# Why does the Carnot cycle require adiabatic processes specifically?

Why does the Carnot cycle need two isentropic adiabatic processes to move between hot and cold reservoirs? Surely as the only requirement for the Carnot cycle is that it is constructed out entirely out of reversible processes and operates between two temperature, could you use a reversible isochoric process to move from hot to cold (and vice versa)?

When I try to derive the Carnot efficiency using isochoric processes for an ideal gas I get: $$\eta = 1 - \frac{T_C(Nk_B\ln(V_f/V_i)-C_V) + C_V T_H}{T_H(Nk_B\ln(V_f/V_i) + C_V) - C_V T_C}$$ where $V_f$ and $V_i$ are the final and initial volumes respectively.

This is decidedly not the true result for the Carnot efficiency except in the case where $C_V = 0$, which is not true for a gas at a temperature above 0K.

• One thing that isn't explicitly mentioned is that Carnot cycle isn't just the most efficient engine, every reversible engine is equally efficient. However, carnot engine is the simplest, as in the simplest to construct and calculate the efficiency for. You'll not be able to construct a simpler working engine with just two heat reservoirs. Apr 15, 2018 at 20:13
• @Isomorphic it is not true that all reversible engines are equally efficient, but it is true that among all engines exchanging heat between temperatures $T_{max}$ and $T_{min}$ the highest possible efficiency is $1-T_{min}/T_{max}$ and this peak efficiency is achieved by the Carnot cycle. But engines that operate also at some in-between temperature(s) the efficiency is actually less than this! Apr 15, 2018 at 20:57
• "Surely as the only requirement for the Carnot cycle is that it […] operates between two temperature, could you use a reversible isochoric process to move from hot to cold (and vice versa)?" But reversible isochoric processes take in or give out heat throughout a continuous range of temperatures, so the cycle is not $operating\ between\ two\ temperatures$ in the meaning of this phrase. Apr 15, 2018 at 21:42
• @hyportnex that's a trivial point. Jun 21, 2018 at 17:37
• @Isomorphic trivial or not you wrote that "One thing that isn't explicitly mentioned is that Carnot cycle isn't just the most efficient engine, every reversible engine is equally efficient." and your claim that all reversible engines are equally is WRONG! Jun 21, 2018 at 17:43