B.4 · IB Physics
Heat engines (Carnot cycle)
One mole of ideal monatomic gas goes round two isotherms and two adiabats. It absorbs heat at the hot temperature, rejects some at the cold temperature, and the enclosed area of the P–V loop is the work done per cycle.
Predict first
Start from the default settings (T_hot = 600 K, T_cold = 300 K). Raise the hot reservoir to 1000 K and leave everything else alone. What happens to the efficiency?
Carnot efficiency depends only on the reservoir temperatures: η = 1 − T_cold ÷ T_hot = 1 − 300 ÷ 1000 = 0.70, so 70 %.
Check it in the model: Move the Hot reservoir, T_hot slider from 600 K to 1000 K and read the Efficiency line.
isotherm at Thot (heat in) isotherm at Tcold (heat out) adiabats
Qin = nRThot ln(V₂÷V₁)
Qout = nRTcold ln(V₂÷V₁)
W = Qin − Qout
adiabat: PV5/3 = constant
n = 1 mol, V₁ = 10.0 L
Try this
- Keep Tcold = 300 K and raise Thot from 400 K to 1000 K. How does the efficiency change, and does the enclosed area grow?
- Use "Small ΔT: poor". Compare Qin and Qout, then raise the expansion ratio to 6.0. What changes and what stays the same?
- Tick Logarithmic axes. What shape do the isotherms and adiabats become, and which pair is steeper?
Exam tip: efficiency limits need temperatures in kelvin. The Carnot value 1 − Tcold/Thot is a ceiling set by the reservoirs, so a real engine between the same two temperatures must do worse.