An ideal gas can be taken form initial state `1` to final state `2` by two different process. Let `DeltaQ` and `W` represent the heat given and work done by the system. Then which quantities is/are same in both process (where `DeltaU =` internal energy of gas)

Options

A. `DeltaQ`
B. `W`
C. `DeltaU`
D. `DeltaQ - W`

Explanation

To solve the problem, we need to analyze the properties of the quantities involved in the processes of an ideal gas transitioning from an initial state (1) to a final state (2) through two different paths. We will focus on the heat (ΔQ), work done (W), internal energy (ΔU), and the relationship between these quantities.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- **Internal Energy (ΔU)**: This is a state function that depends only on the initial and final states of the system, not on the path taken.
- **Heat (ΔQ)** and **Work (W)**: Both are path functions, meaning their values depend on the specific process or path taken between the two states.

2. **Identifying the State Function**:
- Since ΔU is a state function, it will have the same value regardless of the path taken. Therefore, for both processes from state 1 to state 2, ΔU will be the same.

3. **Applying the First Law of Thermodynamics**:
- The First Law of Thermodynamics states that:
\[ \Delta U = \Delta Q - W \]
- Since ΔU is the same for both processes, we can conclude that the difference between heat added (ΔQ) and work done (W) must also be the same for both processes.

4. **Conclusion**:
- The quantities that are the same for both processes are:
- The change in internal energy (ΔU)
- The expression ΔQ - W

### Final Answer:
- The quantities that are the same in both processes are:
- **ΔU (Change in internal energy)**
- **ΔQ - W (Difference between heat added and work done)**