Efficiency of a Carnot Engine

The Carnot engine's efficiency depends only on its two reservoir temperatures, not on the working substance. Using the isothermal-leg heat expressions $Q_H=\mu RT_H\ln(V_2/V_1)$ and $Q_L=\mu RT_L\ln(V_3/V_4)$, and the adiabatic-leg relations $T_HV_2^{\gamma-1}=T_LV_3^{\gamma-1}$ and $T_HV_1^{\gamma-1}=T_LV_4^{\gamma-1}$, one finds $V_2/V_1=V_3/V_4$, so $Q_L/Q_H=T_L/T_H$; substituting into $\eta=1-Q_L/Q_H$ gives the celebrated result $\eta_{Carnot}=1-\dfrac{T_L}{T_H}$ (with $T_H,T_L$ on the absolute Kelvin scale). Three consequences: (1) since $T_L<T_H$ always, $\eta$ is always strictly less than 1, reaching 100% only in the unattainable limit $T_L=0$ K; (2) efficiency depends only on the temperature ratio, not the numerical difference, and not on the working substance at all -- so, as Example 8.26 shows, two Carnot engines with the same temperature difference (150°C-100°C vs. 350°C-300°C) can have very different efficiencies; (3) when $T_H=T_L$, $\eta=0$. Real engines (s …