The work done during the expansion of a gas from a volume of `4 dm^(3)` to 6dm 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`

Answer

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)
The change in volume (ΔV) can be calculated using the formula:
\[ \Delta V = V2 - V1 \]
Substituting the values:
\[ \Delta V = 6 \, \text{dm}^3 - 4 \, \text{dm}^3 = 2 \, \text{dm}^3 \]

### Step 3: Calculate the work done (W)
The work done during the expansion of the gas against a constant external pressure is given by the formula:
\[ W = -P \Delta V \]
Substituting the values:
\[ W = -3 \, \text{atm} \times 2 \, \text{dm}^3 \]
Calculating this gives:
\[ W = -6 \, \text{atm} \cdot \text{dm}^3 \]

### Step 4: Convert the work done to Joules
Since 1 atm·dm³ is equivalent to 101.32 Joules, we can convert the work done:
\[ W = -6 \, \text{atm} \cdot \text{dm}^3 \times 101.32 \, \text{J/(atm} \cdot \text{dm}^3) \]
Calculating this gives:
\[ W = -6 \times 101.32 \, \text{J} = -607.92 \, \text{J} \]
Rounding this, we get:
\[ W \approx -608 \, \text{J} \]

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

---