Accounting for Entropy Class 28.2

1. Accounting for Entropy Class 28.2

2. 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

3. A “natural” process… CO2 H2O Motion Air turbulence Tire deformation Head lights Air conditioning Stereo Hot exhaust Gasoline Air

4. An “unnatural” process… CO2 H2O Motion Air turbulence Tire deformation Head lights Air conditioning Stereo Hot exhaust Gasoline Air Although we all recognize this is impossible, it is still allowed by the first law of thermodynamics (conservation of energy).

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

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

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

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

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

10. 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.

11. Universe System Surroundings 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.

12. 1 Initial state of the system Amount of weight determines friction System boundary

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17. 6 Final state of the system. (Same as initial state.) More weight here causes more weight to be on the floor. This weight controls the amount of irreversibility in the system. Surroundings have changed. (Two weights now on the floor.)

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

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

20. . . . . . . . . . . Steam 100oC Ice Bath 0oC Time passes Copper rod 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.)

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

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

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

24. Expander Irreversible P1 Work Produced P2 P1 , V1 P2 , V2 V1 V2 Reversible Sand P1 P2 P1 , V1 P2 ,V2 V1 V2

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

26. 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 • Irreversible expansion by removing a weight from the piston • Reversible expansion

27. Compressor Irreversible P1 P2 Required Work P1 , V1 P2 , V2 V1 V2 Reversible Sand P1 P2 P1 , V1 P2 ,V2 V1 V2

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

29. 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 • Irreversible compression by adding a weight to the piston • Reversible expansion

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

31. 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.

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

33. 400oF: 1000 Btu 1 lb of 250-psia steam useful work 100oF: 1000 Btu home heating 60oF: 1000 Btu ambient environment Suppose you have 1000 Btu available at 400oF, 100oF, and 60oF. What could you do with it? 1000 Btu 2000 Btu 3000 Btu Observation: Heat flows from higher temperatures to lower temperatures, but becomes less useful as it does so.

34. 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 Qrevindicates the heat associated with a reversible process. Of these possibilities, Rudolf Clausius found the following term was useful Input to the system being studied. which he defined as entropy.

35. System Boundary T Sfinal T Sinitial T Final State Initial State Qrev 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.

36. initial final Energy Accounting (closed system) 0 0 0 0 0 0 DEk + DEp + DU = Win - Wout + Qin - Qout Reversible Expansion Work T T V2 S2 T V1 S1 Qrev

37. Pairs Exercise #3 a. Calculate the entropy change of 1 mole of constant-temperature gas that is reversibly expanded from 1 m3 to 5 m3. b. Calculate the entropy change of 1 mole of constant-temperature gas that is irreversibly expanded from 1 m3 to 5 m3.

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

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

40. final DEk + DEp + DU = Win - Wout + Qin - Qout Energy Accounting (closed system) 0 0 0 0 0 0 Reversible Compression Work T T V2 S2 T V1 S1 Qrev initial

41. Pairs Exercise #4 a. Calculate the entropy change of 1 mole of constant-temperature gas that is reversibly compressed from 5 m3 to 1 m3. b. Calculate the entropy change of 1 mole of constant-temperature gas that is irreversibly compressed from 5 m3 to 1 m3.

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

43. T V2 S2 T V1 S1 initial Cycle – A system that returns to the initial conditions Expansion Compression

44. Pairs Exercise #5 a. Calculate the entropy change of 1 mole of constant-temperature gas that is reversibly expanded from 1 m3 to 5 m3 and then reversibly compressed from 1 m3 to 5 m3. b.Calculate the entropy change of 1 mole of constant-temperature gas that is irreversibly expanded from 1 m3 to 5 m3 and then irreversibly compressed from 1 m3 to 5 m3.

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

46. Reversible Expander Wout Qin Energy Accounting (closed system) 0 0 0 0 0 0 DEk + DEp + DU = Win - Wout + Qin - Qout Sand P1 P2 P1 , V1 P2 ,V2 V1 V2

47. What happens from the perspective of the surroundings? Qin,gas T Wout Qout,surr T Qin(from the perspective of the gas) Qout(from the perspective of the water bath surroundings) Negative because entropy is defined based upon heat input. Here we have output.

48. Reversible Compressor DEk + DEp + DU = Win - Wout + Qin - Qout Energy Accounting (closed system) 0 0 0 0 0 0 Win Sand P1 P2 P1 , V1 P2 ,V2 V1 V2 Qout

49. What happens from the perspective of the surroundings? Win Qin,surr Qout,gas T T Qout(from the perspective of the gas) Qin(from the perspective of the water bath surroundings)

50. Expansion T V2 P2 Compression T V1 P1 initial What happens to the surroundings for a cyclical reversible process?