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

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 formula for work done
The work done (W) during the expansion of a gas can be calculated using the formula:
\[ W = -P \Delta V \]
where:
- \( P \) is the external pressure,
- \( \Delta V \) is the change in volume.

### Step 2: Calculate the change in volume (\( \Delta V \))
Given:
- Initial volume (\( V_1 \)) = 4 dm³
- Final volume (\( V_2 \)) = 6 dm³

We can calculate \( \Delta V \) as follows:
\[ \Delta V = V_2 - V_1 = 6 \, \text{dm}^3 - 4 \, \text{dm}^3 = 2 \, \text{dm}^3 \]

### Step 3: Substitute the values into the work formula
Now, we can substitute the values of \( P \) and \( \Delta V \) into the work formula:
\[ W = -P \Delta V = -3 \, \text{atm} \times 2 \, \text{dm}^3 \]

### Step 4: Convert the units
Since we need the work done in joules, we must convert the units from atm·dm³ to joules. We know that:
1 atm·dm³ = 101.32 J

Thus, we convert:
\[ W = -3 \, \text{atm} \times 2 \, \text{dm}^3 = -6 \, \text{atm·dm}^3 \]
Now converting to joules:
\[ W = -6 \, \text{atm·dm}^3 \times 101.32 \, \text{J/(atm·dm}^3) \]
\[ W = -608 \, \text{J} \]

### Step 5: Final answer
The work done during the expansion of the gas is:
\[ W = -608 \, \text{J} \]