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AP Physics B Lecture Notes

The Laws of Thermodynamics. AP Physics B Lecture Notes. The Laws of Thermodynamics. Topics. 12-01 Work in Thermodynamic Processes. 12-02 The First Law of Thermodynamics. 12-03 Heat Engines and the Second Law of Thermodynamics. 12-04 Entropy. Work in Thermodynamic Processes. F.

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AP Physics B Lecture Notes

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  1. The Laws of Thermodynamics AP Physics B Lecture Notes

  2. The Laws of Thermodynamics Topics 12-01 Work in Thermodynamic Processes 12-02 The First Law of Thermodynamics 12-03 Heat Engines and the Second Law of Thermodynamics 12-04 Entropy

  3. Work in Thermodynamic Processes F DV = A Dx Dx W = -F Dx W= -PA Dx W = -P DV Area During a compression: Work done on a gas is positive. DV During an expansion: Work done by a gas is negative.

  4. The First Law of Thermodynamics Q W System U Environment DU = Q + W First Law of Thermodynamics

  5. The First Law of Thermodynamics Isotherms (lines of constant temperature) Pressure T4 T3 T2 T1 Volume Pressure - Volume Graph P Area under curve represents work Internal energy is proportional to temperature V

  6. The First Law of Thermodynamics Common Processes A. Isobaric B. Isochoric (Isovolumetric) C. Isothermal D. Adiabatic

  7. The First Law of Thermodynamics Po DV Isobaric Expansion (Constant Pressure) During an expansion: Work done by a gas is negative. P DU is positive a) W = -Po DV T4 b) DU increases T3 T2 c) Q = DU - W T1 Heat must be added V First Law of Thermodynamics: DU = Q + W

  8. The First Law of Thermodynamics Po DV Isobaric Compression (Constant Pressure) During a compression: Work done on a gas is positive. P DU is Negative a) W = -Po DV T4 b) DU decreases T3 c) Q = DU - W T2 T1 Heat must be removed V First Law of Thermodynamics: DU = Q + W

  9. The Laws of Thermodynamics 12-01 P o V The process shown on the Pressure-Volume diagram is an (A) adiabatic expansion. (B) isothermal expansion. (C) isometric expansion. (D) isobaric expansion.

  10. The First Law of Thermodynamics Po Pf Isochoric (Constant Volume) Decrease in Pressure P DU is Negative a) W = 0 b) DU decreases c) Q = DU T3 T2 Heat must be removed T1 V First Law of Thermodynamics: DU = Q + W

  11. The First Law of Thermodynamics Pf Po Isochoric (Constant Volume) Increase in Pressure P DU is positive a) W = 0 b) DU increases c) Q = DU T3 T2 Heat must be added T1 V First Law of Thermodynamics: DU = Q + W

  12. The Laws of Thermodynamics 12-02 P o V The process shown on the PV diagram is (A) adiabatic. (B) isothermal. (C) isochoric. (D) isobaric.

  13. The Laws of Thermodynamics 12-03 In an isochoric process, there is no change in (A) pressure. (B) temperature. (C) volume. (D) internal energy.

  14. The First Law of Thermodynamics Po Pf Vo Vf Isothermal (Constant Temperature) Expansion During an expansion: Work done by a gas is negative. P a) W is negative b) DU = 0 T3 c) Q = -W T2 T1 Heat must be added V First Law of Thermodynamics: DU = Q + W

  15. The First Law of Thermodynamics Pf Po Vo Vf Isothermal (Constant Temperature) Compression During a compression: Work done on a gas is positive. P a) W is positive b) DU = 0 T3 c) Q = -W T2 T1 Heat must be removed V First Law of Thermodynamics: DU = Q + W

  16. The Laws of Thermodynamics 12-04 T o V The process shown on the Temperature-Volume graph is an (A) adiabatic compression. (B) isothermal compression. (C) isochoric compression. (D) isobaric compression.

  17. The Laws of Thermodynamics 12-05 When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an isothermal process, (A) ΔU = 0 (B) W = 0 (C) Q = 0 (D) none of the above

  18. The Laws of Thermodynamics 12-06 An ideal gas is compressed to one-half its original volume during an isothermal process. The final pressure of the gas (A) increases to twice its original value. (B) increases to less than twice its original value. (C) increases to more than twice its original value. (D) does not change.

  19. The First Law of Thermodynamics Po Pf Vo Vf Adiabatic (No Heat Exchange) Expansion During an expansion: Work done by a gas is negative. P DU is Negative a) W = Negative b) DU decreases T2 c) Q = 0 W = DU T1 Negative work done by the gas is equal to the decrease in internal energy. V First Law of Thermodynamics: DU = Q + W

  20. The First Law of Thermodynamics Pf Po Vf Vo Adiabatic (No Heat Exchange) Compression During a compression: Work done on a gas is positive. P DU is positive a) W = Positive b) DU increases T2 c) Q = 0 W = DU T1 Positive work done on the gas is equal to the increase in internal energy. V First Law of Thermodynamics: DU = Q + W

  21. The Laws of Thermodynamics 12-07 When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an adiabatic process, (A) ΔU = 0. (B) W = 0. (C) Q = 0. (D) none of the above

  22. The First Law of Thermodynamics (Problem) P Po Pf T Vo Vf V An ideal gas expands to 10 times its original volume, maintaining a constant 440 K temperature. If the gas does 3.3 kJ of work on its surroundings, how much heat does it absorb? First Law of Thermodynamics: Isothermal process: DU = 0 W = -3000 J

  23. The Laws of Thermodynamics 12-08 2P P 3V V 2V 4V A gas is taken through the cycle illustrated here. During one cycle, how much work is done by an engine operating on this cycle? (A) PV (B) 2PV (C) 3PV (D) 4PV

  24. The First Law of Thermodynamics (Problem) 2Po Po Vo 2Vo 3Vo II An ideal gas initially has pressure Po, at volume Vo and absolute temperature To. It then undergoes the following series of processes: III I IV I. Heated, at constant volume to pressure 2Po II. Heated, at constant pressure to pressure 3Vo III. Cooled, at constant volume to pressure Po IV. Cooled, at constant pressure to volume Vo

  25. The First Law of Thermodynamics (Problem) con’t 2Po Po Vo 2Vo 3Vo II 2To 6To III I To IV 3To Find the temperature at each end point in terms of To

  26. The First Law of Thermodynamics (Problem) con’t II 2Po III I Find the net work done by the gas in terms of Po and Vo Po IV Vo 2Vo 3Vo Net work equals net area under curve

  27. The First Law of Thermodynamics (Problem) con’t 2Po Po Vo 2Vo 3Vo II III I Find the net change in internal energy in terms of Po and Vo IV

  28. Heat Engines and the Second Law of Thermodynamics The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one: Heat will flow spontaneously from a hot object to a cold object. It will not flow spontaneously from a cold object to a hot object.

  29. Heat Engines and the Second Law of Thermodynamics Direction of Time

  30. Heat Engines and the Second Law of Thermodynamics High Temp. Th Qh Engine W Qc Low Temp. Tc In a heat engine; mechanical energy can be obtained from thermal energy when heat flows from a higher temperature to a lower temperature. Work done by engine: Thermal efficiency:

  31. Heat Engines and the Second Law of Thermodynamics Th= 550 K = 920 J Qh For the engine Engine W = 630 J Qc Tc Work done by engine each cycle The efficiency of the engine

  32. Heat Engines and the Second Law of Thermodynamics The Carnot Cycle Isothermal Expansion P Th Adiabatic Expansion Adiabatic Compression Tc V Isothermal Compression

  33. Heat Engines and the Second Law of Thermodynamics P High Temp. Th Qh Th Engine W Qc Tc Low Temp. Tc V Work Carnot efficiency:

  34. Heat Engines and the Second Law of Thermodynamics Th= 550 K = 890 J Qh For the engine Engine W = 470 J Qc Tc Work done by engine each cycle The efficiency of the engine

  35. Heat Engines and the Second Law of Thermodynamics Th= 550 K = 890 J Qh Temperature of the cool reservoir Engine W = 420 J = 470 J Qc Tc The engine undergoes 22 cycles per second, its mechanical power output

  36. Heat Engines and the Second Law of Thermodynamics (Problem) Th A carnot engine absorbs 900 J of heat each cycle and provides 350 J of work = 900 J Qh Engine W = 350 J Qc The efficiency of the engine Tc The heat ejected each cycle

  37. Heat Engines and the Second Law of Thermodynamics (Problem) Th A carnot engine absorbs 900 J of heat each cycle and provides 350 J of work = 900 J Qh Engine W = 350 J =550 J The engine ejects heat at 10 oC The temperature of the hot reservoir Qc Tc = 283 K

  38. Heat Engines and the Second Law of Thermodynamics (Problem) Th= 650 K = 400 J Qh Engine W = ? Qc Tc= 300 K A carnot engine operates between a hot reservoir at 650 K and a cold reservoir at 300 K. If it absorbs 400 J of heat at the hot reservoir, how much work does it deliver?

  39. The Laws of Thermodynamics 12-09 If the theoretical efficiency of a Carnot engine is to be 100%, the heat sink must be (A) at absolute zero. (B) at 0°C. (C) at 100°C. (D) infinitely hot.

  40. The Laws of Thermodynamics 12-10 A Carnot cycle consists of (A) two adiabats and two isobars. (B) two isobars and two isotherms. (C) two isotherms and two isomets. (D) two adiabats and two isotherms.

  41. Entropy Definition of the change in entropy S when an amount of heat Q is added: Another statement of the second law of thermodynamics: The total entropy of an isolated system never decreases. When an irreversible process occurs in a closed system, the entropy S of the system always increases: it never decreases.

  42. Entropy Entropy is a measure of the disorder of a system. This gives us yet another statement of the second law: Natural processes tend to move toward a state of greater disorder. Example: If you put milk and sugar in your coffee and stir it, you wind up with coffee that is uniformly milky and sweet. No amount of stirring will get the milk and sugar to come back out of solution.

  43. Entropy Another example: when a tornado hits a building, there is major damage. You never see a tornado approach a pile of rubble and leave a building behind when it passes. Thermal equilibrium is a similar process – the uniform final state has more disorder than the separate temperatures in the initial state.

  44. Entropy Another consequence of the second law: In any natural process, some energy becomes unavailable to do useful work. If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the heat death of the universe.

  45. The Laws of Thermodynamics 12-11 The second law of thermodynamics leads us to conclude that (A) the total energy of the universe is constant. (B) disorder in the universe is increasing with the passage of time. (C) it is theoretically possible to convert heat into work with 100% efficiency. (D) the average temperature of the universe is increasing with the passage of time.

  46. The Law of Thermodynamics Summary First law of thermodynamics: Work done by gas at constant pressure: Efficiency of a heat engine: Carnot efficiency: Isothermal process: temperature is constant. Adiabatic process: no heat is exchanged. Heat engine changes heat into useful work (requires temperature difference).

  47. The Law of Thermodynamics Summary Change in entropy: Second law of thermodynamics: heat flows spontaneously from a hot object to a cold one, but not the reverse Thermal energy cannot be changed entirely to work natural processes tend to increase entropy. Entropy is a measure of disorder. As time goes on, less and less energy is available to do useful work.

  48. END

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