Chapter 12 Thermodynamics

Transcription

1 Chapter 12 Thermodynamics GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: PV diagram efficiency of a heat engine isochoric process Carnot cycle isobaric process refrigerator isothermal process coefficient of performance adiabatic process of a refrigerator heat engine Laws of Thermodynamics State three laws of thermodynamics, and explain the operation of a physical system in terms of these laws. Thermodynamics Problems Solve problems consistent with the laws of thermodynamics. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 5, Energy, and Chapter 10, Temperature and Heat. 107

2 Chapter 12 Thermodynamics OVERVIEW The problem of converting heat energy into useful work energy is an important consideration for an industrial society. The Three Laws of Thermodynamics stated in this chapter express our understanding of how thermal processes operate. As you look over this chapter, please note the importance of the idea of system and the idealized processes by which a system can be changed. SUGGESTED STUDY PROCEDURE As you begin your study of this chapter, carefully read the three Chapter Goals: Definitions, Laws of Thermodynamics, and Thermodynamic Problems. If you need additional assistance with any of the terms listed under Definitions, refer to the next section of this chapter. Next, read chapter sections Be sure to read the examples given at the end of most of these text sections. Answers to most questions posed during your reading are discussed in the second section of this Study Guide chapter. Now read the Chapter Summary and complete Summary Exercises 1-7. Then do Algorithmic Problems 1-5. Now complete Exercises and Problems 1, 3, 5, 6, 11, 12, 17, 19, and 21. For additional work, see the Examples section of this Study Guide. Now you should be prepared to attempt the Practice Test at the end of this Study Guide chapter. If you have difficulty with any of the concepts, check the appropriate section of the text for further help Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions 12.1,12.2,12.7, 1,2, ,5 Laws of 12.4, Thermodynamics Thermodynamics 12.3,12.5,12.6, 7 1,2,3,4, 1,3,6,11, Problems 12.10, ,17,19,

3 DEFINITIONS P-V DIAGRAM Graphical representation of a process using pressure-volume axes. (See Fig. 12.2) The work done during a process is shown as the area under the curve. ISOCHORIC PROCESS Takes place at constant volume. After hot foods are sealed in glass containers during canning, the cooling of a constant volume of material helps to seal the can shut. ISOBARIC PROCESS Takes place at constant pressure. The expansion of a heated gas in a flexible container open to the atmosphere illustrates this process. ISOTHERMAL PROCESS Takes place at constant temperature. The slow compression of a gas can occur at constant temperature. ADIABATIC PROCESS A process that takes place with no change in thermal energy in the system. When we make a rapid change in the pressure of the air in a bicycle tire, we can think of it as an adiabatic process. HEAT ENGINE Absorbs a quantity of energy from higher temperature reservoir, does work, and rejects a quantity of energy to lower temperature reservoir, and returns to its original state. An automobile engine is a heat engine. It makes use of the thermodynamic properties of combustion and gases to convert the chemical energy of gasoline into the kinetic energy of the automobile. EFFICIENCY OF HEAT ENGINE Ratio of useful work output divided by energy input. If we can improve the efficiency of our automobile engines we can save many gallons of gasoline. CARNOT CYCLE Cycle of an ideal engine which includes an isothermal expansion at higher temperature, an adiabatic expansion, an isothermal compression at lower temperature, and an adiabatic compression. REFRIGERATOR A heat engine that operates by the input of work when energy is extracted from a lower temperature reservoir and transferred to a higher temperature reservoir. 109

4 COEFFICIENT OF PERFORMANCE OF A REFRIGERATOR Ratio of heat absorbed to the amount of work supplied to the refrigerator. EXAMPLES THERMODYNAMIC PROBLEMS 1. A container sealed by a moveable piston holds moles of an ideal gas. The gas is originally at a pressure of 1.01 x 10 5 N/m 2, a temperature of 300 ø K and a volume of 22.4 liters. The gas was heated at constant volume until the temperature reached 750 ø K. Then the gas was allowed to expand isothermal until it reached its original pressure. Then the gas was compressed isobarically back to its original state. Draw a P-V diagram for the three processes. Calculate the pressure, volume, and temperature of the gas at the end of each process. Calculate the work done, the change in internal energy and the amount of heat added to or subtracted from the system during each of the processes. What is the efficiency of this cycle of three processes? What Data Are Given? The initial conditions for the moles of gas are given: P 1 = 1.01 x 10 5 N/m 2 ; V 1 = 22.4 x 10 3 cm 3 = 2.24 x 10-2 m 3 ; T 1 = 300 ø K. The highest temperature T 2 = 750 ø K = T 3. The three processes between the initial state, the second state, the third state, and the return to the initial state are all given. What Data Are Implied? The fact that the confined gas is an ideal gas allows us to use equation (12.6), (12.8), and (12.12) to calculate properties of a system where Cv=3/2R; Cp=5/2R. What Physics Principles Are Involved? The first law of thermodynamics can be used in conjunction with the ideal gas laws to solve this problem. What Equations Are to be Used? Process 1 = isochoric compression P 2 = (P 1 T 2 )/T 1 (10.15) ΔU = ΔQ = nc v ΔT (12.6) Process 2 - isothermal expansion P 3 = (P 2 V 2 )/V 3 (10.15) ΔQ = ΔW = PΔV = P 2 V 2 ln (V 3 /V 2 ) (12.12) Process 3 - isobaric compression V 1 = (P 3 V 3 T 1 )/(P 1 T 3 ) (10.15) ΔQ = nc p ΔT (12.8) ΔU = nc v ΔT (12.6) Algebraic Solution 110

5 (a) The P-V diagram - let T 2 be given as some constant times T 1, T 2 = kt 1 ; then P 2 = kp 1 ; V 2 = V 1 Then T 3 = T 2 ; but P 3 = P 1 ; so V 3 = kv 1. ΔQ 12 = nc v (T 2 - T 1 ); ΔQ 23 = P 2 V 2 ln (V 3 /V 2 ) ΔQ 31 = nc p (T 1 - T 3 ); ΔU 12 = ΔQ 12 ; ΔU 23 = 0; ΔU 31 = nc v (T 1 - T 3 ) efficiency = (ΔW 23 + ΔW 31 )/ΔW 23 Numerical Solution P 2 = (1.01 x 10 5 N/m 2 ) (750 ø K/300 ø K) = 2.53 x 10 5 N/m 2 V 2 = V 1 = 2.24 x 10-2 m 3 V 3 = 5.60 x 10-2 m 3 ΔQ 12 = (0.908 mole)(3/2)(8.31 J/mole ø K) x (750 ø K ø K) = 5.09 x 10 3 J ΔW 23 = ΔQ 23 = (2.53 x 10 5 N/m 2 )(2.24 x 10-2 m 3 ) x ln (5.60 x 10-2 )/(2.24 x 10-3 ) = 5.19 x 10 3 J ΔQ 31 = (0.908 moles)(5/2)(8.31 J/mole ø K) x ( ) ø K = x 10 3 J ΔU 31 = (0.908 moles)(3/2)(8.31 J/mole ø K)( ) ø K = x 10 3 J ΔW 31 = P 1 (V 1 - V 3 ) = (1.05 x 10 5 N/m 2 )(2.24 x 10-2 m x 10-2 m 3 ) ΔW 31 = x 10 3 J Efficiency = (5.19 x x 10 3 ) J/ (5.19 x 10 3 J) = 34.0% 111

6 PRACTICE TEST 1. A Carnot engine operates between two temperatures of 600 ø and 300 ø C. a. Calculate this engine's efficiency. b. If the engine absorbs 1000 of heat energy at the higher temperature, how much heat is exhausted at the lower temperature? c. How much mechanical work is produced by the engine per cycle? 2. A heat engine is operated through the cycle process A B C D A as illustrated below. Identify each process as isochoric, isobaric, or isothermal B C C D D A b. Calculate the work done by the system during each cycle and give the net work done per cycle. WA B = WB C = WC D = WD A = Net work done = (per cycle) c. If 60 x 10 5 of heat energy are absorbed during the thermal process A-B, what internal energy change occurs during this process? 3. Is it possible to cool a kitchen during the summer by leaving the door of the refrigerator open? Defend your answer using the Laws of Thermodynamics. ANSWERS: 1. a. 34 b. 660 c a. isochoric, isobaric, isochoric b x 10 6, zero, -3 x 10 5, zero, +7.5 x 10 5 c x No, the heat produced in running the electric motor for cooling exceeds the heat removed from the air inside the refrigerator compartments. This arrangement would be possible only if the system was vented to allow for pumping the heat to the outside. The first law predicts that this is true; the kitchen system has energy added in the form of electrical energy. Thus, when converted to heat, the overall effect is to raise temperature. The second law also predicts the temperature increase; for any system, the entropy either increases or stays constant. Thus, another system (i.e., outdoors) must be involved to cool (or lower the entropy) one system. 112