A monatomic ideal gas follows A(2.0 L, 100 kPa) → B(2.0 L, 250 kPa) → C(5.0 L, 100 kPa) → A, with B→C a straight line. Determine net work by the gas. Calculate temperatures at A, B and C for n = 0.100 mol, then find ΔU and Q on each leg. Use R = 8.31 J mol⁻¹ K⁻¹ and 1 kPa·L = 1 J. State whether the cycle operates as an engine.

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Net work = +225 J. Temperatures: 241 K, 602 K and 602 K. ΔU_AB = +450 J, ΔU_BC = 0, ΔU_CA = −450 J. Heats: +450 J, +525 J and −750 J. It is a heat engine. The clockwise triangular area is W_net = ½(250−100)(5−2) = 225 kPa L = 225 J, positive for work done by the gas. Using PV = nRT gives T_A = 200/[0.100(8.31)] = 241 K, T_B = 500/0.831 = 602 K and T_C = 500/0.831 = 602 K. For a monatomic ideal gas ΔU = (3/2)nRΔT. Thus ΔU_AB = +450 J, ΔU_BC = 0 and ΔU_CA = −450 J, summing to zero over the cycle. Works are W_AB = 0, W_BC = average pressure times ΔV = [(250+100)/2](3) = 525 J, and W_CA = 100(2−5) = −300 J. From Q = ΔU + W_by, heats are +450 J, +525 J and −750 J. Net Q = +225 J equals net work, and positive cycle work means the device acts as a heat engine.

Analyse a triangular gas cycle

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Triangular ideal-gas cycle

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