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

Options

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

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 1: Identify the given values
- Initial volume (V1) = 4 dm³
- Final volume (V2) = 6 dm³
- External pressure (P) = 3 atm

### Step 2: Calculate the change in volume (ΔV)
\[ \Delta V = V2 - V1 = 6 \, \text{dm}^3 - 4 \, \text{dm}^3 = 2 \, \text{dm}^3 \]

### Step 3: Use the formula for work done (W)
The work done by the gas during expansion against a constant external pressure is given by the formula:
\[ W = -P \Delta V \]
Substituting the values we have:
\[ W = -3 \, \text{atm} \times 2 \, \text{dm}^3 \]

### Step 4: Convert dm³ to liters
Since 1 dm³ = 1 L, we can say:
\[ \Delta V = 2 \, \text{dm}^3 = 2 \, \text{L} \]
Thus, the work done becomes:
\[ W = -3 \, \text{atm} \times 2 \, \text{L} = -6 \, \text{L atm} \]

### Step 5: Convert work from L atm to Joules
We know that 1 L atm = 101.32 J. Therefore, we can convert the work done:
\[ W = -6 \, \text{L atm} \times 101.32 \, \text{J/L atm} = -607.92 \, \text{J} \]
Rounding this, we get approximately:
\[ W \approx -608 \, \text{J} \]

### Conclusion
The work done during the expansion of the gas is approximately -608 J. The negative sign indicates that work is done by the system.

---