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A Carnot cycle consists of two reversible isothermal changes, one at each reservoir temperature, joined by two reversible adiabatic processes. It attains the maximum thermal efficiency permitted between those reservoirs, $1-T_c/T_h$.

The total work by the gas is the enclosed triangular area, $W=\frac{1}{2}{p}_0{V}_0$. The positive heat on the sloping leg isAdd $Q_{AB}=48p_0{V}_0/32$, excluding the emitting part of $BC$ and all of $CA$. HenceAs an independent conservation of energy check, the heat emitted on late $BC$ is $p_0{V}_0/32$ and on $CA$ is $80p_0{V}_0/32$; thus $Q_{\mathrm{in}}-Q_{\mathrm{out}}=16p_0{V}_0/32=W$. Using only the net heat on $BC$ would give an incorrect thermal efficiency.

Macroscopically, an adiabatic process exchanges no heat with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore isentropic, since $dS=\delta Q_{\mathrm{rev}}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a heat bath; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a thermodynamic cycle. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the Carnot cycle. Reversibility and zero net entropy change giveThe thermal efficiency is the net work output divided by the absorbed heat:ConsequentlyThe Carnot theorem shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

In the $(p,V)$ plane, the Otto cycle consists of:

It is a clockwise loop, so its enclosed area is the positive net work output. In the $(T,S)$ plane, the two reversible adiabats $A\to B$ and $C\to D$ are vertical constant-entropy segments. Constant-volume heating $B\to C$ moves upward and to the right, while constant-volume cooling $D\to A$ moves downward and to the left.

LetThe two adiabatic relations giveBecause heat is exchanged at constant volume,where $Q_2$ denotes the positive amount rejected. HenceThe thermal efficiency is thereforeFor a fixed gas, $\gamma$ is fixed and the idealized efficiency increases monotonically with the compression ratio $r$, so it is maximized by making $r$ as large as the physical constraints allow.

Heat enters only during the constant-pressure step, so the heat capacity at constant pressure givesThe positive magnitude of the heat rejected during constant-volume cooling isBecause the working gas returns to its initial state, the thermodynamic cycle has zero net change in internal energy and $W=Q_1-Q_2$. Its thermal efficiency is therefore