An ideal gas is allowed to expand from 5 L to 15 L once rapidly and once very slowly. The magnitude of work done in two processes are `W_(1)` and `W_(2)`, they are related as

Options

A. `W_(1) = W_(2)`
B. `W_(1) gt W_(2)`
C. `W_(1) lt W_(2)`
D. Data is insufficient

Answer

To solve the problem, we need to analyze the work done during the expansion of an ideal gas in two different scenarios: one rapid and one slow.

### Step-by-Step Solution:

1. **Understanding Work Done in Gas Expansion**:
The work done (W) by an ideal gas during expansion can be expressed mathematically as:
\[ W = \int P \, dV \]
where \(P\) is the pressure and \(dV\) is the change in volume.

2. **Rapid Expansion (W1)**:
In the case of rapid expansion, the gas expands quickly, and the pressure drops significantly during the process. This can be approximated as an irreversible expansion where the pressure remains constant until the gas reaches the new volume. However, since the expansion is rapid, the work done can be approximated as:
\[ W_1 = P_{initial} \times (V_{final} - V_{initial}) \]
where \(V_{initial} = 5 \, L\) and \(V_{final} = 15 \, L\).

3. **Slow Expansion (W2)**:
In the case of slow expansion, the gas expands gradually, allowing the system to remain in equilibrium. The pressure changes continuously, and the work done can be calculated as:
\[ W_2 = \int_{V_{initial}}^{V_{final}} P \, dV \]
In this case, the work done is greater than in the rapid expansion because the pressure is higher at the initial volume and decreases gradually to the final volume.

4. **Comparison of Work Done**:
Since the work done in the slow process (W2) involves a gradual change in pressure, the area under the pressure-volume curve will be larger than in the rapid process (W1). Therefore, we can conclude:
\[ W_2 > W_1 \]

5. **Final Relationship**:
The relationship between the work done in the two processes is:
\[ W_2 > W_1 \]

### Conclusion:
The magnitude of work done in the two processes is related as \(W_2 > W_1\).

---