Respostas capítulo 19 Halliday, Exercícios de Física

Baixe Respostas capítulo 19 Halliday e outras Exercícios em PDF para Física, somente na Docsity! 1867 Chapter 19 The Second Law of Thermodynamics Conceptual Problems 1 • Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? (a) 25%, (b) 50%, (c) 75%, (d) 100%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. The percentage of the heat of combustion (heat absorbed from the high- temperature reservoir) is the ratio of Qc to Qh. We can use the relationship between W, Qh, and Qc ( ch QQW −= ) to find Qc/ Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: h c h ch h 1 Q Q Q QQ Q W −= − ==ε Solving for Qc/ Qh yields: ε−=1 h c Q Q Substitute for ε to obtain: 75.025.01 h c =−= Q Q and ( )c is correct. 2 • If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? (a) 20%, (b) 25%, (c) 80%, (d) 400%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( ch QQW −= ) to express the efficiency of the heat engine in terms of Qc and W. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: W QQW W Q W cch 1 1 + = + ==ε Chapter 19 1868 Substitute for Qc and W to obtain: 2.0 kJ 100 kJ 4001 1 = + =ε and ( )a is correct. 3 • If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( ch QQW −= ) to express the efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: h c h ch h 1 Q Q Q QQ Q W −= − ==ε Substitute for Qc and Qh to obtain: 2.0 kJ 600 kJ 4801 =−=ε and ( )a is correct. 4 • Explain what distinguishes a refrigerator from a ″heat pump.″ Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its ″natural″ direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter. 5 • [SSM] An air conditioner’s COP is mathematically identical to that of a refrigerator, that is, c AC refCOP COP= = Q W . However a heat pump’s COP is defined differently, as h hpCOP = Q W . Explain clearly why the two COPs are defined differently. Hint: Think of the end use of the three different devices. Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh. The Second Law of Thermodynamics 1871 Substitute numerical values and evaluate COPref, C: 0.3 1 K 300 K 400 1COP C ref, = − = ( )c is correct. 12 •• On a humid day, water vapor condenses on a cold surface. During condensation, the entropy of the water (a) increases, (b) remains constant, (c) decreases, (d) may decrease or remain unchanged. Explain your answer. Determine the Concept When water vapor condenses, its entropy decreases (the liquid state is a more ordered state than is the vapor state) and the entropy of the universe increases. )(a is correct. 13 •• An ideal gas is taken reversibly from an initial state Pi, Vi, Ti to the final state Pf, Vf, Tf. Two possible paths are (A) an isothermal expansion followed by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. For these two paths, (a) ΔEint A > ΔEint B, (b) ΔSA > ΔSB, (c) ΔSA < ΔSB, (d) None of the above. Determine the Concept The two paths are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem. P V iV fV fP iP i f BB A A iT fT (a) Because Eint is a state function and the initial and final states are the same for the two paths, B int,A int, EE Δ=Δ . (b) and (c) S, like Eint, is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs. Thus BA SS Δ=Δ . (d) )(d is correct. 14 •• Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST diagram. Identify this cycle and sketch it on a PV diagram. Chapter 19 1872 Determine the Concept The processes A→B and C→D are adiabatic and the processes B→C and D→A are isothermal. Therefore, the cycle is the Carnot cycle shown in the adjacent PV diagram. P V A B C D 15 •• Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV diagram. Identify the type of engine represented by this diagram. Determine the Concept Note that A→B is an adiabatic expansion, B→C is a constant-volume process in which the entropy decreases, C→D is an adiabatic compression and D→A is a constant-volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). The points A, B, C, and D in Figure 19-13 correspond to points c, d, a, and b, respectively, in Figure 19-3. 16 •• Sketch an ST diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.) Determine the Concept The Otto cycle consists of four quasi-static steps. Refer to Figure 19-3. There a→b is an adiabatic compression, b→c is a constant volume heating, c→d is an adiabatic expansion and d→a is a constant-volume cooling. So, from a to b, S is constant and T increases, from b to c, heat is added to the system and both S and T increase, from c→d S is constant while T decreases, and from d to a both S and T decrease. To determine how S depends on T along b→c and d→a, consider the entropy change of the gas from point b to an arbitrary point on the path b→c where the entropy and temperature of the gas are S and T, respectively: T QS =Δ where, because heat is entering the system, Q is positive. Because Won = 0 for this constant- volume process: ( )bTTCTCQQE −=Δ===Δ VVinint Substituting for Q yields: ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= − =Δ T TC T TTCS bb 1V V The Second Law of Thermodynamics 1873 On path b→c the entropy is given by: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+=Δ+= T TCSSSS b bb 1V The first and second derivatives, dTdS and 22 dTSd , give the slope and concavity of the path. Calculate these derivatives assuming CV is constant. (For an ideal gas CV is a positive constant.): 2V T TC dT dS b= 3V2 2 2 T TC dT Sd b−= These results tell us that, along path b→c, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. Following the same procedure on path d→a gives: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+= T TCSS d d 1V 2V T TC dT dS d= 3V2 2 2 T TC dT Sd d−= These results tell us that, along path d→a, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. An ST diagram for the Otto cycle is shown to the right. S T a b cd 17 •• [SSM] Sketch an SV diagram of the Carnot cycle for an ideal gas. Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4→1 is adiabatic, that is, isentropic, and S is constant while V decreases. Chapter 19 1876 disorder, it is not true that order cannot be brought forth from disorder. What is required is an agent doing work – for example, your friend – on the system in order to reduce the level of chaos and bring about order. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the task. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease. Estimation and Approximation 21 • Estimate the change in COP of your electric food freezer when it is removed from your kitchen to its new location in your basement, which is 8°C cooler than your kitchen. Picture the Problem We can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen. If we further assume that the freezer operates in a Carnot cycle, then we can use the proportion chch TTQQ = to express the ratio of the coefficients of performance in terms of the temperatures in the kitchen, basement, and freezer. The ratio of the coefficients of performance in the basement and kitchen is given by: kitc, kitc, basementc, basementc, kit basement COP COP W Q W Q = Because ch QQW −= for a heat engine or refrigerator: kitc,kith, kitc, basementc,basementh, basementc, kit basement COP COP QQ Q QQ Q − − = Divide the numerators and denominators by Qc,basement and Qc,kit and simplify to obtain: 1 1 1 1 1 1 COP COP basementc, basementh, kitc, kith, kitc, kith, basementc, basementh, kit basement − − = − − = Q Q Q Q Q Q Q Q The Second Law of Thermodynamics 1877 If we assume that the freezer unit operates in a Carnot cycle, then c h c h T T Q Q = and our expression for the ratio of the COPs becomes: 1 1 COP COP basementc, basementh, kitc, kith, kit basement − − = T T T T Assuming that the temperature in your kitchen is 20°C and that the temperature of the interior of your freezer is −5°C, substitute numerical values and evaluate the ratio of the coefficients of performance: 47.1 1 K 268 K 285 1 K 268 K 932 COP COP kit basement = − − = or an increase of %47 in the performance of the freezer! 22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room. Picture the Problem The probability that all the molecules in your bedroom are located in the (open) closet is given by N V Vp ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 2 where N is the number of air molecules in your bedroom and V1 and V2 are the volumes of your bedroom and closet, respectively. We can use the ideal-gas law to find the number of molecules N. We’ll assume that the volume of your room is about 50 m3 and that the temperature of the air is 20°C. If the original volume of the air in your bedroom is V1, the probability p of finding the N molecules, normally in your bedroom, confined to your closet whose volume is V2 is given by: N V Vp ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 2 or, because 110 1 2 VV = , N p ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= 10 1 (1) Use the ideal-gas law to express N: kT PVN = Substitute numerical values and evaluate N: ( )( ) ( )( ) molecules 10252.1 K 293J/K 10381.1 m 50kPa 325.101 27 23 3 ×= × = −N Chapter 19 1878 Substitute for N in equation (1) and evaluate p: 27 27 27 27 10 10252.1 10252.1 10252.1 10 10 10 1 10 1 − ×− × × ≈ ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛=p 23 •• [SSM] Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume γ = 1.4. (The Otto cycle is discussed in Section 19-1. ) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. Express the Carnot efficiency of an engine operating between the temperatures Tc and Th: h c C 1 T T −=ε Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasi- static adiabatic compression from Vc to Vh: 1 hh 1 cc −− = γγ VTVT ⇒ 1 c h 1 c 1 h h c − − − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == γ γ γ V V V V T T Substitute for h c T T to obtain: 1 c h C 1 − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= γ ε V V Express the compression ratio r: h c V Vr = Substituting for r yields: 1C 11 −−= γε r Substitute numerical values for r and γ (1.4 for diatomic gases) and evaluate εC: ( ) %56 0.8 11 14.1C ≈−= −ε 24 •• You are working as an appliance salesperson during the summer. One day, your physics professor comes into your store to buy a new refrigerator. Wanting to buy the most efficient refrigerator possible, she asks you about the efficiencies of the available models. She decides to return the next day to buy the most efficient refrigerator. To make the sale, you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator, and (b) and the highest rate possible for the heat to be released by the refrigerator if the refrigerator uses 600 W of electrical power. The Second Law of Thermodynamics 1881 possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both being on a particular side, or a 50% chance of them both being on either side. Extending this logic, the probability of N molecules all being on one side of the box is P = 2/2N, which means that, if the molecules shuffle 100 times a second, the time it would take them to cover all the combinations and all get on one side or the other is ( )1002 2N t = . In (e) we can apply the ideal gas law to find the number of molecules in 1.0 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K. (a) Evaluate t for N = 10 molecules: ( ) s5s12.5 s 1002 2 1 10 ≈== −t (b) Evaluate t for N = 100 molecules: ( ) y102 s 103.156 y 1s1034.6 s 1002 2 20 7 27 1 100 ×≈ × ××= = −t (c) Evaluate t for N = 1000 molecules: ( )1 1000 s 1002 2 −=t To evaluate 10002 let 1000210 =x and take the logarithm of both sides of the equation to obtain: ( ) 10ln2ln1000 x= ⇒ 301=x Substitute to obtain: ( ) y102 s 103.156 y 1s105.0 s 1002 10 291 7 299 1 301 ×≈ × ××= = −t (d) Evaluate t for N = 1.0 mol =6.022 ×1023 molecules: ( )1 10022.6 s 1002 2 23 − × =t To evaluate 2310022.62 × let 2310022.6210 ×=x and take the logarithm of both sides of the equation to obtain: ( ) 10ln2ln10022.6 23 x=× ⇒ 2310≈x Chapter 19 1882 Substituting for x yields: ( ) y10 s 103.156 y 1 s 1002 10 23 23 10 71 10 ≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × ≈ −t (e) Solve the ideal gas law for the number of molecules N in the gas: kT PVN = Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N: ( )( )( ) ( )( ) molecules1022.3 K300J/K10381.1 L 0.1Pa/torr32.133torr10 7 23 12 ×= × = − − N Evaluate t for N = 3.22×107 molecules: ( )1 1022.3 s 1002 2 7 − × =t To evaluate 71022.32 × let 71022.3210 ×=x and take the logarithm of both sides of the equation to obtain: ( ) 10ln2ln1022.3 7 x=× ⇒ 710≈x Substituting for x yields: ( ) y10 s 103.156 y 1 s 1002 10 7 7 10 71 10 ≈ × ×= −t Express the ratio of this waiting time to the lifetime of the universe tuniverse: 7 7 10 10 10 universe 10 y10 y10 ≈= t t or universe 107 10 tt ≈ Heat Engines and Refrigerators 27 • [SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work during each cycle. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be hQW=ε where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) Because, from conservation of energy, ch QWQ += , we can express the efficiency of the engine in terms of the heat Qc The Second Law of Thermodynamics 1883 released to the cold reservoir during each cycle. (a) Qh absorbed from the hot reservoir during each cycle is given by: J500 0.200 J100 h === ε WQ (b) Use ch QWQ += to obtain: J 400J 100J 500hc =−=−= WQQ 28 • A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does 0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be hQW=ε where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) We can apply conservation of energy to the engine to obtain ch QWQ += and solve this equation for the heat Qc released to the cold reservoir during each cycle. (a) The efficiency of the heat engine is given by: %30 J400 J120 h === Q Wε (b) Apply conservation of energy to the engine to obtain: ch QWQ += ⇒ WQQ −= hc Substitute numerical values and evaluate Qc: J280J 120J 400c =−=Q 29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine. Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output. (a) The efficiency of the heat engine is given by: h c h ch h Q 1 Q Q QQ Q W −= − ==ε Substitute numerical values and evaluate ε: %40 J100 J601 =−=ε Chapter 19 1886 For path 1→2: 0Δ 1212 == VPW and ( ) kJ74.3K300K600 Kmol J8.314ΔΔ 2 3 122 3 12V12 =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st law of thermodynamics: oninintΔ WQE += Because 012 =W : kJ 74.3Δ 1212int, == QE For path 2→3: ( )( ) kJ99.4 atmL J101.325L24.6L49.2atm2.00Δ 2323on −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −−=−=−= VPWW ( ) kJ5.12K600K1200 Kmol J8.314ΔΔ 2 5 232 5 23P23 =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ Apply oninintΔ WQE += to obtain: kJ 5.7kJ 99.4kJ 5.12Δ 23 int, =−=E For path 3→4: 03434 =Δ= VPW and ( ) kJ48.7K0021K600 Kmol J8.314ΔΔΔ 2 3 342 3 34V34int,34 −=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ==== TRTCEQ Apply oninintΔ WQE += to obtain: kJ 48.70kJ 48.7Δ 34 int, −=+−=E For path 4→1: ( )( ) kJ49.2 atmL J101.325L2.94L24.6atm1.00Δ 4141on =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −−=−=−= VPWW and ( ) kJ24.6K600K003 Kmol J8.314ΔΔ 2 5 412 5 41P41 −=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ The Second Law of Thermodynamics 1887 Apply oninintΔ WQE += to obtain: kJ 75.3kJ 49.2kJ 24.6Δ 41 int, −=+−=E For easy reference, the results of the preceding calculations are summarized in the following table: Process onW , kJ inQ , kJ ( )oninintΔ WQE += , kJ 1→2 0 3.74 3.74 2→3 −4.99 12.5 7.5 3→4 0 −7.48 −7.48 4→1 2.49 −6.24 −3.75 (b) The efficiency of the cycle is given by: ( ) 2312 4123 in by QQ WW Q W + −+− ==ε Substitute numerical values and evaluate ε: %15 kJ5.12kJ3.74 kJ2.49kJ4.99 ≈ + − =ε Remarks: Note that the work done per cycle is the area bounded by the rectangular path. Note also that, as expected because the system returns to its initial state, the sum of the changes in the internal energy for the cycle is zero. 32 •• The working substance of an engine is 1.00 mol of a diatomic ideal gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10.0 L to a pressure of 1.00 atm and a volume of 20.0 L, (2) a compression at constant pressure to its original volume of 10.0 L, and (3) heating at constant volume to its original pressure. Find the efficiency of this cycle. Picture the Problem The three steps in the process are shown on the PV diagram. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle. V(L) 2 3 1 P (atm) 2.639 2 1 0 0 10.0 20.0 The pressures and volumes at the end points of the adiabatic expansion are related according to: γγ 2211 VPVP = ⇒ 2 1 2 1 P V VP γ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = Chapter 19 1888 Substitute numerical values and evaluate P1: ( ) atm 639.2atm 00.1 L 0.10 L 0.20 4.1 1 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛=P Express the efficiency of the cycle: hQ W =ε (1) No heat enters or leaves the system during the adiabatic expansion: 012 =Q Find the heat entering or leaving the system during the isobaric compression: ( )( ) Latm35.0 L20.0L10.0atm1.00 ΔΔΔ 2 7 232 7 232 7 23V23 ⋅−= −= === VPTRTCQ Find the heat entering or leaving the system during the constant- volume process: ( )( ) Latm0.41 L10.0atm1.00atm2.639 ΔΔΔ 2 5 312 5 312 5 31V31 ⋅= −= === PVTRTCQ Apply the 1st law of thermodynamics to the cycle ( 0cycle int, =ΔE ) to obtain: Latm6.0 Latm 41.0Latm 35.00 Δ 312312 inininton ⋅= ⋅+⋅−= ++= −=−= QQQ QQEW Substitute numerical values in equation (1) and evaluate ε : %15 Latm41 Latm6.0 = ⋅ ⋅ =ε 33 •• An engine using 1.00 mol of an ideal gas initially at a volume of 24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume, (2) cooling at constant volume to a temperature of 300 K (3) an isothermal compression to its original volume, and (4) heating at constant volume to its original temperature of 400 K. Assume that Cv = 21.0 J/K. Sketch the cycle on a PV diagram and find its efficiency. Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle. The Second Law of Thermodynamics 1891 Use the ideal-gas law to find the temperatures at points 2 and 3: ( )( ) ( ) K601 Kmol J8.314mol1.00 L25.0kPa200 22 32 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = == nR VPTT (b) Find the heat entering the system for the constant-volume process from 1 → 2: ( ) kJ3.74K301K601 Kmol J8.314ΔΔ 2 3 122 3 12V12 =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ Find the heat entering or leaving the system for the isothermal process from 2 → 3: ( ) ( ) kJ46.3 L0.52 L0.05lnK016 Kmol J8.314mol1.00ln 2 3 223 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = V VnRTQ Find the heat leaving the system during the isobaric compression from 3 → 1: ( ) kJ24.6K601K301 Kmol J8.314ΔΔ 2 5 312 5 31P31 −=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ (c) Express the efficiency of the cycle: 2312in QQ W Q W + ==ε (1) Apply the 1st law of thermodynamics to the cycle: kJ0.96 kJ6.24kJ3.46kJ.743 312312 = −+= ++==∑ QQQQW because, for the cycle, 0Δ int =E . Substitute numerical values in equation (1) and evaluate ε : %13 kJ3.46kJ3.74 kJ0.96 = + =ε 35 •• An ideal diatomic gas follows the cycle shown in Figure 19-16. The temperature of state 1 is 200 K. Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition; Chapter 19 1892 using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin. (a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1: 2 22 1 11 T VP T VP = ⇒ 1 2 1 11 22 12 P PT VP VPTT == Substitute numerical values and evaluate T2: ( ) ( ) ( ) K600 atm1.0 atm3.0K2002 ==T Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain: 2 3 2 22 33 23 V VT VP VPTT == Substitute numerical values and evaluate T3: ( ) ( ) ( ) K1800 L100 L300K6003 ==T Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain: 3 4 3 33 44 34 P PT VP VPTT == Substitute numerical values and evaluate T4: ( ) ( ) ( ) K600 atm3.0 atm1.0K18004 ==T (b) The efficiency of the cycle is: inQ W =ε (1) Use the area of the rectangle to find the work done each cycle: ( )( ) Latm400 atm1.0atm3.0L100L300 ΔΔ ⋅= −−= = VPW Apply the ideal-gas law to state 1 to find the product of n and R: ( )( ) atm/KL0.50 K200 L100atm1.0 1 11 ⋅= == T VPnR Noting that heat enters the system between states 1 and 2 and states 2 and 3, express Qin: ( )nRTT TnRTnR TCTCQQQ 232 7 122 5 232 7 122 5 23P12V2312in Δ+Δ= Δ+Δ= Δ+Δ=+= The Second Law of Thermodynamics 1893 Substitute numerical values and evaluate Qin: ( )[ ( )] Latm2600 K atmL50.0K600K1800K200K600 2 7 2 5 in ⋅=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −+−=Q Substitute numerical values in equation (1) and evaluate ε : %15 Latm2600 Latm400 = ⋅ ⋅ =ε 36 ••• Recently, an old design for a heat engine, known as the Stirling engine has been promoted as a means of producing power from solar energy. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume, (3) an isothermal expansion of the gas, and (4) cooling of the gas at constant volume. (a) Sketch PV and ST diagrams for the Stirling cycle. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero. Picture the Problem (a) The PV and ST cycles are shown below. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes. P V T Tc h (1) (2) (3) (4) 1 2 3 4 S T Th cT V = 0 V = 0 Δ Δ (b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle: 41342312cycle ΔΔΔΔΔ SSSSS +++= (1) Express the entropy change for the isothermal process from state 1 to state 2: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 2 12 lnΔ V VnRS Chapter 19 1896 Using an equation for a quasistatic adiabatic process, relate the temperatures Ta and Tb to the volumes Va and Vb: 11 −− = γγ bbaa VTVT ⇒ 1 1 − − = γ γ a b ba V VTT (1) Proceeding similarly, relate the temperatures Tc and Td to the volumes Vc and Vd: 11 −− = γγ ddcc VTVT ⇒ 1 1 − − = γ γ d c cd V VTT (2) Use equations (1) and (2) to eliminate Ta and Td: ( )bc a b b d c c TT V VT V VT − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −= − − − − γ ε γ γ γ γ 1 1 1 1 1 Because Va = Vd: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= −− c b a b c b a c T T V V T T V V 1 1 11 γ ε γγ Noting that Pb = Pc, apply the ideal- gas law to relate Tb and Tc: c b c b V V T T = Substitute for the ratio of Tb to Tc and simplify to obtain: ( )bca bc a b a c a b a c a b a c a b c b a c a c a c c b a b c b a c VVV VV V V V V V V V V V V V V V V V V V V V V V V V V V V V V V V − − −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=⋅ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= − −−−− 1 1111 11 1 1 1 γ γγ γγ γγγγ γ γ γγ ε Second Law of Thermodynamics 39 •• [SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Assume that the heat-engine statement of the second law of thermodynamics is false, and show how a perfect engine working The Second Law of Thermodynamics 1897 with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). Suppose the heat-engine statement of the second law is false. Then a ″perfect″ heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). Then, the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram).This violates the refrigerator statement of the second law. ⇓ ⇓ ⇓ ⇓ ⇓⇓⇓ 500 J Cold reservoir at temperature Tc Hot reservoir at temperature Th 800 J 300 J 300 J 300 J Ordinary refrigerator Perfect refrigerator 500 J 500 J a b c( ) ( ) ( ) Perfect heat engine 40 •• If two curves that represent quasi-static adiabatic processes could intersect on a PV diagram, a cycle could be completed by an isothermal path between the two adiabatic curves shown in Figure 19-18. Show that such a cycle violates the second law of thermodynamics. Determine the Concept The work done by the system is the area enclosed by the cycle, where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. There is no heat transfer in the adiabatic expansion or compression. Thus, we would completely convert heat to mechanical energy, without exhausting any heat to a cold reservoir, in violation of the second law of thermodynamics. Carnot Cycles 41 • [SSM] A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 200 K. (a) What is its efficiency? (b) If it Chapter 19 1898 absorbs 100 J of heat from the hot reservoir during each cycle, how much work does it do each cycle? (c) How much heat does it release during each cycle? (d) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using hc /1 TT−=ε and the work done per cycle from ./ hQW=ε We can apply conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition. (a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs: %3.33 K300 K20011 h c C =−=−= T T ε (b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir: ( )( ) J33.3J1000.333hC === QW ε (c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done: J67 J 66.7J33.3J100hc = =−=−= WQQ (d) Using its definition, express and evaluate the refrigerator’s coefficient of performance: 0.2 J33.3 J66.7COP c === W Q 42 • An engine absorbs 250 J of heat per cycle from a reservoir at 300 K and releases 200 J of heat per cycle to a reservoir at 200 K. (a) What is its efficiency? (b) How much additional work per cycle could be done if the engine were reversible? Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from ,hCQW ε= where εC is the Carnot efficiency. (a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the low- temperature reservoir: %0.20 J250 J2001 1 h c h ch h =−= −= − == Q Q Q QQ Q Wε The Second Law of Thermodynamics 1901 (b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir: ( )( ) J3.74J100743.0hC === QW ε (c) Apply conservation of energy to obtain: J26J74.3J001hc =−=−= WQQ (d) Using its definition, express and evaluate the refrigerator’s coefficient of performance: 35.0 J74.3 J26COP c === W Q 47 •• [SSM] In the cycle shown 1 in Figure 19-19, 1.00 mol of an ideal diatomic gas is initially at a pressure of 1.00 atm and a temperature of 0.0ºC. The gas is heated at constant volume to T2 = 150ºC and is then expanded adiabatically until its pressure is again 1.00 atm. It is then compressed at constant pressure back to its original state. Find (a) the temperature after the adiabatic expansion, (b) the heat absorbed or released by the system during each step, (c) the efficiency of this cycle, and (d) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use TQ Δ= VC and TQ Δ= PC to find the heat entering and leaving during the constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once we’ve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures. (a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1: 3 33 1 11 T VP T VP = or, because P1 = P3, 1 3 13 V VTT = (1) Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1: 2 22 1 11 T VP T VP = ⇒ 12 211 2 TV TVPP = Chapter 19 1902 Because V1 = V2: ( ) atm1.55 K273 K423atm1.00 1 2 12 === T TPP Apply an equation for an adiabatic process to relate the pressures and volumes at points 2 and 3: γγ 3311 VPVP = ⇒ γ 1 3 1 13 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = P PVV Noting that V1 = 22.4 L, evaluate V3: ( ) L30.6 atm1 atm1.55L22.4 1.4 1 3 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V Substitute numerical values in equation (1) and evaluate T3 and t3: ( ) K373 L22.4 L30.6K2733 ==T and C10027333 °=−= Tt (b) Process 1→2 takes place at constant volume (note that γ = 1.4 corresponds to a diatomic gas and that CP – CV = R): ( ) kJ3.12 K273K423 Kmol J8.314 ΔΔC 2 5 122 5 12V12 = −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = == TRTQ Process 2→3 takes place adiabatically: 023 =Q Process 3→1 is isobaric (note that CP = CV + R): ( ) kJ2.91 K373K732 Kmol J8.314 ΔΔC 2 7 122 7 31P31 −= −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = == TRTQ (c) The efficiency of the cycle is given by: inQ W =ε (2) Apply the first law of thermodynamics to the cycle: oninintΔ WQE += or, because 0cycle int, =ΔE (the system begins and ends in the same state) and ingas by theon QWW =−= . Evaluating W yields: kJ0.21kJ2.910kJ3.12 312312 =−+= ++==∑ QQQQW The Second Law of Thermodynamics 1903 Substitute numerical values in equation (2) and evaluate ε : %7.6 kJ3.12 kJ0.21 ==ε (d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K: 35.5% K234 K73211 h c C =−=−= T Tε 48 •• You are part of a team that is completing a mechanical-engineering project. Your team built a steam engine that takes in superheated steam at 270ºC and discharges condensed steam from its cylinder at 50.0ºC. Your team has measured its efficiency to be 30.0%. (a) How does this efficiency compare with the maximum possible efficiency for your engine? (b) If the useful power output of the engine is known to be 200 kW, how much heat does the engine release to its surroundings in 1.00 h? Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1.00 h. (a) The efficiency of the steam engine as a percentage of the maximum possible efficiency is given by: maxmax enginesteam 300.0 εε ε = The efficiency of a Carnot engine operating between temperatures Fc and Th is: %52.40 K543 K32311 h c max =−=−= T T ε Substituting for εmax yields: %05.74 4052.0 300.0 max enginesteam == ε ε or maxenginesteam 740.0 εε = (b) Relate the heat Qc discharged to the engine’s surroundings to Qh and the efficiency of the engine: h ch h Q QQ Q W − ==ε ⇒ ( ) hc 1 QQ ε−= Chapter 19 1906 Substituting for COP yields: tP TT TQ Δ ch c c ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = Substitute numerical values and evaluate Qc: ( ) MJ 30.0kJ 303 min s 60min00.1W370 K 273K 293 K 273 c ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =Q (b) If the COP is 70% of the efficiency of an ideal pump: ( )( ) MJ10.2kJ 30370.0c =='Q 51 • A refrigerator is rated at 370 W. (a) What is the maximum amount of heat it can absorb for the food compartment in 1.00 min if the temperature in the compartment is 0.0ºC and it releases heat into a room at 35ºC? (b) If the COP of the refrigerator is 70% of that of a reversible pump, how much heat can it absorb from the food compartment in 1.00 min? Is the COP for the refrigerator greater when the temperature of the room is 35ºC or 20ºC? Explain. Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and Δt. (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP: ( ) ( ) tP WQ Δ= = COP COPc Express the COP in terms of Th and Tc and simplify to obtain: ch c h c h h h cc 1 1 1 111 COP TT T T T Q WQ Q Q W Q − =− − = −= − = − === εε ε εε Substituting for COP yields: tP TT TQ Δ ch c c ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = The Second Law of Thermodynamics 1907 Substitute numerical values and evaluate Qc: ( ) MJ 71.0kJ 731 min s 60min00.1W370 K 273K 083 K 732 c ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =Q (b) If the COP is 70% of the efficiency of an ideal pump: ( )( ) MJ12.0kJ 73170.0c =='Q Because the temperature difference increases when the room is warmer, the COP decreases. 52 ••• You are installing a heat pump, whose COP is half the COP of a reversible heat pump. You will use the pump on chilly winter nights to increase the air temperature in your bedroom. Your bedroom’s dimensions are 5.00 m × 3.50 m × 2.50 m. The air temperature should increase from 63°F to 68°F. The outside temperature is 35°F , and the temperature at the air handler in the room is 112°F. If the pump’s electric power consumption is 750 W, how long will you have to wait in order for the room’s air to warm (take the specific heat of air to be 1.005 kJ/(kg·°C)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings and floors. Also assume that the heat capacity of the floor, ceiling, walls and furniture are negligible. Picture the Problem We can use the definition of the coefficient of performance of a heat pump and the relationship between the work done per cycle and the pump’s power consumption to find your waiting time. The coefficient of performance of the heat pump is defined as: tP Q W Q Δ COP hh HP == ⇒ ( )P Qt HP h COP Δ = where Qh is the heat required to raise the temperature of your bedroom, P is the power consumption of the heat pump, and Δt is the time required to warm the bedroom. We’re given that the coefficient of performance of the heat pump is half the coefficient of performance of an ideal heat pump: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = == ch h 2 1 max2 1h HP COPCOP TT T W Q Substituting for HPCOP yields: P TT T Qt ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ch h h2Δ Chapter 19 1908 The heat required to warm the room is related to the volume of the room, the density of air, and the desired increase in temperature: TVcTmcQ ΔΔh ρ== where ρ is the density of air and c is its specific heat capacity. Substitute for Qh to obtain: P TT T TVct ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ch h Δ2Δ ρ Substitute numerical values and evaluate Δt: ( ) ( ) s 56 W750 K 275K 317 K 317 F 9 C 5F 5 Ckg J1005m 2.50m 3.50m 00.5 m kg293.12 Δ 3 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ° ×°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ °⋅ ××⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =t Entropy Changes 53 • [SSM] You inadvertently leave a pan of water boiling away on the hot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas? Picture the Problem Because the water absorbed heat in the vaporization process its change in entropy is positive and given by T Q S OHby absorbed OH 2 2 Δ = . See Table 18-2 for the latent heat of vaporization of water. The change in entropy of the water is given by: T Q S OHby absorbed OH 2 2 Δ = The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization: vv OHby absorbed 2 VLmLQ ρ== Substituting for OHby absorbed 2 Q yields: T VLS v OH2 Δ ρ = The Second Law of Thermodynamics 1911 Substitute numerical values and evaluate ΔSu: ( ) ( ) J/K40.2 K263 K263K273 Kkg J2100 kg J105.333 K273 K263ln Kkg J2100 K273 kg J105.333 kg0500.0Δ 3 3 u = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ +× + ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ + × −=S and, because ΔSu > 0, the entropy of the universe increases. 56 • In this problem, 2.00 mol of an ideal gas at 400 K expand quasi- statically and isothermally from an initial volume of 40.0 L to a final volume of 80.0 L. (a) What is the entropy change of the gas? (b) What is the entropy change of the universe for this process? Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS for the ideal gas as a function of its initial and final volumes. (a) The entropy change of the gas is given by: T QS =gasΔ (1) Apply the first law of thermodynamics to the isothermal process to express Q in terms of Won: onintΔ WEQ −= or, because ΔEint = 0 for an isothermal expansion of a gas, onWQ −= The work done on the gas is given by: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = f i on ln V VnRTW ⇒ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= f iln V VnRTQ Substitute for Q in equation (1) to obtain: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= f i gas lnΔ V VnRS Substitute numerical values and evaluate ΔS: ( ) K J11.5 L80.0 L40.0ln Kmol J8.314mol2.00Δ gas =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −=S Chapter 19 1912 (b) Because the process is reversible: 0u =ΔS Remarks: The entropy change of the environment of the gas is −11.5 J/K. 57 •• [SSM] A system completes a cycle consisting of six quasi-static steps, during which the total work done by the system is 100 J. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat from a reservoir at temperature T3. (During steps 2, 4 and 6 the system undergoes adiabatic processes in which the temperature of the system changes from one reservoir’s temperature to that of the next.) (a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible, what is the temperature T3? Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. Because the entropy change for the complete cycle is the sum of the entropy changes for each process, we can find the temperature T3 from the entropy changes during the 1st two processes and the heat released during the third. (a) Because S is a state function of the system, and because the system’s final state is identical to its initial state: 0Δ cyclecomplete 1 system =S (b) Relate the entropy changes for each of the three heat reservoirs and the system for one complete cycle of the system: 0ΔΔΔΔ system321 =+++ SSSS or 00 3 3 2 2 1 1 =+++ T Q T Q T Q Substitute numerical values. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir: 0J400 K400 J200 K300 J300 3 =+ − + − T Solving for T3 yields: K2673 =T 58 •• In this problem, 2.00 mol of an ideal gas initially has a temperature of 400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to twice its initial volume. What is (a) the entropy change of the gas and (b) the entropy change of the universe? The Second Law of Thermodynamics 1913 Picture the Problem The initial and final temperatures are the same for a free expansion of an ideal gas. Thus, the entropy change ΔS for a free expansion from Vi to Vf is the same as ΔS for an isothermal process from Vi to Vf. We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS for the ideal gas as a function of its initial and final volumes. (a) The entropy change of the gas is given by: T QS =gasΔ (1) Apply the first law of thermodynamics to the isothermal process to express Q: onintΔ WEQ −= or, because ΔEint = 0 for a free expansion of a gas, onWQ −= The work done on the gas is given by: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = f i on ln V VnRTW ⇒ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= f iln V VnRTQ Substitute for Q in equation (1) to obtain: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= f i gas lnΔ V VnRS Substitute numerical values and evaluate ΔS: ( ) K J11.5 L80.0 L40.0ln Kmol J8.314mol2.00Δ gas =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −=S (b) The change in entropy of the universe is the sum of the entropy changes of the gas and the surroundings: gssurroundingasu ΔΔΔ SSS += For the change in entropy of the surroundings we use the fact that, during the free expansion, the surroundings are unaffected: 00Δ rev gssurroundin === TT QS The change in entropy of the universe is the change in entropy of the gas: K J5.11Δ u =S 59 •• A 200-kg block of ice at 0.0ºC is placed in a large lake. The temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very Chapter 19 1916 Remarks: The result that ΔSu > 0 tells us that this process is irreversible. 61 •• [SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe. (a) Use the equation for the entropy change during a constant-pressure process to express the entropy change of the copper block: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ i f CuCuCu ln T TcmS (1) Apply conservation of energy to obtain: 0 i i =∑Q or 0waterwarmingblockcopper =+QQ Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water: ( ) ( ) ( ) ( ) 0K273 Kkg kJ 4.18 L kg1.00L4.00 K373 Kkg kJ0.386kg1.00 f f =−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛+ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ T T Solve for Tf to obtain: K275.26 f =T Substitute numerical values in equation (1) and evaluate CuΔS : ( ) K J117 K373 K275.26ln Kkg kJ0.386kg1.00Δ Cu −=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =S (b) The entropy change of the water is given by: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ i f waterwaterwater ln T TcmS The Second Law of Thermodynamics 1917 Substitute numerical values and evaluate waterΔS : ( ) K J138 K273 K26.275ln Kkg kJ18.4kg00.4Δ water =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =S (c) Substitute for CuΔS and waterΔS and evaluate the entropy change of the universe: K J02 K J138 K J117ΔΔΔ waterCuu = +−=+= SSS Remarks: The result that ΔSu > 0 tells us that this process is irreversible. 62 •• If a 2.00-kg piece of lead at 100ºC is dropped into a lake at 10ºC, find the entropy change of the universe. Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead, the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10°C. We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead, the water in the lake, and the universe. Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake: wPbu SSS Δ+Δ=Δ (1) Using the equation for the entropy change during a constant-pressure process, express and evaluate the entropy change of the lead: ( ) K J69.70 K373 K283ln Kkg kJ0.128kg2.00lnΔ i f PbPbPb −=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = T TcmS The entropy change of the water in the lake is given by: w PbPbPb w Pb w w w ΔΔ T Tcm T Q T QS === Chapter 19 1918 Substitute numerical values and evaluate ΔSw: ( ) ( ) J/K41.81 K283 K90 Kkg kJ0.128kg2.00 Δ w = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =S Substitute numerical values in equation (1) and evaluate ΔSu: K J11 K J81.41 K J70.69Δ u =+−=S Entropy and ″Lost″ Work 63 •• [SSM] A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs. (a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−= +−=Δ+Δ=Δ ch ch chu 11 TT Q T Q T QSSS Substitute numerical values and evaluate ΔSu: ( ) J/K0.42 K300 1 K400 1J500Δ u = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−=S (b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs: hmaxQW ε= The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures: h c Cmax 1 T T −== εε The Second Law of Thermodynamics 1921 (a) The efficiency of the engine is given by: h c h ch h 1 Q Q QQQ Q W −= − ==ε Substitute numerical values and evaluate ε: %7.16%67.16 J150 J1251 ==−=ε (b) Find the efficiency of a Carnot engine operating between the same reservoirs: %45.21 K 373 K 29311 h c C =−=−= T Tε Express the ratio of the two efficiencies: 777.0 %45.21 %67.16 C == ε ε 67 • [SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs. (a) What is the efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle. (a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −== h c C 185.085.0 T Tεε Substitute numerical values and evaluate ε : %51510.0 K500 K200185.0 ==⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=ε (b) Use the definition of efficiency to find the work done in each cycle: ( )( ) MJ0.10 kJ 102kJ200.5100h = === QW ε (c) Apply the first law of thermodynamics to the cycle to obtain: kJ89 kJ021kJ002cycleh,cyclec, = −=−= WQQ 68 • Estimate the change in entropy of the universe associated with an Olympic diver diving into the water from the 10-m platform. Chapter 19 1922 Picture the Problem Assume that the mass of the diver is 75 kg and that the temperature of the water in the pool is 25°C. The energy added to the water in the pool is the change in the gravitational potential energy of the diver during the dive. The change in entropy of the universe associated with a dive is given by: water water toadded wateru ΔΔ T QSS == where water toaddedQ is the energy entering the water as a result of the kinetic energy of the diver as he enters the water. The energy added to the water is the change in the gravitational potential energy of the diver: water uΔ T mghS = Substitute numerical values and evaluate uΔS : ( )( )( ) ( ) K J25 K27325 m 10m/s 81.9kg 75Δ 2 u ≈ + =S 69 • To maintain the temperature inside a house at 20ºC, the electric power consumption of the electric baseboard heaters is 30.0 kW on a day when the outside temperature is –7ºC. At what rate does this house contribute to the increase in the entropy of the universe? Picture the Problem The change in entropy of the universe is the change in entropy of the house plus the change in entropy of the environment. We can find the change in entropy of the house by exploiting the given information that the temperature inside the house is maintained at a constant temperature. We can find the change in entropy of the surrounding by dividing the heat added by the temperature. Entropy is a state function, and the state of the house does not change. Therefore the entropy of the house does not change: gssurroundinhouseu ΔΔΔ SSS += or, because 0Δ house =S , gssurroundinu ΔΔ SS = Heat is absorbed by the surroundings at the same rate R that energy is delivered to the house: gssurroundingssurroundin gssurroundin ΔΔ T tR T QS == The Second Law of Thermodynamics 1923 Substitute for ΔSsurroundings yields: gssurroundin u ΔΔ T tRS = ⇒ gssurroundin u Δ Δ T R t S = Substitute numerical values and evaluate ΔSu/Δt: K W131 K266 kW30.0 Δ Δ u == t S 70 •• Calvin Cliffs Nuclear Power Plant, located on the Hobbes River, generates 1.00 GW of power. In this plant, liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine. Heat is absorbed by the liquid sodium in the core, and released by the liquid sodium (and into the superheated steam) in the heat exchanger. The temperature of the superheated steam is 500 K. Heat is released into the river, and the water in the river flows by at a temperature of 25ºC. (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river every second? (c) How much heat must be released by the core to supply 1.00 GW of electrical power? (d) Assume that new environmental laws have been passed to preserve the unique wildlife of the river. Because of these laws, the plant is not allowed to heat the river by more than 0.50ºC. What is the minimum flow rate that the water in the Hobbes River must have? Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1.00 GW of power and, from this value, the power that is wasted. The rate at which heat is being released to the river is related to the requisite flow rate of the river by .dtdVTcdtdQ ρΔ= (a) The Carnot efficiency of a plant operating between temperatures Tc and Th is given by: h c Cmax 1 T T −== εε Substitute numerical values and evaluate εC: 404.0 K500 K2981max =−=ε (c) The power that must be supplied, at 40.4% efficiency, to produce an output of 1.00 GW is given by: GW48.2 0.404 GW00.1 max output supplied = == ε P P (b) Relate the wasted power to the power generated and the power supplied: generatedsuppliedwasted PPP −= Chapter 19 1926 Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle, its efficiency is that of a Carnot engine operating between the temperatures of its isotherms. The Carnot efficiency of the cycle is given by: h c C 1 T T −=ε Substitute numerical values and evaluate εC: %0.60 K750 K3001C =−=ε 73 •• [SSM] (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K, or (2) A reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process? Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process. (a) For process (2): inCrecoveredmax,2 QWW ε== The efficiency of a Carnot engine operating between temperatures Th and Tc is given by: h c C 1 T T −=ε and hence in h c recovered 1 Q T TW ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= Substitute for Cε to obtain: ( ) J250kJ1.00 K 400 K 3001recovered =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=W or 750 J are lost. Process (2) is more wasteful of available work. (b) Find the change in entropy of the universe for process (1): J/K1.67 K300 J500ΔΔ 1 === T QS The Second Law of Thermodynamics 1927 Express the change in entropy of the universe for process (2): ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −Δ= Δ + Δ −=Δ+Δ=Δ hc ch ch2 11 TT Q T Q T QSSS Substitute numerical values and evaluate ΔS2: ( ) J/K833.0 K400 1 K300 1kJ1.00Δ 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=S 74 •• Helium, a monatomic gas, is initially at a pressure of 16 atm, a volume of 1.0 L, and a temperature of 600 K. It is quasi-statically expanded at constant temperature until its volume is 4.0 L and is then quasi-statically compressed at constant pressure until its volume and temperature are such that a quasi-static adiabatic compression will return the gas to its original state. (a) Sketch this cycle on a PV diagram. (b) Find the volume and temperature after the compression at constant pressure. (c) Find the work done during each step of the cycle. (d) Find the efficiency of the cycle. Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being the initial state. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures, volumes, and pressures at points 1, 2, and 3. To find the work done during each cycle, we can use the equations for the work done during isothermal, isobaric, and adiabatic processes. Finally, we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion. (a) The PV diagram of the cycle is shown to the right. (b) Apply the ideal-gas law to the isothermal expansion 1→2 to find P2: ( ) atm4.0 L4.0 L1.0atm16 2 1 12 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == V VPP Chapter 19 1928 Apply an equation for an adiabatic process to relate the pressures and volumes at 1 and 3: γγ 3311 VPVP = ⇒ γ1 3 1 13 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = P PVV Substitute numerical values and evaluate V3: ( ) L2.3 L2.294 atm4.0 atm16L1.0 1.671 3 = =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V Apply an equation for an adiabatic process (γ =1.67) to relate the temperatures and volumes at 1 and 3: 1 11 1 33 −− = γγ VTVT ⇒ 1 3 1 13 − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = γ V VTT Substitute numerical values and evaluate T3: ( ) K103.4 K 344 L2.294 L1.0K600 2 11.67 3 ×= =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − T (c) Express the work done each cycle: 312312 WWWW ++= (1) For the process 1→2: ( )( ) Latm22.18 L1.0 L4.0lnL1.0atm16 lnln 1 2 11 1 2 112 ⋅= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = V VVP V VnRTW For the process 2→3: ( )( ) Latm824.6 L4.00L2.294atm4.0 Δ 23223 ⋅−= −= = VPW For the process 3→1: ( ) ( ) ( )( ) ( )( )[ ] Latm0.241 L2.294atm4.0L1.0atm16 Δ 2 3 33112 3 312 3 31V31 ⋅−= −−= −−=−−=−= VPVPTTnRTCW Substitute numerical values in equation (1) and evaluate W: Latm5Latm5.116 Latm10.24 Latm6.824Latm18.22 ⋅=⋅= ⋅− ⋅−⋅=W The Second Law of Thermodynamics 1931 leaks through the walls of the freezer at a rate of 50 W. Find the power of the motor that is needed to maintain the temperature in the freezer. Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer. Using the definition of the COP, relate the heat that must be removed from the freezer to the work done by the motor: W QcCOP = ⇒ COP cQW = Differentiate this expression with respect to time to express the power of the motor: COP c dtdQ dt dWP == Express the maximum COP of the motor: T T Δ = c maxCOP Substitute for COPmax to obtain: c c T T dt dQP Δ = Substitute numerical values and evaluate P: ( ) W10 K250 K50W50 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =P 78 •• In a heat engine, 2.00 mol of a diatomic gas are taken through the cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed or released by the gas in each segment of this cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the constant- pressure, adiabatic, and isothermal processes of the cycle. Chapter 19 1932 (a) Apply the ideal-gas law to find the volume of the gas at A: ( ) ( ) L19.7 L 69.19 m 10 L 1 m 10969.1 atm kPa101.325atm5.00 K600 Kmol J8.314mol2.00 33 32 A A A = =××= × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅= = − − P nRTV (b) We’re given that AB 2VV = . Hence: ( ) L39.4L 38.39L19.692B ===V Apply the ideal-gas law to this constant-pressure process to obtain: ( ) K1200 2K600 A A A B AB = == V V V VTT (c) Because the process C→A is isothermal: K600AC == TT (d) Apply an equation for an adiabatic process (γ = 1.4) to find the volume of the gas at C: 1 CC 1 BB −− = γγ VTVT ⇒ 1 1 C B BC − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = γ T TVV Substitute numerical values and evaluate VC: ( ) L223 L 77.222 K600 K1200L39.38 11.4 1 C = =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − V (e) The work done by the gas during the constant-pressure process AB is given by: ( ) ( ) AA AAAABAAB 2 VP VVPVVPW = −=−= Substitute numerical values and evaluate WAB: ( )( ) kJ9.98J 109754.9 Latm J101.325Latm98.45 L19.69atm5.00 3 AB =×= ⋅ ×⋅= =W The Second Law of Thermodynamics 1933 Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion BC: BC2 5 BCVBC int, BC int,BC in,BCint,BC 0 TnR TncE EQEW Δ−= Δ−=Δ= −Δ=−Δ= Substitute numerical values and evaluate WBC: ( ) ( ) kJ24.9 J 10494.2K1200K600 Kmol J8.314mol2.00 4 2 5 BC = ×=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −=W The work done by the gas during the isothermal compression CA is: ( ) ( ) kJ2.24kJ20.24 L222.77 L19.69lnK600 Kmol J8.314mol2.00ln C A CCA −=−= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = V VnRTW (f) The heat absorbed during the constant-pressure expansion AB is: ( ) ( ) kJ9.34kJ92.34 K600K1200 Kmol J8.314mol2.00ΔΔ 2 7 BA2 7 BAPAB == −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === −− TnRTncQ The heat absorbed during the adiabatic expansion BC is: 0BC =Q Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA: kJ2.24 CACA int,CACA −= =Δ+= WEWQ because 0CA int, =ΔE for an isothermal process. 79 •• [SSM] In a heat engine, 2.00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. Chapter 19 1936 (a) Apply the ideal-gas law to find the volume of the gas at A: ( ) ( ) L19.7L19.69 atm kPa101.325atm5.00 K600 Kmol J8.314mol2.00 A A A == × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅= = P nRTV (b) We’re given that: ( ) L39.4 L 39.39L19.6922 AB = === VV Apply the ideal-gas law to this isobaric process to find the temperature at B: ( ) K1200 2K600 A A A B AB = == V V V VTT (c) Because the process CA is isothermal: K600AC == TT (d) Apply an equation for an adiabatic process (γ = 5/3) to express the volume of the gas at C: 1 CC 1 BB −− = γγ VTVT ⇒ 1 1 C B BC − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = γ T TVV Substitute numerical values and evaluate VC: ( ) L111L 4.111 K600 K1200L39.39 2 3 C == ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V (e) The work done by the gas during the isobaric process AB is given by: ( ) ( ) AA AAAABAAB 2 VP VVPVVPW = −=−= Substitute numerical values and evaluate WAB: ( )( ) kJ9.98kJ 975.9 Latm J101.325Latm98.45 L19.69atm5.00AB == ⋅ ×⋅= =W The Second Law of Thermodynamics 1937 Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC: ( ) BC2 3 BCVBC int, BC int, BC in,BC int,BC 0 TnR TncE E QEW Δ−= Δ−=Δ= −Δ= −Δ= Substitute numerical values and evaluate WBC: ( ) ( ) kJ0.51 kJ 97.14K1200K600 Kmol J8.314mol2.002 3 BC = =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −=W The work done by the gas during the isothermal compression CA is: ( ) ( ) kJ3.17kJ29.17 L4.111 L19.69lnK600 Kmol J8.314mol2.00ln C A CCA −−= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = V VnRTW (f) The heat absorbed during the isobaric expansion AB is: ( ) ( ) kJ9.24 K600K1200 Kmol J8.314mol2.00ΔΔ 2 5 AB2 5 ABPAB in, = −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TnRTncQ Express and evaluate the heat absorbed during the adiabatic expansion BC: 0BC =Q Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA: kJ3.17 Δ CACA int,CACA −= =+= WEWQ because ΔEint = 0 for an isothermal process. 81 •• In a heat engine, 2.00 mol of a monatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. Chapter 19 1938 (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB: ( ) kPa253kPa 3.253 atm1 kPa101.325atm2.50 2 atm00.5 A A B A AB == ×= == V V V VPP (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC: BB CC BC VP VPTT = (1) Use the ideal-gas law to find the volume at B: ( ) ( ) L39.39 kPa253.3 K600 Kmol J8.314mol2.00 B B B = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅= = P nRTV Use the equation of state for an adiabatic process and γ = 5/3 to find the volume occupied by the gas at C: ( ) L26.86 atm1.00 atm2.50L39.39 531 C B BC = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = γ P PVV Substitute numerical values in equation (1) and evaluate TC: ( ) ( )( ) ( )( ) K416K 9.415 L39.39atm2.50 L26.86atm1.00K600C == =T (c) The work done by the gas in one cycle is given by: DACDBCAB WWWWW +++= (2) The Second Law of Thermodynamics 1941 The efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle is given by: c a T T −=1Carnotε Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures: 1 1 1 Otto Carnot > − − = b a c a T T T T ε ε because Tc > Tb. Hence, OttoCarnot εε > 83 ••• [SSM] A common practical cycle, often used in refrigeration, is the Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric (constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric compression back to the original state. Assume the system begins the adiabatic compression at temperature T1, and transitions to temperatures T2, T3 and T4 after each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the efficiency of the overall cycle is given by ε = 1− T4 − T1( ) T3 − T2( ) . (c) Show that this efficiency, can be written as ε = 1− r 1−γ( ) γ , where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the engine during the isobaric transitions. (a) The Brayton heat engine cycle is shown to the right. The paths 1→2 and 3→4 are adiabatic. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1. 1 2 3 4 P V ⇓ ⇓ hQ cQ (b) The efficiency of a heat engine is given by: in ch in Q QQ Q W − ==ε (1) Chapter 19 1942 During the constant-pressure expansion from state 1 to state 2 heat enters the system: ( )23PPh23 Δ TTnCTnCQQ −=== During the constant-pressure compression from state 3 to state 4 heat enters the system: ( )41PPc41 Δ TTnCTnCQQ −−=−=−= Substituting in equation (1) and simplifying yields: ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )23 14 23 4123 23P 41P23P 1 TT TT TT TTTT TTnC TTnCTTnC − − −= − −+− = − −−−− =ε (c) Given that, for an adiabatic transition, constant 1 =−γTV , use the ideal-gas law to eliminate V and obtain: constant 1 =−γ γ P T Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2: 1 high 2 1 low 1 −− = γ γ γ γ P T P T ⇒ 2 1 high low 1 T P PT γ γ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Similarly, for the adiabatic transition from state 3 to state 4: 3 1 high low 4 T P PT γ γ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Subtract T1 from T4 and simplify to obtain: ( )23 1 high low 2 1 high low 3 1 high low 14 TT P P T P PT P PTT −⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =− − −− γ γ γ γ γ γ The Second Law of Thermodynamics 1943 Dividing both sides of the equation by T3 − T2 yields: γ γ 1 high low 23 14 − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − P P TT TT Substitute in the result of Part (b) and simplify to obtain: ( ) γ γ γ γ γ γ ε − −− −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= 1 1 low high 1 high low 1 11 r P P P P where low high P P r = 84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as a refrigerator in your kitchen. In this case, the cycle begins at temperature T1 and expands at constant pressure until its temperature T4. Then the gas is adiabatically compressed, until its temperature is T3. And then it is compressed at constant pressure until its temperature T2. Finally, it adiabatically expands until it returns to its initial state at temperature T1. (a) Sketch this cycle on a PV diagram. (b) Show that the coefficient of performance isCOPB= T4 − T1( ) T3 − T2 − T4 + T1( ) . (c) Suppose your ″Brayton cycle refrigerator″ is run as follows. The cylinder containing the refrigerant (a monatomic gas) has an initial volume and pressure of 60 mL and 1.0 atm. After the expansion at constant pressure, the volume and temperature are 75 mL and –25°C. The pressure ratio r = Phigh/Plow for the cycle is 5.0. What is the coefficient of performance for your refrigerator? (d) To absorb heat from the food compartment at the rate of 120 W, what is the rate at which electrical energy must be supplied to the motor of this refrigerator? (e) Assuming the refrigerator motor is actually running for only 4.0 h each day, how much does it add to your monthly electric bill. Assume 15 cents per kWh of electric energy and thirty days in a month. Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio of the heat that enters the system to the work done to operate the refrigerator. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the refrigerator during the isobaric transitions. Chapter 19 1946 Because the gas is monatomic, RC 2 5 P = . Substitute for n and CP to obtain: ( ) ( ) ( ) 4B 14442 5 B 14 4 44 2 5 COP COP T TTVfP TTR RT VPf dt dW − = − = Substitute numerical values and evaluate dW/dt: ( )( ) ( ) ( )( ) kW 21.0 W207 K 2481.11 K 198K 248 L m 10mL 75kPa 325.101s 60 33 1 2 5 = = −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = − − dt dW (e) The monthly cost of operation is given by monthper days ofnumber nconsumptiodaily rate nConsumptioPower Power ofPer Unit Cost CostMonthly ××= ×= Substitute numerical values and evaluate the monthly cost of operation: 4$d 03 d h 4.0 W207 kWh 15.0$CostMonthly ≈×××= 85 ••• Using ( ) ( )v 2 1 2 1ln lnΔ = −S C T T nR V V (Equation 19-16) for the entropy change of an ideal gas, show explicitly that the entropy change is zero for a quasi-static adiabatic expansion from state (V1, T1) to state (V2, T2). Picture the Problem We can use VP CCnR −= , VP CC=γ , and 1−γTV = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1,T1) to state (V2,T2) is zero. Express the entropy change for a general process that proceeds from state 1 to state 2: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ 1 2 1 2 V lnln V VnR T TCS For an adiabatic process: 1 2 1 1 2 − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = γ V V T T The Second Law of Thermodynamics 1947 Substitute for 1 2 T T and simplify to obtain: ( ) ( )[ ]V 1 2 2 1 2 1 V 1 2 1 2 1 2 1 V 1 2 1 2 1 2 1 V 1ln ln ln1 ln ln ln lnlnln CnR V V V V V VC nR V V V V V VC nR V V V VnR V VCS −−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ − − γ γ γ γ Use the relationship between CP and CV to obtain: VP CCnR −= Substituting for nR and γ and simplifying yields: 0 1ln V V p VP 1 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ C C C CC V VS 86 ••• (a) Show that if the refrigerator statement of the second law of thermodynamics were not true, then the entropy of the universe could decrease. (b) Show that if the heat-engine statement of the second law were not true, then the entropy of the universe could decrease. (c) A third statement of the second law is that the entropy of the universe cannot decrease. Have you just proved that this statement is equivalent to the refrigerator and heat-engine statements? Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal mount of heat is transferred to the hot reservoir and W = 0. The entropy change of the universe is then ΔSu = Qc/Th − Qc/Tc. Because Th > Tc, Su < 0, i.e., the entropy of the universe would decrease. (b) In this case, is heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir; that is, Qc = 0, then the entropy change of the universe is ΔSu = −Qh/Th + 0, which is negative. Again, the entropy of the universe would decrease. Chapter 19 1948 (c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir, but these statements do not specify the minimum amount of heat rejected or work that must be done. The statement ΔSu ≥ 0 is more restrictive. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to ΔSu ≥ 0. 87 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine as shown in Figure 19-22. The efficiencies of the engines are ε1 and ε2, respectively. Show that the net efficiency of the combination is given by 2121net εεεεε −+= . Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Express the net efficiency of the two heat engines connected in series: h 21 net Q WW + =ε Express the efficiencies of engines 1 and 2: h 1 1 Q W =ε and m 2 2 Q W =ε Solve for W1 and W2 and substitute to obtain: 2 h m 1 h m2h1 net εεεεε Q Q Q QQ += + = Express the efficiency of engine 1 in terms of Qm and Qh: h m 1 1 Q Q −=ε ⇒ 1 h m 1 ε−= Q Q Substitute for Qm/Qh and simplify to obtain: ( ) 2121211net 1 εεεεεεεε −+=−+= 88 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine, as shown in Figure 19-22. Suppose that each engine is an ideal reversible heat engine. Engine 1 operates between temperatures Th and Tm and Engine 2 operates between Tm and Tc, where Th > Tm > Tc. Show that the net efficiency of the combination is given by c net h 1ε = − T T . (Note that this result means that two