1. Use the ideal gas relation on a $P$-$T$ diagram

For an ideal gas, $PV=nRT \,\Rightarrow\, V=\frac{nRT}{P}.$

On a $P$-$T$ graph:

• Horizontal line $\Rightarrow P=$ constant (isobaric)
• Vertical line $\Rightarrow T=$ constant (isothermal)
• Straight line through origin $P\propto T$ $\Rightarrow V=$ constant (isochoric)

We need the work done on the gas in going from $D$ to $A$.

2. Identify the process $D \to A$

From the given cycle on the $P$-$T$ diagram, the segment $D \to A$ is an isobaric compression at $P=2\text{ atm}$ with temperature changing from $T_D=345\text{ K} \quad \text{to} \quad T_A=138\text{ K}.$

So, $\Delta T = T_A-T_D = 138-345=-207\text{ K}.$

3. Work done by the gas in an isobaric process

For an isobaric process, $W_{\text{by}}=P\Delta V.$ Using $PV=nRT$ at constant $P$, $P\Delta V=nR\Delta T.$ Hence, $W_{\text{by}}=nR\Delta T.$

Given $n=2$ moles, $W_{\text{by}}=2R(-207)=-414R.$

4. Convert to work done on the gas

Work done on the gas is the negative of work done by the gas: $W_{\text{on}}=-W_{\text{by}}=+414R.$

5. Match with options

Thus the required answer is $\boxed{+414R}$ which corresponds to Option A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.