chapter 3.pdf

Chapter 3 Volumetric Properties of Pure Fluids The equations of the preceding chapter provide the means for calculation of the heat and work quantities associated with various processes, but they are useless without knowledge of property values for internal energy or enthalpy. Such properties differ from one substance to another, and the laws of thermodynamics themselves do not provide any description or model of material behavior. Property values come from experiment, or from the correlated results of experiment, or from models grounded in and validated by experiment. Because there are no internal-energy or enthalpy meters, indirect measurement is the rule. For fluids, the most com- prehensive procedure requires measurements of molar volume in relation to temperature and pressure. The resulting pressure/volume/temperature (PVT ) data are most usefully correlated by equations of state, through which molar volume (or density), temperature, and pressure are functionally related. In this chapter we: ∙ Present the phase rule, which relates the number of independent variables required to fix the thermodynamic state of a system to the number of chemical species and phases present ∙ Describe qualitatively the general nature of PVT behavior of pure substances ∙ Provide a detailed treatment of the ideal-gas state ∙ Treat equations of state, which are mathematical formulations of the PVT behavior of fluids ∙ Introduce generalized correlations that allow prediction of the PVT behavior of fluids for which experimental data are lacking 3.1 THE PHASE RULE As indicated in Section 2.5, the state of a pure homogeneous fluid is fixed whenever two intensive thermodynamic properties are set at specific values. In contrast, when two phases of the same pure species are in equilibrium, the state of the system is fixed when only a sin- gle property is specified. For example, a system of steam and liquid water in equilibrium at 68 3.1. The Phase Rule 69 101.33 kPa can exist only at 100°C. It is impossible to change the temperature without also changing the pressure, if equilibrium between vapor and liquid phases is to be maintained. There is a single independent variable. For a multiphase system at equilibrium, the number of independent variables that must be arbitrarily fixed to establish its intensive state is called the number of degrees of freedom of the system. This number is given by the phase rule of J. Willard Gibbs.1 It is presented here without proof in the form applicable to nonreacting systems:2 F = 2 − π + N (3.1) where F is the number of degrees of freedom, π is the number of phases, and N is the number of chemical species present in the system. The intensive state of a system at equilibrium is established when its temperature, pressure, and the compositions of all phases are fixed. These are the variables of the phase rule, but they are not all independent. The phase rule gives the number of variables from this set that must be specified to fix all remaining intensive variables, and thus the intensive state of the system. A phase is a homogeneous region of matter. A gas or a mixture of gases, a liquid or a liquid solution, and a crystalline solid are examples of phases. An abrupt change in properties always occurs at the boundary between phases. Various phases can coexist, but they must be in equilibrium for the phase rule to apply. A phase need not be continuous; examples of discontinuous phases are a gas dispersed as bubbles in a liquid, a liquid dispersed as droplets in another liquid with which it is immiscible, and solid crystals dispersed in either a gas or a liquid. In each case a dispersed phase is distributed throughout a continuous phase. As an example, the phase rule may be applied to an aqueous solution of ethanol in equi- librium with its vapor. Here N = 2, π = 2, and F = 2 − π + N = 2 − 2 + 2 = 2 This is a system in vapor/liquid equilibrium, and it has two degrees of freedom. If the system exists at specified T and P (assuming this is possible), its liquid- and vapor-phase composi- tions are fixed by these conditions. A more common specification is of T and the liquid-phase composition, in which case P and the vapor-phase composition are fixed. Intensive variables are independent of the size of the system and of the individual phases. Thus, the phase rule gives the same information for a large system as for a small one and for different relative amounts of the phases. Moreover, the phase rule applies only to individual-phase compositions, and not to the overall composition of a multiphase system. Note also that for a phase only N − 1 compositions are independent, because the mole or mass fractions of a phase must sum to unity. The minimum number of degrees of freedom for any system is zero. When F = 0, the system is invariant; Eq. (3.1) becomes π = 2 + N. This value of π is the maximum number of phases that can coexist at equilibrium for a system containing N chemical species. When N = 1, this limit is reached for π = 3, characteristic of a triple point (Sec. 3.2). For example, the 1Josiah Willard Gibbs (1839–1903), American mathematical physicist, who deduced it in 1875. See http://en.wiki- pedia.org/wiki/Willard_Gibbs 2The theoretical justification of the phase rule for nonreacting systems is given in Sec. 12.2, and the phase rule for reacting systems is considered in Sec. 14.8. 70 CHAPTER 3. Volumetric Properties of Pure Fluids triple point of H2O, where liquid, vapor, and the common form of ice exist together in equilib- rium, occurs at 0.01°C and 0.0061 bar. Any change from these conditions causes at least one phase to disappear. Example 3.1 How many phase-rule variables must be specified to fix the thermodynamic state of each of the following systems? (a) Liquid water in equilibrium with its vapor. (b) Liquid water in equilibrium with a mixture of water vapor and nitrogen. (c) A three-phase system of a saturated aqueous salt solution at its boiling point with excess salt crystals present. Solution 3.1 (a) The system contains a single chemical species existing as two phases (one liquid and one vapor), and F = 2 − π + N = 2 − 2 + 1 = 1 This result is in agreement with the fact that for a given pressure water has but one boiling point. Temperature or pressure, but not both, may be specified for a system comprised of water in equilibrium with its vapor. (b) Two chemical species are present. Again there are two phases, and F = 2 − π + N = 2 − 2 + 2 = 2 The addition of an inert gas to a system of water in equilibrium with its vapor changes the characteristics of the system. Now temperature and pressure may be independently varied, but once they are fixed the system described can exist in equilibrium only at a particular composition of the vapor phase. (If nitrogen is considered negligibly soluble in water, the liquid phase is pure water.) (c) The three phases (π = 3) are crystalline salt, the saturated aqueous solution, and vapor generated at the boiling point. The two chemical species (N = 2) are water and salt. For this system, F=2−3+2=1 3.2 PVT BEHAVIOR OF PURE SUBSTANCES Figure 3.1 displays the equilibrium conditions of P and T at which solid, liquid, and gas phases of a pure substance exist. Lines 1-2 and 2-C represent the conditions at which solid and liq- uid phases exist in equilibrium with a vapor phase. These vapor pressure versus temperature 3.2. PVT Behavior of Pure Substances 71 lines describe states of solid/vapor (line 1-2) and liquid/vapor (line 2-C) equilibrium. As indicated in Ex. 3.1(a), such systems have but a single degree of freedom. Similarly, solid/liquid equilibrium is represented by line 2-3. The three lines display conditions of P and T at which two phases may coexist, and they divide the diagram into single-phase regions. Line 1-2, the sublimation curve, separates the solid and gas regions; line 2-3, the fusion curve, separates the solid and liquid regions; line 2-C, the vaporization curve, separates the liquid and gas regions. Point C is known as the critical point; its coordinates Pc and Tc are the highest pressure and highest temperature at which a pure chemical species is observed to exist in vapor/liquid equilibrium. The positive slope of the fusion line (2-3) represents the behavior of the vast majority of substances. Water, a very common substance, has some very uncommon properties, and exhibits a fusion line with negative slope. The three lines meet at the triple point, where the three phases coexist in equilibrium. According to the phase rule the triple point is invariant (F = 0). If the system exists along any of the two-phase lines of Fig. 3.1, it is univariant (F = 1), whereas in the single-phase regions it is divariant (F = 2). Invariant, univariant, and divariant states appear as points, curves, and areas, respectively, on a PT diagram. Changes of state can be represented by lines on the PT diagram: a constant-T change by a vertical line, and a constant-P change by a horizontal line. When such a line crosses a phase boundary, an abrupt change in properties of the fluid occurs at constant T and P; for example, vaporization for the transition from liquid to vapor. 3 A Fluid region Pc Liquid region C Fusion curve Pressure Vaporization curve Figure 3.1: PT diagram B for a pure substance. Solid region Gas region Triple Vapor 2 point region 1 Sublimation curve Tc Temperature Water in an open flask is obviously a liquid in contact with air. If the flask is sealed and the air is pumped out, water vaporizes to replace the air, and H2O fills the flask. Though the pressure in the flask is much reduced, everything appears unchanged. The liquid water resides at the bottom of the flask because its density is much greater than that of water vapor (steam), and the two phases are in equilibrium at conditions represented by a point on curve 2-C of Fig. 3.1. Far from point C, the properties of liquid and vapor are very different. However, if 72 CHAPTER 3. Volumetric Properties of Pure Fluids the temperature is raised so that the equilibrium state progresses upward along curve 2-C, the properties of the two phases become more and more nearly alike; at point C they become identical, and the meniscus disappears. One consequence is that transitions from liquid to vapor may occur along paths that do not cross the vaporization curve 2-C, i.e., from A to B. The transition from liquid to gas is gradual and does not include the usual vaporization step. The region existing at temperatures and pressures greater than Tc and Pc is marked off by dashed lines in Fig. 3.1; these do not represent phase boundaries, but rather are limits fixed by the meanings accorded the words liquid and gas. A phase is generally considered a liquid if vaporization results from pressure reduction at constant temperature. A phase is considered a gas if condensation results from temperature reduction at constant pressure. Since neither process can be initiated in the region beyond the dashed lines, it is called the fluid region. The gas region is sometimes divided into two parts, as indicated by the dotted vertical line of Fig. 3.1. A gas to the left of this line, which can be condensed either by compression at constant temperature or by cooling at constant pressure, is called a vapor. A fluid existing at a temperature greater than Tc is said to be supercritical. An example is atmospheric air. PV Diagram Figure 3.1 does not provide any information about volume; it merely displays the bounda- ries between single-phase regions. On a PV diagram [Fig. 3.2(a)] these boundaries in turn become regions where two phases—solid/liquid, solid/vapor, and liquid/vapor—coexist in equilibrium. The curves that outline these two-phase regions represent single phases that are in equilibrium. Their relative amounts determine the molar (or specific) volumes within the two-phase regions. The triple point of Fig. 3.1 here becomes a triple line, where the three phases with different values of V coexist at a single temperature and pressure. Figure 3.2(a), like Fig. 3.1, represents the behavior of the vast majority of substances wherein the transition from liquid to solid (freezing) is accompanied by a decrease in specific volume (increase in density), and the solid phase sinks in the liquid. Here again water dis- plays unusual behavior in that freezing results in an increase in specific volume (decrease in density), and on Fig. 3.2(a) the lines labeled solid and liquid are interchanged for water. Ice therefore floats on liquid water. Were it not so, the conditions on the earth’s surface would be vastly different. Figure 3.2(b) is an expanded view of the liquid, liquid/vapor, and vapor regions of the PV diagram, with four isotherms (paths of constant T) superimposed. Isotherms on Fig. 3.1 are vertical lines, and at temperatures greater than Tc do not cross a phase boundary. On Fig. 3.2(b) the isotherm labeled T > Tc is therefore smooth. The lines labeled T1 and T2 are for subcritical temperatures and consist of three seg- ments. The horizontal segment of each isotherm represents all possible mixtures of liquid and vapor in equilibrium, ranging from 100% liquid at the left end to 100% vapor at the right end. The locus of these end points is the dome-shaped curve labeled BCD, the left half of which (from B to C) represents single-phase liquids at their vaporization (boiling) temperatures and the right half (from C to D) single-phase vapors at their condensation temperatures. Liquids and vapors represented by BCD are said to be saturated, and coexisting phases are connected by the horizontal segment of the isotherm at the saturation pressure specific to the isotherm. Also called the vapor pressure, it is given by a point on Fig. 3.1 where an isotherm (vertical line) crosses the vaporization curve. 3.2. PVT Behavior of Pure Substances 73 Solid/liquid Tc Fluid C C Liquid Pc Pc Pc Q N P P Liquid Solid Gas Vap or T Tc Liquid/vapor Vapor Tc Tc J K Liquid/vapor T1 Tc Solid/vapor B T2 Tc D Vc Vc V V (a) (b) Figure 3.2: PV diagrams for a pure substance. (a) Showing solid, liquid, and gas regions. (b) Showing liquid, liquid/vapor, and vapor regions with isotherms. The two-phase liquid/vapor region lies under dome BCD; the subcooled-liquid region lies to the left of the saturated-liquid curve BC, and the superheated-vapor region lies to the right of the saturated-vapor curve CD. Subcooled liquid exists at temperatures below, and superheated vapor, at temperatures above the boiling point for the given pressure. Isotherms in the subcooled-liquid region are very steep because liquid volumes change little with large changes in pressure. The horizontal segments of the isotherms in the two-phase region become progressively shorter at higher temperatures, being ultimately reduced to a point at C. Thus, the critical iso- therm, labeled Tc, exhibits a horizontal inflection at the critical point C at the top of the dome, where the liquid and vapor phases become indistinguishable. Critical Behavior Insight into the nature of the critical point is gained from a description of the changes that occur when a pure substance is heated in a sealed upright tube of constant volume. The dotted vertical lines of Fig. 3.2(b) indicate such processes. They may also be traced on the PT diagram of Fig. 3.3, where the solid line is the vaporization curve (Fig. 3.1), and the dashed lines are constant-volume paths in the single-phase regions. If the tube is filled with either liquid or vapor, the heating process produces changes that lie along the dashed lines of Fig. 3.3, for example, by the change from E to F (subcooled-liquid) and by the change from G to H (super- heated-vapor). The corresponding vertical lines on Fig. 3.2(b) are not shown, but they lie to the left and right of BCD respectively. If the tube is only partially filled with liquid (the remainder being vapor in equilibrium with the liquid), heating at first causes changes described by the vapor-pressure curve (solid 74 CHAPTER 3. Volumetric Properties of Pure Fluids V2l Vc Liquid C V2v V1l Q Figure 3.3: PT diagram for a pure fluid showing P F N the vapor-pressure curve and constant-volume ( J, K ) V1v lines in the single-phase regions. H G E Vapor T line of Fig. 3.3). For the process indicated by line JQ on Fig. 3.2(b), the meniscus is initially near the top of the tube (point J), and the liquid expands sufficiently upon heating to fill the tube (point Q). On Fig. 3.3 the process traces a path from (J, K) to Q, and with further heating departs from the vapor-pressure curve along the line of constant molar volume V 2l .  The process indicated by line KN on Fig. 3.2(b) starts with a meniscus level closer to the bottom of the tube (point K), and heating vaporizes liquid, causing the meniscus to recede to the bottom (point N). On Fig. 3.3 the process traces a path from (J, K) to N. With further heating the path continues along the line of constant molar volume V  2v  . For a unique filling of the tube, with a particular intermediate meniscus level, the heating process follows a vertical line on Fig. 3.2(b) that passes through the critical point C. Physically, heating does not produce much change in the level of the meniscus. As the critical point is approached, the meniscus becomes indistinct, then hazy, and finally disappears. On Fig. 3.3 the path first follows the vapor-pressure curve, proceeding from point (J, K) to the critical point C, where it enters the single-phase fluid region, and follows Vc, the line of constant molar volume equal to the critical volume of the fluid.3 PVT Surfaces For a pure substance, existing as a single phase, the phase rule tells us that two state variables must be specified to determine the intensive state of the substance. Any two, from among P, V, and T, can be selected as the specified, or independent, variables, and the third can then be regarded as a function of those two. Thus, the relationship among P, V, and T for a pure substance can be represented as a surface in three dimensions. PT and PV diagrams like those illustrated in Figs. 3.1, 3.2, and 3.3 represent slices or projections of the three-dimensional PVT surface. Fig. 3.4 presents a view of the PVT surface for carbon dioxide over a region including liquid, vapor, and supercritical fluid states. Isotherms are superimposed on this 3A video illustrating this behavior is available in the online learning center at http://highered.mheducation.com/ sites/1259696529. 3.2. PVT Behavior of Pure Substances 75 surface. The vapor/liquid equilibrium curve is shown in white, with the vapor and liquid portions of it connected by the vertical segments of the isotherms. Note that for ease of visualization, the molar volume is given on a logarithmic scale, because the vapor volume at low pressure is several orders of magnitude larger than the liquid volume. 5 10 4 10 V (cm3/mol) 3 10 2 10 1 10 350 0 300 20 40 250 60 T (K) 80 200 100 P (bar) Figure 3.4: PVT surface for carbon dioxide, with isotherms shown in black and the vapor/liquid equi- librium curve in white. Single-Phase Regions For the regions of the diagram where a single phase exists, there is a unique relation connect- ing P, V, and T. Expressed analytically, as f (P, V, T) = 0, such a relation is known as a PVT equation of state. It relates pressure, molar or specific volume, and temperature for a pure homogeneous fluid at equilibrium. The simplest example, the equation for the ideal-gas state, PV = RT, has approximate validity for the low-pressure gas region and is discussed in detail in the following section. An equation of state may be solved for any one of the three quantities P, V, or T, given values for other two. For example, if V is considered a function of T and P, then V = V(T, P), and ( ∂T ) (∂P) ∂V ∂V dV =   ___    dT + ___       dP (3.2) P T 76 CHAPTER 3. Volumetric Properties of Pure Fluids The partial derivatives in this equation have definite physical meanings and are related to two properties, commonly tabulated for liquids, and defined as follows: ∙ Volume expansivity: V( ∂T )P 1 ∂V β ≡ __ ___       (3.3) ∙ Isothermal compressibility: V(∂P)T 1 ∂V κ ≡ − __ ___       (3.4) Combining Eqs. (3.2) through (3.4) yields: dV ___ = β dT − κ dP (3.5) V The isotherms for the liquid phase on the left side of Fig. 3.2(b) are very steep and closely spaced. Thus both (∂ V / ∂ T)Pand (∂ V / ∂ P)T are small. Hence, both β and κ are small. This char- acteristic behavior of liquids (outside the critical region) suggests an idealization, commonly employed in fluid mechanics and known as the incompressible fluid, for which both β and κ are zero. No real fluid is truly incompressible, but the idealization is useful, because it pro- vides a sufficiently realistic model of liquid behavior for many practical purposes. No equation of state exists for an incompressible fluid, because V is independent of T and P. For liquids, β is almost always positive (liquid water between 0°C and 4°C is an excep- tion), and κ is necessarily positive. At conditions not close to the critical point, β and κ are weak functions of temperature and pressure. Thus for small changes in T and P little error is introduced if they are assumed constant. Integration of Eq. (3.5) then yields: V2 ln ___ = β(T2− T1)− κ(P2− P1) (3.6) V1 This is a less restrictive approximation than the assumption of an incompressible fluid. Example 3.2 For liquid acetone at 20°C and 1 bar, β = 1.487 × 10−3°C−1 κ = 62 × 10−6bar−1 V = 1.287 cm3⋅g−1 For acetone, find: (a) The value of (∂ P / ∂ T)  V at 20°C and 1 bar. (b) The pressure after heating at constant V from 20°C and 1 bar to 30°C. (c) The volume change when T and P go from 20°C and 1 bar to 0°C and 10 bar. 3.3. Ideal Gas and Ideal-Gas State 77 Solution 3.2 (a) The derivative (∂ P / ∂ T)V is determined by application of Eq. (3.5) to the case for which V is constant and dV = 0: β dT − κ dP = 0 (const V) or ( ∂T ) ∂P β 1.487 × 10−3         = __ ___ = __________    = 24 bar ⋅ ° C−1 κ 62 × 10−6 V (b) If β and κ are assumed constant in the 10°C temperature interval, then for con- stant volume Eq. (3.6) can be written: β P2= P1+ __ (T2− T1)= 1 bar + 24 bar ⋅ ° C−1× 10° C = 241 bar κ (c) Direct substitution into Eq. (3.6) gives: V2 = (1.487 × 10−3)(−20)− (62 × 10−6)(9)= −0.0303 ln ___ V1       V2 ___ = 0.9702 and V2= (0.9702)(1.287)= 1.249cm3⋅g−1 V1 Then, ΔV = V2− V1= 1.249 − 1.287 = −0.038 cm3⋅g−1 The preceding example illustrates the fact that heating a liquid that completely fills a closed vessel can cause a substantial rise in pressure. On the other hand, liquid volume decreases very slowly with rising pressure. Thus, the very high pressure generated by heating a subcooled liquid at constant volume can be relieved by a very small volume increase, or a very small leak in the constant volume container. 3.3 IDEAL GAS AND IDEAL-GAS STATE In the 19th century, scientists developed a rough experimental knowledge of the PVT behavior of gases at moderate conditions of temperature and pressure, leading to the equation PV = RT, wherein V is molar volume and R is a universal constant. This equation adequately describes PVT behavior of gases for many practical purposes near ambient conditions of T and P. How- ever, more precise measurements show that for pressures appreciably above, and temperatures appreciably below, ambient conditions, deviations become pronounced. On the other hand, deviations become ever smaller as pressure decreases and temperature increases. The equation PV = RT is now understood to define an ideal gas and to represent a model of behavior more or less approximating the behavior of real gases. It is called the ideal gas law, but is in fact valid only for pressures approaching zero and temperatures approaching 78 CHAPTER 3. Volumetric Properties of Pure Fluids infinity. Thus, it is a law only at limiting conditions. As these limits are approached, the mol- ecules making up a gas become more and more widely separated, and the volume of the mol- ecules themselves becomes a smaller and smaller fraction of the total volume occupied by the gas. Furthermore, the forces of attraction between molecules become ever smaller because of the increasing distances between them. In the zero-pressure limit, molecules are separated by infinite distances. Their volumes become negligible compared with the total volume of the gas, and the intermolecular forces approach zero. The ideal gas concept extrapolates this behavior to all conditions of temperature and pressure. The internal energy of a real gas depends on both pressure and temperature. Pressure dependence results from intermolecular forces. If such forces did not exist, no energy would be required to alter intermolecular distances, and no energy would be required to bring about pressure and volume changes in a gas at constant temperature. Thus, in the absence of inter- molecular forces, internal energy would depend on temperature only. These observations are the basis for the concept of a hypothetical state of matter designated the ideal-gas state. It is the state of a gas comprised of real molecules that have negligible molecular volume and no intermolecular forces at all temperatures and pressures. Although related to the ideal gas, it presents a different perspective. It is not the gas that is ideal, but the state, and this has practical advantages. Two equations are fundamental to this state, namely the “ideal-gas law” and an expression showing that internal energy depends on temperature alone: ∙ The equation of state: PVig= RT (3.7) ∙ Internal energy: Uig= U(T) (3.8) The superscript ig denotes properties for the ideal-gas state. The property relations for this state are very simple, and at appropriate conditions of T and P they may serve as suitable approximations for direct application to the real-gas state. However, they have far greater importance as part of a general three-step procedure for calcu- lation of property changes for real gases that includes a major step in the ideal-gas state. The three steps are as follows: 1. Evaluate property changes for the mathematical transformation of an initial real-gas state into the ideal-gas state at the same T and P. 2. Calculate property changes in the ideal-gas state for the T and P changes of the process. 3. Evaluate property changes for the mathematical transformation of the ideal-gas state back to the real-gas state at the final T and P. This procedure calculates the primary property-value changes resulting from T and P changes by simple, but exact, equations for the ideal-gas state. The property-value changes for transitions between real and ideal-gas states are usually relatively minor corrections. These transition calculations are treated in Chapter 6. Here, we develop property-value calculations for the ideal-gas state. 3.3. Ideal Gas and Ideal-Gas State 79 Property Relations for the Ideal-Gas State The definition of heat capacity at constant volume, Eq. (2.15), leads for the ideal-gas state to ig the conclusion that CVis a function of temperature only: ( ∂T ) ∂ Uig dUig( T)       = ______ ig ig CV ≡  ____   = CV(T) (3.9) dT V The defining equation for enthalpy, Eq. (2.10), applied to the ideal-gas state, leads to the conclusion that Higis also a function only of temperature: H  ig≡ Uig+ PVig= Uig(T)+ RT = Hig(T) (3.10) ig ig The heat capacity at constant pressure CP, defined by Eq. (2.19), like CV, is a function of temperature only: ( ∂T ) ∂ Hig dHig( T)       = ___________ = CP(T) ig ig CP ≡  ____   (3.11) dT P ig ig A useful relation between C P  and C V  for the ideal-gas state comes from differentiation of Eq. (3.10): dHig dUig CP≡  _____     = ____ ig ig + R = CV+ R (3.12) dT dT ig ig This equation does not mean that CP and CV are themselves constant for the ideal-gas state, but only that they vary with temperature in such a way that their difference is equal to R. For any change in the ideal-gas state, Eqs. (3.9) and (3.11) lead to: ∫ V ig ig dU  ig= CVdT (3.13a) ΔUig=    C   dT  (3.13b) ∫ ig ig dHig= CPdT (3.14a) ΔHig=    CP  dT  (3.14b) ig Because both Uigand CVfor the ideal-gas state are functions of temperature only, ΔUig for the ideal-gas state is always given by Eq. (3.13b), regardless of the kind of process causing the change. This is illustrated in Fig. 3.5, which shows a graph of internal energy as a function of Vigat two different temperatures. The dashed line connecting points a and b represents a constant-volume process for which the temperature increases from T1 to T2 and the internal energy changes by Δ Uig= U  2ig  − U  1ig  . ig This change in internal energy is given by Eq. (3.13b) as Δ Uig= ∫CV  dT. The dashed lines connecting points a and c and points a and d represent other processes not occurring at constant volume but which also lead from an initial temperature T1 to a final temperature T2. The graph shows that the change in Uigfor these processes is the same as for the constant-volume process, ig and it is therefore given by the same equation, namely, ΔUig= ∫CVdT. However, ΔUig is not equal to Q for these processes, because Q depends not only on T1 and T2 but also on the path of the process. An entirely analogous discussion applies to the enthalpy Higin the ideal-gas state. 80 CHAPTER 3. Volumetric Properties of Pure Fluids b c d Figure 3.5: Internal energy changes for the U2 T2 ideal-gas state. Because Uigis independent of Vig, the plot of Uigvs. Vigat constant U temperature is a horizontal line. For different U1 T1 temperatures, Uighas different values, with a a separate line for each temperature. Two such lines are shown, one for temperature T1 and one for a higher temperature T2. V ig Process Calculations for the Ideal-Gas State Process calculations provide work and heat quantities. The work of a mechanically reversible closed-system process is given by Eq. (1.3), here written: dW = −P dVig (1.3) For the ideal-gas state in any closed-system process, the first law as given by Eq. (2.6) written for a unit mass or a mole, may be combined with Eq. (3.13a) to give: ig dQ + dW = CV dT Substitution for dW by Eq. (1.3) and solution for dQ yields an equation valid for the ideal-gas state in any mechanically reversible closed-system process: ig dQ = CVdT + PdVig (3.15) This equation contains the variables P, Vig, and T, only two of which are independent. Working equations for dQ and dW depend on which pair of these variables is selected as independent; i.e., upon which variable is eliminated by Eq. (3.7). We consider two cases, eliminating first P, and second, Vig. With P = RT / Vig, Eqs. (3.15) and (1.3) become: dVig dVig dQ = CV  dT + RT ____ ig (3.16) d W = −RT ____ ig (3.17) V ig V R dP For Vig= RT / P, dVig=  __ (dT − T ___ ). Substituting for dVigand for CV= CP− R transforms ig ig P P Eqs. (3.15) and (1.3) into: dP dP dQ = CP  dT − RT ___ (3.18) dW = −RdT + RT ___ ig (3.19) P P These equations apply to the ideal-gas state for various process calcula- tions. The assumptions implicit in their derivation are that the system is closed and the process is mechanically reversible. 3.3. Ideal Gas and Ideal-Gas State 81 Isothermal Process By Eqs. (3.13b) and (3.14b), ΔUig= ΔHig= 0 (const T) ig V P1 2ig = RT ln ___ By Eqs. (3.16) and (3.18), Q = RT ln ___ V1 P2 ig V P2 1ig = RT ln ___ By Eqs. (3.17) and (3.19), W = RT ln ___ V2 P1 Because Q = −W, a result that also follows from Eq. (2.3), we can write in summary: ig V P1 2ig = RT ln _ Q = −W = RT ln _ (const T) (3.20) V1 P2 Isobaric Process By Eqs. (3.13b) and (3.19) with dP = 0, ∫ ig ΔUig =     CV  dT  and W = −R(T2− T1) By Eqs. (3.14b) and (3.18), ∫ ig Q = ΔHig=     CP  dT  (const P) (3.21) Isochoric (Constant-V) Process With dVig= 0, W = 0, and by Eqs. (3.13b) and (3.16), ∫ ig Q = ΔUig=     CV  dT  (const Vig) (3.22) Adiabatic Process; Constant Heat Capacities An adiabatic process is one for which there is no heat transfer between the system and its surroundings; i.e., dQ = 0. Each of Eqs. (3.16) and (3.18) may therefore be set equal to zero. ig ig Integration with CV and CP constant then yields simple relations among the variables T, P, and V , valid for mechanically reversible adiabatic compression or expansion in the ideal-gas ig state with constant heat capacities. For example, Eq. (3.16) becomes: dT ___ R dVig = − ___ ig ____ ig T CV V 82 CHAPTER 3. Volumetric Properties of Pure Fluids ig Integration with CVconstant gives: ig ig R/CV T1 ( Vig ) 2 T V ___   1     =  ___ 2 Similarly, Eq. (3.18) leads to: ig R/CP T1 ( P1) T2 ___ P2 =  ___       These equations may also be expressed as: ig) γ − 1= const (3.23a) T ( V T ( 1 − γ)/γ= const P (3.23b) P ig) γ= const (3.23c) ( V where Eq. (3.23c) results by combining Eqs. (3.23a and (3.23b) and where by definition,4 ig C γ ≡ _ Pig (3.24) CV Equations (3.23) apply for the ideal-gas state with constant heat capaci- ties and are restricted to mechanically reversible adiabatic expansion or compression. The first law for an adiabatic process in a closed system combined with Eq. (3.13a) yields: ig dW = dU = CVdT ig For constant CV, ig W = ΔUig= CV ΔT (3.25) ig Alternative forms of Eq. (3.25) result if CVis eliminated in favor of the heat-capacity ratio γ: ig ig C CV+ R R R γ ≡ ___ Pig = _____ ig = 1 + ___ ig or CV= ___ ig CV CV CV γ − 1 and RΔT W = CVΔT = ____ ig γ−1 ig ig 4If C   and C   are constant, γ is necessarily constant. The assumption of constant γ is equivalent to the assumption p v ig ig that the heat capacities themselves are constant. This is the only way that the ratio Cp  /Cv  and the difference ig ig ig ig Cp  -Cv   = R can both be constant. Except for the monatomic gases, both Cp   and Cv   actually increase with temperature, but the ratio γ is less sensitive to temperature than the heat capacities themselves. 3.3. Ideal Gas and Ideal-Gas State 83 ig ig Because R T1= P1V1  and R T2= P2V2  , this expression may be written: ig ig P2V − P1V1 RT2− RT1 ___________ W = _______ 2 =    (3.26) γ−1 γ−1 Equations (3.25) and (3.26) are general for adiabatic compression and expansion processes in a closed system, whether reversible or not, because P, Vig, and T are state functions, ig independent of path. However, T2 and V2  are usually unknown. Elimination of V  2ig  from Eq. (3.26) by Eq. (3.23c), valid only for mechanically reversible processes, leads to the expression: γ − 1 [( P 1 ) ] γ − 1 [( P 1 ) ] ig (γ − 1)/γ (γ − 1)/γ P1V P 2 RT1 ___  2 P 1 ___ W = _____       − 1   = ___       − 1 (3.27) The same result is obtained when the relation between P and Vig given by Eq. (3.23c) is used for the integration, W = −∫ Pd Vig. Equation (3.27) is valid only for the ideal-gas state, for constant heat capacities, and for adiabatic, mechanically reversible, closed-system processes. When applied to real gases, Eqs. (3.23) through (3.27) often yield satisfactory approxi- mations, provided the deviations from ideality are relatively small. For monatomic gases, γ = 1.67; approximate values of γ are 1.4 for diatomic gases and 1.3 for simple polyatomic gases such as CO2, SO2, NH3, and CH4. Irreversible Processes All equations developed in this section have been derived for mechanically reversible, closed-system processes for the ideal-gas state. However, the equations for property changes— dUig, dHig, ΔUig, and ΔHig—are valid for the ideal-gas state regardless of the process. They apply equally to reversible and irreversible processes in both closed and open systems, because changes in properties depend only on initial and final states of the system. On the other hand, an equation for Q or W, unless it is equal to a property change, is subject to the restrictions of its derivation. The work of an irreversible process is usually calculated by a two-step procedure. First, W is determined for a mechanically reversible process that accomplishes the same change of state as the actual irreversible process. Second, this result is multiplied or divided by an efficiency to give the actual work. If the process produces work, the absolute value for the reversible process is larger than the value for the actual irreversible process and must be multiplied by an efficiency. If the process requires work, the value for the reversible process is smaller than the value for the actual irreversible process and must be divided by an efficiency. Applications of the concepts and equations of this section are illustrated in the examples that follow. In particular, the work of irreversible processes is treated in Ex. 3.5. 84 CHAPTER 3. Volumetric Properties of Pure Fluids Example 3.3 Air is compressed from an initial state of 1 bar and 298.15 K to a final state of 3 bar and 298.15 K by three different mechanically reversible processes in a closed system: (a) Heating at constant volume followed by cooling at constant pressure. (b) Isothermal compression. (c) Adiabatic compression followed by cooling at constant volume. These processes are shown in the figure. We assume air to be in its ideal-gas state, ig ig and assume constant heat capacities, C P= 29.100 J·mol−1·K−1. V= 20.785and C Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process. 6 c 4 P bar 2 a 2 b 1 0 5 10 15 20 25 V ig 103 m3 Solution 3.3 Choose the system as 1 mol of air. The initial and final states of the air are identi- cal with those of Ex. 2.7. The molar volumes given there are V1ig = 0.02479m3 V2ig = 0.008263 m3 Because T is the same at the beginning and end of the process, in all cases, ΔUig= ΔHig= 0 (a) The process here is exactly that of Ex. 2.7(b), for which: Q = −4958 J and W = 4958 J (b) Equation (3.20) for isothermal compression applies. The appropriate value of = 8.314 J·mol−1·K−1. R here (from Table A.2 of App. A) is R P1 1 Q = −W = RT ln ___ = (8.314)(298.15)ln __ = −2723 J P2 3 (c) The initial step of adiabatic compression takes the air to its final volume of 0.008263 m3. By Eq. (3.23a), the temperature at this point is: 3.3. Ideal Gas and Ideal-Gas State 85 γ−1 ( 0.008263 ) ig 0.4 ( V2 ) V 0.02479   1ig    T′ = T1 ___   = (298.15) ________         = 462.69 K For this step, Q = 0, and by Eq. (3.25), the work of compression is: ig ig W = CVΔT = CV(T′ − T1)  = (20.785)(462.69 − 298.15)= 3420 J For the constant-volume step, no work is done; the heat transfer is: ig Q = ΔUig= CV(T2− T′  )  = 20.785(298.15 − 462.69)= −3420 J Thus for process (c), W = 3420 J and Q = −3420 J Although the property changes ΔUig and ΔHig are zero for each process, Q and W are path-dependent, and here Q = −W. The figure shows each process on a PVig diagram. Because the work for each of these mechanically reversible processes is given by W = −∫ PdVig, the work for each process is proportional to the total area below the paths on the PVig diagram from 1 to 2. The relative sizes of these areas correspond to the numerical values of W. Example 3.4 A gas in its ideal-gas state undergoes the following sequence of mechanically reversible processes in a closed system: (a) From an initial state of 70°C and 1 bar, it is compressed adiabatically to 150°C. (b) It is then cooled from 150 to 70°C at constant pressure. (c) Finally, it expands isothermally to its original state. Calculate W, Q, ΔUig, and ΔHig for each of the three processes and for the entire cycle. ig ig Take CV= 12.471and CP= 20.785 J·mol−1·K−1. Solution 3.4 Take as a basis 1 mol of gas. (a) For adiabatic compression, Q = 0, and ig ΔUig= W = CVΔT = (12.471)(150 − 70)= 998 J      ig ΔHig= CPΔT = (20.785)(150 − 70)= 1663 J Pressure P2 is found from Eq. (3.23b): γ/(γ − 1) 150 + 273.15 2.5 ( T1) ( 70 + 273.15 ) T2 P2= P1 ___         = (1) ___________          = 1.689 bar 86 CHAPTER 3. Volumetric Properties of Pure Fluids 3 b 2 70°C 150 °C P a c 70°C 1 V ig (b) For this constant-pressure process, ig Q = ΔHig= CPΔT = (20.785)(70 − 150)= −1663 J            ΔU = CigΔT = (12.471)(70 − 150)= −998 J V W = ΔUig− Q = −998 − (−1663)= 665 J (c) For this isothermal process, ΔUig and ΔHig are zero; Eq. (3.20) yields: P3 P2 1.689 Q = −W = RT ln ___ = RT ln ___ = (8.314)(343.15)ln _____ = 1495 J P1 P1 1 For the entire cycle, Q = 0 − 1663 + 1495 = −168 J W = 998 + 665 − 1495 = 168 J             ΔUig = 998 − 998 + 0 = 0 ΔHig = 1663 − 1663 + 0 = 0 The property changes ΔUig and ΔHig both are zero for the entire cycle because the initial and final states are identical. Note also that Q = −W for the cycle. This follows from the first law with ΔUig = 0. Example 3.5 If the processes of Ex. 3.4 are carried out irreversibly but so as to accomplish exactly the same changes of state—the same changes in P, T, Uig, and Hig—then different values of Q and W result. Calculate Q and W if each step is carried out with a work efficiency of 80%. Solution 3.5 If the same changes of state as in Ex. 3.4 are carried out by irreversible processes, the property changes for the steps are identical with those of Ex. 3.4. However, the values of Q and W change. 3.3. Ideal Gas and Ideal-Gas State 87 (a) For mechanically reversible, adiabatic compression, the work is Wrev = 998 J. If the process is 80% efficient compared with this, the actual work is larger, and W = 998/0.80 = 1248 J. This step cannot here be adiabatic. By the first law, Q = ΔUig− W = 998 − 1248 = −250 J (b) The work required for the mechanically reversible cooling process is 665 J. For the irreversible process, W = 665/0.80 = 831 J. From Ex. 3.4(b), ΔUig = −998 J, and Q = ΔUig− W = −998 − 831 = −1829 J (c) As work is done by the system in this step, the irreversible work in absolute value is less than the reversible work of −1495 J, and the actual work done is: W = (0.80)(−1495)= −1196 J     Q = ΔUig− W = 0 + 1196 = 1196 J For the entire cycle, ΔUig and ΔHig are zero, with Q = −250 − 1829 + 1196 = −883 J     W = 1248 + 831 − 1196 = 883 J A summary of these results and those for Ex. 3.4 is given in the following table; values are in joules. Mechanically reversible, Ex. 3.4 Irreversible, Ex. 3.5 ΔUig ΔHig Q W ΔUig ΔHig Q W (a) 998 1663 0 998 998 1663 −250 1248 (b) −998 −1663 −1663 665 −998 −1663 −1829 831 (c) 0 0 1495 −1495 0 0 1196 −1196 Cycle 0 0 −168 168 0 0 −883 883 The cycle is one which requires work and produces an equal amount of heat. The striking feature of the comparison shown in the table is that the total work required when the cycle consists of three irreversible steps is more than five times the total work required when the steps are mechanically reversible, even though each irreversible step is assumed to be 80% efficient. Example 3.6 Air flows at a steady rate through a horizontal pipe to a partly closed valve. The pipe leaving the valve is enough larger than the entrance pipe that the kinetic-energy change of the air as it flows through the valve is negligible. The valve and connecting pipes are well insulated. The conditions of the air upstream from the valve are 20°C and 6 bar, and the downstream pressure is 3 bar. If the air is in its ideal-gas state, what is the temperature of the air some distance downstream from the valve? 88 CHAPTER 3. Volumetric Properties of Pure Fluids Solution 3.6 Flow through a partly closed valve is known as a throttling process. The system is insulated, making Q negligible; moreover, the potential-energy and kinetic- energy changes are negligible. No shaft work is accomplished, and Ws = 0. Hence, Eq. (2.31) reduces to: ig ig ΔHig= H2 − H1 = 0. Because Hig is a function of temperature only, this requires that T 2 = T 1. The result that ΔH ig = 0 is general for a throttling process, because the assumptions of negligible heat transfer and potential- and kinetic- energy changes are usually valid. For a fluid in its ideal-gas state, no temperature change occurs. The throttling process is inherently irreversible, but this is immaterial to the calculation because Eq. (3.14b) is valid for the ideal-gas state whatever the process.5 Example 3.7 If in Ex. 3.6 the flow rate of air is 1 mol·s–1 and if both upstream and downstream pipes have an inner diameter of 5 cm, what is the kinetic-energy change of the air and ig what is its temperature change? For air, CP= 29.100 J·mol−1 and the molar mass is ℳ = 29 g·mol−1. Solution 3.7 By Eq. (2.23b), ∙ ∙ n nVig u =  ___=  ____   Aρ A where D2= (_ )(5 × 10−2)2= 1.964 × 10−3m2 π π A = __ 4 4 The appropriate value here of the gas constant for calculation of the upstream molar volume is R = 83.14 × 10−6 bar·m3·mol−1·K−1. Then ( 83.14 × 10 )(293.15 K) −6 RT1 ___________________ V1 = ___ ig =       = 4.062 × 10−3m3·mol−1 P1 6 bar Then, 1 mol·s−1) ( 4.062 × 10−3m3·mol−1) ( u1= ___________________________        = 2.069 m·s−1 1.964 × 10−3m2 If the downstream temperature is little changed from the upstream temperature, then to a good approximation: V2ig = 2V1ig and u2= 2u1= 4.138 m·s−1 5The throttling of real gases may result in a relatively small temperature increase or decrease, known as the Joule/ Thomson effect. A more detailed discussion is found in Chapter 7. 3.4. Virial Equations of State 89 The rate of change in kinetic energy is therefore: (2 ) (2 ) 1 1 m  Δ _ u2 = nℳ Δ _ u2 ∙ ∙                (4.138  2− 2.0692)m2·s−2 = (1 × 29 × 10−3kg·s−1)___________________    2 = 0.186 kg·m2·s−3= 0.186 J·s−1 In the absence of heat transfer and work, the energy balance, Eq. (2.30), becomes: ( 2 ) (2 ) 1 1 Δ Hig+ _ u2 m= mΔ Hig+ m  Δ _ u2 = 0 ∙ ∙ ∙          (2 ) (2 ) 1 1 (m/ ℳ)CPΔT + m Δ _ u2 = nCPΔT + m Δ _ ig ig u2 = 0 ∙ ∙ ∙ ∙ Then (2 ) 1 (1)(29.100)ΔT = −m Δ _ u2 = −0.186 ∙ and ΔT = −0.0064 K Clearly, the assumption of negligible temperature change across the valve is jus- tified. Even for an upstream pressure of 10 bar and a downstream pressure of 1 bar and for the same flow rate, the temperature change is only −0.076 K. We conclude that, except for very unusual conditions, ΔHig = 0 is a satisfactory energy balance. 3.4 VIRIAL EQUATIONS OF STATE Volumetric data for fluids are useful for many purposes, from the metering of fluids to the sizing of tanks. Data for V as a function of T and P can of course be given as tables. However, expression of the functional relation f(P, V, T) = 0 by equations is much more compact and convenient. The virial equations of state for gases are uniquely suited to this purpose. Isotherms for gases and vapors, lying to the right of the saturated-vapor curve CD in Fig. 3.2(b), are relatively simple curves for which V decreases as P increases. Here, the product PV for a given T varies much more slowly than either of its members, and hence is more easily represented analytically as a function of P. This suggests expressing PV for an isotherm by a power series in P: PV = a + bP + cP2+... If we define, b ≡ aB′ , c ≡ aC′, etc., then, PV = a(1 + B′P + C′P2+ D′P3+...) (3.28) where B′, C′, etc., are constants for a given temperature and a given substance. 90 CHAPTER 3. Volumetric Properties of Pure Fluids Ideal-Gas Temperature; Universal Gas Constant Parameters B′, C′, etc., in Eq. (3.28) are species-dependent functions of temperature, but param- eter a is found by experiment to be the same function of temperature for all chemical species. This is shown by isothermal measurements of V as a function of P for various gases. Extrapola- tions of the PV product to zero pressure, where Eq. (3.28) reduces to PV = a, show that a is the same function of T for all gases. Denoting this zero-pressure limit by an asterisk provides: (PV)*= a = f (T) This property of gases serves as a basis for an absolute temperature scale. It is defined by arbi- trary assignment of the functional relationship f (T) and the assignment of a specific value to a single point on the scale. The simplest procedure, the one adopted internationally to define the Kelvin scale (Sec. 1.4): ∙ Makes (PV)* directly proportional to T, with R as the proportionality constant: (PV)*= a ≡ RT (3.29) ∙ Assigns the value 273.16 K to the temperature of the triple point of water (denoted by subscript t): (PV)  *t    = R × 273.16 K (3.30) Division of Eq. (3.29) by Eq. (3.30) gives: ( PV)* T K = 273.16 ______*  (3.31) (PV)  t  This equation provides the experimental basis for the ideal-gas temperature scale throughout the temperature range for which values of (PV)* are experimentally accessible. The Kelvin temperature scale is defined so as to be in as close agreement as possible with this scale. The proportionality constant R in Eqs. (3.29) and (3.30) is the universal gas constant. Its numerical value is found from Eq. (3.30): ( PV)  *t  R =  ________   273.16 K The accepted experimental value of (PV)  *t  is 22,711.8 bar·cm3 · mol−1, from which6 22,711.8 bar·cm3·mol−1 R = ___________________   = 83.1446 bar·cm3·mol−1·K−1 273.16 K Through the use of conversion factors, R may be expressed in various units. Commonly used values are given in Table A.2 of App. A. Two Forms of the Virial Equation A useful auxiliary thermodynamic property is defined by the equation: PV V Z ≡ _ = _ (3.32) RT Vig 6See http://physics.nist.gov/constants. 3.4. Virial Equations of State 91 This dimensionless ratio is called the compressibility factor. It is a measure of the deviation of the real-gas molar volume from its ideal-gas value. For the ideal-gas state, Z = 1. At moderate temperatures its value is usually 1. Figure 3.6 shows the compressibility factor of carbon dioxide as a function of T and P. This figure presents the same information as Fig. 3.4, except that it is plotted in terms of Z rather than V. It shows that at low pressure, Z approaches 1, and at moderate pressures, Z decreases roughly linearly with pressure. 1 0.8 0.6 Z 0.4 0.2 0 350 0 300 20 40 250 60 T (K) 80 P (bar) 200 100 Figure 3.6: PZT surface for carbon dioxide, with isotherms shown in black and the vapor/liquid equilibrium curve in white. With Z defined by Eq. (3.32) and with a = RT [Eq. (3.29)], Eq. (3.28) becomes: Z = 1 + B′P + C′P2+ D′P3+... (3.33) An alternative expression for Z is also in common use:7 B C _ D Z = 1 + _ + _ + +... (3.34) V V2 V3 Both of these equations are known as virial expansions, and the parameters B′, C′, D′, etc., and B, C, D, etc., are called virial coefficients. Parameters B′ and B are second virial coeffi- cients; C′ and C are third virial coefficients, and so on. For a given gas the virial coefficients are functions of temperature only. 7Proposed by H. Kamerlingh Onnes, “Expression of the Equation of State of Gases and Liquids by Means of Series,” Communications from the Physical Laboratory of the University of Leiden, no. 71, 1901. 92 CHAPTER 3. Volumetric Properties of Pure Fluids The two sets of coefficients in Eqs. (3.33) and (3.34) are related as follows: B C−B 2 D − 3BC + 2B3 ′= ___ B ′= _____ (3.35a) C ′= __________ (3.35b) D    (3.35c) RT ( R 2 T) T)3 ( R To derive these relations, we set Z = PV/RT in Eq. (3.34) and solve for P. This allows elimina- tion of P on the right side of Eq. (3.33). The resulting equation reduces to a power series in 1/V which may be compared term by term with Eq. (3.34) to yield the given relations. They hold exactly only for the two virial expansions as infinite series, but they are acceptable approxima- tions for the truncated forms used in practice. Many other equations of state have been proposed for gases, but the virial equations are the only ones firmly based on statistical mechanics, which provides physical significance to the virial coefficients. Thus, for the expansion in 1/V, the term B/V arises on account of interactions between pairs of molecules; the C/V2 term, on account of three-body interactions; etc. Because, at gas-like densities, two-body interactions are many times more common than three-body interactions, and three-body interactions are many times more numerous than four-body interactions, the contributions to Z of the successively higher-ordered terms decrease rapidly. 3.5 APPLICATION OF THE VIRIAL EQUATIONS The two forms of the virial expansion given by Eqs. (3.33) and (3.34) are infinite series. For engineering purposes their use is practical only where convergence is very rapid, that is, where two or three terms suffice for reasonably close approximations to the values of the series. This is realized for gases and vapors at low to moderate pressures. Figure 3.7 shows a compressibility-factor graph for methane. All isotherms originate at Z = 1 and P = 0, and are nearly straight lines at low pressures. Thus the tangent to an isotherm at P = 0 is a good approximation of the isotherm from P → 0 to some finite pressure. Differ- entiation of Eq. (3.33) for a given temperature gives: (∂P) ∂Z ___        = B′+2C′P + 3D′P2+... T from which, (∂P) ∂Z ___         = B′ T; P=0 Thus the equation of the tangent line is Z = 1 + B′P, a result also given by truncating Eq. (3.33) to two terms. A more common form of this equation results from substitution for B′ by Eq. (3.35a): PV BP Z = _ = 1 + _ (3.36) RT RT 3.5. Application of the Virial Equations 93 500 K 1.00 350 K 300 K Figure 3.7: Compressibility-factor graph 0.75 for methane. Shown are isotherms of the Z = PV/RT 250 K compressibility factor Z, as calculated 175 K from PVT data for methane by the defining 5K equation Z = PV/RT. They are plotted vs. 22 0.50 pressure for a number of constant temperatures, 0K and they show graphically what the virial 20 expansion in P represents analytically. 0.25 0 50 100 150 200 P (bar) This equation expresses direct linearity between Z and P and is often applied to vapors at subcritical temperatures up to their saturation pressures. At higher temperatures it often pro- vides a reasonable approximation for gases up to a pressure of several bars, with the pressure range increasing as the temperature increases. Equation (3.34) as well may be truncated to two terms for application at low pressures: PV B Z = ___ = 1 + __ (3.37) RT V However, Eq. (3.36) is more convenient in application and is normally at least as accurate as Eq. (3.37). Thus when the virial equation is truncated to two terms, Eq. (3.36) is preferred. The second virial coefficient B is substance dependent and a function of temperature. Experimental values are available for a number of gases.8 Moreover, estimation of second virial coefficients is possible where no data are available, as discussed in Sec. 3.7. For pressures above the range of applicability of Eq. (3.36) but below the critical pres- sure, the virial equation truncated to three terms often provides excellent results. In this case Eq. (3.34), the expansion in 1/V, is far superior to Eq. (3.33). Thus when the virial equation is truncated to three terms, the appropriate form is: PV B C Z = _ = 1 + _ + _ (3.38) RT V V2 This equation is explicit in pressure but is cubic in volume. Analytic solution for V is possible, but solution by an iterative scheme, as illustrated in Ex. 3.8, is often more convenient. Values of C, like those of B, depend on the gas and on temperature. Much less is known about third virial coefficients than about second virial coefficients, though data for a num- ber of gases are found in the literature. Because virial coefficients beyond the third are rarely known and because the virial expansion with more than three terms becomes unwieldy, its use is uncommon. 8J. H. Dymond and E. B. Smith, The Virial Coefficients of Pure Gases and Mixtures, Clarendon Press, Oxford, 1980. 94 CHAPTER 3. Volumetric Properties of Pure Fluids 100 4000 C 2000 0 0 B 2 1 2000 C/cm 6 mol B/cm 3 mol 100 4000 Figure 3.8: Virial coefficients B and C for nitrogen. 200 300 0 100 200 300 400 T/K Figure 3.8 illustrates the effect of temperature on the virial coefficients B and C for nitrogen; although numerical values are different for other gases, the trends are similar. The curve of Fig. 3.8 suggests that B increases monotonically with T; however, at temperatures much higher than shown, B reaches a maximum and then slowly decreases. The temperature dependence of C is more difficult to establish experimentally, but its main features are clear: C is negative at low temperatures, passes through a maximum at a temperature near the critical temperature, and thereafter decreases slowly with increasing T. Example 3.8 Reported values for the virial coefficients of isopropanol vapor at 200°C are: B = − 388 cm3· mol−1 C = − 26,000 cm6· mol−2 Calculate V and Z for isopropanol vapor at 200°C and 10 bar: (a) For the ideal-gas state; (b) By Eq. (3.36); (c) By Eq. (3.38). Solution 3.8 The absolute temperature is T = 473.15 K, and the appropriate value of the gas constant is R = 83.14 bar·cm3·mol−1·K−1. (a) For the ideal-gas state, Z = 1, and RT (83.14)(473.15) Vig= ___ = ____________ −    = 3934 cm3·mol 1 P 10 (b) From the second equality of Eq. (3.36), we have RT V = ___ + B = 3934 − 388 = 3546 cm3·mol−1 P 3.6. Cubic Equations of State 95 and PV V V 3546 Z = ___ = _____ = ___ = _____ = 0.9014 RT RT / P Vig 3934 (c) If solution by iteration is intended, Eq. (3.38) may be written: P( Vi Vi2) RT B C Vi+1= ___ 1 + _ + _ where i is the iteration number. Iteration is initiated with the ideal-gas state value Vig. Solution yields: V = 3488 cm3·mol−1 from which Z = 0.8866. In comparison with this result, the ideal-gas-state value is 13% too high, and Eq. (3.36) gives a value 1.7% too high. 3.6 CUBIC EQUATIONS OF STATE If an equation of state is to represent the PVT behavior of both liquids and vapors, it must encompass a wide range of temperatures, pressures, and molar volumes. Yet it must not be so complex as to present excessive numerical or analytical difficulties in application. Poly- nomial equations that are cubic in molar volume offer a compromise between generality and simplicity that is suitable to many purposes. Cubic equations are in fact the simplest equations capable of representing both liquid and vapor behavior. The van der Waals Equation of State The first practical cubic equation of state was proposed by J. D. van der Waals9 in 1873: RT a P = _ − _ (3.39) V − b V2 Here, a and b are positive constants, specific to a particular species; when they are zero, the equation for the ideal-gas state is recovered. The purpose of the term a/V2 is to account for the attractive forces between molecules, which make the pressure lower than that which would be exerted in the ideal-gas state. The purpose of constant b is to account for the finite size of molecules, which makes the volume larger than in the ideal-gas state. Given values of a and b for a particular fluid, one can calculate P as a function of V for various values of T. Figure 3.9 is a schematic PV diagram showing three such isotherms. 9Johannes Diderik van der Waals (1837–1923), Dutch physicist who won the 1910 Nobel Prize for physics. See: http://www.nobelprize.org/nobel_prizes/physics/laureates/1910/waals-bio.html. 96 CHAPTER 3. Volumetric Properties of Pure Fluids Superimposed is the “dome” representing states of saturated liquid and saturated vapor.10 For the isotherm T1 > Tc, pressure decreases with increasing molar volume. The critical isotherm (labeled Tc) contains the horizontal inflection at C characteristic of the critical point. For the isotherm T2 < Tc, the pressure decreases rapidly in the subcooled-liquid region with i ncreasing V; after crossing the saturated-liquid line, it goes through a minimum, rises to a maximum, and then decreases, crossing the saturated-vapor line and continuing downward into the superheated-vapor region. C Figure 3.9: PV isotherms as given by a cubic equation of T1 > Tc P state for T above, at, and below the critical temperature. The Tc superimposed darker curve P sat shows the locus of saturated vapor and liquid volumes. T2 < Tc V sat(liq) V sat(vap) V Experimental isotherms do not exhibit the smooth transition from saturated liquid to saturated vapor characteristic of equations of state; rather, they contain a horizontal segment within the two-phase region where saturated liquid and saturated vapor coexist in varying proportions at the saturation (or vapor) pressure. This behavior, shown by the dashed line in Fig. 3.9, cannot be represented by an equation of state, and we accept as inevitable the unrealistic behavior of equations of state in the two-phase region. Actually, the PV behavior predicted in the two-phase region by proper cubic equations of state is not wholly fictitious. If pressure is decreased on a saturated liquid devoid of vapor nucleation sites in a carefully controlled experiment, vaporization does not occur, and liquid persists alone to pressures well below its vapor pressure. Similarly, raising the pressure on a saturated vapor in a suitable experiment does not cause condensation, and vapor persists 10Though far from obvious, the equation of state also provides the basis for calculation of the saturated liquid- and vapor-phase volumes that determine the location of the “dome.” This is explained in Sec. 13.7. 3.6. Cubic Equations of State 97 alone to pressures well above the vapor pressure. These nonequilibrium or metastable states of superheated liquid and subcooled vapor are approximated by those portions of the PV isotherm which lie in the two-phase region adjacent to the states of saturated liquid and saturated vapor.11 Cubic equations of state have three volume roots, of which two may be complex. Physically meaningful values of V are always real, positive, and greater than constant b. For an isotherm at T > Tc, reference to Fig. 3.9 shows that solution for V at any value of P yields only one such root. For the critical isotherm (T = Tc), this is also true, except at the critical pressure, where there are three roots, all equal to Vc. For isotherms at T < Tc, the equation may exhibit one or three real roots, depending on the pressure. Although these roots are real and positive, they are not physically stable states for the portion of an isotherm lying between saturated liquid and saturated vapor (under the “dome”). Only for the saturation pressure Psat are the roots, V sat(liq) and V sat(vap), stable states, lying at the ends of the horizontal portion of the true isotherm. For any pressure other than P sat, there is only a single physically meaningful root, corresponding to either a liquid or a vapor molar volume. A Generic Cubic Equation of State A mid-20th-century development of cubic equations of state was initiated in 1949 by publica- tion of the Redlich/Kwong (RK) equation:12 RT a( T) P = ____ − ______ (3.40) V − b V(V + b) Subsequent enhancements have produced an important class of equations, represented by a generic cubic equation of state: RT a( T) P = _ − ___________    (3.41) V − b (V + εb)(V + σb) The assignment of appropriate parameters leads not only to the van der Waals (vdW) equation and the Redlich/Kwong (RK) equation, but also to the Soave/Redlich/Kwong (SRK)13 and the Peng/Robinson (PR) equations.14 For a given equation, ɛ and σ are pure numbers, the same for all substances, whereas parameters a(T) and b are substance dependent. The temper- ature dependence of a(T) is specific to each equation of state. The SRK equation is identical to the RK equation, except for the T dependence of a(T). The PR equation takes different values for ɛ and σ, as indicated in Table 3.1. 11The heating of liquids in a microwave oven can lead to a dangerous condition of superheated liquid, which can “flash” explosively. 12Otto Redlich and J. N. S. Kwong, Chem. Rev., vol. 44, pp. 233–244, 1949. 13G. Soave, Chem. Eng. Sci., vol. 27, pp. 1197–1203, 1972. 14D.-Y. Peng and D. B. Robinson, Ind. Eng. Chem. Fundam., vol. 15, pp. 59–64, 1976. 98 CHAPTER 3. Volumetric Properties of Pure Fluids Determination of Equation-of-State Parameters The parameters b and a(T) of Eq. (3.41) can in principle be found from PVT data, but suf- ficient data are rarely available. They are in fact usually found from values for the critical constants Tc and Pc. Because the critical isotherm exhibits a horizontal inflection at the critical point, we may impose the mathematical conditions: (∂V) ( ∂ V ) ∂P ∂2P  ___      = 0 (3.42)  ____2      = 0 (3.43)   T;cr T;cr Subscript “cr” denotes the critical point. Differentiation of Eq. (3.41) yields e xpressions for both derivatives, which may be equated to zero for P = Pc, T = Tc, and V = Vc. The equation of state may itself be written for the critical conditions. These three equa- tions contain five constants: Pc, Vc, Tc, a(Tc), and b. Of the several ways to treat these equa- tions, the most suitable is elimination of Vc to yield ex

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