3.2 Energy Conservationeach step. Such changes often occur rapidly and give rise to flows and “friction"effects. After an irreversible change the system cannot be brought back to its orig-inal thermodynamic state without causing a change in the thermodynamic stateof the universe.3.2EnergyConservationThere is a store of energy in a thermodynamic system, called the internal energy U. Infinitesimal changes du in internal energy content can occur by causingthe system to do mechanical work,dW or by adding heat, dQ, to the system.Theinternal energy can also be changed by adding particles of types j = 1,2,.,vto the system causing infinitesimal changes in the internal energy equal toZj1M, dN, whereu,isthe chemical potentialof particles oftypejand dN,istheinfinitesimal change in the number of typej particles. We use the notation,d W,toindicate thatthedifferential dW is not exact (see Appendix B).Thequantities dQand dw are not exact differentials because they depend on the path taken (onthe way in which heat is added or work is done).The quantities dN, are exactdifferentials.The change in the internal energy that results from these thermal, mechanical,andchemicalprocessesisgivenbyVdu=dQ-dw+ZudN,.(3.1)合The work,dW,may be due to changes in any relevant extensive"mechanical"variable. In general it can be writtendW=PdV-JdL-adA-E.dP-H.dM-@de,(3.2)wheredU,dV,dL,dA,dP,dM,andde areexactdifferentials.ThemagnetizationM and the magnetic field H are related to the magnetic flux density B by theequationB=μoH+M,whereμo is thepermeabilityoffreespace.Thedefinitionofthefirst fiveterms in Eq. (3.2)was discussed in Section 3.1.The term,-@ de,isthe work the system does if it has an electric potential, @, and increases its chargeby an amount,de.Wemay think of-P,J,o,E,H and as generalizedforces,andwe may think of dV,dL,dA,dP,dM, and de as generalized displacements.It is useful to introduce ageneralizedmechanical force,Y,which denotesquan-titiessuchas,-P,J,o,E,H,and@,andageneralizeddisplacement,X,whichdenotes the corresponding displacements, V,L,A,P,M, and e, respectively.Thendw=-YdX and the first law ofthermodynamics can bewritten in theformVEu,dN,.du=dQ+Ydx+(3.3)j-1
3.2 Energy Conservation 29 each step. Such changes often occur rapidly and give rise to flows and “friction” effects. After an irreversible change the system cannot be brought back to its original thermodynamic state without causing a change in the thermodynamic state of the universe. 3.2 Energy Conservation There is a store of energy in a thermodynamic system, called the internal energy U. Infinitesimal changes dU in internal energy content can occur by causing the system to do mechanical work, d∕W or by adding heat, d∕Q, to the system. The internal energy can also be changed by adding particles of types j = 1, 2, . , ν to the system causing infinitesimal changes in the internal energy equal to ∑ν j=1 μj dNj, where μj is the chemical potential of particles of type j and dNj is the infinitesimal change in the number of type j particles.We use the notation,d∕W, to indicate that the differential d∕W is not exact (see Appendix B). The quantities d∕Q and d∕W are not exact differentials because they depend on the path taken (on the way in which heat is added or work is done). The quantities dNj are exact differentials. The change in the internal energy that results from these thermal, mechanical, and chemical processes is given by dU = d∕Q − d∕W + ∑ν j=1 μj dNj . (3.1) The work, d∕W, may be due to changes in any relevant extensive “mechanical” variable. In general it can be written d∕W = P dV − J dL − σ dA − E ⋅ dP − H ⋅ dM − φ de , (3.2) where dU, dV, dL, dA, dP, dM, and de are exact differentials. The magnetization M and the magnetic field H are related to the magnetic flux density B by the equation B = μ0H + M, where μ0 is the permeability of free space. The definition of the first five terms in Eq. (3.2) was discussed in Section 3.1. The term, −φ de, is the work the system does if it has an electric potential, φ, and increases its charge by an amount, de. We may think of −P, J, σ, E, H and φ as generalized forces, and we may think of dV, dL, dA, dP, dM, and de as generalized displacements. It is useful to introduce a generalized mechanical force, Y, which denotes quantities such as, −P, J, σ, E, H, and φ, and a generalized displacement, X, which denotes the corresponding displacements, V, L, A, P, M, and e, respectively. Then d∕W = −Y dX and the first law of thermodynamics can be written in the form dU = d∕Q + Y dX + ∑ν j=1 μj dNj . (3.3)
3ThermodynamicsNote that μ, is a chemical force and dN, is a chemical displacement. Note alsothat the pressure,P, has a different sign from the other generalized forces. If weincrease the pressure, the volume increases, whereas if we increase theforce, Y,for all other cases, the extensive variable, X, decreases.3.3EntropyThe second law of thermodynamics is of immense importance for many rea-sons[24,47,183].Wecanuseittocomputethemaximumpossibleefficiencyofan engine that transforms heat into work.It also enables us to introduce a newstate variable, theentropyS, which is conjugate to the temperature.The entropygives us a measure of the degree of thermal disorder in a system and also gives usameansfor determining the stabilityofequilibrium states.In addition,it providesan important link between reversible and irreversible processes.3.3.1CarnotEngineThe second law is most easilydiscussed in terms of a universal heat engine firstintroduced byCarnot.Theconstruction ofall heatenginesisbasedontheobser-vation that,ifheat is allowedtoflowfroma hightemperaturetoalowertempera-ture,part of the heat can be turned into work. Carnot observed that temperaturedifferences can disappear spontaneouslywithout producing work.Therefore,themost efficient heat engines must be those whose cycles consist only of reversiblesteps, thereby eliminating wasteful heat flows.Thereare many ways to constructreversibleheatengines,and theygenerallyhave different levels ofefficiency.How-ever, Carnot found themost efficient of all possible heat engines.The Carnot heatengine is universal because,notonly is itthemostefficientofallheatengines,buttheeficiencyoftheCarnotengineisindependentofthematerialsused to run it.The Carnot engine consists of the four steps shown in Figure 3.1.Theseinclude:1.Isothermal (constanttemperature)absorptionofheatQ2fromareservoiratahightemperatureTh(weusetoindicateafiniteratherthananinfinitesimalamount of heat) (the process 1 → 2).2.Adiabatic (constant heatcontent) change in temperaturefrom Tyto thelowervalue T, (the process 2 → 3).3.Isothermal expulsionofheatQ intoareservoirattemperatureT(theprocess 3 → 4).Adiabatic return of the state at temperature T,to the state at temperatureTh4.(the process 4→1).The work done by the engine during one complete cycle can be found by inte-grating the differential element of work Y dX about the entire cycle and, there-
30 3 Thermodynamics Note that μj is a chemical force and dNj is a chemical displacement. Note also that the pressure, P, has a different sign from the other generalized forces. If we increase the pressure, the volume increases, whereas if we increase the force, Y, for all other cases, the extensive variable, X, decreases. 3.3 Entropy The second law of thermodynamics is of immense importance for many reasons [24, 47, 183]. We can use it to compute the maximum possible efficiency of an engine that transforms heat into work. It also enables us to introduce a new state variable, the entropy S, which is conjugate to the temperature. The entropy gives us a measure of the degree of thermal disorder in a system and also gives us a means for determining the stability of equilibrium states. In addition, it provides an important link between reversible and irreversible processes. 3.3.1 Carnot Engine The second law is most easily discussed in terms of a universal heat engine first introduced by Carnot. The construction of all heat engines is based on the observation that, if heat is allowed to flow from a high temperature to a lower temperature, part of the heat can be turned into work. Carnot observed that temperature differences can disappear spontaneously without producing work. Therefore, the most efficient heat engines must be those whose cycles consist only of reversible steps, thereby eliminating wasteful heat flows. There are many ways to construct reversible heat engines, and they generally have different levels of efficiency. However, Carnot found the most efficient of all possible heat engines. The Carnot heat engine is universal because, not only is it the most efficient of all heat engines, but the efficiency of the Carnot engine is independent of the materials used to run it. The Carnot engine consists of the four steps shown in Figure 3.1. These include: 1. Isothermal (constant temperature) absorption of heat ΔQ12 from a reservoir at a high temperature Th (we use Δ to indicate a finite rather than an infinitesimal amount of heat) (the process 1 → 2). 2. Adiabatic (constant heat content) change in temperature from Th to the lower value Tc (the process 2 → 3). 3. Isothermal expulsion of heat ΔQ43 into a reservoir at temperature Tc (the process 3 → 4). 4. Adiabatic return of the state at temperature Tc to the state at temperature Th (the process 4 → 1). The work done by the engine during one complete cycle can be found by integrating the differential element of work Y dX about the entire cycle and, there-
3.3 Entropyisothermal (Th)Y1.AQ12adiabaticsgaoaisothermal (Tc)SGX(a)(b)XFigure 3.1 (a) A Carnot engine which runscess takes place reversibly. (b) An arbitraryonasubstancewithstatevariables,XandY.reversible heat engine is composed of manyThe heat absorbed is Qμz and the heat eject-infinitesimal Carnot engines. The area en-ed is Q. The shaded area is equal to theclosed by the curve is equal to the work donework done during the cycle. The whole pro-by the heat engine.fore, the net work Wot done by the engine is given by the shaded area in Figure 3.la.Theeficiency ofany heatengine is givenbythe ratio ofthe net work done Wtotto heat absorbed Qabs so, in general, the efficiency of a heat engine is given by=(△Wtot)/(Qabs).For the Carnot engine, heat is only absorbed during the process 1 -→ 2, so theefficiencyof theCarnot engine (CE)canbe writtenWtot(3.4)ICE=△Q12The internal energyU is a state variable and, therefore,the total change utot foronecompletecycleof theenginemustbezerobecausecompletion of thecyclereturns the system to the thermodynamic state it started in.The first law thenenablesustowrite(3.5)Utot=Qtot-Wtot=0and thus(3.6)△Wtot=AQtot=AQ12+4Q34=AQ12-△Q43If we combine Eqs. (3.4)and (3.6),we can write the eficiencyofthe Carnot engineintheformAQ43(3.7)IcE = 1.△Q12A 100% efficient engine is one which converts all the heat it absorbs into work.However,as we shall see,no such engine can exist in nature because real enginesdonot operate in a completely reversiblemanner.Carnot engines can operate using anyofa variety of substances (examples areleft as problems).In Exercise3.1,wecompute theefficiency of a Carnot enginewhich uses an ideal gas as an operating substance. However, regardless of the op-erating substance,all Carnotengineshavethe same efficiency
3.3 Entropy 31 Figure 3.1 (a) A Carnot engine which runs on a substance with state variables, X and Y. The heat absorbed is ΔQ12 and the heat ejected is ΔQ43. The shaded area is equal to the work done during the cycle. The whole process takes place reversibly. (b) An arbitrary reversible heat engine is composed of many infinitesimal Carnot engines. The area enclosed by the curve is equal to the work done by the heat engine. fore, the net work ΔWtot done by the engine is given by the shaded area in Figure 3.1a. The efficiency η of any heat engine is given by the ratio of the net work done ΔWtot to heat absorbed ΔQabs so, in general, the efficiency of a heat engine is given by η = (ΔWtot)∕(ΔQabs ). For the Carnot engine, heat is only absorbed during the process 1 → 2, so the efficiency of the Carnot engine (CE) can be written ηCE = ΔWtot ΔQ12 . (3.4) The internal energy U is a state variable and, therefore, the total change ΔUtot for one complete cycle of the engine must be zero because completion of the cycle returns the system to the thermodynamic state it started in. The first law then enables us to write ΔUtot = ΔQtot − ΔWtot = 0 (3.5) and thus ΔWtot = ΔQtot = ΔQ12 + ΔQ34 = ΔQ12 − ΔQ43 . (3.6) If we combine Eqs. (3.4) and (3.6), we can write the efficiency of the Carnot engine in the form ηCE = 1 − ΔQ43 ΔQ12 . (3.7) A 100% efficient engine is one which converts all the heat it absorbs into work. However, as we shall see, no such engine can exist in nature because real engines do not operate in a completely reversible manner. Carnot engines can operate using any of a variety of substances (examples are left as problems). In Exercise 3.1, we compute the efficiency of a Carnot engine which uses an ideal gas as an operating substance. However, regardless of the operating substance, all Carnot engines have the same efficiency
3ThermodynamicsThe Kelvin temperature scale T was introduced by W.Thomson (Lord Kelvin)and is a universal temperature scale because it is based on the universality of theCarnot engine. It is defined asAQ = T(3.8)AQ12Thwhere T, (T,) is the hottest (coldest) temperature reached by the Carnot engine.The units of degree Kelvin (K) are the same as degree Celsius (°C).The ice pointandboilingpoints of water,at one atmosphere pressure aredefined tobeOC and100°C,respectively.Thetriple point of water is 0.01°C and occurs at a pressureof 611.73 Pa.The relation between degree Celsius, t,and degree Kelvin,T, isT = (t。+ 273.15). The triple point of water is fixed at T = 273.16 K.WecanusetheCarnot engineto define anewstatevariable called theentropy.All Carnot engines have an efficiencyA0=1-(3.9)ICE=1△Q12Thregardless ofoperating substance.Using Eq.(3.9),wecan writethefollowingre-lation fora Carnot cycle:AQ12 + AQu = 0(3.10)ThT.(notethatQ34=-△Q43)Equation (3.10) can be generalized to the case of an arbitrary reversible heatenginebecausewe can consider suchan engineasbeing composed of a sum ofmanyinfinitesimal Carnot cycles (cf.Figure3.1b).For anarbitraryreversibleheatengine we havefdo2=0，(3.11)Tand,therefore,dods=(3.12)Tis an exact differential.The quantity S,is called the entropy and is a state variablebecausetheintegralofdSaboutaclosedpathgiveszeroExercise 3.1口Compute the efficiency of a Carnot cycle which uses a monatomic ideal gas asan operating substance.Use the equation ofstate ofa monatomic ideal gas PV=nRTandtheinternalenergyU=(3/2)nRT
32 3 Thermodynamics The Kelvin temperature scale T was introduced by W. Thomson (Lord Kelvin) and is a universal temperature scale because it is based on the universality of the Carnot engine. It is defined as ΔQ43 ΔQ12 = Tc Th , (3.8) where Th (Tc) is the hottest (coldest) temperature reached by the Carnot engine. The units of degree Kelvin (K) are the same as degree Celsius (◦C). The ice point and boiling points of water, at one atmosphere pressure are defined to be 0 ◦C and 100 ◦C, respectively. The triple point of water is 0.01 ◦C and occurs at a pressure of 611.73 Pa. The relation between degree Celsius, tc, and degree Kelvin, T, is T = (tc + 273.15). The triple point of water is fixed at T = 273.16 K. We can use the Carnot engine to define a new state variable called the entropy. All Carnot engines have an efficiency ηCE = 1 − ΔQ43 ΔQ12 = 1 − Tc Th (3.9) regardless of operating substance. Using Eq. (3.9), we can write the following relation for a Carnot cycle: ΔQ12 Th + ΔQ34 Tc = 0 (3.10) (note that ΔQ34 = −ΔQ43). Equation (3.10) can be generalized to the case of an arbitrary reversible heat engine because we can consider such an engine as being composed of a sum of many infinitesimal Carnot cycles (cf. Figure 3.1b). For an arbitrary reversible heat engine we have ∮ d∕Q T = 0 , (3.11) and, therefore, dS ≡ d∕Q T (3.12) is an exact differential. The quantity S, is called the entropy and is a state variable because the integral of dS about a closed path gives zero. Exercise 3.1 Compute the efficiency of a Carnot cycle which uses a monatomic ideal gas as an operating substance. Use the equation of state of a monatomic ideal gas PV = 𝔫RT and the internal energy U = (3∕2)𝔫RT
3.3 EntropyThe Carnot cycle for an ideal gas is shown in thefigure belowAnswer:RT--1TnRaPATciP3/1VV2V4Vs(a) Let us first considerthe isothermal paths1-→2and 3-→4. Since the temperature is constant alongthesepaths,dT=Oand du =(3/2)nRdT=0.Thus,alongthe path 1 -2, dQ= dW = nRT,(dV/V).The heat absorbed/ejected along thepaths1→2and3-→4aredvand AQ34 = nRT,lnAQ12=nRTh=nRT,lVVirespectively.SinceV>Vi,Q12>0 and heat is absorbed alongthe path1→2Since V>V, Q34<0 and heat is ejected along the path 3→ 4.(b) Let us next consider the adiabatic paths 2 -→ 3 and 4 - 1. Along the adi-abatic path, dQ=0=du +PdV =(3/2)nRdT+PdV.If wemake use ofthe equation of state,wefind (3/2)dT/T=-dV/V.We now integrate to findT3/2V=constantforan adiabaticprocess.Thus,along thepaths2-→3and4 1, respectively, we have T.V2/3 = T, V2/3 and T,V2/ = T, V2/3 which givesV/V,=V,/V.Because U is a state variable, for the entire cycle we can writeUtot=QtotWtot=0.ThusWtot=Qtot=Q12+Q34.Theefficiencyof the Carnot cycle is△Wtot =1+AQ34=1-T。 In(Vs/V4)T.=1-1=AQ12Q12T, In(V2/V)ThNoheat engine, operating between the samehighand lowtemperatures, can bemore efficientthan a Carnotengine.Thus,an enginewhichruns between the sametwo reservoirs but contains spontaneous or irreversible processes in some part ofthe cycle will have a lower efficiency. Therefore, for an irreversible heat engine wecanwriteAQ3>3T.AQ12AQ43<0.and(3.13)ThT.△Q12Th
3.3 Entropy 33 Answer: The Carnot cycle for an ideal gas is shown in the figure below. (a) Let us first consider the isothermal paths 1 → 2 and 3 → 4. Since the temperature is constant along these paths, dT = 0 and dU = (3∕2)nR dT = 0. Thus, along the path 1 → 2, d∕Q = d∕W = nRTh(dV∕V). The heat absorbed/ejected along the paths 1 → 2 and 3 → 4 are ΔQ12 = nRTh V2 ∫ V1 dV V = nRTh ln (V2 V1 ) and ΔQ34 = nRTc ln (V4 V3 ) , respectively. Since V2 > V1, ΔQ12 > 0 and heat is absorbed along the path 1 → 2. Since V3 > V4, ΔQ34 < 0 and heat is ejected along the path 3 → 4. (b) Let us next consider the adiabatic paths 2 → 3 and 4 → 1. Along the adiabatic path, d∕Q = 0 = dU + P dV = (3∕2)nR dT + P dV. If we make use of the equation of state, we find (3∕2) dT∕T = −dV∕V. We now integrate to find T3∕2V = constant for an adiabatic process. Thus, along the paths 2 → 3 and 4 → 1, respectively, we have TcV2∕3 3 = ThV2∕3 2 and TcV2∕3 4 = ThV2∕3 1 which gives V3∕V4 = V2∕V1. Because U is a state variable, for the entire cycle we can write ΔUtot = ΔQtot − ΔWtot = 0. Thus ΔWtot = ΔQtot = ΔQ12 + ΔQ34. The efficiency of the Carnot cycle is η = ΔWtot ΔQ12 = 1 + ΔQ34 ΔQ12 = 1 − Tc Th ln(V3∕V4) ln(V2∕V1) = 1 − Tc Th . No heat engine, operating between the same high and low temperatures, can be more efficient than a Carnot engine. Thus, an engine which runs between the same two reservoirs but contains spontaneous or irreversible processes in some part of the cycle will have a lower efficiency. Therefore, for an irreversible heat engine we can write ΔQ43 ΔQ12 > Tc Th and ΔQ12 Th − ΔQ43 Tc < 0 . (3.13)