Along segment CD of the same cyclical process, the gas is compressed at a constant pressure of 2.00 x 10^5 N/m^2 while its volume decreases by 5.00 x 10^-4 m^3. What is the work done by the gas along this segment?

1. Segment CD is still an isobaric process, so work is again W = P x ΔV, but here the volume decreases. 2. A decreasing volume makes ΔV negative, so ΔV_CD = -5.00 x 10^-4 m^3. 3. Substitute the values with units: W_CD = (2.00 x 10^5 N/m^2)(-5.00 x 10^-4 m^3). 4. Multiplying: 2.00 x 5.00 = 10.0, and 10^5 x 10^-4 = 10^1, giving -10.0 x 10^1 N*m = -100 J. 5. The negative sign matters here: it shows the gas has work done on it during this segment rather than doing work itself. 6. Option A gives the right size but the wrong sign, ignoring that a shrinking volume must produce negative work in this convention. 7. Option B would only be correct for an isochoric segment where volume does not change at all, which is not the case here. 8. Option D applies the wrong pressure value, mixing up this segment's 2.00 x 10^5 N/m^2 with the 1.50 x 10^6 N/m^2 used on segment AB. _Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.2 The First Law of Thermodynamics and Some Simple Processes_