The Carnot cycle and maximum efficiency

A reversible step is one that could be retraced, leaving the gas and everything around it as they were. Heat crossing a temperature gap is the commonest way to fail that test: let 1.00 kJ leak straight from 800 K to 300 K and the first law notices nothing, yet 625 J of work that an engine could have taken out of it has gone for good. So while the gas is in contact with a reservoir it must sit at that reservoir's own temperature, which makes those stages isothermal. Travelling between 800 K and 300 K then has to be done with the gas insulated, which makes the connecting stages adiabatic. No third kind of stage is available, so the cycle reads isotherm, adiabat, isotherm, adiabat — and any engine that exchanges heat reversibly with only two reservoirs is a Carnot engine, whatever it looks like from outside.

Take 1.00 mol of a monatomic ideal gas, T_H = 600 K, T_C = 300 K, and let the hot isotherm expand the gas by a factor of 3.00. On a→b the temperature never changes, so ΔU = 0 and the heat in equals the work out: Q_H = nRT_H ln 3.00 = 8.314 × 600 × 1.0986 = 5.48 kJ. On b→c the gas is insulated, Q = 0, and it does work at the expense of its own internal energy, nC_V(T_H − T_C) = 1.5 × 8.314 × 300 = 3.74 kJ, cooling to 300 K as it expands. On c→d the gas is compressed at 300 K through the same volume ratio 3.00, rejecting Q_C = nRT_C ln 3.00 = 2.74 kJ. On d→a the surroundings put back exactly the 3.74 kJ the gas released on b→c. The adiabatic pair therefore cancels, the net work is 5.48 − 2.74 = 2.74 kJ, and η = 2.74/5.48 = 0.500.

The claim that both isotherms run through the same volume ratio is the hinge of the derivation, and the adiabats supply it. On a reversible adiabat of an ideal gas T V^(γ−1) stays constant, so b→c gives T_H V_b^(γ−1) = T_C V_c^(γ−1) and d→a gives T_H Vₐ^(γ−1) = T_C V_d^(γ−1). Divide the first by the second and both temperatures disappear: (V_b/Vₐ)^(γ−1) = (V_c/V_d)^(γ−1), hence V_b/Vₐ = V_c/V_d. Now set the two isothermal heats side by side. Q_H = nRT_H ln(V_b/Vₐ) and Q_C = nRT_C ln(V_c/V_d) share an identical logarithm, so Q_C/Q_H = T_C/T_H and η = 1 − T_C/T_H. The amount of gas, the volumes and γ have all cancelled, which is the first hint that the answer is bigger than the ideal gas it came from.

Suppose an engine X between 800 K and 300 K really did convert 70% of its heat, beating the Carnot value 1 − 300/800 = 0.625. Feed X 1000 J: it delivers 700 J of work and drops 300 J into the cold reservoir. Now spend that 700 J driving a Carnot engine backwards, which being reversible it permits. Run in reverse it still obeys Q_C/Q_H = 0.375, so 700 J of work returns 700/0.625 = 1120 J to the hot reservoir while lifting 420 J out of the cold one. Add the two machines together: no work has crossed the boundary, the hot reservoir has gained 1120 − 1000 = 120 J, and the cold reservoir has lost 420 − 300 = 120 J. Heat has moved from cold to hot with nothing paid for it. Since that is impossible, no X can beat Carnot; and running the argument both ways with two reversible engines shows that every reversible engine between the same reservoirs has exactly the same efficiency, whatever its working substance.

Differentiate η = 1 − T_C/T_H and the asymmetry is plain: ∂η/∂T_H = T_C/T_H² while ∂η/∂T_C = −1/T_H, so a one-kelvin drop in the cold reservoir is worth T_H/T_C times a one-kelvin rise in the hot one. Take a steam plant at T_H = 810 K against a cooling tower at T_C = 300 K, where the ceiling is 0.630. Ten kelvin more in the boiler lifts it to 0.634; ten kelvin less in the tower lifts it to 0.642, a gain 2.7 times larger, which is the ratio T_H/T_C the derivatives predict. The catch is ownership: the cold reservoir is the river or the air, and nobody gets to choose its temperature, so every engineering effort goes into raising T_H, which metallurgy caps. A gas turbine entering at 1700 K against the same 300 K has a ceiling of 0.824, and that is why combined-cycle plants exist.

Reversible means quasi-static, and a quasi-static engine produces zero power: every stage takes forever. Real heat exchangers must move heat at a finite rate, which needs a finite temperature drop across them, so the working fluid never quite reaches 810 K or 300 K and the loop it actually traverses is narrower than the reservoirs allow. Model just that one compromise — an internally reversible engine coupled to its reservoirs through finite conductances — and maximise the power rather than the efficiency: the optimum sits at η = 1 − √(T_C/T_H) = 1 − √(300/810) = 0.391. Large coal plants deliver about 0.40, so this finite-time figure, not the 0.630 ceiling, is the honest yardstick. Friction, turbulence, throttling through valves, incomplete combustion and heat leaking past the working fluid then take their share of what is left.