[figure] An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A B C A as shown in the diagram above. The total work done in the process is J.
Answer
200
Solution
Work on the gas: w = - P dV (1 kPa dm³ = 1 J)
A → B: P = 10 kPa, V 10 → 30 dm³ ⇒ w = −10 × 20 = −200 J
B → C: V 30 → 10 dm³, P 10 → 30 kPa ⇒ w = −(average P)ΔV = −20 × (−20) = +400 J
C → A: ΔV = 0 ⇒ w = 0
∴ Total w = −200 + 400 + 0 = +200 J (= area of triangle ABC = ½ × 20 × 20)
Answer: 200