2. Accounting for Entropy Class 28.2

3. Objectives Qualitatively understand reversibility/irreversibility Quantitatively understand reversibility/irreversibility Understand entropy Perform simple calculations involving entropy Know how to account for entropy Quantitatively state the second law of thermodynamics

4. Gasoline Air CO 2 H 2 O Motion Air turbulence Tire deformation Head lights Air conditioning Stereo Hot exhaust A “natural” process…

5. Gasoline Air CO 2 H 2 O Motion Air turbulence Tire deformation Head lights Air conditioning Stereo Hot exhaust An “unnatural” process… Although we all recognize this is impossible, it is still allowed by the first law of thermodynamics (conservation of energy).

6. We need another law… The second law of thermodynamics i.e., naturally occurring processes are directional

7. How do we quantify the second law of thermodynamics? Entropy

8. Entropy is closely tied to… Reversible processes Irreversible processes – do not generate entropy – do generate entropy

9. A reversible process… Frictionless pulley If a movie of this process were run backwards, you could not tell.

10. An irreversible process… If a movie of this process were run backwards, you could tell.

11. Imagining a movie running forwards or backwards is a useful method for thinking about reversibility, but what would we do for a process that we are not familiar with? We need a better way to determine if a process is reversible or not.

12. Better Approach: Return the system to its initial state, i.e., run a “cycle.” The more change in the surroundings, the more irreversible the process. Universe Surroundings System

13. 1 Initial state of the system Amount of weight determines friction System boundary

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18. 6 Final state of the system. (Same as initial state.) Surroundings have changed. (Two weights now on the floor.) More weight here causes more weight to be on the floor. This weight controls the amount of irreversibility in the system.

19. Energy changes of this process: Potential energy internal energy heat } } “Ordered” energy“Disordered” energy Potential Kinetic Work Internal energy Heat

20. Observation: Irreversibilities occur when ordered energy is converted to disordered energy.

21. .......... Steam 100 o C Ice Bath 0 o C Time passes This is an irreversible process. Heat will not spontaneously flow from the ice bath to regenerate the steam. (A movie run backwards would look funny.) Copper rod

22. Observation: Heat transfer from a high-temperature body to a low-temperature body is an irreversible process.

23. T + dT V + dV TVTV T + dT V + dV TVTV Reversible heat transfer… Perfect insulation

24. Observation: Systems with differential driving forces are reversible. Corollary: Systems with differential driving forces are infinitely slow.

25. Expander P 1, V 1 P 2, V 2 V1V1 P1P1 P2P2 V2V2 Work Produced Irreversible V1V1 P1P1 P2P2 V2V2 P 1, V 1 P 2, V 2 Sand Reversible

26. Generalized Observation: A reversible process produces more work than an irreversible process.

27. Pairs Exercise #1 The initial conditions for 1 mol of air in a piston/cylinder are 5 atm and 300 K. The piston decreases the pressure to final conditions of 1 atm and 300 K. Calculate the work (J) produced from the gas using a.Irreversible expansion by removing a weight from the piston b.Reversible expansion

28. Compressor V1V1 P1P1 P2P2 V2V2 Required Work Irreversible Reversible V1V1 P1P1 P2P2 V2V2 P 1, V 1 P 2, V 2 P 1, V 1 P 2, V 2 Sand

29. Generalized Observation: A reversible process requires less work than an irreversible process.

30. Pairs Exercise #2 The initial conditions for 1 mol of air in a piston/cylinder are 1 atm and 300 K. The piston increases the pressure to final conditions of 5 atm and 300 K. Calculate the work (J) required to compress the gas using a.Irreversible compression by adding a weight to the piston b.Reversible expansion

31. Note: For the reversible case, the work produced by the expansion was identical to the work required by the compression.

32. Generalized Observation: A reversible process that has a given work output when run in the forward direction requires the same work input when run in the reverse direction.

33. Work Heat Work Heat Expansion Compression P V P V Many irreversible paths, but only one reversible path. Each path has its own work and heat.

34. Suppose you have 1000 Btu available at 400 o F, 100 o F, and 60 o F. What could you do with it? 400 o F: 1000 Btu 1 lb of 250-psia steam useful work 100 o F: 1000 Btu home heating 60 o F: 1000 Btu ambient environment Observation: Heat flows from higher temperatures to lower temperatures, but becomes less useful as it does so. 1000 Btu 2000 Btu 3000 Btu

35. How can we quantify the notion that heat available at a higher temperature is more useful than heat available at a lower temperature? The following combinations of heat and temperature may be proposed: where Q rev indicates the heat associated with a reversible process. Of these possibilities, Rudolf Clausius found the following term was useful which he defined as entropy. Input to the system being studied.

36. System Boundary Initial State T S initial T Final State Q rev T S final State quantity Path quantity State quantity Rule 9, page 490: An algebraic combination of a well-defined path quantity with a state quantity is a state quantity. This is why it is important to specify reversible path.

37. Work Reversible Expansion T T Q rev T S1S1 S2S2 V1V1 V2V2  E k +  E p +  U = W in - W out + Q in - Q out 0 0 000 0 Energy Accounting (closed system) initial final

38. Pairs Exercise #3 a. Calculate the entropy change of 1 mole of constant-temperature gas that is reversibly expanded from 1 m 3 to 5 m 3. b. Calculate the entropy change of 1 mole of constant-temperature gas that is irreversibly expanded from 1 m 3 to 5 m 3.

39. The entropy of the system is a state quantity and does not depend upon the path, whether reversible or irreversible. Observation

40. The entropy increases when the volume increases. In the larger volume the gas is more “disordered” so more entropy corresponds to more disorder. Observation

41. Work Reversible Compression T T Q rev T S1S1 S2S2 V1V1 V2V2 initial  E k +  E p +  U = W in - W out + Q in - Q out 0 0 0 0 0 0 Energy Accounting (closed system) final

42. Pairs Exercise #4 a. Calculate the entropy change of 1 mole of constant-temperature gas that is reversibly compressed from 5 m 3 to 1 m 3. b. Calculate the entropy change of 1 mole of constant-temperature gas that is irreversibly compressed from 5 m 3 to 1 m 3.

43. The entropy decreases when the volume decreases. In the smaller volume the gas is less “disordered” so less entropy corresponds to less disorder. Observation

44. Cycle – A system that returns to the initial conditions TS1S1 V1V1 TS2S2 V2V2 initial Expansion Compression

45. Pairs Exercise #5 a. Calculate the entropy change of 1 mole of constant-temperature gas that is reversibly expanded from 1 m 3 to 5 m 3 and then reversibly compressed from 1 m 3 to 5 m 3. b.Calculate the entropy change of 1 mole of constant-temperature gas that is irreversibly expanded from 1 m 3 to 5 m 3 and then irreversibly compressed from 1 m 3 to 5 m 3.

46. For a cycle, the system entropy does not change, regardless of whether the path is reversible or irreversible. Observation

47. Reversible Expander  E k +  E p +  U = W in - W out + Q in - Q out 0 0 000 0 Energy Accounting (closed system) V1V1 P1P1 P2P2 V2V2 P 1, V 1 P 2, V 2 Sand W out Q in

48. What happens from the perspective of the surroundings? Q out (from the perspective of the water bath surroundings) Q in,gas T W out Q out,surr T Q in (from the perspective of the gas) Negative because entropy is defined based upon heat input. Here we have output.

49. Reversible Compressor  E k +  E p +  U = W in - W out + Q in - Q out 0 0 0 0 0 0 Energy Accounting (closed system) V1V1 P1P1 P2P2 V2V2 P 1, V 1 P 2, V 2 Sand W in Q out

50. What happens from the perspective of the surroundings? Q in (from the perspective of the water bath surroundings) Q out,gas T W in Q in,surr T Q out (from the perspective of the gas)

51. What happens to the surroundings for a cyclical reversible process? TP2P2 V2V2 TP1P1 V1V1 initial Compression Expansion

52. For a reversible cycle, the entropy of the surroundings does not change. Observation

53. What happens to the universe for a cyclical reversible process? TP2P2 V2V2 TP1P1 V1V1 initial Compression Expansion

54. For a reversible cycle, the entropy of the universe does not change. Observation

55. Irreversible Expander  E k +  E p +  U = W in - W out + Q in - Q out 0 0 000 0 Energy Accounting (closed system) V1V1 P1P1 P2P2 V2V2 P 1, V 1 P 2, V 2 T T W out Q in T

56. What happens from the perspective of the surroundings? Q out (from the perspective of the water bath surroundings) Q in,gas T W out Q out,surr T Q in (from the perspective of the gas) Negative because entropy is defined based upon heat input. Here we have output.

57. Irreversible Compressor  E k +  E p +  U = W in - W out + Q in - Q out 0 0 0 0 0 0 Energy Accounting (closed system) V1V1 P1P1 P2P2 V2V2 P 1, V 1 P 2, V 2 W in T T Q out T

58. What happens from the perspective of the surroundings? Q in (from the perspective of the water bath surroundings) Q out,gas T W in Q in,surr T Q out (from the perspective of the gas)

59. What happens to the surroundings for a cyclical irreversible process? TP2P2 V2V2 TP1P1 V1V1 initial Compression Expansion Positive

60. For an irreversible cycle, the entropy of the surroundings always increases. Observation

61. What happens to the universe for a cyclical irreversible process? TP2P2 V2V2 TP1P1 V1V1 initial Compression Expansion Positive

62. For a irreversible cycle, the entropy of the universe increases. Observation

63. Restatement of the second law of thermodynamics… For any process that occurs in nature,

64. Entropy Accounting 0