The work done by a weightless piston in causing an expansing `Delta V` (at constant temperature), when the opposing pressure P is variable, is given by :

Options

A. `W = - int p Delta V`
B. W = 0
C. `W = -P Delta V`
D. None

Answer

To solve the question regarding the work done by a weightless piston during an expansion at constant temperature with variable opposing pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Work Done by a Piston**:
The work done \( W \) by a piston in a thermodynamic process can be expressed as the integral of pressure \( P \) with respect to volume \( V \). The general formula for work done is:
\[ W = -\int_{V_1}^{V_2} P \, dV \]
where \( V_1 \) is the initial volume and \( V_2 \) is the final volume.

2. **Expressing Change in Volume**:
The change in volume \( \Delta V \) is defined as:
\[ \Delta V = V_2 - V_1 \]
Therefore, we can rewrite the work done in terms of \( \Delta V \):
\[ W = -\int_{V_1}^{V_2} P \, dV = -P \Delta V \]
Here, we assume that \( P \) is the average pressure during the expansion.

3. **Considering Constant Temperature**:
Since the process occurs at constant temperature, we can assume that the gas behaves ideally. However, the pressure \( P \) is variable, so we cannot simplify it further without additional information about how \( P \) changes with \( V \).

4. **Final Expression for Work Done**:
The final expression for the work done by the weightless piston during the expansion at constant temperature is:
\[ W = -P \Delta V \]
This indicates that the work done by the system (the gas) is negative since the system is expanding against an external pressure.

5. **Evaluating the Options**:
- **Option A**: \( W = -P \Delta V \) (Correct)
- **Option B**: \( W = 0 \) (Incorrect, as work is done)
- **Option C**: \( W = -P \Delta V \) (Incorrect, lacks the integral)
- **Option D**: None (Incorrect, since we have a correct answer)

### Conclusion:
The correct answer is **Option A**: \( W = -P \Delta V \).