# Supplementary Notes on Entropy and the Second Law of Thermodynamics

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1 ME 4- hermodynamics I Supplementary Notes on Entropy and the Second aw of hermodynamics Reversible Process A reversible process is one which, having taken place, can be reversed without leaving a change in either the system or the surroundings. Entropy Consider a closed system such as a piston-cylinder assembly. et us say that a gas is contained in the cylinder, and that it consists of our system. We take our system through a closed cycle by adding (or removing) heat. We ensure that the path is reversible. If we think of our cycle as consisting of a series of infinitesimally small links as shown in Figure, and we measure the infinitesimal heat transfer to the system in that link, δ i, and the temperature of our system, i, during that infinitesimal transfer, we can evaluate the sum: N i δ i i where i is the ith link, and N is the number of links. If the links were made truly infinitesimally small, we would in effect be evaluating the cyclic integral δ cycle If we did an experiment of this type, we would find that this cyclic integral would always come out to be zero. Since we know that the change in a property of a system over a closed cycle is zero, we postulate that there must exist a property of the system, S, such that: ds δ for a reversible process () he property S is called Entropy. It is an extensive property whose units are J/K. he corresponding specific property is denoted by s. Its units are J/(kg K). he temperature appearing in Equation is the absolute temperature of the system. he change in entropy between two states and may be evaluated using: S S S δ rev () It is important to understand what Equation means. Since S is a property of the system, S S is independent of the path connecting the two points and ; it is the same for all paths connecting the two points as long as the end points are fixed. owever, we can evaluate S S directly from δ only if the path connecting the two is reversible. 3 Entropy Change for an Irreversible Process If the path connecting points and is irreversible, it is found that ds δ ds gen (3)

2 i link i Figure : Cycle Integral Evaluation for Entropy ere ds gen is the entropy generated by irreversibilities such as friction, heat transfer across finite temperature differences or uncontrolled expansions, and is always positive. Irreversibility always increases the entropy. he entropy change between two states and is given by: S S S δ S gen (4) Again, it is important to interpret Equation 4 correctly. Once the points and are fixed, the entropy change between them is fixed, regardless of the process connecting them. owever, this change S is only equal to δ if the process is reversible. If the process is irreversible, the entropy change would turn out to be greater than δ because of entropy generation due to irreversibilities. Alternatively, if we fixed δ for the two paths, the irreversible path would end up with a different end state than the reversible path. he end state would have a higher value of S than the reversible path because of the extra entropy generation due to irreversibilities. From Equation 4 we see that the entropy of a system can change because of two agents:(i) heat transfer to/from a system, and (ii) irreversibilities. eat transfer to the system tends to increase its entropy; heat transfer from a system tends to decrease its entropy. Irreversibilities always tend to increase a system s entropy. herefore, the entropy of a system can either increase or decrease overall; a decrease can occur if the system loses enough heat to make up for entropy increases due to irreversibilities. Another important point is that the work done by/on the system does not appear in the entropy change expression. Work does not change the entropy directly. he doing of work may indirectly lead to friction or other irreversibilities which can generate entropy. For an isolated system, δ is zero. herefore ds ds gen or S S S gen 0 (5) hus, the entropy of an isolated system can only increase (or stay the same if all processes within it are reversible). Since the universe, which is the sum of the system and its surroundings, is the ultimate isolated system, we can also say: S universe S system S surroundings 0 (6) his is probably the most useful operational statement of the Second aw of thermodynamics, and one which we shall use most often in our problem solving.

3 W net Figure : A eat Engine Which Violates the Kelvin-Planck Statement 4 he Isentropic Process Consider Equation4. If the process - is reversible, S gen δ 0. If the process is adiabatic, 0. hus, for a reversible, adiabatic process, S S. Such a process is called an isentropic process. Remember that all reversible processes are not isentropic. Similarly, all adiabatic processes are not isentropic. It is only a reversible and adiabatic process which is isentropic. 5 Equivalence with the Kelvin-Planck and Clausius Statements of the Second aw he Kelvin-Planck statement of the Second aw states: It is impossible to construct a device that will operate in a cycle and produce no effect other than new work and the exchange of heat with a single reservoir. et us consider a heat engine that violates the Kelvin-Planck statement, as shown in Figure. ere we have a heat engine that receives at temperature, and does net work W net. Consider one cycle of operation. Applying the First aw and realizing that E is zero in a closed cycle W net he total entropy change of the universe may be written as: S universe S system S δ where S system is the entropy change of the heat engine itself and S is the entropy change of the high-temperature reservoir. Because S is a property of the system, S system 0 in one complete cycle. Further, since the heat transfer occurs in an isothermal process, δ S since is removed from the reservoir. herefore S universe 0 0 hus, an engine which violates the Kelvin-Planck statement of the Second aw also violates the rule that S universe 0. 3

4 W W net Figure 3: A eat Engine Operating Between wo Reservoirs Why does heat rejection to fix the problem? hat is, why is the engine shown in Figure 3 legal whereas the one shown in Figure illegal? Consider the net entropy change of the universe for the case shown in Figure 3. S universe S system S S From the First aw, since E=0 for one closed cycle of operation, W net where W net 0 is the work done by the system, and is a positive number. As before, S system =0 in a closed cycle. herefore S universe W net net (7) he first term in the last equation of Equation 7 is always positive; the second term (which is itself positive) appears with a negative sign. So the sum can be negative. owever, if W net were small enough, we could get a net increase in entropy of the universe. Removing from decreases the entropy of the universe. Adding to increases it. So if we add enough to to make up for the decrease due to the removal of, we can build a system for which the Second aw is not violated. It is clear we cannot do this unless there is a non-zero. he Clausius statement of the Second aw says: It is impossible to construct a device that operates in a cycle and produces no other effect than the transfer of heat from a cooler body to a hotter body. hat is, a heat engine of the type shown in Figure 4 is impossible. et us consider the entropy changes associate with an engine that violates the Clausius statement. he net entropy change 4

5 Figure 4: A eat Engine Which Violates the Clausius Statement of the universe for such an engine is S universe Since, S universe 0. herefore such an engine is impossible. We see that even though we cannot prove the Kelvin-Planck or the Clausius statements of the Second aw, they are consistent with our operational statement S universe 0. 6 eat ransfer and Irreversibility eat transfer between two bodies is considered reversible if it occurs across infinitesimally small temperature differences. It is irreversible if it occurs across a finite temperature difference. et us consider the entropy changes associated with heat transfer between two bodies at and respectively, as shown in Figure 5. ere S universe If (i.e. the heat transfer is across infinitesimally small differences in temperature), S universe 0. hat is, the heat transfer is reversible. If, S universe 0. So heat transfer from a hot body to a cold one causes a net increase in the entropy of the universe. If, S universe 0. his is, of course, impossible. 7 he Carnot Cycle and Entropy Change Consider the Carnot cycle shown in Figure 3. Recall that the Carnot cycle is a reversible cycle which involves four reversible processes: 5

6 - -3: 3-4: 4-: S Figure 5: eat ransfer Between Reservoirs and : Isothermal expansion due to heat addition from a high temperature reservoir at. Adiabatic expansion. Isothermal compression due to heat loss to a low temperature reservoir at. Adiabatic compression. Consider one closed cycle of operation. he entropy change of the universe is given by: S universe 0 system S S he universe is an isolated system, so it is adiabatic. Since the Carnot cycle is reversible and all the heat transfer interactions are reversible, all processes in the universe are reversible. So the universe undergoes a reversible, adiabatic process, i.e., an isentropic process: S universe 0 herefore Recall that we had assumed this relationship in deriving the thermal efficiency of the Carnot cycle in terms of temperature. It is valid as long as the cycle is reversible, and the heat addition and heat rejection happen isothermally at and respectively. 6

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