Interpret the $P$–$V$ graph

The process is:

• $D \to E$: linear path
• $E \to F$: isobaric compression back to the original volume

The work done by the gas is the area under the $P$–$V$ curve.

Use coordinates from the graph

From the figure, the states are:

• $D:(V=1\,\text{m}^3,\; P=300\,\text{N/m}^2)$
• $E:(V=4\,\text{m}^3,\; P=600\,\text{N/m}^2)$
• $F:(V=1\,\text{m}^3,\; P=600\,\text{N/m}^2)$

Work done from $D$ to $E$

Since $D \to E$ is a straight line, work equals the area of the trapezium:

$W_{DE}=\frac{P_D+P_E}{2}(V_E-V_D)$

$W_{DE}=\frac{300+600}{2}\times (4-1)$

$W_{DE}=\frac{900}{2}\times 3=450\times 3=1350\,\text{J}$

Work done from $E$ to $F$

This is an isobaric process at $P=600\,\text{N/m}^2$ with volume decreasing from $4$ to $1\,\text{m}^3$:

$W_{EF}=P\,(V_F-V_E)$

$W_{EF}=600(1-4)=600(-3)=-1800\,\text{J}$

Total work done

$W_{\text{total}}=W_{DE}+W_{EF}$

$W_{\text{total}}=1350-1800=-450\,\text{J}$

Check options

The total work done by the gas is:

$\boxed{-450\,\text{J}}$

So the correct option is A.