Step 1: Understanding the Concept:
The process described is an isothermal expansion against a constant external pressure.
This means the process is an irreversible thermodynamic process.
Step 2: Key Formula or Approach:
The work done in an isothermal expansion against a constant external pressure is given by \(W = -P_{\text{ext}} \Delta V = -P_{\text{ext}}(V_2 - V_1)\).
Use the conversion factor \(1 \text{ L} \cdot \text{atm} = 101.3 \text{ J}\) (noting that \(1 \text{ dm}^3 = 1 \text{ L}\)).
Step 3: Detailed Explanation:
Identify the given values from the problem statement:
Constant external pressure, \(P_{\text{ext}} = 1 \text{ atm}\).
Initial volume, \(V_1 = 15.5 \text{ dm}^3\).
Final volume, \(V_2 = 20 \text{ dm}^3\).
Calculate the change in volume, \(\Delta V\):
\[ \Delta V = 20 \text{ dm}^3 - 15.5 \text{ dm}^3 = 4.5 \text{ dm}^3 \]
Now, calculate the work done in units of \(\text{atm} \cdot \text{dm}^3\):
\[ W = - (1 \text{ atm}) \times (4.5 \text{ dm}^3) = -4.5 \text{ atm} \cdot \text{dm}^3 \]
Next, convert the work to Joules using the standard conversion factor (\(1 \text{ atm} \cdot \text{dm}^3 = 101.3 \text{ J}\)):
\[ W = -4.5 \times 101.3 \text{ J} \]
\[ W = -455.85 \text{ J} \]
Rounding to the nearest whole number to match the given options, we get \(-456 \text{ J}\).
Step 4: Final Answer:
The work done is approximately \(-456 \text{ J}\).