Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where E is internal energy of the system)

Options

A. `-DeltaP//P`
B. Zero
C. `-V DeltaP`
D. `-DeltaE`

Explanation

To solve the question regarding the pressure-volume (PV) work done by an ideal gaseous system at constant volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Work Done**:
- The work done (W) by a gas during expansion or compression can be expressed mathematically as:
\[ W = -\int P \, dV \]
- Here, \(P\) is the pressure and \(dV\) is the change in volume.

2. **Identify the Condition**:
- The question specifies that the process occurs at constant volume. This means that the volume does not change (\(dV = 0\)).

3. **Evaluate the Work Done**:
- Since the volume is constant, there is no change in volume (\(dV = 0\)). Therefore, substituting this into the work done equation gives:
\[ W = -\int P \, dV = -P \cdot 0 = 0 \]
- Thus, the work done by the system at constant volume is zero.

4. **Conclusion**:
- The work done by an ideal gaseous system at constant volume is:
\[ W = 0 \]

### Final Answer:
The work done by an ideal gaseous system at constant volume is **0**.