The process `Delta U=0`, for an ideal gas can be best represented in the form of a graph

Explanation

To solve the question regarding the representation of the process where \( \Delta U = 0 \) for an ideal gas, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept of Internal Energy (\( \Delta U \))**:
- For an ideal gas, the change in internal energy (\( \Delta U \)) is given by the formula:
\[ \Delta U = nC_v \Delta T \]
- Here, \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature.

2. **Setting \( \Delta U = 0 \)**:
- When \( \Delta U = 0 \), it implies that:
\[ nC_v \Delta T = 0 \]
- Since \( n \) and \( C_v \) are constants for a given gas, this means that \( \Delta T = 0 \).

3. **Interpreting \( \Delta T = 0 \)**:
- A change in temperature of zero indicates that the temperature remains constant throughout the process. This is characteristic of an **isothermal process**.

4. **Graphical Representation of an Isothermal Process**:
- In a Pressure-Volume (P-V) diagram, an isothermal process for an ideal gas is represented by a hyperbolic curve. This is because, according to the ideal gas law \( PV = nRT \), if the temperature (T) is constant, the product \( PV \) will also remain constant.
- Therefore, the graph will show a hyperbola, which reflects the inverse relationship between pressure and volume at constant temperature.

5. **Conclusion**:
- The process \( \Delta U = 0 \) for an ideal gas can be best represented as an isothermal process on a P-V diagram, which is depicted as a hyperbola.

### Final Answer:
The process \( \Delta U = 0 \) for an ideal gas is best represented by a hyperbolic curve on a Pressure-Volume (P-V) graph, indicating an isothermal process.

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