An ideal gas sample of 2.0 mol undergoes the cyclic process ABCA as...

In the given cyclic process $ABCA$, the graph is a $T-V$ (Temperature-Volume) diagram.

1. Process $C \rightarrow A$:
The line is vertical, which means the volume is constant ($V = \text{constant}$). This is an isochoric process.
Work done, $W_{CA} = 0$.

2. Process $A \rightarrow B$:
The line passes through the origin, which means $T \propto V$. From the ideal gas equation $PV = nRT$, this implies that the pressure $P$ is constant. This is an isobaric process.
The work done is given by:
$W_{AB} = P \Delta V = nR \Delta T = nR(T_B - T_A)$
From the graph, $T_A = 300 \text{ K}$. The point $B$ is at the same temperature level as $C$ to close the cycle at the given upper temperature, so $T_B = T_C = 500 \text{ K}$.
Given $n = 2.0 \text{ mol}$ and using $R = 8.3 \text{ J/(mol K)}$:
$W_{AB} = 2.0 \times 8.3 \times (500 - 300) = 2.0 \times 8.3 \times 200 = 3320 \text{ J}$

3. Cyclic Process $ABCA$:
For a complete cycle, the change in internal energy is zero ($\Delta U = 0$). According to the first law of thermodynamics:
$Q_{\text{cycle}} = W_{\text{cycle}}$
We are given that $1200 \text{ J}$ of heat is withdrawn from the sample, so $Q_{\text{cycle}} = -1200 \text{ J}$.
Therefore, the total work done in the cycle is $W_{\text{cycle}} = -1200 \text{ J}$.

The total work done is the sum of the work done in each process:
$W_{\text{cycle}} = W_{AB} + W_{BC} + W_{CA}$
$-1200 = 3320 + W_{BC} + 0$
$W_{BC} = -1200 - 3320 = -4520 \text{ J}$

Answer: $-4520 \text{ J}$