Q.An ideal gas is taken through a cyclic process as shown in the figure: a closed loop on a $P(\text{Pa})$ vs $V(\text{m}^3)$ diagram through three labelled states A, B, C, with the pressure axis marked at 400 Pa and 600 Pa and the volume axis marked at 3 m$^3$ and 6 m$^3$. Calculate

Step 1. The figure's axes are marked at $P=400\ \text{Pa}$ and $600\ \text{Pa}$, and $V=3\ \text{m}^3$ and $6\ \text{m}^3$, with three labelled states A, B, C forming a closed loop. The leg $A(V=6,P=400)\to B(V=3,P=400)$ is an isobaric compression: $W_{AB}=P\Delta V=400\times(3-6)=-1200\ \text{J}=-1.2\ \text{kJ}$ (work done ON the gas, matching part (b) exactly).

Step 2. The leg $B(V=3,P=400)\to C(V=3,P=600)$ is isochoric (volume fixed at 3 m³, only pressure rises): $W_{BC}=0.$

Step 3. The leg $C(V=3,P=600)\to A(V=6,P=400)$ closes the loop with volume increasing back to 6 m³; taking this as a straight-line path on the P-V diagram, the work done is the trapezoidal area under it: average pressure $\times$ change in volume $=\dfrac{600+400}{2}\times(6-3)=500\times3=1500\ \text{J}=+1.5\ \text{kJ}$ (work done BY the gas, matching part (a) exactly). …

Identify which leg is isobaric (rectangular area), which is isochoric (zero work), and which is the diagonal closing leg (trapezoidal …

• Adding the raw magnitudes of the two non-zero legs (1.2 kJ + 1.5 kJ) instead of correctly taking the signed sum (which correctly gives the much smaller …