Determine the work done by an ideal gas undergoing a cyclic process from 1 → 4 → 3 → 2 → 1 . Given P 1 = 10 5 Pa , P 0 = 3 × 10 5 Pa , P 3 = 4 × 10 5 Pa and V 2 - V 1 = 10 L .

Step-by-step solution

From figure V 4 - V 3 V 2 - V 1 = P 3 - P 0 P 0 - P 1 ⇒ V 4 - V 3 1 0 = 4 × 1 0 5 - 3 × 1 0 5 3 × 1 0 5 - 1 0 5       V 4 -   V 3   = 5   L Now, work done W = 1 2 × 1 × 2 × 1 0 5 - 1 2 × 5 × 1 × 1 0 5 × 1 0 - 3 = 7 5 0 J