The work done during the expansion of a gas from a volume of `4dm^(3)` to `6dm^(3)` againts a constant external presuure of `3atm` is `(1atm-L=101.32J)`

Options

A. `-6J`
B. `-608J`
C. `+304J`
D. `-304J`

Explanation

To solve the problem of calculating the work done during the expansion of a gas from a volume of 4 dm³ to 6 dm³ against a constant external pressure of 3 atm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Volumes**:
- Initial Volume (V1) = 4 dm³
- Final Volume (V2) = 6 dm³

2. **Calculate the Change in Volume (ΔV)**:
- ΔV = V2 - V1
- ΔV = 6 dm³ - 4 dm³ = 2 dm³

3. **Convert the Change in Volume to Litres**:
- Since 1 dm³ = 1 L, ΔV = 2 L.

4. **Identify the External Pressure**:
- External Pressure (P) = 3 atm

5. **Use the Work Done Formula**:
- The work done (W) during expansion against a constant external pressure is given by:
\[ W = -P \Delta V \]
- Substituting the values:
\[ W = -3 \text{ atm} \times 2 \text{ L} \]

6. **Calculate the Work Done in atm-L**:
- W = -6 atm-L

7. **Convert Work Done to Joules**:
- We know that 1 atm-L = 101.32 J.
- Therefore, to convert -6 atm-L to Joules:
\[ W = -6 \text{ atm-L} \times 101.32 \text{ J/atm-L} \]
- Calculating this gives:
\[ W = -607.92 \text{ J} \]

8. **Round the Answer**:
- The work done can be approximated to -608 J.

### Final Answer:
The work done during the expansion of the gas is approximately **-608 J**.

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