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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.

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    $\begingroup$ 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. $\endgroup$ – Isomorphic Apr 15 '18 at 20:13
  • $\begingroup$ @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! $\endgroup$ – hyportnex Apr 15 '18 at 20:57
  • $\begingroup$ "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. $\endgroup$ – Philip Wood Apr 15 '18 at 21:42
  • $\begingroup$ @hyportnex that's a trivial point. $\endgroup$ – Isomorphic Jun 21 '18 at 17:37
  • $\begingroup$ @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! $\endgroup$ – hyportnex Jun 21 '18 at 17:43
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The idea of the Carnot cycle is that it is the most efficient possible thing you could do, given a high-temperature and low-temperature reservoir.

You always want to draw heat from the high-temperature reservoir when your engine at the same temperature, because transferring heat through a finite temperature difference will generate entropy. The same goes for dumping heat into the low-temperature reservoir. That means that you should not have any heat transfer while you change your engine's temperature between these two, so the intermediate processes must be adiabatic.

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