A gas undergoes a cyclic process ABCA as shown. Find the work done by the gas for A ⎯→ B ⎯→ C.

The correct option is 1200 J.

To find the total work done by the gas during the process A ⎯→ B ⎯→ C, we analyze the pressure-volume ($P-V$) diagram provided in the image:
- The vertical axis representing pressure ($P$) is in units of $N/m^2$ and shows values of 300 and 900.
- The horizontal axis representing volume ($V$) is in units of $m^2$ and shows values of 3 and 7.
- From the diagram, the coordinates of the states are:
State A: $(V_A,P_A)=(3,900)$
State B: $(V_B,P_B)=(7,300)$
State C: $(V_C,P_C)=(3,300)$

The total work done $W_{ABC}$ is the sum of the work done in path A ⎯→ B and path B ⎯��� C:
$W_{ABC}=W_{AB}+W_{BC}$

Step 1: Work done during the path A ⎯→ B ($W_{AB}$)
The path A ⎯→ B is a straight line on the diagram. The work done is equal to the area under the line AB, which is a trapezoid bounded by $V=3$ and $V=7$:
$W_{AB}=\frac{1}{2}\times (P_A+P_B)\times (V_B-V_A)$
Substituting the values from the graph:
$W_{AB}=\frac{1}{2}\times (900+300)\times (7-3)$
$W_{AB}=\frac{1}{2}\times 1200\times 4=2400\text{J}$

Step 2: Work done during the path B ⎯→ C ($W_{BC}$)
The path B ⎯→ C is an isobaric compression at a constant pressure of $P=300\text{N}/m^2$. The volume decreases from $V_B=7$ to $V_C=3$:
$W_{BC}=P\times (V_C-V_B)$
Substituting the values from the graph:
$W_{BC}=300\times (3-7)$
$W_{BC}=300\times (-4)=-1200\text{J}$

Step 3: Total work done ($W_{ABC}$)
Summing the work done in both steps:
$W_{ABC}=2400\text{J}+(-1200\text{J})=1200\text{J}$