Select the incorrect statement

Options

A. For isothermal process of ideal gas, `DeltaU = 0`
B. For isochoric process, `W = 0`
C. For adiabatic process, `DeltaU = -DeltaW`
D. For cyclic process, `DeltaW = 0`

Answer

To solve the question of selecting the incorrect statement regarding thermodynamic processes, we will analyze each statement one by one.

### Step-by-Step Solution:

1. **Statement 1: For isothermal process, ΔU = 0.**
- In an isothermal process, the temperature of the system remains constant (T1 = T2). The change in internal energy (ΔU) for an ideal gas is given by the formula:
\[ ΔU = mC_V(T2 - T1) \]
Since T1 = T2, we have:
\[ ΔU = mC_V(0) = 0 \]
- **Conclusion:** This statement is **correct**.

2. **Statement 2: For isochoric process, W = 0.**
- An isochoric process is one where the volume remains constant. The work done (W) in a thermodynamic process is given by:
\[ W = P \Delta V \]
Since the volume does not change (ΔV = 0), the work done is:
\[ W = P(0) = 0 \]
- **Conclusion:** This statement is **correct**.

3. **Statement 3: For adiabatic process, ΔQ = 0.**
- An adiabatic process is defined as one in which there is no heat transfer into or out of the system (ΔQ = 0). According to the first law of thermodynamics:
\[ ΔQ = ΔU + W \]
If ΔQ = 0, then:
\[ 0 = ΔU + W \implies ΔU = -W \]
- **Conclusion:** This statement is **correct**.

4. **Statement 4: For cyclic process, ΔW = 0.**
- In a cyclic process, the system returns to its initial state, which means that the change in internal energy (ΔU) over one complete cycle is zero. However, the work done (W) is not necessarily zero. The work done in a cyclic process is represented by the area enclosed in the P-V diagram. Therefore, the work done can be non-zero.
- **Conclusion:** This statement is **incorrect**.

### Final Answer:
The incorrect statement is **Statement 4: For cyclic process, ΔW = 0.**