A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be :

Question

G1 Physics · Accepted Answer

Image Work done from A to B is the area under the straight-line process: W(AB) = ½(P(A) + P(B))ΔV = ½(8000 + 4000) dyne/cm² × 4 m³ = 6000 dyne/cm² × 4 m³ Work done from B to C: W(BC) = -4000 dyne/cm² × 4 m³ Therefore, W(total) = (6000 - 4000) dyne/cm² × 4 m³ = 2000 dyne/cm² × 4 m³ Since, 1 dyne/cm² = 0.1 N/m² W(total) = 200 N/m² × 4 m³ = 800 J Official answer in the source paper: BONUS. The calculated answer is 800 J, which was not present in the original options. For clean upload, Option A has…