# 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. Commented 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! Commented 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. Commented Apr 15, 2018 at 21:42
• @hyportnex that's a trivial point. Commented Jun 21, 2018 at 17:37
• @hyportnex Do you understand this that when we compare efficiency of two engines, we’ve to make them operate for same source and sinks. Or is this thing too non-trivial for you? Commented Jul 29, 2018 at 20:14

The laws of Thermodynamics give the following relation between Heat absorbed by the system, the work done by the system, and the change in the energy of the system- $$dQ=dU+dW$$ Now take the first step of the cycle-Adiabatic expansion. By definition, an adiabatic process is one where no heat is absorbed. Any work done by the system in this leg of the cycle is a result of the drop in internal energy, which can also be observed by the equation- $$dW=-dU$$ The reason for using Adiabatic expansion is that the system does not absorb heat, thereby improving the efficiency. Now, the same process could be used to regain the state but that would leave the cycle redundant as no net work would be done. To keep the entire process cyclic the state needs to be changed. The most efficient way to do so is to convert all the heat that the system absorbs into work-therefore the isothermal expansion.