The full cyclical process ABCDA consists of isobaric segment AB doing 750 J of work, isochoric segment BC doing no work, isobaric segment CD doing -100 J of work, and isochoric segment DA doing no work. What is the total work done by the gas over the complete cycle?

1. The total work in a cyclical process is the sum of the work done along each segment of the path. 2. Add the four segments: W = W_AB + W_BC + W_CD + W_DA = 750 J + 0 J + (-100 J) + 0 J. 3. Carrying out the addition: 750 J - 100 J = 650 J. 4. This positive total matches the area enclosed by the rectangle traced on the pressure-volume diagram, found independently as (P_AB - P_CD) x ΔV = (1.50 x 10^6 - 2.00 x 10^5)(5.00 x 10^-4 m^3) = 650 J. 5. Option A gives the correct size but a negative sign, which would only occur if the loop were traversed counter-clockwise instead. 6. Option B and option C each drop one of the nonzero segments from the sum rather than combining all four correctly. _Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.2 The First Law of Thermodynamics and Some Simple Processes_