# 4.5 The carnot cycle

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• Describe the Carnot cycle with the roles of all four processes involved
• Outline the Carnot principle and its implications
• Demonstrate the equivalence of the Carnot principle and the second law of thermodynamics

In the early 1820s, Sadi Carnot (1786−1832), a French engineer, became interested in improving the efficiencies of practical heat engines. In 1824, his studies led him to propose a hypothetical working cycle with the highest possible efficiency between the same two reservoirs, known now as the Carnot cycle    . An engine operating in this cycle is called a Carnot engine    . The Carnot cycle is of special importance for a variety of reasons. At a practical level, this cycle represents a reversible model for the steam power plant and the refrigerator or heat pump. Yet, it is also very important theoretically, for it plays a major role in the development of another important statement of the second law of thermodynamics. Finally, because only two reservoirs are involved in its operation, it can be used along with the second law of thermodynamics to define an absolute temperature scale that is truly independent of any substance used for temperature measurement.

With an ideal gas as the working substance, the steps of the Carnot cycle, as represented by [link] , are as follows.

1. Isothermal expansion. The gas is placed in thermal contact with a heat reservoir at a temperature ${T}_{\text{h}}.$ The gas absorbs heat ${Q}_{\text{h}}$ from the heat reservoir and is allowed to expand isothermally, doing work ${W}_{1}.$ Because the internal energy ${E}_{\text{int}}$ of an ideal gas is a function of the temperature only, the change of the internal energy is zero, that is, $\text{Δ}{E}_{\text{int}}=0$ during this isothermal expansion. With the first law of thermodynamics, $\text{Δ}{E}_{\text{int}}=Q-W,$ we find that the heat absorbed by the gas is
${Q}_{\text{h}}={W}_{1}=nR{T}_{\text{h}}\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{V}_{N}}{{V}_{M}}.$
2. Adiabatic expansion . The gas is thermally isolated and allowed to expand further, doing work ${W}_{2}.$ Because this expansion is adiabatic, the temperature of the gas falls—in this case, from ${T}_{\text{h}\phantom{\rule{0.2em}{0ex}}}\text{to}\phantom{\rule{0.2em}{0ex}}{T}_{\text{c}}.$ From $p{V}^{\gamma }=\phantom{\rule{0.2em}{0ex}}\text{constant}$ and the equation of state for an ideal gas, $pV=nRT$ , we have
$T{V}^{\text{γ}\phantom{\rule{0.2em}{0ex}}\text{−}\phantom{\rule{0.2em}{0ex}}\text{1}}=\text{constant},$

so that
${T}_{\text{h}}{V}_{N}{}^{\gamma -1}={T}_{\text{c}}{V}_{O}{}^{\gamma -1}.$
3. Isothermal compression . The gas is placed in thermal contact with a cold reservoir at temperature ${T}_{\text{c}}$ and compressed isothermally. During this process, work ${W}_{3}$ is done on the gas and it gives up heat ${Q}_{\text{c}}$ to the cold reservoir. The reasoning used in step 1 now yields
${Q}_{\text{c}}=nR{T}_{\text{c}}\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{V}_{O}}{{V}_{P}},$

where ${Q}_{\text{c}}$ is the heat dumped to the cold reservoir by the gas.
4. Adiabatic compression . The gas is thermally isolated and returned to its initial state by compression. In this process, work ${W}_{4}$ is done on the gas. Because the compression is adiabatic, the temperature of the gas rises—from ${T}_{\text{c}}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{T}_{\text{h}}$ in this particular case. The reasoning of step 2 now gives
${T}_{\text{c}}{V}_{P}{}^{\gamma -1}={T}_{\text{h}}{V}_{M}{}^{\gamma -1}.$

The total work done by the gas in the Carnot cycle is given by
$W={W}_{1}+{W}_{2}-{W}_{3}-{W}_{4}.$

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