Abstract
Much of the formulation of the first law of thermodynamics has been based on empirical observations of conversion of work into heat energy. But the reader needs to be cautioned that the insignia of the first law as displayed in (3.7) is not an ordinary mathematical equation because it does not guarantee that the reverse—i.e., the conversion of heat energy into work—can also be fully accomplished. Indeed, such a reverse process is a vexed undertaking whose analysis lies at the very heart of thermodynamics. Not only do Carnot’s ideas about perfect heat engines—and how they produce work with maximum possible efficiency that is related to the temperatures of the warm and cold external heat energy—provide a basis for the second law, they also help identify an important thermodynamic state function, the entropy. These ideas, their relevant subject matter, and many associated issues are treated in the form of solved problems. After some introductory remarks about ideal gas heat engines in Sect. 4.1, a perfect Carnot engine that runs on ideal gas as its working substance is described in Sect. 4.2. The Kelvin description of the absolute temperature is discussed in Sect. 4.3. Infinitesimal and finite Carnot cycles are studied in Sect. 4.4. Carnot version of the second law and calculation of the entropy are discussed in Sect. 4.5. Perfect Carnot engine with arbitrary working substance is studied in Sect. 4.6. Different statements of the second law are enunciated in Sect. 4.7. The fact that entropy increases in all spontaneous processes is noted and analyzed in Sects. 4.8 and 4.9. Kelvin–Planck version of the second law and its prediction that entropy always increases in irreversible–adiabatic processes is discussed in Sect. 4.10. Sections 4.10.1 and 4.10.2 respectively deal with the fact that while entropy increases in irreversible adiabats, it remains constant in reversible adiabats. In Sect. 4.11 non-Carnot heat engines and the Clausius inequality are studied. Integral and differential forms of Clausius inequality are reported in Sects. 4.11.1 and 4.11.2, respectively. Section 4.12 contains many solved problems that relate to thermal contact and the entropy change. Carnot refrigerator is described in Sect. 4.13. Section 4.14 deals with idealized version of realistic engine cycles, including the Stirling cycle, perfect Carnot cycle, diesel cycle, Otto and Joule cycles. Some cursory remarks about negative temperature are appended at the end of the chapter in Sect. 4.15.
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Notes
- 1.
A perpetual machine of the first kind was described in the preceding chapter.
- 2.
The operational details include such things as the nature of the cycle of operation and the quality of the engineering, etc.
- 3.
As we know from the introductory chapter, a reversible process is necessarily quasistatic. There are, however, two additional points to note. First, the entropy of the universe remains unchanged during all reversible processes. This means that any loss in the entropy of the working substance is exactly counterbalanced by the increase in the entropy of the reservoir, or vice versa. Thus, globally speaking, a reversible process is isentropic. Second, despite the fact that all reversible processes are quasistatic, the reverse is not always true. One can imagine a quasistatic process that is not reversible in the sense that globally it is not isentropic. However, for our purposes in this book, essentially all of the quasistatic processes—unless otherwise stated—are likely to be reversible. Therefore, the statement quasistatic may be considered synonymous with the word reversible.
- 4.
The similarity between the two notations lies in the fact that both \(Q^{\mathrm{rev}}(T_{\mathrm{H}})\) and \(Q^{\mathrm{rev}}(T_{\mathrm{C}})\) refer to the amounts of heat energy transferred “into” the working substance. The latter, occurring at \(T_{\mathrm{C}}\), is of course negative.
- 5.
As noted earlier, the heat energy, \(Q^{\mathrm{rev}}(T_{\mathrm{C}})\), transferred (supposedly) out of the dump and into the working substance is actually negative!
- 6.
Remember that the two adiabatic legs, \(2\to 3\) and \(4\to 1\), do not entail any heat energy transfer. Also do not forget that positive heat energy is actually discarded out of the working substance into the cold dump. Therefore, \(Q^{\mathrm{rev}}(T_{\mathrm{C}})\), as described here, is in fact a negative quantity.
- 7.
Indeed, in my experience, what we have dubbed here as the Carnot version of the second law, is as simple—if not simpler—to comprehend by a beginning student as any of the usual statements of the second law.
- 8.
Meaning, very large compared to the very tiny Carnot cycles mentioned above.
- 9.
Recall that for the given three state variables, \(P\), \(V\), and \(T\), a simple thermodynamic system is completely specified by any of the three pairs. Furthermore, we have seen that the entropy \(S\) is also a state variable and thus can be represented by any of the given three pairs of variables. As a result, any thermodynamic function of a simple system can be represented in terms of \(S\) and one of the above mentioned three variables, say the temperature \(T\).
- 10.
Constancy of entropy is ensured if the process occurs adiabatically and quasistatically.
- 11.
This can be arranged by letting the working substance adiabatically perform some work.
- 12.
Although, one might imagine that the following applies only to two separate systems that are in physical contact, the argument equally well applies to a single thermodynamic system because it can be “imagined” to have two macroscopic parts.
- 13.
This is so because \(T_{\mathrm{H}}>T_{\mathrm{C}}\).
- 14.
Note that the entropy at point 3, i.e., \(S_{3}\), can be arranged to be exactly equal to \(S_{0}\) because we are free to choose the amount of heat energy, \(Q^{\mathrm{rev}}(T_{2})\), to be added.
- 15.
To check this, write it as \(1-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\geq 1+ \frac{\mathrm{d} Q_{\mathrm{i}}'(T_{\mathrm{C}})}{\mathrm{d} Q_{\mathrm{i}}' (T_{\mathrm{H}})}\). Then subtract 1 from both sides, and multiply both sides by \(\frac{\mathrm{d} Q_{\mathrm{i}}'(T_{\mathrm{H}})}{T_{\mathrm{C}}}\) to get \(-\frac{\mathrm{d} Q_{\mathrm{i}}'(T_{\mathrm{H}})}{T_{\mathrm{H}}} \geq \frac{\mathrm{d} Q_{\mathrm{i}}'(T_{\mathrm{C}})}{T_{\mathrm{C}}}\). Now transfer \(-\frac{\mathrm{d} Q_{\mathrm{i}}'(T_{\mathrm{H}})}{T_{\mathrm{H}}}\) to the right-hand side.
- 16.
- 17.
As mentioned before, while reversible heat energy transfers and work output/input have been denoted with the superscript “rev,” the corresponding irreversible—or, partially irreversible—processes are denoted with a “prime” (′).
- 18.
The equality holds only if the path \(\mathrm{a}\to \mathrm{b}\) is traversed quasistatically.
- 19.
It is not absolutely necessary that the heat energy must flow out of the reservoir quasistatically. Because the reservoir is infinitely large, all heat flows out at exactly the same temperature \(T\), namely the temperature of the reservoir. Thus, even if the entropy change occurs via irreversible processes, one can use reversible paths for its computation, and we can correctly calculate the entropy loss of the reservoir by assuming it occurred quasistatically.
- 20.
Notice that the entropy gain by the hot object is negative because it cools down from \(T_{\mathrm{h}}\) to \(T_{\mathrm{f}}\).
- 21.
Although our choice of constants \(A,B,D,E\) is arbitrary, in crystalline solids the occurrence of the \(ET^{3}\) term at low temperatures is owed to temperature induced lattice vibrations. Similarly, the term linear in temperature—\(BT\)—is the electronic contribution to the specific heat in metals. In amorphous solids, the contribution to low temperature specific heat can sometime be quadratic in form, e.g., \(DT^{2}\).
- 22.
This increase is actually negative because the reservoir loses positive amount of heat energy in the process.
- 23.
Of course, in reality, it is a decrease because the object has yielded heat energy in this process.
- 24.
Note that an object being cooled inside a refrigerator is a finite cold reservoir.
- 25.
Note that (4.184) below is exactly equivalent to the perfect Carnot equality recorded in (4.2), namely
$$ \frac{Q^{\mathrm{rev}}(T_{\mathrm{C}})}{Q^{\mathrm{rev}}(T_{\mathrm{H}})} = - \biggl(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}} \biggr) \, . $$(4.183)The appropriate translation of (4.184) in terms of the notation used in (4.183) is as follows: Because \(|Q_{\mathrm{hot}}|\) is rejected to the hot reservoir, an equal amount of heat energy is extracted from the working substance at temperature \(T_{\mathrm{H}}\). Thus the heat energy reversibly added to the working substance at temperature \(T_{\mathrm{H}}\) is \(Q^{\mathrm{rev}}(T_{\mathrm{H}})=-|Q_{\mathrm{hot}}|\). On the other hand, because \(|Q_{\mathrm{cold}}|\) is extracted from the cold object/reservoir, it is equal to \(Q^{\mathrm{rev}}(T_{\mathrm{C}})\) added to the working substance at temperature \(T_{\mathrm{C}}\). Thus \(Q^{\mathrm{rev}}(T_{\mathrm{C}})=+|Q_{\mathrm{cold}}|\).
- 26.
Note that entropy is an extensive variable. So any increase in entropy is proportional to the system size. Here, a convenient measure of the size of the slice is its mass \(\varDelta M\). Therefore, the increase in the entropy of the slice \(\varDelta S\) is proportional to the mass \(\varDelta M\) of the slice.
- 27.
The \(\epsilon \) used here is not to be confused with its earlier usage relating to engine efficiency.
- 28.
Note that whether we use \(+1\) or −1, the expansion in powers of \(\epsilon \) for the entropy change is the same.
- 29.
Note that the universe here consists only of the isolated rod!
- 30.
\(R_{\mathrm{comp}}\) is defined as the ratio of the largest to the smallest volumes achieved during the cycle. Here that would be equal to \(\frac{V_{2}}{V_{1}}=\frac{V_{3}}{V_{1}}=\frac{V_{2}}{V_{4}}= \frac{V_{3}}{V_{4}}\).
- 31.
This mixture is the working substance and admittedly ideal gas is a rather crude approximation to it.
- 32.
Of course, the frictional loss is somewhat reduced by the presence of engine oil-lubricants.
- 33.
Mitsubishi, GDI engine, according to the manufacturer, achieves a compression ratio of 11.5. This requires clever engineering. For instance, part of the fuel is injected during the compression and in the process the mixture is cooled by the fuel spray.
- 34.
This is true despite the fact—as previously concluded—that “A violation of the Carnot version of the second law results in a violation of the Clausius statement of the second law.”
- 35.
Namely, “Without assistance it is impossible to withdraw positive amount of heat energy from a colder object and transfer the same to a warmer object.” In other words, heat energy does not spontaneously get transferred from a colder object to a warmer one.
References
F.W. Sears, G.L. Salinger, Thermodynamics, Kinetic Theory, and Statistical Thermodynamics, 3rd edn. (Addison Wesley, Reading, 1975)
D. ter Haar, H. Wergeland, Elements of Thermodynamics (Addison-Wesley, Reading, 1966)
N.F. Ramsey, Thermodynamics and statistical mechanics at negative absolute temperature. Phys. Rev. 103, 20 (1956)
H.B. Callen, See Callen’s Rule in Chap. 8 (8.16), also Callen’s scaling principle
E.M. Purcell, R.V. Pound, Phys. Rev. 81, 279 (1951)
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Tahir-Kheli, R. (2020). The Second Law. In: General and Statistical Thermodynamics. Graduate Texts in Physics. Springer, Cham. https://doi.org/10.1007/978-3-030-20700-7_4
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