Trap 1 — Celsius in Carnot: η = 1 − $T_{2}$/$T_{1}$ requires kelvin. Temperatures in °C give wrong answers. Always add 273. Example: 127°C = 400 K, 527°C = 800 K; η = 1 − 400/800 = 50%, not 1 − 127/527 = 75.9%.

Trap 2 — Adiabatic = constant temperature: Q = 0 means no heat exchange, NOT constant temperature. Temperature changes significantly in adiabatic processes. Constant temperature is isothermal.

Trap 3 — $\Delta U$ ≠ 0 in isothermal: Students apply $\Delta U$ = nCᵥ$\Delta T$ and forget $\Delta T$ = 0 for isothermal. For ideal gas isothermal: $\Delta U$ = 0 always.

Trap 4 — Adiabatic curve not steeper: Students draw the adiabatic and isothermal as the same shape. Adiabatic is always steeper because PV^γ = const and γ > 1 (compared to PV = const for isothermal).

Trap 5 — v_mp is largest speed: v_mp is the most probable speed (peak of distribution), not the maximum speed. Order: v_mp < v_avg < v_rms. RMS is always the largest of the three characteristic speeds.

Trap 6 — COP ≤ 1: COP of refrigerator = $T_{2}$/($T_{1}$ − $T_{2}$) can far exceed 1. It is not constrained to be ≤ 1 unlike engine efficiency.

Trap 7 — Using diatomic γ for $CO_{2}$: $CO_{2}$ looks like a diatomic problem (C and $O_{2}$), but $CO_{2}$ is a triatomic molecule. For linear polyatomic: f = 5; for nonlinear: f = 6. Context determines which to use.

Trap 8 — W = 0 in isothermal: Students confuse $\Delta U$ = 0 with W = 0. In isothermal expansion, W = nRT ln($V_{2}$/$V_{1}$) ≠ 0. It is isochoric (W = 0), not isothermal.

Key correction: For adiabatic work, W = nCᵥ($T_{1}$ − $T_{2}$) = (P_{1}$$V_{1} − P_{2}$$V_{2})/(γ − 1). Students sometimes add a minus sign error by using $\Delta T$ in the wrong order.