Two processes on a system are illustrated in the figure. The...

From the given $p-T$ graph, both processes A and B are represented by horizontal lines, which means they are isobaric (constant pressure) processes.

Let the pressure of process A be $p_A$ and that of process B be $p_B$. From the graph, it is clear that $p_B > p_A$.

The work done by the system in an isobaric process is given by $\Delta W = p \Delta V = p(V_f - V_i)$.

It is given that the initial volumes for both processes are identical ($V_{i,A} = V_{i,B} = V_i$) and the final volumes are also identical ($V_{f,A} = V_{f,B} = V_f$). Therefore, the change in volume $\Delta V = V_f - V_i$ is the same for both processes.

From the graph, the temperature $T$ increases in both processes. For an ideal gas at constant pressure, $V \propto T$, so the volume must increase, implying $\Delta V > 0$.

The work done in process A is $\Delta W_1 = p_A \Delta V$.

The work done in process B is $\Delta W_2 = p_B \Delta V$.

Since $p_B > p_A$ and $\Delta V > 0$, we have $p_B \Delta V > p_A \Delta V$.

Therefore, $\Delta W_2 > \Delta W_1$, which can be written as $\Delta W_1 < \Delta W_2$.