Step 1: Understanding the Concept:
Work done in a chemical process under constant external pressure is called pressure-volume work ($PV$ work). When a gas is compressed, work is done on the system, so the sign of work is positive ($W > 0$).
Step 2: Formula Application:
The formula for work done is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
To convert the result from $bar \cdot dm^3$ to Joules, we use the conversion factor: $1 \text{ bar} \cdot dm^3 = 100 \text{ J}$.
Step 3: Explanation:
Given: $P_{ext} = 3$ bar, $V_1 = 24$ dm$^3$, $V_2 = 13$ dm$^3$.
$\Delta V = V_2 - V_1 = 13 - 24 = -11$ dm$^3$.
$W = -(3 \text{ bar}) \times (-11 \text{ dm}^3) = 33 \text{ bar} \cdot dm^3$.
Converting to Joules: $W = 33 \times 100 \text{ J} = 3300 \text{ J}$.
Step 4: Final Answer:
The work done is 3300 J.