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A monoatomic ideal gas cycle of isothermal, isochoric and adiabatic processes: heat in and out, temperature and pressure ratios and the efficiency

A gas cylinder runs the cycle: isothermal expansion on a hot bath, isochoric cooling on a cold bath, adiabatic compression. For V2/V1 = 8 the run measures T_a = 4 T_c, heat out 1.125 nRT_a below heat in 2.08 nRT_a, and the same efficiency at 300 K and 600 K; P_a is 8 P_b, not 4 - answer (A), (B), (C).

Sub-topics covered

• Isothermal process: dU = 0, Q = W = nRT ln(V2/V1), P proportional to 1/V
• Isochoric process: W = 0, Q = n C_V dT
• Adiabatic process: Q = 0, T V^(gamma - 1) constant; gamma = 5/3 for a monoatomic gas
• Efficiency of a cycle, eta = (Q_in - Q_out)/Q_in

What it teaches

• The adiabat gives T_a/T_c = (V2/V1)^(2/3) = 4 for V2/V1 = 8 - statement (C)
• Heat absorbed on ab = nRT_a ln 8 = 2.1 nRT_a exceeds the heat released on bc = (3/2) nR (3/4) T_a = 1.125 nRT_a - statement (A)
• Both heats scale with T_a, so eta = 1 - (3/2)(1 - r^(-2/3))/ln r depends only on r - statement (B)
• On the isotherm P_a/P_b = V2/V1 = 8, not 4 - statement (D) false

Simulation ID

• ADV-2026-P1-PHY-Q07 · revision 1

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