Chapter 6.

1. Chapter 6. The second law of Thermodynamics

2. Ranque and his tube: No moving parts compressed air 2 kmoles, 4 atm., 294K COLD HOT 1 kmole, 1 atm., 333K 1 kmole, 1 atm., 255K hollow tube Is this a crazy idea or what?

3. In earlier chapters we alluded to aspects of thermodynamics which require further discussion or analysis. 1. Many processes do not occur in nature even though they are not forbidden by the first law. Is there some other law stating that these processes cannot occur? 2. Is there some general way to decide if a process is reversible or irreversible? 3. Can we write the first law solely in terms of state variables? We examine these questions in this chapter. We can convert to a state variable by multiplying it by the integrating factor (1/P): where V is a state variable đW đW Similarly if we multiply by an integrating factor which turns out to be (1/T) we obtain an exact differential. We write đQ đQ S=entropy (reversible processes)

4. The r in the above equations emphasizes that we are dealing with reversible processes. Hence dU= đQ- đW The boxed equation is the first law in terms of state variables. It connects any two neighboring equilibrium states. Of course we must still prove that dS is an exact differential. The boxed formula is probably the most important equation in thermodynamics.

5. Irreversible processes. A reversible process is one whose direction can be reversed by an infinitesimal change in some property. It is a quasi-static process in which no dissipative forces are present. All reversible processes must be quasi-static, but not all quasi-static processes are reversible (e.g., slowly letting the air out of a tire). In a reversible process everything (including the surroundings) can be restored to its initial state. All natural processes are, in actual fact, irreversible. How can we determine if a process is reversible or irreversible? On the other hand any process involving dissipative work, such as rubbing two solids together, is irreversible.

6. Dissipative work is done on the system. As an example consider a system consisting of a stirrer and fluid. Work done by rotating the stirrer is converted to “heat” energy. To reverse the process, the same amount of heat would need to be extracted from the fluid to perform the original amount of work, with no other changes. This does not violate the first law. A new law, the second law of thermodynamics, will be introduced which forbids this reverse process. Consider a gas in one portion of a chamber which is then allowed to undergo a free expansion into an evacuated section of the chamber. The gas will not compress itself back to the original volume. Again any process which attempts to restore the gas to its original volume, with no other changes, would violate the second law. Consider heat flowing from a high-T body to a lower-T body in the absence of other effects. The reverse process does not take place. It is impossible to create a device to simply reverse the process. Again the second law would be violated.

7. The impossibility of the above reverse processes occurring is contained in the second law of thermodynamics. Clausius statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body. Kelvin-Planck statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat with a single reservoir. Carnot’s theorem No engine operating between two reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. Remember that a Carnot engine is an ideal reversible engine and hence can be run as a refrigerator.

8. R2=Carnot refrigerator (so reversible) I= Engine (not an ideal reversible one) High T In the drawing all quantities are magnitudes. The diagram is of a self-contained device operating between two reservoirs. One part of the device is a Carnot refrigerator and the other part is an engine. The system is adjusted so that the work output from I is used to drive R2. I R2 Low T Let us assume that the efficiency of I is greater than R2. We would then have and so The heat extracted from the low temperature reservoir is By our assumption, this is positive.

9. The heat deposited to the high-T reservoir is: Therefore the net result of this cyclic system is to extract some heat from the low temperature reservoir and deposit it into the high-T reservoir, without any work being done by the surroundings. This violates the Clausius statement of the second law. Hence our original assumption is false and the efficiency of I must be less than, or at most equal to, the efficiency of the Carnot cycle.

10. Suppose now that I is replaced by a Carnot engine R1. There is no reason that we cannot do this. The analysis proceeds as before and, based on the above analysis, The nice thing about our Carnot devices is that they are reversible. Hence we can reverse the engines so that R1 is now the refrigerator and R2 is the heat engine. Again the analysis proceeds as before and we obtain Comparing equations (1) and (2) we conclude that All Carnot engines (operating between the same temperatures) have the same efficiency. In the above analysis no particular working substance was assumed. The efficiency of the Carnot cycle is independent of the working substance.

11. Breaking news! Two devils have invaded our laboratory from their den at the center of the Earth. The red devil is capable of violating the Clausius Statement of the 2nd Law. That is, she can take energy from a low T reservoir and deposit it into a reservoir at higher T, without any other changes, in some cyclic process. (She argues that it’s OK because the 1st Law is not violated.) The blue devil is capable of violating the Kelvin-Planck Statement of the 2nd Law by extracting energy from a single reservoir and performing some work in a cyclic process without depositing any energy into a reservoir at lower T. (He argues that it’s OK because the 1st Law is not violated.)

12. We wish to demonstrate that, if the Clausius Statement of the 2nd Law is violated, then the Kelvin-Planck statement is also violated. We begin with an ordinary engine (blue). It takes some energy from the high T reservoir, does some work and then dumps some waste energy into the low T reservoir. But the dreaded red devil (Violator of the Clausius statement) sneaks into the system. The devil takes the energy that the engine deposited into the low T reservoir and puts it back into the high T reservoir. High T This complete system then takes energy from the high T reservoir and performs some work with engine no change in the low T reservoir. This violates the Kelvin-Planck statement of the 2nd Law! Low T

13. In other words, the Kelvin-Planck statement says that it is impossible to construct any cyclical system (no matter how complex!) which takes energy from a single reservoir and perform useful work. But this statement is violated if we permit a violation of the Clausius statement (with the aid of the red devil). A violation of one statement of the 2nd Law leads to a violation of the other statement of the 2nd Law. Hence they are equivalent.

14. We wish to demonstrate that, if the Kelvin-Plank Statement of the 2nd Law is violated, then the Clausius statement is also violated. A device (R) would like to extract some energy from the low T reservoir and deposit the same amount of energy into the high T reservoir. This does not violate the 1st Law. We know that this is impossible as it would cause Herr Clausius to be very unhappy. Enter the blue devil (which violates the Kelvin-Planck Statement.) He extracts some energy from the high T reservoir and uses it to perform an equal amount of work without depositing energy into the low T reservoir. High T R Low T He uses this work to drive the refrigerator. The result of this system is to take a certain amount of energy from the low T reservoir and deposit the same amount into the high T reservoir, without any influence from, or change of, the environment external to the system.

15. In other words, the Clausius Statement says that it is impossible to construct any cyclical system (no matter how complex!) which takes energy from a low T reservoir and transport it to a high T reservoir with no change in the system or envionment. But this statement is violated if we permit a violation of the Kelvin-Planck statement (with the aid of the blue devil). Again a violation of one statement of the 2nd Law leads to a violation of the other statement of the 2nd Law. Hence they are equivalent. We are now going to study entropy in some detail. In preparation for this discussion we prove a theorem given on the next slide.

16. To emphasize that our development is completely general we introduce a generalized force Y (which could be P) and a generalized displacement X (which could be V) and we consider some reversible process if as shown on the diagram. isotherm adiabat reversible process Y i a b f adiabat X

17. We now demonstrate the following: Any arbitrary reversible process, in which the temperature may change in any manner, can be replaced by two reversible adiabatic processes connected by a reversible isothermal process in such a way that the heat exchanged over the isothermal process equals that exchanged over the original arbitrary process. In the above diagram, adiabats are drawn passing through i and f.

18. Curve ab represents an isothermal process. It is selected so that the area under iabf is equal to the area under the original curve. This ensures that the work done in the two processes is equal. Since the two paths connect the same states, the change in internal energy must be the same also. Therefore the heat exchanged must also be the same. However, for the curve iabf, the heat exchange takes place only during the isothermal part. This completes the proof.

19. adiabats . f . T2 Now we consider some reversible cycle as shown below. reversible cycle P i T1 V We draw a large number of adiabats, dividing the cycle into a large number of narrow strips, one of which is shown on the diagram. The isotherms (T1 and T2) are drawn as discussed above. We now have a Carnot cycle.

20. Heat energy Q2 enters the system during the isothermal process at T2 Heat energy Q1 leaves the system during the isothermal process at T1 We have previously shown, for a Carnot cycle, Now we revert to a notation in which heat entering a system is positive and heat leaving the system is negative. Hence and we have {Remember that “heat energy” means an energy transfer by the heating process.}

21. Continuing with the other Carnot cycles we have The Q’s represent infinitesimal heat transfers. Taking the limit, the sum becomes an integral and we have đQ Clausius Theorem: The integral of around any reversible cycle equals zero. đQ/T

22. We can use the Clausius Theorem to show the following: Consider 2 points (i, f) on any reversible cycle giving đQ đQ đQ đQ đQ đQ/T is independent of the path taken. It depends only on the state of the system at the initial and final points. Hence we can introduce a state function S, called the entropy, by đQ Reversible process!

23. The main points so far in this chapter are: First Law: Second Law: Clausius statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body. Kelvin-Planck statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat with a single reservoir. A Carnot engine is the most efficient engine. đQ đQ Clausius Theorem

24. In preparation for the next point we review the efficiency. Carnot !! Remember that is negative. Hence is negative. Hence the smaller (more negative) is, the smaller the efficiency. For example, if the efficiency is zero, while if , the efficiency is 0.5

25. Now consider an irreversible cycle between the same two reservoirs. Since I < R this gives or đQ đQ or, for an infinitesimal heat transfer, đQ As before (!!!!!!!!!) In this expression /T is not an entropy. đQ Combining our result with the expression for a reversible process we have đQ Clausius Inequality. This is sometimes taken as a statement of the second law. Now we have a way to decide whether or not a process is reversible!

26. Now we wish to examine, again, the change in entropy of an irreversible process. Consider going between two equilibrium states if by some irreversible process. We complete the cycle by fi by some reversible process. Because part of the cycle is irreversible đQ đQ đQ đQ đQ đQ đQ đQ The change in entropy is greater than the integral going from the initial state to the final state in the irreversible process!

27. đQ Combining with the reversible case: đQ For isentropic processes: dS = 0 and so For an adiabatic process đQ = 0 and so Now consider an isolated system. Then đQ =0 and so The entropy of an isolatedsystem increases in any irreversible process and is unaltered in any reversible process. It is the system’s entropy that governs the direction of spontaneous change.

28. Let us again consider an isolatedsystem. Suppose we start at some initial state which has an entropy Si .Now we begin some process. As the process proceeds, the entropy can only increase or, at best, remain unaltered. If now, at some stage, we reverse the direction of the process, the entropy must continue to increase, or remain the same. If the entropy continues to increase, the system can never regain the initial state with entropy Si . Hence it is obvious that , for a process to be reversible, the entropy cannot change. We now can state when a process is irreversible: In an isolated system, an irreversible process is one in which the entropy increases.

29. Now suppose the isolated system is in equilibrium, so no changes are taking place in the system. Then If , this implies a change towards equilibrium. Since the entropy of the system can only increase, this means that the equilibrium state is the state of maximum possible entropy. This is true regardless of the type of process taking place in the system. Thus, at equilibrium, the entropy of an isolated system has its maximum value with respect to all possible variations and the condition for this maximum is that An example of an isolated system is a universe. We must then have NOTE: One of these two terms can be negative!!!

30. This principle of increasing entropy gives a direction for the sequence of natural events. The laws of mechanics are second order equations in time and are unaltered by the transformation t  -t. According to mechanics all physical processes can run backwards. This is not so. The process can only go in a direction which increases the entropy. Entropy provides the “arrow of time” for physical processes. As will be discussed later, increasing entropy is associated with increasing disorder.

31. Entropy and available energy. What is the significance of the statement that entropy is created in an irreversible process? Of course, it tells us whether or not a process is possible. For example a chemist might wish to know if two substances will react under certain conditions. If the reaction would decrease the entropy of the universe, it is not possible. Something about the energy is irretrievably lost during an irreversible process, even though the energy is conserved. This is often stated as the “availability of the energy to do useful work”. Consider a simple situation:- the transfer of Q from a high-T reservoir to a low-T reservoir through a conductor. This reduces the capacity of Q to be converted to do useful work. Let us consider this in more detail. Suppose an even lower temperature reservoir exists. Consider a Carnot engine operating between and and extracts energy Q from the high-T reservoir.

32. Q same Q Q W C C Increasing T of reservoirs

33. Now suppose that the heat Q had been transferred to the reservoir at and we now operate the same Carnot engine between and extracting energy Q from the T1 reservoir, then The difference is the loss in work that Q can do as a result of it being transferred from the reservoir at to the reservoir at the lower temperature . is the increase in entropy which has resulted from the energy transfer. Although we have chosen to discuss a simple case, the result is true in general. Although the energy is not lost, it has been degraded in that it is less available do useful work.

34. To summarize: All natural processes are irreversible Every irreversible process makes energy less available to do external work. All natural processes continually make energy less available to do useful work.

35. EXAMPLE: We consider an irreversible engine. Because the process is irreversible, we cannot draw the cycle. We consider one complete cycle. Since S is a state variable: I W Must have Since

36. EXAMPLE: A 0.5kg sample of water (the system) at 90oC cools to room temperature (surroundings) at 293K. Calculate the entropy changes. Since the entropy is a state variable, the change in entropy is the same as if the water were cooled reversibly from đ P is constant, so The heat enters the room (surroundings), which is at constant T and so its entropy increases (the entropy change must be positive). The energy by the heating process that leaves the water is đ

37. The energy that enters the surroundings is

38. EXAMPLE: Two reservoirs are separated by a diathermic wall. We consider an interval of time over which some heat is exchanged, the amount being sufficiently small so that the temperatures of the reservoirs do not change appreciably. adiabatic walls diathermic wall Can the heat flow be in the opposite direction? Why not?

39. EXAMPLE (Problem 6.5):A thermally insulated 50 ohm resistor • carries a current of 1A for 1 sec. The initial temperature of the • resistor is 100C, its mass is 5 grams and the specific heat capacity • at constant pressure is 850 J/(kgK). • What is the entropy change of the resistor? • (b) What is the entropy change of the universe? đ đ (a) đ đ

40. (b) Surroundings isolated, so

41. đ Remember: (reversible process) into system, the entropy increases đ đ out of system, the entropy decreases For any closed cycle (because U and S are state variables): đ

42. EXAMPLE: Consider one cycle of this reversible process. The heat capacities are: P(Pa) b c (a) Q=? đ đ a d đ đ For the cycle (b) W=area enclosed

43. đ đ (c) đ because S is a state variable (like U)

44. EXAMPLE (Problem 6.1): A Carnot engine operates on 1 kg of methane which we will consider to be an ideal gas. The value of the ratio of specific heats is 1.35. If the ratio of the maximum to the minimum volume is 4 and the efficiency is 0.25, find the entropy increase of the methane during the isothermal expansion. b c P adiabat adiabat a isothermal expansion: d V

45. adiabatic expansion Using (1) From (2) Energy enters the engine during this part of the cycle so the entropy increases.

46. EXAMPLE (Problem 5.11): A reversible engine is connected to 3 • reservoirs as shown below. During some integral number of complete • cycles the engine absorbs 1200J from the reservoir at 400K and • performs 200J of mechanical work. • Find the quantities of energy exchanged with the other reservoirs, • and state whether the reservoir gives up or receives energy • (b) Find the change in entropy of each reservoir. (a) W=200J after n cycles R For n complete reversible cycles, the entropy change of the engine=0 (see slide 20)

47. From (1) Placing in (2) (Out of engine) From (1) (Into engine) (b) For the reservoirs where Q is for the engine ΔS1+ ΔS2+ ΔS3=0 (Of course) It is the requirement that the entropy change be zero that heat is transferred into the reservoir which has the middle temperature.

48. EXAMPLE: A mass of liquid at a temperature T1 is mixed with an equal mass of the same liquid at a temperature T2. The system is thermally isolated. Calculate the entropy change of the universe in terms of cp and the temperatures and show that it is necessarily positive. (Assume reversible process connecting states.) đ đ (universe)

49. One way to show that this is an increase in entropy is to show that Hence the minimum value of x=1 (temperatures the same) and so the minimum change in entropy is zero A second way is as follows:

50. EXAMPLE: The value for for some substances can be reasonably represented by • Find the heat absorbed and the increase in entropy of a mass m of the substance when its temperature is increased at constant pressure from T1 to T2. • (b) From a plot for the specific heat for Cu (see Figure 4.1) find the change in the molal specific entropy for Cu, when the temperature is increased at constant pressure, from 500 K to 1200 K. (a) đ đ