Interpret the processes on the $P$-$V$ diagram

From the figure (linear path $A \to B$ followed by isobaric compression $B \to C$), we read:

• State $A: (V_A, P_A)$
• State $B: (V_B, P_B)$
• State $C: (V_A, P_B)$ since $BC$ is isobaric and volume is reduced back to the original value.

From the diagram values:

$P_A = 200\,\text{N/m}^2, \quad P_B = 600\,\text{N/m}^2$ $V_A = 2\,\text{m}^3, \quad V_B = 6\,\text{m}^3$

Work done from $A$ to $B$

Since $A \to B$ is a straight line on a $P$-$V$ graph, the work done is the area under the line, i.e. trapezium area:

$W_{AB} = \frac{P_A + P_B}{2}(V_B - V_A)$

Substituting:

$W_{AB} = \frac{200 + 600}{2}(6-2)$ $W_{AB} = \frac{800}{2}\times 4 = 400 \times 4 = 1600\,\text{J}$

Work done from $B$ to $C$

Process $B \to C$ is isobaric at pressure $P_B = 600\,\text{N/m}^2$.

Volume decreases from $6$ to $2\,\text{m}^3$, so

$W_{BC} = P_B (V_C - V_B) = 600(2-6)$ $W_{BC} = 600(-4) = -2400\,\text{J}$

Negative sign means work is done on the gas during compression.

Total work done by the gas

$W_{\text{total}} = W_{AB} + W_{BC}$ $W_{\text{total}} = 1600 + (-2400) = -800\,\text{J}$

Match with options

The total work done by the gas is $-800\,\text{J}$. Its magnitude is $800\,\text{J}$, and the listed option corresponds to:

$\boxed{800\,\text{J}}$

So the intended answer is Option A.