One mole of an ideal monatomic gas undergoes a cyclic process shown in the figure (rectangle a→ b→ c→ d in the P–V plane, with P in N m⁻² between 100 and 300, and V in m³ between 2 and 5). The total heat supplied to the gas is:

Solution & Explanation

For any complete cyclic process the gas returns to its initial state, so ΔU = 0 over the cycle. First law for a cycle: Q = ΔU + W = 0 + W, so the net heat supplied equals the net work done. The net work in a P–V cycle is the area enclosed by the loop. The rectangle has: Width ΔV = 5 − 2 = 3 m³. Height ΔP = 300 − 100 = 200 N m⁻². Enclosed area = ΔV × ΔP = 3 × 200 = 600 J. Since the loop is traversed clockwise (work done by the gas is positive), W = +600 J. Therefore Q = W = 600 J, matching option C.