The area under the graph in a PV diagram is equal in magnitude to

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1 a volume V and exerts a uniform pressure P on the cylinder walls and the piston. The gas is compressed slowly enough so the system remains essentially in thermodynamic equilibrium at all times. As the piston is pushed downward by an external force F through a distance y, the work done on the gas is W F y PA y Chapter 12 _12_p /3/04 9:13 AM Page 388 where we have set the magnitude F of the external force equal to PA, possible because 12.1 WORK IN THERMODYNAMIC PROCESSES the pressure is the same everywhere in the system (by the assumption of equilibrium). Note that if the piston is pushed downward, y y f y i is negative, so we need an explicit negative sign in the expression for W to make the work positive. The change in volume of the gas is V A y, which leads to the following definition: Energy can be transferred to a system by heat and by work done on the system. The work W done on a gas at constant pressure is given by W P V [12.1] where P is the pressure throughout the gas and V is the change in volume of the gas during the process. 388 Chapter 12 The Laws of Thermodynamics If the gas is compressed as in Active Figure 12.1b, V is negative and the work done on the gas is positive. If the gas expands, V is positive and the work done on the gas is P negative. The isobaric work done process by the gas on its environment, W env, is simply the negative of the work done on the gas. In the absence of a change in volume, the work is zero. EXAMPLE 12.1 Goal P f i Work Done by an Expanding Gas Apply the definition of work at constant pressure. A Log into PhysicsNow at and go to Active Figure 12.1 y to move the piston and see the resulting work done on P the gas. V (a) (b) The pressure vs. volume ACTIVE FIGURE graph, 12.1 or PV diag P A (a) V y (b) ACTIVE FIGURE 12.1 (a) A gas in a cylinder occupying a volume V at a pressure P. (b) Pushing the piston down compresses the gas. Equation 12.1 can be used to calculate t the pressure of the gas remains constant d A process in which the pressure remains Figure The curve on such a graph is c and final states, with the arrow indicating th case from smaller to larger volume. The are Problem In a system similar to that shown in Active Figure 12.1, the gas in the cylinder is at a pressure of Pa and the piston has an Varea of m 2. As energy is slowly added to the by heat, the piston is V pushed up a distance f V of 4.00 cm. i Calculate the work work done by the done expanding on gas on the surroundings, gas W env, assuming the pressure remains constant. Figure 12.2 The PV diagram for a gas being compressed at constant pressure. The shaded area represents The area under the graph in a PV diagram is equal in magnitude to Area P(V f The area under the graph in a PV diagram on the gas. Strategy The work done on the environment is the negative of the work done on the gas given in Equation Compute the change in volume and multiply by the pressure.

2 This is true in general, whether or not the process proceeds at constant pressure. Just draw the PV diagram of the process, find the area underneath the graph (and above the horizontal axis), and that area will be the equal to the magnitude of the work done on the gas. If the arrow on the graph points toward larger volumes, the work done on the gas is negative. If the arrow on the graph points toward smaller volumes, the work done on the gas is positive. Whenever negative work is done on a system, positive work is done by the system on its environment. The negative work done on the system represents a loss of energy from the system the cost of doing positive work on the environment.

3 12.2 THE FIRST LAW OF THERMODYNAMICS If a system undergoes a change from an initial state to a final state, where Q is the energy transferred to the system by heat and W is the work done on the system, the change in the internal energy of the system, U, is given by U U f U i Q W [12.2] The quantity Q is positive when energy is transferred into the system by heat and negative when energy is transferred out of the system by heat. The quantity W is positive when work is done on the system and negative when the system does work on its environment. TIP 12.1 Dual Sign Conventions Many physics and engineering textbooks present the first law as U Q W, with a minus sign between the heat and the work. The reason is that work is defined in these treatments as the work done by the gas rather than on the gas, as in our treatment. This form of the first law represents the original interest in applying it to steam engines, where the primary concern is the work extracted from the engine.

4 From Chapter 10 U 3 2 nrt molar specific heat at constant volume monatomic ideal U 3 2 nr T tant volume of a C v 3 2 R n ideal gas ca U nc v T A gas with a larger molar specific heat requires more energy to realize a given temperature change. The size of the molar specific heat depends on the structure of the gas molecule and how many different ways it can store energy. is always valid, e

5 Isobaric Processes pressure remains constant An expanding gas does work on its environment. first law. The work done by the gas on its environment must come at the expense of the change in its internal energy, U. Because the change in the internal energy of an ideal gas is given by U nc v T, the temperature of an expanding gas must decrease as the internal energy decreases. Expanding volume and decreasing temperature means the pressure must also decrease, in conformity with the ideal gas law, PV nrt. Consequently, the only way such a process can remain at constant pressure is if thermal energy Q is transferred into the gas by heat. Rearranging the first law, we obtain Q U W U P V Q 3 2 nr T nr T 5 2 nr T ss this transfer by heat is

6 Adiabatic Processes no energy enters or leaves the system by heat A sufficiently rapid process may be considered approximately adiabatic because there isn t time for any significant transfer of energy by heat. For adiabatic processes Q 0, so the first law becomes U W he work done during an adiabatic process can be calc PV constant where C p is called the adiabatic index of the gas. Values of the adiabat C v adiabatic index of the gas

Isovolumetric Processes constant volume If the volume doesn t change, no work is done on or by the system, so W=0, and the first law of thermodynamics reads U Q (isovolumetric

7 Isovolumetric Processes constant volume If the volume doesn t change, no work is done on or by the system, so W=0, and the first law of thermodynamics reads U Q (isovolumetric process) This result tells us that in an isovolumetric process, the change in internal energy of a system in an process, the cha equals the energy transferred to the system by heat. Q nc v T

8 Isothermal Processes temperature of a system doesn t change. ideal gas the internal ene U 0 because T 0. W Q (isothermal process) P(10 5 Pa) 1.00 P nrt V 0.25 adiabatic isothermal W env nrt ln V f V i V(m3 )

9 TABLE 12.2 The First Law and Thermodynamic Processes (Ideal Gases) Process U Q W Isobaric nc v T nc p T P V Adiabatic nc v T 0 U Isovolumetric nc v T U 0 Isothermal 0 W nrt ln V f V i General nc v T U W (PV Area) 1. Isovolumetric 2. Isobaric 3.Isothermal

13 ACTIVE FIGURE 12.9 A schematic representation of a heat engine. The engine receives energy Q h from the hot reservoir, expels 12.3 energy HEAT ENGINES Q c to the AND cold THE reservoir, SECOND and LAW OF THERMODYNAMICS does work W. Log into PhysicsNow at and A heat go to engine Active Figure takes 12.9 in to select energy by heat and partially converts it to other forms, the efficiency of the engine and observe the transfer of energy. P Hot reservoir at T h Q h Area = W eng V Engine Figure The PV diagram for an arbitrary cyclic process. The area enclosed by the curve equals the net work done. Q c EXAMPLE W eng net energy absorbed by the engine. As we can see from Active Figure the engine is carried through a cycle in w Q net Q h Q h Q c. Therefore, boiler and then expands against a piston. A ing water, it returns to the boiler, and the pr W eng W eng It s Q h useful Qto c draw a heat engine schema[ gine absorbs energy Q h from the hot reser Engine Ordinarily, a transfer of thermal energy ergy QQ can c to be the either cold reservoir. positive or (Note negative, that n that W W eng.) Because the working sub use of absolute value signs makes the signs of Q turning to its h and Q initial thermodynamic c explicit. state, i If the working Q c substance is a gas, then the work done by the engine for a equal, so U 0. From the first law of therm process is the area enclosed by the curve representing the process on diagram. This area is shown for an arbitrary cyclic process U in 0Figure Q W : The thermal efficiency e of a heat Cold reservoir at T The engine last is equation defined shows as the that work the done workby Wt c gine, W eng, divided by the energy absorbed net energy during absorbed one cycle: by the engine. As Q h Q c. Therefore, ACTIVE FIGURE 12.9 A schematic representation of a heat engine. The engine receives energy thermal efficiency Q h from the hot reservoir, expels energy Q c to the cold reservoir, Qand h does work W. Ordinarily, a transfer of thermal energy Q c We can think of thermal efficiency use as of the absolute ratio of value the signs benefit makes received the signs (woo the cost Log into incurred PhysicsNow (energy at transfer at If the the higher working temperature). substance is a gas, Equation then th shows and Wthat go to eng a Active heat Figure Q h engine 12.9 Qhas to select c 100% efficiency process is (ethe 1) area only enclosed if Q c by 0 mean the curv the efficiency of the engine and energy observe is expelled the transfer to of the energy. cold reservoir. diagram. In other This words, area is shown a heat for engine an arbitrary with p efficiency would have to expel all the input The thermal energy by efficiency doing mechanical e of a heat engine work P isn t possible, as will be seen in Section gine, W eng, divided by the energy absorbed Area = W eng The Efficiency of an Engine e W eng e W eng Q h W eng Q h [ Q h Q h Cold reservoir at T c We can think of thermal efficiency as the the cost incurred (energy transfer at the Goal ACTIVE Apply FIGURE the efficiency 12.9 formula to a heat engine. shows that a heat engine has 100% efficienc V energy is expelled to the cold reservoir. In o Problem During one cycle, an engine Figure extracts The PV 10diagram for the work done an by arbitrary the cyclic engine process. for 3 J of energy from The area efficiency a hot a cyclic process is would reservoir the area have enclosed to and expel transfers all the 1.50 input to a cold reservoir. (a) Find the thermal efficiency enclosed by the of curve the engine. equals the net (b) How isn t much possible, work as does will this be seen engine Section do in 12.4 one (c) How much power does by the the engine curve generate work representing done. if it goes the through process four on a cycles PV diagram in 2.50 s? Q net Q h Q c Q h 1 Q c Q h Strategy Apply Equation to obtain the thermal efficiency, then use the first law, adapted to engines (

14 The Second Law of Thermodynamics No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the performance of an equal amount of work. J-L. Charmet/SPL/Photo Researchers, Inc. LORD KELVIN, British Physicist and Mathematician ( ) To summarize, the first law says we can t get a greater amount of energy out of a cyclic process than we put in, and the second law says we can t break even. Reversible and Irreversible Processes In a reversible process, every state along the path is an equilibrium state, so the system can return to its initial conditions by going along the same path in the reverse direction. A process that doesn t satisfy this requirement is irreversible.

15 ch Engineer the founder of mics. Some of eath indicate ognize the relaheat. J-L. Charmet/SPL/Photo Rese The cycle consists of two adiabatic and two isothermal processes, all reversible: 1. The process A : B is an isothermal expansion at temperature T h in which the gas is placed in thermal contact with a hot reservoir (a large oven, for example) at temperature T h (Active Fig a). During the process, the gas absorbs energy Q h from the reservoir 44337_12_p and does work W AB 11/3/04 in raising 9:14 the AM piston. Page In the process B : C, the base of the cylinder is replaced by a thermally nonconducting wall and the gas expands adiabatically, so no energy enters or 44337_12_p /3/04 9:14 AM Page Chapter 12 The Laws of Thermodynamics The Carnot Engine J-L. Charmet/SPL/Photo Researchers, Inc. viewpoint. He showed that a heat engine operating in an ideal, reversible cycle now called a Carnot cycle between two energy reservoirs is the most efficient engine possible. Such an engine establishes an upper limit on the efficiencies of all real engines. Carnot s theorem can be stated as follows: the system by heat (Active No real Fig. engine operating 12.15b). between two During energy reservoirs can the be more process, efficient than a Carnot engine operating between the same two reservoirs. the temperalls from T h to T c and the gas does work W In a Carnot cycle, an ideal gas is contained in BC in raising the piston. a cylinder with a movable piston at one end. The temperature of the gas varies between T c and T h. The cylinder walls rocess C : D, the gas is placed in thermal contact with a cold reservoir and the piston are thermally nonconducting. Active Figure shows the four stages of the Carnot cycle, and Active Figure is the PV diagram for the cycle. The cycle consists of two adiabatic and two isothermal processes, all reversible: perature T c (Active Fig c) and is compressed 1. The process A : B is an isothermal expansion at temperature T h isothermally at temre T c. During this time, the ple) at temperature gas expels T h (Active Fig a). energy During the Qprocess, c to the gas the absorbsreservoir and the in which the gas is placed in thermal contact with a hot reservoir (a large oven, for exam- energy Q h from the reservoir and does work W AB in raising the piston. one on the 2. In the process B : C, the base of the cylinder is replaced by a thermally nonconducting wall and the gas expands adiabatically, so no energy enters or SADI CARNOT, gas French is WEngineer CD. ( ) nal process, D : A, the base of the cylinder is again replaced by a ther- Carnot is considered to be the founder of the science of thermodynamics. Some of his notes found after his death indicate onconducting that he was the first to wall recognize the relationship between work and heat. (Active Fig d) and the gas is compressed adialy. The temperature of the gas increases A B (a) to TIsothermal expansion h, and the work done on the DA. work done on the gas is W CD. 4. In the final process, D : A, the base of the cylinder is aga mally nonconducting wall (Active Fig d) and the ga No real engine operating 12.3 Heat between Engines and two the energy Second reservoirs Law of Thermodynamics can be more effi-40cient batically. than a Carnot The engine temperature operating between of the the same gas two reservoirs. increases to T h, and gas is W DA. For a Carnot engine, the following relationship between the fers and the absolute temperatures can be derived: leaves the system by heat (Active Fig b). During the process, the temperature falls from T h to T c and the gas does work W BC in raising the piston. A P A B Isothermal 3. In the expansion process C : D, the gas is placed in thermal contact with a cold reservoir at temperature T c (Active Fig c) and is compressed isothermally at temperature T c. During this time, the gas expels energy Q c to the reservoir and the work done on the gas is W CD. 4. In the final process, D : A, the base of the cylinder is again replaced by a thermally nonconducting wall (Active Fig d) and the gas is compressed adia- B Q h Q c Q h T Q h c batically. The temperature of the gas increases to T h, and the work done on the B Energy reservoir at gas T is W DA. W h Q h T eng T h h C For a Carnot engine, the following relationship between the thermal energy transfers and the absolute temperatures can be derived: W D Q T c c eng V T ACTIVE FIGURE h The PV diagram C for the Carnot cycle. The net work done, W eng, equals the net energy received D Q Tby heat in one cycle, Q h c c Q c. Substituting this expression into Equation 12.12, we find D A B C Q c Q h T c [12.15] Adiabatic Adiabatic Cycle Q ot engine, the following relationship compression between Energy reservoir at T h expansion the thermal energy h T transe absolute temperatures can be derived: h ciency of (a) Substituting a Carnot this expression engine into Equation 12.12, is we find that the thermal efficiency of a Carnot engine is 5 D A B C ocess A : B, Adiabatic Adiabatic mally while in Q = 0 Cycle compression expansion Q = 0 Q at T h. In c ACTIVE FIGURE T c e C 1 [12.15] T c [12.16] T expands adiacess C : D, contact with a reservoir at T h. The Carnot cycle. In process A : B, the gas expands isothermally while in Q = 0 Q = 0 Q In h T h h process B : C, the gas expands adiabatically (Q 0). In process C : D, othermally where T must be in kelvins. From this result, we see that all Carnot engines operating reversibly between the same two temperatures have the same efficiency. (d) (b) cycle, Q eservoir at the gas is compressed isothermally (d) (b) g this expression A, the gas is while in contact with a reservoir at T c T h. into Equation 12.12, we find that the thermal effi- h Q c. In process D : A, the gas is Equation can be applied to any working substance operating in a Carnot compressed adiabatically. The upward ly. The upward arrows on the piston indicate the Carnot engine is removal of sand during the expansions, and the downward arrows Isothermal C D cycle between two energy reservoirs. According to that equation, the efficiency is icate the the expand arrows Isothermal indicate the addition of sand during C D compression zero if T c T h. The efficiency increases as T c is lowered and as T h is increased. the compressions. sand during compression Log into PhysicsNow at Q c and go to Active Figure to observe the motion of e [12.16] the piston in the Carnot cycle while C 1 T The efficiency can be one (100%), however, only if T c 0 K. According to the third law of thermodynamics, it s impossible to lower the temperature of a system c to absolute zero in a finite number of steps, so such reservoirs are not available, Energy reservoir at T c you also observe the cycle on the PV T diagram of Active Figure h and the maximum All efficiency real engines is always less operate than one. irreversibly, In most practical due to cases, the (c) cold reservoir is near Q c friction room temperature, and the brevity about 300 of their K, so increasing cycles, and the are efficiency requires raising the temperature of the hot reservoir. All real engines operate irreversibly, that due all to Carnot friction therefore and engines the less brevity efficient of operat- their than cycles, the and Carnot are therefore engine. less effi- to Active the ust motion be ofin kelvins. From this result, we see t cycle while Energy reservoir cient at Tthan the Carnot engine. c le bly on the between PV the same two temperatures have the same efficiency. e (c) n can be applied to any working substance operating in a Carnot een two energy reservoirs. According to Quick that equation, Quiz 12.3 the efficiency is T h. The efficiency increases as T c is lowered and as T h is increased. P 12.3 Heat Engines and the Second Law of Thermodynamics 407 A V ACTIVE FIGURE Log into PhysicsNow at The PV diagram for the Carnot cycle. The net work done, W eng, equals the net energy received by heat in one e C 1 T c and go to Active Figure to observe the Carnot cycle on the PV diagram while you also observe the motion of the piston in Active Figure Log into PhysicsNow at where T must be in kelvins. From this result, and go TIP to 12.2 we Active Don t see Shop for that a all C Figure to observe Carnot the Engine Carnot ing reversibly between the same cycle two on the temperatures PV diagram while youhave the sa The Carnot engine is only an idealization. If a Carnot engine were also observe the motion developed of the an piston effort maximize Equation can be applied in Active to Figure any working substance efficiency, it would have zero power output, because, in order for all of the processes to be reversible, the cycle between two energy reservoirs. According that equa engine would have to run infinitely slowly. zero if T c T h. The efficiency increases as T c is lowered an Three engines operate between reservoirs separated in temperature by 300 K. The The efficiency reservoir temperatures can are as follows: be one (100%), however, only if T c 0 T h

16 12.4 ENTROPY Let Q r be the energy absorbed or expelled during a reversible, constant temperature process between two equilibrium states. Then the change in entropy during any constant temperature process connecting the two equilibrium states is defined as S Q r T [12.17] SI unit: joules/kelvin ( J/K) the entropy of the Universe increases in all natural processes Although the entropy of the Universe increases in all natural processes, the entropy of a system can decrease e TIP 12.4 Don t Confuse the W s The symbol W used here is a probability, not to be confused with the same symbol used for work. Entropy and Disorder S k B ln W a disorderly arrangement is much more probable than an orderly one if the laws of nature are allowed to act without interference. isolated systems tend toward greater disorder, and entropy is a measure of that disorder. The second law of thermodynamics is really a statement of what is most probable rather than of what must be

Chapter 12. The Laws of Thermodynamics

Chapter 12. The Laws of Thermodynamics Chapter 12 The Laws of Thermodynamics First Law of Thermodynamics The First Law of Thermodynamics tells us that the internal energy of a system can be increased by Adding energy to the system Doing work