A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. Its volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be:

The system starts at state A, goes to state B via a linear process, and then from state B to state C via an isobaric process.
From the graph:
State A: $P_A = 8000 \, \text{Dyne/cm}^2$, $V_A = 3 \, \text{m}^3$
State B: $P_B = 4000 \, \text{Dyne/cm}^2$, $V_B = 7 \, \text{m}^3$
State C: The process B to C is isobaric, so $P_C = P_B = 4000 \, \text{Dyne/cm}^2$. The volume at C is reduced to the original value from A, so $V_C = V_A = 3 \, \text{m}^3$.

We know that $1 \, \text{Dyne/cm}^2 = 0.1 \, \text{N/m}^2 = 0.1 \, \text{Pa}$.
$P_A = 8000 \, \text{Dyne/cm}^2 = 8000 \times 0.1 \, \text{Pa} = 800 \, \text{Pa}$
$P_B = 4000 \, \text{Dyne/cm}^2 = 4000 \times 0.1 \, \text{Pa} = 400 \, \text{Pa}$
$P_C = 400 \, \text{Pa}$

The process A to B is a linear process on a P-V diagram. The work done is the area of the trapezoid under the line segment AB.

Substitute the values:

The process B to C is an isobaric process (constant pressure). The work done is given by $W = P \Delta V$.

Substitute the values:

The total work done is the sum of the work done in each process.

The final answer is $\text{800 J}$.