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Introduction to the Thermodynamicsof MaterialsThird EditionDavid R. GaskellPreliminaries‡ [email protected]::spellD‡ Physical Constants Needed for Problemsü Heat CapacitiesThe generic heat capcityc 105bTCp a ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ;T2103The heat capacities of various elements and compounds areCpAgs Cp ê. 8a Ø 21.30, b Ø 8.54, c Ø 1.51 ;CpAgl Cp ê. 8a Ø 30.50, b Ø 0, c Ø 0 ;20.75 T2CpAl Cp ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. 8a Ø 31.38, b Ø -16.4, c Ø -3.6 ;106—CpAll Cp ê. 8a Ø 31.76, b Ø 0, c Ø 0 ;General::spell1 : Possible spelling error:new symbol name "CpAll" is similar to existing symbol "CpAl".CpAl2O3 Cp ê. 8a Ø 117.49, b Ø 10.38, c Ø -37.11 ;CpCaO Cp ê. 8a Ø 50.42, b Ø 4.18, c Ø -8.49 ;CpCaTiO3 Cp ê. 8a Ø 127.39, b Ø 5.69, c Ø -27.99 ;CpCord Cp ê. 8a Ø 626.34, b Ø 91.21, c Ø -200.83 ;2.26 T2CpCr Cp ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. 8a Ø 21.76, b Ø 8.98, c Ø -0.96 ;106CpCr2O3 Cp ê. 8a Ø 119.37, b Ø 9.30, c Ø -15.65 ;

2Notes on Gaskell Text—CpCO Cp ê. 8a Ø 28.41, b Ø 4.10, c Ø -0.46 ;General::spell1 : Possible spelling error:new symbol name "CpCO" is similar to existing symbol "CpCaO".CpCO2 Cp ê. 8a Ø 44.14, b Ø 9.04, c Ø -8.54 ;9.47 T2CpCu Cp ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. 8a Ø 30.29, b Ø -10.71, c Ø -3.22 ;106CpDiamond Cp ê. 8a Ø 9.12, b Ø 13.22, c Ø -6.19 ;17.38 T2CpGraphite Cp - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. 8a Ø 0.11, b Ø 38.94, c Ø -1.48 ;106CpH2Og Cp ê. 8a Ø 30.00, b Ø 10.71, c Ø -0.33 ;N2 over range 298-2500KCpN2 Cp ê. 8a Ø 27.87, b Ø 4.27, c Ø 0 ;O2 over range 298-3000K—CpO2 Cp ê. 8a Ø 29.96, b Ø 4.18, c Ø -1.67 ;General::spell : Possible spelling error: newsymbol name "CpO2" is similar to existing symbols 8CpCO2, CpN2 .Si3N4 over range 298-900K27.07 T2CpSi3N4 Cp - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. 8a Ø 76.36, b Ø 109.04, c Ø -6.53 ;106SiO2 (alpha quartz) for 298-847KCpSiO2Q Cp ê. 8a Ø 43.93, b Ø 38.83, c Ø -9.69 ;CpTiO2 Cp ê. 8a Ø 73.35, b Ø 3.05, c Ø -17.03 ;CpZra Cp ê. 8a Ø 22.84, b Ø 8.95, c Ø -0.67 ;—CpZrb Cp ê. 8a Ø 21.51, b Ø 6.57, c Ø 36.69 ;General::spell1 : Possible spelling error: newsymbol name "CpZrb" is similar to existing symbol "CpZra".CpZraO2 Cp ê. 8a Ø 69.62, b Ø 7.53, c Ø -14.06 ;—CpZrbO2 Cp ê. 8a Ø 74.48, b Ø 0, c Ø 0 ;General::spell1 : Possible spelling error: newsymbol name "CpZrbO2" is similar to existing symbol "CpZraO2".ü Enthalpies at 298K and Enthalpies of TransitionsHere are some enthalpies at 298. For compounds, these are enthalpies for formation from elements. The enthalpiesof pure elements are taken, by convention to be zero.HAl2O3 -1675700;

Notes on Gaskell Text3HAlmelt 10700;HCaO -634900;HCaTiO3 -1660600;HCH4 -74800;HCr2O3 -1134700;HCO2 -393500;HDiamond 1500;HH2Og -241800;HO2 0;HSi3N4 -744800;HSiO2Q -910900;HTiO -543000;HTiO2 -944000;HTi2O3 -1521000;HTi3O5 -2459000;Transformation Zr(a) to Zr(b)DHZratob 3900;Transformation Zr(a)O(2) to Zr(b)O2DHZrO2atob 5900;Formation of Zr(a)O(2)HZraO2 -1100800;ü Entropies at 298KThere are absolute entropies of some elements at compounds at 298KSCaO 38.1;SCaTiO3 93.7;SN2 191.5;SO2 205.1;SSi3N4 113.0;SSiO2Q 41.5;

4Notes on Gaskell TextSTiO 34.7;STiO2 50.6;STi2O3 77.2;STi3O5 129.4;SZra 39.0;SZraO2 50.6;ü Molecular WeightsmassAl 26.98;massAu 196.97;massCr 52.;massCu 63.55;massFe 55.85;massH 1.008;massMg 24.31;massN 14.007;massO 16;massC 12;massCa 40.08;massSi 28.04;massTi 47.88;massMn 54.94;massF 19 ;massZn 65.38 ;ü Vapor Pressurevapor -A ê T B [email protected] CAC - ÅÅÅÅ B [email protected] for the range 298-630KlnvapHgl vapor ê. 8A - 7611 , B - -0.795, C - 17.168 ;

Notes on Gaskell Text5lnvapSiCl4 vapor ê. 8A - 3620 , B - 0, C - 10.96 ;lnvapCO2s vapor ê. 8A - 3116 , B - 0, C - 16.01 ;lnvapMn vapor ê. 8A - 33440 , B - -3.02, C - 37.68 ;lnvapFe vapor ê. 8A - 45390 , B - -1.27, C - 23.93 ;lnvapZn vapor ê. 8A - 15250 , B - -1.255, C - 21.79 ;Chapter 1: Introduction and Definition of Terms‡ HistoryThermodynamics began with the study of heat and work effects and relations between heat and work. Some earlythermodynamics problems were for very practical problems. For example, in a steam engine heat is supplied towater to create steam. The steam is then used to turn an engine which does work. Finally, the water is exhasted tothe environment or in a cyclic engine it can be condensed and recyled to the heating chamber or boilerBoiler at TExhaustat T21EngineWorkDoneCondenserSteam power plant or steam engineAn early goal for thermodynamics was to analyze the steam engine and to figure out the maximum amount ofwork that could be done for an engine operating between the input temperature T1 and the output temperature T2 .Some of the most important work on thermodynamics of heat engines was done by Nicholas Carnot around 1810.He was a French engineer and wrote one paper, Reflections on the Motive Power of Heat, that introduced the“Carnot” cycle and helped explain the maximum efficiency of heat engines. It is interesting to note that the firststeam engines were invented in 1769. Thus the practical engineering was done without knowledge ofthermodynamics and well before the theory of the heat engine was developed. It can be said that the invention ofthe steam engine spawned the development of thermodynamics or that the steam engine did much more forthermodynamics than thermodynamics ever did for the steam engine.Although analysis of devices like steam engines, combustion engines, refrigerators, etc., are important,thermodynamics has much wider applicability. In material science, one is normally not that interested in heat andwork, but interested more the state of matter and how things might change when mixed, heated, pressurized, etc.Some important effects are chemical reactions (such as oxidation), formation of solutions, phase transformations.

6Notes on Gaskell TextOther issues might include response of materials to stress, strain, electrical fields, or magnetic fields. In otherwords, the changes in the matter are more interesting than the heat and work effects.‡ System and SurroundingsThe universe is divided into the System and the Surroundings. The system is any collection of objects that wechoose to analyze. The surroudings is the rest of the universe, but in more practical terms is the environment of thesystem. Our interest is in understanding the system. The system and surroundings interact be exchanging heat andwork. The surroundings can supply heat to the system or do work on the system. Alternatively, the system maygive off heat (supply heat to the surroundings) or do work on the surroundings.Some examples of material science type systems are a metalllic alloy in a crucible, a multi-component, multiphaseceramic, a blend of polymer molecules, a semiconductor alloy, or a mixutre of gases in a container. In materialscience, our main interest in such systems is the equilibrium state of the system, will the components react, willthey mix or phase separate, will there by phase transitions, and how will they respond to externally applied stimulisuch as pressure, temperature, stress, strain, electrical field, or magnetic filed.Thermodynamics is concerned only with the equilibrium state of matter and not in the rate at which matter reachesthe equlibrium state. Early thermodynamics was on heat (thermo) and work (dynamics) effects. In heat engineswith gases and liquids, equilibrium is often reached very fast and the rate of reaching equilibrium is very fast. The“dynamics” part refers to work effects and not to rates of processes. The study of the rates of processes is knownas “kinetics.”In material science, particularly problems dealing with solids or condensed matter, it is possible to deviate fromequilbrium for long times. For example, a polymer glass well below its glass transition is a non-equilibriumstructure. A detailed thermodynamic analysis of glass polymers (a difficult problem) would predict that thepolymer should exist in a different state than it actually does. At sufficient low temperatures, the polymer,however, will remain in the non-equilibrium glassy state; the equilibrium state will not be realized on any practicaltime scale.‡ Concept of StateMatter contains elementary particles such as atoms and molecules. The state of a system can be defined byspecifying the masses, velocities, positions, and all modes of motion (e.g., accelerations) of all of the particles inthe system. Such a state is called the microscopic state of the system. Given the microscopic state, we coulddeduce all the properties of the system. Normally, however, we do not have such detailed knowledge becausethere will always be a large number of particles (e.g. 1023 molecules in 1 mole of molecules). Fortunately suchdetailed knowlege is not required. Instead, it is possible to define a macroscopic state of the system by specifyingonly a few macroscopic and measurable variables such as pressure, volume, and temperature. It is found that whenonly a few of these variables are fixed, the entire state of the system is also fixed. Thus, the thermodynamic stateof a system is uniquely fixed when a small number of macroscopic, independent variables are fixed.For example, consider a gas or a liquid of constant composition such as a pure gas or liquid. The three keyvariables are pressure, P, temperature, T, and volume, V. It has been observed that when P and T are fixed that Valways has a unique value. In other words, P and T are the independent variables and V is a function of P and T:Volume [email protected], TD ;Such an equation is called an equation of state. Once P and T are known, V (and all other properties in this simpleexample) are determined. P, V, and T are all known as state variables; they only depend on the current state andnot the path the system took to reach the current state.

Notes on Gaskell Text7The use of P and T as the independent variables is simply a matter of choice and is done usually because P and Tare easy to control and measure. It would be equally acceptable to define V and T as the independent variables anddefine the system by an equation of state for pressure:Pressure [email protected], TD ;or to use P and V is independent variables and define the system by an equation of state for temperature:Temperature [email protected], VD ;V . ; P .; T .;ü More than Two Independent VariablesPure gases and liquids are particularly simple because their state depends only on two independent variables.Other systems require more variables, but the number required is always relatively small. For example, the volumeof a mixture of two gases will depend on the P and T and the compositions of the two gases orVolume [email protected], T, n1 , n2 D ;where n1 and n2 are the number of moles of the two gases. The volume of the system will depend not only on Pand T, but also on which gases are present. As above, this new equation of state could be done instead as anequation for P in terms of V, T, and composition:Pressure [email protected], T, n1 , n2 D ;or similarly as an equation for T in terms of P, V, and composition.Pressure or volume are all that are needed to define mechanical stimuli on a gas or a liquid. For solids, however,the matter might experience various states of stress and strain. For a pure solid, the natural variables aretemperature, stress s (instead of P), and strain e (instead of V). Unlike P and V which are scalar quantities, stressand strain are tensors with 6 independent coordinates. In general, the strain components are a function of T and thestress componentsStrainComponent εi @T, si D ;where εi and si are components of stress and strain. Alternatively, stress can be written as a function oftemperature and strainStressComponent si @T, εi D ;These equations of state are the thermomechanical stress-strain relations for a material. If the material is not a purematerial, such as a composite material, the stress-strain relations will also depend on the compositions of thematerial and typically on the geometry of the structure.For interactions of matter with other stimuli suich as electric or magnetic fields, the equations of state will alsodepend on the intensity of those fields.Thus, in summary, the thermodynamic state can also be expressed as an equation of state that is a function of arelatively small number of variables. For most problems encountered in thermodynamics, the variables are limitedto P, T, V, εi , si , composition, and applied fields. The simplest examples involve only two variables. Morecomplicated systems require more variables.

8Notes on Gaskell Textü Multivariable MathematicsAn equation of state is a function that defines one variable in terms of several other variables. Thus equations ofstate follow the rules of mutlivariable mathematics. In thermodynamics, we are often concered with howsomething changes as we change the independent variables. A general analysis of such a problem can be writtendown purely in mathematical terms. Let f @x1 , x2 , . xn D be a function of n variables x1 to xn . The totaldifferential in f (df) is given byn fdf ‚ J ÅÅÅÅÅÅÅÅÅÅÅÅ N dxi ; xii 1where the partial derivative is taken with all x j π xi being held constant. In Mathematica notation, this totaldifferential is written asndf ‚ xi f dxi ;i 1where xi fmeans the partial derivative of f with respect to xi while all other variables (here x j π xi ) are heldconstant. This Mathematica notation will be used throughout these notes which were prepared in a Mathematicanotebook.ü Example: V[P,T]For example, the equation of state V[P,T] for a pure gas depends on only two variables and has the totaldifferentialdV P [email protected], TD dP T [email protected], TD dTdT VH0,1L @P, TD dP VH1,0L @P, TDNote: blue text is these notes is Mathematica output after evaluating an input expression in red. Many inputexpressions are followed be semicolons which simple supresses uninteresting Mathematica output.Any change in volume due to a change in T and P can be calculated by integrating dV:fDeltaV ‡ „ V ;iwhere i and f are the initial and final values of T and P.This expression for dV is simply treating V[P,T] as an mathematical function of P and T. In thermodynamics weare usually dealing with physical quantities. In general, the partial derviatives for the total differentials themselvesoften have physical significance. In other words, they often correspond to measurable quanties. In the dVexpression, T [email protected], TD is the change in volume per degree at constant pressure which is thermal expansion ofthe matter. Thermal expansion coefficient is normalized to give T [email protected],, TDα ÄÄÄÄÄÄÄÄÄÄÄ[email protected],, TDVH0,1L @P, ÅÅÅÅÅÅ[email protected], TD

Notes on Gaskell Text9Likewise, P [email protected], TD is the change in volume due to pressure at constant temperature which is thecompressibility of the matter. After normalizing and adding a minus sign to make it positive, compressibility is P [email protected],, TDβ - ÄÄÄÄÄÄÄÄÄÄÄ[email protected],, TDVH1,0L @P, TD- ÅÅÅÅÅ[email protected], TDIn terms of thermal expansion and compressibility, the total differential for volume becomes:a . ; b . ; dV - b [email protected], TD dP a [email protected], TD dTdT a [email protected], TD - dP b [email protected], TDMany thermodynamic relations involve writing total differentials functions and then evaluating the physicalsignificance of the terms. Sometimes the physical significance is not clear. In such problems, the partial derivativeis defined as having having physical significance or it becomes a new thermodynamic quantity. One good exampleto be encountered later in this course is chemical potential.ü State VariablesA state variable is a variable that depends only on the state of a system and not on how the system got to that thatstate. For example V is a state variable. It depends only on the independent variables (P, T, and perhaps others)and not on the path taken to get to the variables. There are many thermodynamic state variables and they are veryimportant in thermodynamics.There are some thermodynamic quantities that are not state variables. The two most important are heat and work.The heat supplied to a system or the work done by a system depend on the path taken between states and thus bydefinition, heat and work are not state variables.ü EquilibriumAs stated before, thermodynamics always deals with the equilibrium state of matter. The previous sections defineequations of state for matter. Equilibrium is the state of the system when the variable reaches the value it shouldhave as defined by the equation of state. For example, a pure gas has an equation of state V[P,T]. Equilibrium isreached when after changing P and T to some new values, the volume becomes equal to the V[P,T] defined by theequation of state.All systems naturally proceed towards equilibrium. They are driven there by natural tendencies to minimizeenergy and to maximize entropy. These concepts will be discussed later. Although all systems tend towardsequilibrium, thermodynamics says nothing about the rate at which they will reach equilibrium. Some systems,particularly condensed solids as encountered in material science, may not approach equilibrium on a pratical timescale.ü Equation of State of an Ideal GasCharles’s law is that volume is proportional to temperature (which is true no matter what temperature scale isused) at constant pressure. In other words dV/dT is constant at constant pressure. If we take Tc as the temperatureon the centigrate scale and let V0 a0 dV/dT, where V0 and a0 are the volume and thermal expansion coefficientat 0 C, then volume at any other temperature on the centigrate scale is found by integration

10Notes on Gaskell TextTcV CollectAV0 ‡V0 a0 „ T , V0E0H1 a0 TcL V0But, this result implies that the volume will become zero [email protected] 0 , TcD199Tc Ø - ÅÅÅÅÅÅÅ a0and become negative if Tc drops lower. It is physcially impossible to have negative volume, thusw Tc -1/a01must define the lowest possible temperature or absolute zero. In 1802, Guy-Lussac measured a0 to be ÅÅÅÅÅÅÅÅÅ or2671ÅÅÅÅÅÅÅÅorabsoloutezeroabsolute zero to be at -267 C. More accurate experiments later (and today) show that a0 ÅÅÅÅÅÅÅÅ273.15to be at -273.15. These observations lead to the absolute or Kelvin temperature T defined by11T Tc ÅÅÅÅÅÅÅ ê. a0 - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅa0273.15273.15 TcOn the absolute scale1T . ; V SimplifyAV ê. Tc - T - ÅÅÅÅÅÅÅ Ea0a0 T V0Thus the volume is zero at T 0 and increases linearly with T (as observed experimentally).Boyle found that at constant T that V is inversely proportional to P. Combining the laws of Boyle and Charles, anideal gas can be defined byVV .; constant P ÅÅÅÅÅÅTPVÅÅÅÅÅÅÅÅTThe constant for one mole of gas is defined as the gas constant R. Thus, the equation of state for V for n moles ofgas isTV n R ÅÅÅÅÅÅPnRTÅÅÅÅÅÅÅÅÅÅÅÅPThe thermal expansion coefficient of an ideal gas is

Notes on Gaskell Text11 T Va ÅÅÅÅÅÅÅÅÅÅÅV1ÅÅÅÅTThe compressibility of an ideal gas is P Vb - ÅÅÅÅÅÅÅÅÅÅÅV1ÅÅÅÅPThus for the special case of an ideal gas, we can writeV . ; dV a V dT - b V dPdP VdT V- ÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅPTEquations of state for P and T can be solved by simple rearrangementTV . ; Solve A V n R ÅÅÅÅÅÅ , PEPnRT99P Ø ÅÅÅÅÅÅÅÅÅÅÅÅ VTSolve A V n R ÅÅÅÅÅÅ , TEPPV99T Ø ÅÅÅÅÅÅÅÅ nRP .; V .; T .;‡ Units of Work and EnergyP V has units of Force/Area X Volume Force X length. These are the units of work or energy. Thus, R musthave units of energy/degree/mole. When R was first measured, P was measured in atm and V in liters; thus P V orwork or energy has units liter-atm. In these units, R isRla 0.082057 ;with units liter-atm/(degree mole).SI units for energy is Joules. Also, in SI units, 1 atm isoneatm 101325. ;Nê m2 .Because 1 liter is 1000 cm3 or 10-3 m3 , 1 liter-atm is

12Notes on Gaskell Textonela oneatm 10-3101.325Joules. Then, in SI units of J/(degree mole), the gas constant isRSI Rla onela101.325 RlaIn cgs units with energy units of egs 10-7 J, the gas constant isRerg RSI 1071.01325 µ 109 RlaFinally, there are .239 cal/J. The gas constant using calories as the energy unit isRcal RSI .23924.2167 RlaNote that in early studies of work and heat, calories were used for heat energy and Joules (or an equivalent F Xlength) for work or mechanical energy. The first law of thermodynamics connects the two energy units and allowsone to relate heat and work energy or to relate calories and Joules.‡ Extensive and Intensive PropertiesProperties (or state variables) are extensive or intensive. Extensive variables depend on the size of the system suchas volume or mass. Intensive variables do not depend on the size such as pressure and temperature. Extensivevariables can be changed into intensive variables by dividing them by the mass or number of moles. Suchintensive variables are often called specific or molar quantities. For example, the volume per mole or molarvolume is an intensive variable of a system. Similarly, mass is an extensive property, by mass per unit volume ordensity is an intensive property.‡ Phase Diagrams and Thermodynamics ComponentsA Phase diagram is a 2D representation that plots the state of a system as a function of two independent variables.Systems are characterized by the number of components and the type of phase diagrams depend on the number ofcomponents. Examples are one-component (unary), two-component (binary), three-component (ternary), fourcomponent (quarternary), etc.In each zone, one state is the most stable state. On lines, two phases can coexist. At triple points, three phases cancoexist. Example of unary is water phase diagram. Unary diagrams usually use two variables like P and T.Binary diagrams add composition as a third variable. Binary diagrams are usually for one variable (T, P, or V)together with the composition variable. The complete phase space is 3D. Thus, 2D binary plots are sections of the3D curves. Zones can be single phase solutions or two-phase regions. The relative proportions of phases in twophase regions are given by the lever rule.Choice of components is arbitrary.

Notes on Gaskell Text13‡ OverviewZeroth law of thermodynamics defines temperature. First law connected heat and work and clarified conservationof energy in all systems. The key new energy term that developed from the first law is internal energy. Internalenergy often has a nice physical significance; sometimes, it significance is less apparant. The first law says energyis conserved, but it makes no statement about the possible values of heat and work. The second law defines limitson heat and work in processes. It was used to define the efficiency of heat engines. The second law also lead to thedefinition of entropy. Entropy was slow to be accepted, because it has less apparant physical significance thaninternal energy. Rougly speaking, entropy is the degree of mixed-upedness. Some thermodynamic problemsrequire an absolute value of entropy, the third law of thermodynamics defines the entropy of a pure substance atabsolute zero to be zero.The principles of thermodynamics is are nearly fully defined after defining the laws of thermodynamics, internalenergy, and entropy. The rest of the study of thermodynamics is application of those principles to variousproblems. All systems try to minimize energy and maximize entropy. Most problems we ever encounter can besolved from these basic principles. It turns out, however, that direct use of internal energy and entropy can bedifficult. Instead, we define new functions called free energy - Gibbs free energy or Helmholz free energy. Thesenew energies perform the same function as other thermodynamics functions, but that are physcially much morerelevant to typical problems of chemistry and material science. In particular, Gibbs free energy is the mostcommon term needed for chemical and material science problems that are typically encounted in various states ofapplied temperature and pressure.Chapter 2: The First Law of Thermodynamics‡ Ideal Gas Change of Stateü Change in Internal Energy U UÅÅÅÅ L 0 for an ideal gas an H ÅÅÅÅÅÅÅÅ L n cv for an ideal gas, the total differential for internal energy forBecause H ÅÅÅÅ V T T Vany change of state of an ideal gas is dU n cv dT. The total change in internal energy is thuys always given by:T2DU ‡n cv „ TT1-n cv T1 n cv T2which can be rewritten ascvDU ÅÅÅÅÅÅÅ n R DT ;Rwhere DT T2 - T1 . For an ideal gas, n R(T2 - T1 ) P2 V2 - P1 V1 DHPVL. Thus internal energy can also bewritten ascvDU ÅÅÅÅÅÅÅ D HPVL ;R

14Notes on Gaskell Textü Change in EnthalpyOnce the change in internal energy is known, the change in enthalpy is easily found fromcvDH DU D HPVL J ÅÅÅÅÅÅÅÅ 1N D HPVLRccvpBut, for an ideal gas c p - cv R which leads to H ÅÅÅÅRÅÅ 1L ÅÅÅÅRÅÅ . The total change in enthalpy can be written twoways as:cpDH ÅÅÅÅÅÅÅ D HPVL ; DH RcpÅÅÅÅÅÅÅ n R DT;Rü Heat and Work in Various ProcessesThe previous sections gave results for DU and DH for any change of state in a ideal gas. The values for heat andwork during a change of state, however, will depend on path. This section gives some results for heat and workduring some common processes:1. Adiabatic ProcessThe definition of an adiabatic process is that q 0; thus all the change in U is caused by work or:q 0 ; w -DU ;2. Isometric ProcessIn an isometric process volume is constant which means w 0. Heat and work are thus:q DU ;w 0 ;3. Isobaric ProcessThe definition of enthalpy is the it is equal to the heat during a constant pressure or isobaric process; thus q DH. Work is found thethe first law as w q - DU; thusq DH ; w D HPVL ;4. Isothermal ProcessBecause U is a function only of T for an ideal gas, DU DH 0 for an isothermal process. These results alsofollow from the general results by using DT D(PV) 0 for an isothermal process. In general, all that can be saidabout q and w for an isothermal process isq w ;w q ;The actually value of q and w will depend on whether the process is conducted reversibly or irreversibly. For areversible process q and w can be calculated from P dV work asV2q w ‡P „V ;V1which using the ideal gas equation of state becomesV2q w ‡V1nRTÅÅÅÅÅÅÅÅÅÅÅÅ „ VV-n R T [email protected] D n R T [email protected] D

Notes on Gaskell Text15or because PV constant, we can writeV2q w n R T LogA ÅÅÅÅÅÅÅ E ; q w V1P1n R T LogA ÅÅÅÅÅÅÅ E ;P25. Any ProcessesFor any other process, w can be calculated for the P dV integral and q from the first law of thermodynamics.Thus, we can writeV2q DU ‡V1V2P „V ; w ‡P „V ;V1To do these calculations, we need to know P as a function of V throughout the process. This result applies for bothreversible and irreversible processes; P, however, will be given by an equation of state only for reversibleprocesses.‡ Numerical ExamplesV1 liters or and ideal gas at T1 and P1 are expanded (or compressed) to a new pressure P2 . Here are some constantsdefined in a table used to get numerical results:nums 8V1 - 10 , T1 - 298 ,P1 - 10 , P2 - 1 , R - 8.3144 , Rla - 0.082057 ;The number of moles can be calculated from the starting state:P 1 V1nmols ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. nums ;Rla T1subs [email protected], n - nmolsD8V1 Ø 10, T1 Ø 298, P1 Ø 10, P2 Ø 1,R Ø 8.3144, Rla Ø 0.082057, n Ø 4.08948 Finally, this constant will convert liter-atm energy units to Joule energy units. All results are given in Joules:laToJ 101.325 ;ü 1. Reversible, Isothermal ProcessIn an isothermal process for an ideal gas,DU 0 ; DH 0 ;thus heat and work are equal and given by:P2q w n R T1 LogA ÅÅÅÅÅÅÅ E J ê. subsP1-23330.9 J

16Notes on Gaskell Textü 2. Reversible Adiabatic ExpansionIn an adiabatic expansionq 0;and PV g isa constant. Thus the final state has1êggP2 V2i P1 V1 zy; T2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. g - 5 ê 3V2 jj ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ zn Rlak P2 {5ê33ê5P1 V 1P2 I ÅÅÅÅÅÅÅÅP2 ÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅn RlaFor an ideal gas cv 3R/2; thus3DU ÅÅÅÅ n R HT2 - T1 L ê. subs2-9147.99or we can use3DU ÅÅÅÅ HP2 V2 - P1 V1 L laToJ ê. [email protected], g - 5 ê 3D2-9148.02For some numeric results, the final temperature and volumes weread2 [email protected] , T2 ê. [email protected], g - 5 ê 3DD839.8107, 118.636 The work done isdw -DU9148.02For an ideal gas c p 5R/2; thus the enthalpy change is5DH ÅÅÅÅ HP2 V2 - P1 V1 L laToJ ê. [email protected], g - 5 ê 3D2-15246.7or

Notes on Gaskell Text175DH ÅÅÅÅ n R HT2 - T1 L ê. subs2-15246.7For numerical results in the subsequent examples, the initial and final states for the adiabatic process areV2 . ; T2 . ;sub2 [email protected], 8V2 - [email protected]@1DD , T2 - [email protected]@2DD, g - [email protected] ê 3D D8V1 Ø 10, T1 Ø 298, P1 Ø 10, P2 Ø 1, R Ø 8.3144, Rla Ø 0.082057,n Ø 4.08948, V2 Ø 39.8107, T2 Ø 118.636, g Ø 1.66667 ü Altenate Paths to End of Adiabatic Expansion(i) Get to P2 V2 T2 by isothermal process followed by constant volume process. DU for isothermal step is zero(because of the ideal gas). The constant volume step has the total DU which is3DU ÅÅÅÅ n R HT2 - T1 L ê. sub22-9147.99(ii) Get to P2 V2 T2 by isometric process followed by isothermal process. DU for isothermal step is zero (becauseof the ideal gas). The constant volume step is same as above and thus obviously gives the same result.(iii) Get to P2 V2 T2 by isothermal process followed by constant pressure process. DU for isothermal step is zero(because of the ideal gas). The enthalpy change for the constant pressure step is simply the same as before3DU ÅÅÅÅ n R HT2 - T1 L ê. sub22-9147.99(iv) Get to P2 V2 T2 by isometric process followed by constant pressure process. For isometric process, we onlyneed to know the intermediate temperature given byP2 V 1Ti ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ;n RlaThus, the first step has3DUi ÅÅÅÅ n R HTi - T1 L ê. sub22-13678.8The internal energy change in the second step is

18Notes on Gaskell Text3DUii ÅÅÅÅ n R HT2 - Ti L ê. sub224530.84Thus total energy change isDU DUi DUii-9147.99(v) Get to P2 V2 T2 by constant pressure process followed