constant.

4. **Analyzing Temperature Changes**:
From the equation \( nRT V^{0.25} = \text{constant} \), we can see that if the volume \( V \) increases, the temperature \( T \) must also increase to maintain the constant. Conversely, if the volume decreases, the temperature must decrease.

**Hint**: Consider how changes in volume affect temperature and internal energy.

5. **Applying the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
For an ideal gas, the change in internal energy \( \Delta U \) is given by:
\[ \Delta U = nC_V\Delta T \]
Since \( C_V \) for a monatomic gas is \( \frac{3}{2}R \), if the temperature remains constant, \( \Delta U = 0 \).

**Hint**: Remember that for an isothermal process (constant temperature), the internal energy change is zero.

6. **Work Done by the Gas**:
The work done by the gas during expansion or compression can be defined as:
\[ W = \int P \, dV \]
If the volume increases, the work done is positive; if the volume decreases, the work done is negative.

**Hint**: Relate the sign of work done to the changes in volume.

7. **Conclusion on Heat Transfer**:
Since \( \Delta U = 0 \), the first law simplifies to:
\[ Q = W \]
If the volume increases (work done on the gas is negative), then heat must be absorbed by the gas. If the volume decreases (work done by the gas is negative), then heat is released.

**Hint**: This leads to the conclusion that the heat absorbed or released depends on the direction of the volume change.

### Final Answer:
Based on the analysis, the correct conclusion is that if the volume increases, the gas absorbs heat; if the volume decreases, the gas releases heat.

Thus, the option stating that "Heat is absorbed by the gas if the volume is increased" is correct.