Thermodynamic change in the system from state (P_1, V_1) to (P_2,...

The work done by the system during a thermodynamic process is equal to the area under the $P-V$ graph projected on the volume axis.

From the given graph, the process from state A to state D forms a trapezium with the volume axis.

The initial state A has coordinates $P_1 = 1 \times 10^5\ \text{N/m}^2$ and $V_1 = 1\ \text{m}^3$.

The final state D has coordinates $P_2 = 3 \times 10^5\ \text{N/m}^2$ and $V_2 = 5\ \text{m}^3$.

The work done $W$ is the area of this trapezium:

$W = \dfrac{1}{2} (P_1 + P_2) (V_2 - V_1)$

Substituting the values:

$W = \dfrac{1}{2} (1 \times 10^5 + 3 \times 10^5) (5 - 1)$

$W = \dfrac{1}{2} (4 \times 10^5) (4)$

$W = 2 \times 10^5 \times 4$

$W = 8 \times 10^5\ \text{J}$

Answer: $8 \times 10^5\ \text{J}$