Chapter 6 - Thermodynamic Properties of Fluids PDF

Chapter 6 Thermodynamic Properties of Fluids Application of thermodynamics to practical problems requires numerical values of thermody- namic properties. A very simple example is calculation of the work required for a steady-state gas compressor. If designed to operate adiabatically with the purpose of raising the pressure of a gas from P1 to P2, this work can be determined by an energy balance [Eq. (2.32)], wherein the small kinetic- and potential-energy changes of the gas are neglected: Ws= ΔH = H2− H1 The shaft work is simply ΔH, the difference between inlet and outlet values for the enthalpy of the gas. The necessary enthalpy values must come from experimental data or by estimation. It is our purpose in this chapter to: ∙ Develop from the first and second laws the fundamental property relations which underlie the mathematical structure of applied thermodynamics for systems of constant composition ∙ Derive equations that allow calculation of enthalpy and entropy values from PVT and heat-capacity data ∙ Illustrate and discuss the types of diagrams and tables used to present property values for convenient use ∙ Develop generalized correlations that provide estimates of property values in the absence of complete experimental information 6.1 FUNDAMENTAL PROPERTY RELATIONS Equation (2.6), the first law for a closed system of n moles of a substance, may be written for the special case of a reversible process: d(nU) = dQrev + dWrev 210 6.1. Fundamental Property Relations 211 Equations (1.3) and (5.1) as applied to this process are: = − P d(nV) dWrev = T d(nS) dQrev These three equations combine to give: d(nU)= T d(nS)− P d(nV) (6.1) where U, S, and V are molar values of the internal energy, entropy, and volume. All of the primitive thermodynamic properties—P, V, T, U, and S—are included in this equation. It is a fundamental property relation connecting these properties for closed PVT systems. All other equations relating properties of such systems derive from it. Additional thermodynamic properties, beyond those appearing in Eq. (6.1), are defined as a matter of convenience, in relation to the primary properties. Thus, the enthalpy, defined and applied in Chap. 2, is joined here by two others. The three, all with recognized names and useful applications, are: Enthalpy H ≡ U + PV (6.2) Helmholtz energy A ≡ U − TS (6.3) Gibbs energy G ≡ U + PV − TS = H − TS (6.4) The Helmholtz energy and Gibbs energy,1 with their origins here, find application in phase- and chemical-equilibrium calculations and in statistical thermodynamics. Multiplication of Eq. (6.2) by n, followed by differentiation, yields the general expression d(nH)= d(nU)+ P d(nV)+ nV dP Substitution for d(nU) by Eq. (6.1) reduces this result to d(nH)= T d(nS)+ nV dP (6.5) The differentials of nA and nG are obtained similarly: d(nA)= − nS dT − P d(nV) (6.6) d(nG)= − nS dT + nV dP (6.7) Equations (6.1) and (6.5) through (6.7) are equivalent fundamental property relations. They are derived for a reversible process. However, they contain only properties of the system. which depend only on the state of the system, and not the path by which it reached that state. These equations are therefore not restricted in application to reversible processes. However, the restrictions placed on the nature of the system cannot be relaxed. Application is to any process in a closed PVT system resulting in a differ- ential change from one equilibrium state to another. The system may consist of a single phase (a homogeneous system), or it may comprise several phases (a heterogeneous system); it may be chemically inert, or it may undergo 1These have traditionally been called Helmholtz free energy and the Gibbs free energy. The word free originally had the connotation of energy available to perform useful work, under appropriate conditions. However, in current usage, the word free adds nothing, and is best omitted. Since 1988, the IUPAC-recommended terminology omits the word free. 212 CHAPTER 6. Thermodynamic Properties of Fluids chemical reaction. The choice of which equation to use in a particular application is dictated by convenience. However, the Gibbs energy G is special, because of its unique functional relation to T and P, the variables of primary interest by virtue of the ease of their measurement and control. An immediate application of these equations is to one mole (or to a unit mass) of a homogeneous fluid of constant composition. For this case, n = 1, and they simplify to: dU = T dS − P dV (6.8) dH = T dS + V dP (6.9) dA = −S dT − P dV (6.10) dG = −S dT + V dP (6.11) Implicit in each of these equations is a functional relationship that expresses a molar (or unit mass) property as a function of a natural or special pair of independent variables: U = U(S, V) H = H(S, P) A = A(T, V) G = G(T, P) These variables are said to be canonical,2 and a thermodynamic property known as a function of its canonical variables has a unique characteristic: All other thermodynamic properties may be evaluated from it by simple mathematical operations. Equations (6.8) through (6.11) lead to another set of property relations because they are exact differential expressions. In general, if F = F(x, y), then the total differential of F is defined as: ( ∂x ) ( ∂y ) ∂F ∂F dF ≡ ___       dx + ___       dy y x or dF = M dx + N dy (6.12) ( ∂x ) ( ∂y ) ∂F ∂F where M ≡  ___    N ≡ ___       y x ( ∂y ) ∂y ∂x ( ∂x ) ∂x ∂y ∂M ∂2F ∂N ∂2F Then  ___    = _____  ___    = _____ x y The order of differentiation in mixed second derivatives is immaterial, and these equations combine to give: ( ∂y ) ( ∂x ) ∂M ∂N  ___    = ___       (6.13) x y Because F is a function of x and y, the right side of Eq. (6.12) is an exact differential expression, and Eq. (6.13) correctly relates the partial derivatives. This equation serves in fact as the criterion of exactness. For example, y dx + x dy is a simple differential expression which is exact because the derivatives of Eq. (6.13) are equal, i.e., 1 = 1. The function that produces this expression is clearly F(x, y) = xy. On the other hand, y dx – x dy is an equally simple differential expression, but the derivatives of Eq. (6.13) are not equal, i.e., 1 ≠ −1, and it is not exact. There is no function of x and y whose differential yields the original expression. 2Canonical here means that the variables conform to a general rule that is both simple and clear. 6.1. Fundamental Property Relations 213 The thermodynamic properties U, H, A, and G are known to be functions of the canonical variables on the right sides of Eqs. (6.8) through (6.11). For each of these exact differential expressions, we may write the relationship of Eq. (6.13), producing the Maxwell relations.3 (∂V) ( ∂S ) (∂P) ( ∂S ) ∂T ∂P ∂T ∂V ___       = − ___       (6.14)       = ___ ___       (6.15) S V S P (∂V) ( ∂T ) (∂P) ( ∂T ) ∂S ∂P ∂S ∂V ___       = ___       (6.16) − ___       = ___       (6.17) T V    T P Expression of U, H, A, and G as functions of their canonical variables does not preclude the validity of other functional relationships for application to particular systems. Indeed, Axiom 3 of Sec. 2.5, applied to homogeneous PVT systems of constant composition, asserts their dependence on T and P. The restrictions exclude heterogeneous and reacting systems, except for G, for which T and P are the canonical variables. A simple example is a system comprised of a pure liquid in equilibrium with its vapor. Its molar internal energy depends on the relative amounts of liquid and vapor present, and this is in no way reflected by T and P. However, the canonical variables S and V also depend on the relative amounts of the phases, giving U = U(S, V) its greater generality. On the other hand, T and P are the canonical variables for the Gibbs energy, and G = G(T, P) is general. Thus G is fixed for given T and P, regardless of the relative amounts of the phases, and provides the fundamental basis for the working equations of phase equilibria. Equations (6.8) through (6.11) lead not only to the Maxwell relations but also to many other equations relating thermodynamic properties. The remainder of this section develops those most useful for evaluation of thermodynamic properties from experimental data. Enthalpy and Entropy as Functions of T and P In engineering practice, enthalpy and entropy are often the thermodynamic properties of inter- est, and T and P are the most common measurable properties of a substance or system. Thus, their mathematical connections, expressing the variation of H and S with changes in T and P, are needed. This information is contained in the derivatives (∂H/∂T)P, (∂S/∂T)P, (∂H/∂P)T, and (∂S/∂P)T, with which we can write: ( ∂T ) ( ∂P ) (∂T) (∂P) ∂H ∂H ∂S ∂S dH = ___       dT + ___       dP dS = ___       dT + ___       dP P T P T Our goal here is to express these four partial derivatives in terms of measurable properties. The definition of heat capacity at constant pressure is: ( ∂T ) ∂H ___       = CP (2.19) P Another expression for this quantity is obtained by applying Eq. (6.9) to changes with respect to T at constant P: ( ∂T ) (∂T) ∂H ∂S ___       = T ___       P P 3After James Clerk Maxwell (1831–1879). See http://en.wikipedia.org/wiki/James_Clerk_Maxwell. 214 CHAPTER 6. Thermodynamic Properties of Fluids Combination of this equation with Eq. (2.19) gives: (∂T) ∂S CP  ___    = ___ (6.18) T P The pressure derivative of entropy results directly from Eq. (6.17): (∂P) ( ∂T ) ∂S ∂V ___       = − ___       (6.19) T P The corresponding derivative for enthalpy is found by applying Eq. (6.9) to changes with respect to P at constant T: ( ∂P ) (∂P) ∂H ∂S ___       = T ___       + V T T As a result of Eq. (6.19) this becomes: ( ∂P ) ( ∂T ) ∂H ∂V ___       = V − T ___       (6.20) T P With expressions for the four partial derivatives given by Eqs. (2.19) and (6.18) through (6.20), we can write the required functional relations as: [ ( ∂T ) ] ∂V dH = CP dT + V − T ___       dP (6.21) P T ( ∂T ) dT ∂V dS = CP _ − ___       dP (6.22) P These are general equations relating enthalpy and entropy to temperature and pressure for homogeneous fluids of constant composition. Eqs. (6.19) and (6.20) illustrate the utility of the Maxwell relations, particularly Eqs. (6.16) and (6.17), which relate changes in entropy that are not experimentally accessible to PVT data that are experimentally measurable. The Ideal-Gas State The coefficients of dT and dP in Eqs. (6.21) and (6.22) are evaluated from heat-capacity and PVT data. The ideal-gas state (denoted by superscript ig) provides an example of PVT behavior: ( ∂T ) P ∂ Vig R       = __ PVig= RT ____ P 6.1. Fundamental Property Relations 215 Substituting these equations into Eqs. (6.21) and (6.22) reduces them to: dT ___ dP ___ dH  ig= C  ig P  dT (6.23) d Sig= Cig P T − R P (6.24) These are restatements of equations for the ideal-gas state presented in Secs. 3.3 and 5.5. Alternative Forms for Liquids Alternative forms of Eqs. (6.19) and (6.20) result when (∂V/∂T)P is replaced by βV [Eq. (3.3)]: (∂P) ( ∂P ) ∂S ∂H ___       = − βV (6.25) ___       = (1 − βT)V (6.26) T T These equations incorporating β, although general, are usually applied only to liquids. How- ever, for liquids at conditions far from the critical point, both the volume and β are small. Thus at most conditions pressure has little effect on the properties of liquids. The important ideali- zation of an incompressible fluid (Sec. 3.2) is considered in Ex. 6.2. Replacing (∂V/∂T)P in Eqs. (6.21) and (6.22) with βV yields: dT dH = CP  dT + (1 − βT)V dP (6.27) dS = CP ___ − βV dP (6.28) T Because β and V are weak functions of pressure for liquids, they are usually assumed constant at appropriate average values for integration of the final terms. Internal Energy as a Function of P Internal energy is related to enthalpy by Eq. (6.2) as U = H – PV. Differentiation yields: ( ∂P ) ( ∂P ) (∂P) ∂U ∂H ∂V ___       = ___       − P ___       − V T T T Then by Eq. (6.20), ( ∂P ) ( ∂T ) (∂P) ∂U ∂V ∂V ___       = − T ___       − P ___       T P T An alternative form results if the derivatives on the right are replaced by βV [Eq. (3.3)] and –κV [Eq. (3.4)]: ( ∂P ) ∂U ___       = (− βT + κP)V (6.29) T 216 CHAPTER 6. Thermodynamic Properties of Fluids Example 6.1 Determine the enthalpy and entropy changes of liquid water for a change of state from 1 bar and 25°C to 1000 bar and 50°C. Data for water are given in the following table. t ºC P/bar CP/J · mol−1 · K−1 V/cm3 · mol−1 β/K−1 25 1 75.305 18.071 256 × 10−6 25 1000...... 18.012 366 × 10−6 50 1 75.314 18.234 458 × 10−6 50 1000...... 18.174 568 × 10−6 Solution 6.1 For application to the change of state described, Eqs. (6.27) and (6.28) require integration. Enthalpy and entropy are state functions, and the path of integration is arbitrary; the path most suited to the given data is shown in Fig. 6.1. Because the data indicate that CP is a weak function of T and that both V and β change relatively slowly with P, integration with arithmetic means is satisfactory. The integrated forms of Eqs. (6.27) and (6.28) that result are: ΔH = ⟨CP  ⟩(T  2 − T 1) + (1 − ⟨β⟩T 2)⟨ V⟩(P 2 − P 1)      T2   ΔS = ⟨ CP⟩ln ___     −⟨β⟩⟨ V⟩ (P2− P1) T1 1 H1 and S1 at 1 bar, 25°C ∫CP dT at 1 bar ∫CP dTT Figure 6.1: Calculation path for Ex. 6.1. ∫ V(1 T)dP at 50°C 1 bar, 50°C ∫ V dP 2 H2 and S2 at 1,000 bar, 50°C For P = 1 bar, 75.305 + 75.314 ⟨CP  ⟩= ______________      = 75.310 J·mol−1·K−1 2 For t = 50°C, 18.234 + 18.174 ⟨V⟩= ______________       = 18.204 cm3·mol−1 2 6.1. Fundamental Property Relations 217 and 458 + 568 〈β〉 = ________ × 10−6= 513 × 10−6K−1 2 Substitution of these numerical values into the equation for ΔH gives: ΔH = 75.310(323.15 − 298.15)J⋅mol−1 [1 − (513 × 10−6)(323.15)](18.204)(1000 − 1)bar⋅cm3⋅mol−1         + _________________________________________________                       10 bar⋅cm3⋅J−1 = 1883 + 1517 = 3400 J⋅mol−1 Similarly for ΔS, 323.15 ΔS = 75.310 ln ______ J⋅mol−1⋅K−1 298.15 (513 × 10−6)(18.204)(1000− 1)bar⋅cm3⋅mol−1⋅K−1           − __________________________________________          10 bar ⋅cm3⋅J−1 = 6.06 − 0.93 = 5.13 J⋅mol−1⋅K−1 Note that the effect of a pressure change of almost 1000 bar on the enthalpy and entropy of liquid water is less than that of a temperature change of only 25°C. Internal Energy and Entropy as Functions of T and V In some circumstances, temperature and volume may be more convenient independent var- iables than temperature and pressure. The most useful property relations are then for inter- nal energy and entropy. Required here are the derivatives (∂U/∂T )V, (∂U/∂V)T, (∂S/∂T )V, and (∂S/∂V)T, with which we can write: ( ∂T ) ( ∂V ) (∂T) (∂V) ∂U ∂U ∂S ∂S dU = ___       dT + ___       dV dS = ___       dT + ___       dV V T V T The partial derivatives of U follow directly from Eq. (6.8): ( ∂T ) (∂T) ( ∂V ) (∂V) ∂U ∂S ∂U ∂S ___       = T ___       ___       = T ___       − P V V T T Combining the first of these with Eq. (2.15) and the second with Eq. (6.16) gives: (∂T) ( ∂V ) ( ∂T ) ∂S CV ∂U ∂P       = ___ ___ (6.30) ___       = T ___       − P (6.31) T V T V 218 CHAPTER 6. Thermodynamic Properties of Fluids With expressions for the four partial derivatives given by Eqs. (2.15), (6.31), (6.30), and (6.16) we can write the required functional relations as: [ ( ∂ T )V ] ∂P dU = CV dT + T ___       − P dV (6.32) T ( ∂T ) dT ∂P dS = CV _ + ___       dV (6.33) V These are general equations relating the internal energy and entropy of homogeneous fluids of constant composition to temperature and volume. Equation (3.5) applied to a change of state at constant volume becomes: ( ∂T ) ∂P β       = __ ___ (6.34) κ V Alternative forms of Eqs. (6.32) and (6.33) are therefore: (κ ) β CV β dU = CV dT + _ T − P dV (6.35) dS = ___ dT + __ dV (6.36) T κ Example 6.2 Develop the property relations appropriate to the incompressible fluid, a model fluid for which both β and κ are zero (Sec. 3.2). This is an idealization employed in fluid mechanics. Solution 6.2 Equations (6.27) and (6.28) written for an incompressible fluid become: dH = CP dT + V dP (A) dT dS = CP ___ T The enthalpy of an incompressible fluid is therefore a function of both temperature and pressure, whereas the entropy is a function of temperature only, independent of P. With κ = β = 0, Eq. (6.29) shows that the internal energy is also a function of temperature only, and is therefore given by the equation, dU = CV dT. Equation (6.13), the criterion of exactness, applied to Eq. (A), yields: ( ∂P ) ( ∂T ) ∂ CP ∂V ____       = ___       T P 6.1. Fundamental Property Relations 219 However, the definition of β, given by Eq. (3.3), shows that the derivative on the right equals βV, which is zero for an incompressible fluid. This means that CP is a function of temperature only, independent of P. The relation of CP to CV for an incompressible fluid is of interest. For a given change of state, Eqs. (6.28) and (6.36) must give the same value for dS; they are therefore equated. The resulting expression, after rearrangement, is: βT (CP − CV )dT = βTV dP + ___ dV κ Upon restriction to constant V, this reduces to: ( ∂T ) ∂P CP − CV = βTV ___       V Elimination of the derivative by Eq. (6.34) yields: ( κ) β CP − CV = βTV _ (B) Because β = 0, the right side of this equation is zero, provided that the indeter- minate ratio β/κ is finite. This ratio is indeed finite for real fluids, and a contrary presumption for the model fluid would be irrational. Thus the definition of the incompressible fluid presumes this ratio is finite, and we conclude for such a fluid that the heat capacities at constant V and at constant P are identical: CP = CV = C The Gibbs Energy as a Generating Function The fundamental property relation for G = G(T, P), dG = V dP − S dT (6.11) has an alternative form. It follows from the mathematical identity: ( RT ) RT G 1 G d _ ≡ ___ dG − ____ 2 dT RT Substitution for dG by Eq. (6.11) and for G by Eq. (6.4) gives, after algebraic reduction: ( RT ) RT G V H d _ = _ dP − _ 2 dT (6.37) RT The advantage of this equation is that all terms are dimensionless; moreover, in contrast to Eq. (6.11), the enthalpy rather than the entropy appears on the right side. 220 CHAPTER 6. Thermodynamic Properties of Fluids Equations such as Eqs. (6.11) and (6.37) are most readily applied in restricted form. Thus, from Eq. (6.37) [ ] [ ] V ∂ (G / RT) H ∂ (G / RT) ___= ________       (6.38) ___= − T ________       (6.39) RT ∂P RT ∂T P T Given G/RT as a function of T and P, V/RT and H/RT follow by simple differentiation. The remaining properties follow from defining equations. In particular, S __ H G U H PV = ___ − ___ ___= ___ − ___ R RT RT RT RT RT The Gibbs energy, G or G/RT, when given as a function of its canonical variables T and P, serves as a generating function for the other thermo- dynamic properties through simple mathematics, and implicitly repre- sents complete property information. Just as Eq. (6.11) leads to expressions for all thermodynamic properties, so Eq. (6.10), the fundamental property relation for the Helmholtz energy A = A(T, V), leads to equations for all thermodynamic properties from knowledge of A as a function of T and V. This is particularly useful in connecting thermodynamic properties to statistical mechanics because closed systems at fixed volume and temperature are often most amenable to treatment by both theoretical methods of statistical mechanics and computational methods of molecular simulation based on statistical mechanics. 6.2 RESIDUAL PROPERTIES Unfortunately, no experimental method for the measurement of numerical values of G or G/RT is known, and the equations which relate other properties to the Gibbs energy are of little direct practical use. However, the concept of the Gibbs energy as a generating function for other thermodynamic properties carries over to a closely related property for which numer- ical values are readily obtained. By definition, the residual Gibbs energy is: GR ≡ G − Gig, where G and Gig are the actual and the ideal-gas-state values of the Gibbs energy at the same temperature and pressure. Other residual properties are defined in an analogous way. The residual volume, for example, is: RT VR≡ V − Vig= V − ___ P Because V = ZRT/P, the residual volume and the compressibility factor are related: RT VR= ___ (Z − 1) (6.40) P The generic residual property4 is defined by: MR≡ M − Mig (6.41) 4Sometimes referred to as a departure function. Generic here denotes a class of properties with the same characteristics. 6.2. Residual Properties 221 where M and Mig are actual and ideal-gas-state properties at the same T and P. They represent molar values for any extensive thermodynamic property, e.g., V, U, H, S, or G. The underlying purpose of this definition is more easily understood when it is written as: M = Mig+ MR From a practical perspective this equation divides property calculations into two parts: first, simple calculations for properties in the ideal-gas state; second, calculations for the residual properties, which have the nature of corrections to the ideal-gas-state values. Properties for the ideal-gas state reflect real molecular configurations but presume the absence of intermolecular interactions. Residual properties account for for the effect of such interactions. Our purpose here is to develop equations for the calculation of residual properties from PVT data or from their representation by equations of state. Equation (6.37), written for the ideal-gas state, becomes: ( RT ) RT Gig Vig Hig d _ = ___ dP − ____ 2 dT RT Subtracting this equation from Eq. (6.37) itself gives: ( RT ) RT GR VR HR d _ = _ dP − _ 2 dT (6.42) RT This fundamental residual-property relation applies to fluids of constant composition. Useful restricted forms are: [ ] [ ] V R ___ ( ) ∂ GR/ RT H R ( ) ∂ GR/ RT = _________        (6.43) ___ = − T _________        (6.44) RT ∂P T RT ∂T P Equation (6.43) provides a direct link between the residual Gibbs energy and experiment. Written ( RT ) RT GR VR d _ = ___ dP (const T ) it may be integrated from zero pressure to arbitrary pressure P, yielding: RT ( RT ) GR GR P VR ∫ 0 RT ___       +   ___ = ___ dP  (const T ) P=0 For convenience, define: ( RT ) GR ___       ≡ J P=0 With this definition and elimination of VR by Eq. (6.40), GR P ∫0 ___ dP = J +   (Z − 1)___   (const T ) (6.45) RT P 222 CHAPTER 6. Thermodynamic Properties of Fluids As explained in the Addendum to this chapter, J is a constant, independent of T, and the derivative of this equation in accord with Eq. (6.44) gives: ∫ 0 (∂T) P HR P ∂Z dP _= − T   ___       _   (const T ) (6.46) RT P The defining equation for the Gibbs energy, G = H – TS, may also be written for the ideal-gas state, Gig = Hig − TSig; by difference, GR = HR − TSR, and SR HR ___ ___ GR = ___ − (6.47) R RT RT Combining this equation with Eqs. (6.45) and (6.46) gives: ( ) SR P ∂Z P ∫ 0 ∂T dP ∫0 ___ dP = − T   _    ___ − J −   (Z − 1)  ___ (const T ) R P P P In application, entropy always appears in differences. In accord with Eq. (6.41), we write: S = Sig + SR for two different states. Then by difference: ΔS ≡ S2− S1= (S2 − S1 )+ (S2R− S1R) ig ig Because J is constant, it cancels from the final term, and its value is of no consequence. Con- stant J is therefore arbitrarily set equal to zero, and the working equation for SR becomes: ∫ 0 (∂T) P ∫ 0 SR P ∂Z dP P dP _= − T   ___         _ −   (Z − 1)_   (const T ) (6.48) R P P and Eq. (6.45) is written: GR P RT ∫ 0 dP _=   (Z − 1)_   (const T ) (6.49) P Values of the compressibility factor Z = PV/RT and of (∂Z/∂T )P may be calculated from experimental PVT data, with the two integrals in Eqs. (6.46), (6.48), and (6.49) evaluated by numerical or graphical methods. Alternatively, the two integrals may be expressed analytically with Z given as a function of T and P by a volume-explicit equation of state. This direct con- nection with experiment allows evaluation of the residual properties HR and SR for use in the calculation of enthalpy and entropy values. Enthalpy and Entropy from Residual Properties General expressions for Hig and Sig are found by integration of Eqs. (6.23) and (6.24) from an ideal-gas state at reference conditions T0 and P0 to an ideal-gas state at T and P:5 T T P ∫ T0 ∫ T0 ig ig ig ig dT Hig= H0 +   C  P    dT    − R ln ___ Sig= S0 +   CP ___ T P 0 5Thermodynamic properties for organic compounds in the ideal-gas state are given by M. Frenkel, G. J. Kabo, K. N. Marsh, G. N. Roganov, and R. C. Wilhoit, Thermodynamics of Organic Compounds in the Gas State, Thermo- dynamics Research Center, Texas A & M Univ. System, College Station, Texas, 1994. For many compounds, these data are also available via the NIST Chemistry Webbook, http://webbook.nist.gov. 6.2. Residual Properties 223 Because H = Hig + HR and S = Sig + SR: T ∫ T0 ig ig H = H0 +   CP  dT + HR (6.50) T ∫ T0 ig ig dT P S = S0 +   CP ___   − R ln ___ + SR (6.51) T P0 Recall (Secs. 4.1 and 5.5) that for purposes of computation the integrals in Eqs. (6.50) and (6.51) are represented by: T ∫ T 0 ig   CP  dT  = R × ICPH(T  0, T; A, B, C, D)      T   ∫ T 0 ig___ dT   CP        = R × ICPS(T  0, T; A, B, C, D) T Equations (6.50) and (6.51) have alternative forms when the integrals are replaced by equivalent terms that include the mean heat capacities introduced in Secs. 4.1 and 5.5: ig ig H = H0 + ⟨C  P  ⟩  H(T − T0)+ HR (6.52) ig ig T P   P  ⟩  S ln ___ S = S0 + ⟨C − R ln ___ + SR (6.53) T0 P0 In Eqs. (6.50) through (6.53), HR and SR are given by Eqs. (6.46) and (6.48). Again, for com- putational purposes, the mean heat capacities are represented by: ig ⟨C   P  ⟩  H = R × MCPH(T  0, T; A, B, C, D)     ig   ⟨C  P  ⟩  S = R × MCPS(T  0, T; A, B, C, D) Applications of thermodynamics require only differences in enthalpy and entropy, and these do not change when the scale of values is shifted by a constant amount. The reference-state conditions T0 and P0 are therefore selected for convenience, and values ig ig are assigned to H 0 and S 0 arbitrarily. The only information needed for application of Eqs. (6.52) and (6.53) are ideal-gas-state heat capacities and PVT data. Once V, H, and S are known at given conditions of T and P, the other thermodynamic properties follow from defining equations. The great practical value of the ideal-gas state is now evident. It provides the base for calculation of real-gas properties. Residual properties have validity for both gases and liquids. However, the advantage of Eqs. (6.50) and (6.51) in application to gases is that HR and SR, the terms which contain all the complex calculations, are residuals that are usually small. They act as corrections to the major terms, Hig and Sig. For liquids, this advantage is largely lost because HR and SR must include the large enthalpy and entropy changes of vaporization. Property changes of liquids are usually calculated by integrated forms of Eqs. (6.27) and (6.28), as illustrated in Ex. 6.1. 224 CHAPTER 6. Thermodynamic Properties of Fluids Example 6.3 Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information: 1. Table 6.1 gives compressibility-factor data (values of Z) for isobutane vapor. 2. The vapor pressure of isobutane at 360 K is 15.41 bar. ig ig 3. Set H0 = 18,115.0 J · mol−1 and S0 = 295.976 J· mol−1· K−1 for the reference state at 300 K and 1 bar. [These values are in accord with the bases adopted by R. D. Goodwin and W. M. Haynes, Nat. Bur. Stand. (U.S.), Tech. Note 1051, 1982.] 4. The ideal-gas-state heat capacity of isobutane vapor at temperatures of interest is: ig CP/ R = 1.7765 + 33.037 × 10−3T (T K) Table 6.1: Compressibility Factor Z for Isobutane P bar 340 K 350 K 360 K 370 K 380 K 0.10 0.99700 0.99719 0.99737 0.99753 0.99767 0.50 0.98745 0.98830 0.98907 0.98977 0.99040 2.00 0.95895 0.96206 0.96483 0.96730 0.96953 4.00 0.92422 0.93069 0.93635 0.94132 0.94574 6.00 0.88742 0.89816 0.90734 0.91529 0.92223 8.00 0.84575 0.86218 0.87586 0.88745 0.89743 10.0 0.79659 0.82117 0.84077 0.85695 0.87061 12.0...... 0.77310 0.80103 0.82315 0.84134 14.0............ 0.75506 0.78531 0.80923 15.41............ 0.71727 Solution 6.3 Calculation of HR and SR at 360 K and 15.41 bar by application of Eqs. (6.46) and (6.48) requires evaluation of two integrals: ( ) P ∂Z P ∫ 0 ∂T dP ∫0 dP   ___         ___     ( Z − 1)___   P P P Graphical integration requires simple plots of (∂Z/∂T )P/P and (Z −1)/P vs. P. Values of (Z−1)/P are found from the compressibility-factor data at 360 K. The quantity (∂Z/∂T )P/P requires evaluation of the partial derivative (∂Z/∂T)P, given by the slope of a plot of Z vs. T at constant pressure. For this purpose, separate plots are made of Z vs. T for each pressure at which compressibility-factor data are given, and a slope is determined at 360 K for each curve (for example, by construction of a tangent line at 360 K). Data for the required plots are shown in Table 6.2. 6.2. Residual Properties 225 Table 6.2: Values of the Integrands Required in Ex. 6.3 Values in parentheses are by extrapolation. P bar [(∂ Z / ∂ T)P/ P]× 104 K−1⋅bar−1 [− (Z − 1)/ P]× 102 bar−1 0.00 (1.780) (2.590) 0.10 1.700 2.470 0.50 1.514 2.186 2.00 1.293 1.759 4.00 1.290 1.591 6.00 1.395 1.544 8.00 1.560 1.552 10.0 1.777 1.592 12.0 2.073 1.658 14.0 2.432 1.750 15.41 (2.720) (1.835) The values of the two integrals, as determined from the plots, are: ∫ 0 (∂T) P P ∂Z dP P ∫0 dP   ___         ___ = 26.37 × 10−4 K−1    ( Z − 1)___   = − 0.2596 P P By Eq. (6.46), HR ___ = − (360) (26.37 × 10−4) = − 0.9493 RT By Eq. (6.48), SR ___ = − 0.9493 − (− 0.2596)= − 0.6897 R For R = 8.314 J⋅mol−1⋅K−1, HR= (−0.9493)(8.314)(360)= −2841.3 J⋅mol−1      R S = (−0.6897)(8.314)= −5.734 J⋅mol−1⋅K−1 Values of the integrals in Eqs. (6.50) and (6.51), with parameters from the given ig equation for CP/ R, are: 8.314 × ICPH(300, 360; 1.7765, 33.037 × 10  −3, 0.0, 0.0) = 6324.8 J⋅mol  −1        8.314 × ICPS(300, 360; 1.7765, 33.037 × 10  −3, 0.0, 0.0) = 19.174 J⋅mol  −1⋅K   −1 Substitution of numerical values into Eqs. (6.50) and (6.51) yields: H = 18,115.0 + 6324.8 − 2841.3 = − 21,598.5 J⋅mol−1       S = 295.976 + 19.174 − 8.314 ln 15.41 − 5.734 = 286.676 J⋅mol−1⋅K−1 Although calculations are here carried out for just one state, enthalpies and entropies can be evaluated for any number of states, given adequate data. After having completed a set of calculations, one is not irrevocably committed to the 226 CHAPTER 6. Thermodynamic Properties of Fluids ig ig particular values of H0 and S0 initially assigned. The scale of values for either the enthalpy or the entropy can be shifted by addition of a constant to all values. In this way one can give arbitrary values to H and S for some particular state so as to make the scales convenient for some particular purpose. The calculation of thermodynamic properties is an exacting task, seldom required of an engineer. However, engineers do make practical use of thermodynamic properties, and an understanding of methods by which they are calculated should suggest that some uncertainty is associated with every property value. Inaccuracy derives partly from experimental error in the data, which are also frequently incomplete and must be extended by interpolation and extrapolation. Moreover, even with reliable PVT data, a loss of accuracy occurs in the differ- entiation process required in the calculation of derived properties. Thus data of high accuracy are required to produce enthalpy and entropy values suitable for engineering calculations. 6.3 RESIDUAL PROPERTIES FROM THE VIRIAL EQUATIONS OF STATE The numerical or graphical evaluation of integrals, as in Eqs. (6.46) and (6.48), is often tedious and imprecise. An attractive alternative is analytical evaluation through equations of state. The procedure depends on whether the equation of state is volume explicit, i.e., expresses V (or Z) as a function of P at constant T, or pressure explicit, i.e., expresses P (or Z) as a function of V (or ρ) at constant T.6 Equations (6.46) and (6.48) are directly applicable only for a volume- explicit equation, such as the two-term virial equation in P [Eq. (3.36)]. For pressure-explicit equations, such as the virial expansions in reciprocal volume [Eq. (3.38)], Eqs. (6.46), (6.48), and (6.49) require reformulation. The two-term virial equation of state, Eq. (3.36), is volume explicit, Z − 1 = BP/RT. Differentiation yields (∂Z/∂T )P. We therefore have the required expressions for substitution into Eqs. (6.46) and (6.48). Direct integration gives HR/RT and SR/R. An alternative procedure is to evaluate GR/RT by Eq. (6.49): GR BP ___ = ___ (6.54) RT RT From this result HR/RT is found from Eq. (6.44), and SR/R is given by Eq. (6.47). Either way, we find: R( T dT ) HR P _ ___ B dB SR P dB = __ − _ (6.55) ___ = − __ ___ (6.56) RT R R dT Evaluation of residual enthalpies and residual entropies by Eqs. (6.55) and (6.56) is straightforward for given values of T and P, provided one has sufficient information to evalu- ate B and dB/dT. The range of applicability of these equations is the same as for Eq. (3.36), as discussed in Sec. 3.5. 6The ideal-gas equation of state is both pressure and volume explicit. 6.3. Residual Properties from the Virial Equations of State 227 Equations (6.46), (6.48), and (6.49) are incompatible with pressure-explicit equations of state and must be transformed such that P is no longer the variable of integration. In carrying out this transformation, the molar density ρ is a more convenient variable of integration than V, because ρ goes to zero, rather than to infinity, as P goes to zero. Thus, the equation PV = ZRT is written in alternative form as P = ZρRT (6.57) Differentiation at constant T gives: dP = RT(Zdρ + ρdZ) (const T ) Dividing this equation by Eq. (6.57) gives: dP dρ ___ ___ dZ = ____ + (const T ) P ρ Z Upon substitution for dP/P, Eq. (6.49) becomes: ρ GR RT ∫ 0 dρ _=   (Z − 1)_   + Z − 1 − ln Z (6.58) ρ where the integral is evaluated at constant T. Note also that ρ → 0 when P → 0. Solving Eq. (6.42) for its final term and substituting for VR by Eq. (6.40) yields: ( RT ) HR ____ dP GR 2 dT = (Z − 1)___ − d _ R T P Applying this for changes with respect to T at constant ρ gives: P ( ∂ T )ρ [ ] HR Z − 1 ___ ____ ∂P ∂ ( GR/ RT) 2 = ____       − _________       R T ∂T ρ Differentiation of Eq. (6.57) provides the first derivative on the right, and differentiation of Eq. (6.58) provides the second. Substitution leads to: ∫ 0 ( ∂ T )ρ ρ ρ HR ∂ Z dρ _= − T   ___         _ + Z − 1 (6.59) RT The residual entropy is found from Eq. (6.47) in combination with Eqs. (6.58) and (6.59): ∫ 0 ( ∂ T )ρ ρ ∫ 0 ρ ρ SR ∂ Z dρ dρ _= ln Z − T   ___         _ −   (Z − 1)_   (6.60) R ρ We now apply this to the pressure-explicit three-term virial equation: Z − 1 = Bρ + C ρ2 (3.38) Substitution into Eqs. (6.58) through (6.60) leads to: GR ___ 3 = 2Bρ + __ C ρ2− ln Z (6.61) RT 2 228 CHAPTER 6. Thermodynamic Properties of Fluids [( T dT ) ( T 2 dT ) ] HR ___ B dB C 1_ dC = T _ − _ ρ + _ − _ ρ2 (6.62) RT [( T dT ) 2 ( T dT ) ] SR ___ B dB 1 C _ dC = ln Z − T _ − _ ρ + _ _ + ρ2 (6.63) R Application of these equations, useful for gases up to moderate pressures, requires data for both the second and third virial coefficients. 6.4 GENERALIZED PROPERTY CORRELATIONS FOR GASES Of the two kinds of data needed for evaluation of thermodynamic properties, heat capacities and PVT data, the latter are most frequently missing. Fortunately, the generalized methods developed in Sec. 3.7 for the compressibility factor are also applicable to residual properties. Equations (6.46) and (6.48) are put into generalized form by substitution of the relations: P = Pc Pr T = Tc Tr    dP = Pc d Pr dT = Tc d Tr The resulting equations are: ∫ 0 ( ∂ Tr)P  Pr Pr HR ____ ∂Z d Pr = − Tr2   ___         ___ (6.64) RTc r ∫ 0 ( ∂ Tr)P  Pr ∫ 0 Pr Pr SR ___ ∂Z d Pr d Pr = − Tr   ___         ___ −   (Z − 1)___   (6.65) R Pr r The terms on the right sides of these equations depend only on the upper limit Pr of the integrals and on the reduced temperature at which they are evaluated. Thus, values of HR/RTc and SR/R may be determined once and for all at any reduced temperature and pressure from generalized compressibility-factor data. The correlation for Z is based on Eq. (3.53): Z = Z0+ ω Z1 Differentiation yields: ( ∂ Tr) ( ∂ Tr) ( ∂ Tr) ∂Z ∂ Z0 ∂ Z1 ___       = ____       + ω ____       P r  Pr Pr Substitution for Z and (∂ Z / ∂ T )Pr in Eqs. (6.64) and (6.65) gives: ∫ 0 ( ∂ Tr) Pr ∫ 0 ( ∂ Tr) Pr Pr Pr HR ____ ∂ Z0 d Pr ∂ Z1 d Pr         ___ = − Tr2   ____         ___ − ω Tr2   ____ R Tc P r P r ∫ 0 [ ( ∂ Tr) ∫ 0 [ ( ∂ Tr) Pr ] Pr Pr ] Pr SR ∂ Z0 d Pr ∂ Z1 d Pr ___= −   Tr ____       + Z0− 1  ___− ω   Tr ____       + Z1  ___ R P r P r 6.4. Generalized Property Correlations for Gases 229 The first integrals on the right sides of these two equations may be evaluated n umerically or graphically for various values of Tr and Pr from the data for Z0 given in Tables D.1 and D.3 of App. D, and the integrals which follow ω in each equation may be similarly evaluated from the data for Z1 given in Tables D.2 and D.4. If the first terms on the right sides of the preceding equations (including the minus signs) are represented by (HR)0/RTc and (SR)0/R and if the terms which follow ω, together with the preceding minus signs, are represented by (HR)1/RTc and (SR)1/R, then: R R 0 H (H ) R 1 (H ) R 0 S (S ) R R 1 (S ) ____ = _____ + ω _____ (6.66) ___ = _____ + ω _____ (6.67) RTc RTc RTc R R R Calculated values of the quantities (HR)0/RTc, (HR)1/RTc, (SR)0/R, and (SR)1/R as deter- mined by Lee and Kesler are given as functions of Tr and Pr in Tables D.5 through D.12. These values, together with Eqs. (6.66) and (6.67), allow estimation of residual enthalpies and entropies on the basis of the three-parameter corresponding-states principle as developed by Lee and Kesler (Sec. 3.7). Tables D.5 and D.7 for (HR)0/RTc and Tables D.9 and D.11 for (SR)0/R, used alone, pro- vide two-parameter corresponding-states correlations that quickly yield rough estimates of the residual properties. The nature of these correlations is indicated by Fig. 6.2, which shows a plot of (HR)0/RTc vs. Pr for six isotherms. 5 0.7 Compressed liquids Tr 0.9 4 3 1.0 1.2 ( HR 0 ) Two-phase C 1.5 RTc region Gases 2 1 4.0 0 0.05 0.1 0.2 0.5 1.0 2.0 5.0 10.0 Pr Figure 6.2: The Lee/Kesler correlation for (HR)0/ RTc as a function of Tr and Pr. 230 CHAPTER 6. Thermodynamic Properties of Fluids As with the generalized compressibility-factor correlation, the complexity of the functions (HR)0/RTc, (HR)1/RTc, (SR)0/R, and (SR)1/R precludes their general representation by simple equa- tions. However, the generalized second-virial-coefficient correlation forms the basis for analyti- cal correlations of the residual properties at low pressures. Recall Eqs. (3.58) and (3.59): B Pc Bˆ= ____ = B0+ ω B1 R Tc ˆ, B0, and B1 are functions of Tr only. Hence, Quantities B dBˆ d B0 d B1  ____  = ____ + ω ____ d Tr d Tr d Tr Equations (6.55) and (6.56) may be written: ( d Tr) HR dBˆ S R dB ˆ = Pr Bˆ− Tr____ _       ___ = − P____     r R Tc R d Tr Combining each of these equations with the previous two equations yields: [ ( d Tr)] HR ____ d B0 d B1 = Pr B0− Tr_ + ω B1− Tr_ (6.68) R Tc d Tr ( d Tr d Tr) SR ___ d B0 d B1 = − Pr _ + ω _ (6.69) R The dependence of B0 and B1 on reduced temperature is given by Eqs. (3.61) and (3.62). Differentiation of these equations provides expressions for dB0/dTr and dB1/dTr. Thus the equations required for application of Eqs. (6.68) and (6.69) are: 0.422 0.172 B0= 0.083 − _____ 1.6 (3.61) B1= 0.139 − _____ 4.2 (3.62) Tr Tr 0 d B _____ 0.675 d B1 _____ 0.722 ____ = (6.70) ____ = 5.2 (6.71) d Tr T2.6 r d Tr Tr Figure 3.10, drawn specifically for the compressibility-factor correlation, is also used as a guide to the reliability of the correlations of residual properties based on generalized s econd virial coefficients. However, all residual-property correlations are less accurate than the compressibility-factor correlations on which they are based and are, of course, least reliable for strongly polar and associating molecules. The generalized correlations for HR and SR, together with ideal-gas heat capacities, allow calculation of enthalpy and entropy values of gases at any temperature and pres- sure by Eqs. (6.50) and (6.51). For a change from state 1 to state 2, write Eq. (6.50) for both states: T2 T1 ∫ T0 ∫ T0 ig ig ig ig H2= H0 +   CPdT  + H2R H1= H0 +   CPdT  + H1R 6.4. Generalized Property Correlations for Gases 231 The enthalpy change for the process, ΔH = H2 − H1, is the difference between these two equations: T2 ∫ T1 ig ΔH =   CPdT  + H2R− H1R (6.72) Similarly, by Eq. (6.51), T2 P2 ∫ T1 ig dT ΔS =   CP___   − R ln ___ + S2R− S1R (6.73) T P1 Written in alternative form, these equations become: ΔH = ⟨CP⟩  (T2− T1)+ H2R− H1R ig (6.74) H T2 P2 ΔS = ⟨CP⟩  ln ___ ig − R ln ___ + S2R− S1R (6.75) S T1 P1 Just as we have given names to functions used in evaluation of the integrals in Eqs. (6.72) and (6.73) and the mean heat capacities in Eqs. (6.74) and (6.75), we also name functions useful for evaluation of HR and SR. Equations (6.68), (3.61), (6.70), (3.62), and (6.71) together pro- vide a function for the evaluation of HR/RTc, named HRB(Tr, Pr, OMEGA):7 HR ____ = HRB (Tr, Pr, OMEGA) RTc A numerical value of HR is therefore represented by: RTc × HRB (Tr, Pr, OMEGA) Similarly, Eqs. (6.69) through (6.71) provide a function for the evaluation of SR/R, named SRB(Tr, Pr, OMEGA): SR ___ = SRB (Tr, Pr, OMEGA) R A numerical value of SR is therefore represented by: R × SRB (Tr, Pr, OMEGA) The terms on the right sides of Eqs. (6.72) through (6.75) are readily associated with steps in a computational path leading from an initial to a final state of a system. Thus, in Fig. 6.3, the actual path from state 1 to state 2 (dashed line) is replaced by a three-step computational path: ∙ Step 1 → 1ig: A hypothetical process that transforms a real gas into an ideal gas at T1 and P1. The enthalpy and entropy changes for this process are: ig ig H1 − H1= − H1R S1 − S1= − S1R ∙ Step 1ig → 2ig: Changes in the ideal-gas state from (T1, P1) to (T2, P2). For this process, T2 ∫ T1 ig ig ig ΔHig= H2 − H1 =   CPdT  (6.76) 7Sample programs and spreadsheets for evaluation of these functions are available in the online learning center at http://highered.mheducation.com:80/sites/1259696529. 232 CHAPTER 6. Thermodynamic Properties of Fluids T2, P2 (real) 2 T1 , P1 S (real) H R 1 H2 Figure 6.3: Computational path for property H R1 changes ΔH and ΔS. S2R S R1 1 ig H ig S ig 2ig T1 , P1 T2, P2 (ideal) (ideal) T2 P2 ∫ T1 ig ig ig dT ΔSig= S2 − S1 =   CP___   − R ln ___ (6.77) T P1 ∙ Step 2ig → 2: Another hypothetical process that transforms the ideal gas back into a real gas at T2 and P2. Here, ig ig H2− H2 = H2R S2− S2 = S2R Equations (6.72) and (6.73) result from addition of the enthalpy and entropy changes for the three steps. Example 6.4 Supercritical CO2 is increasingly used as an environmentally friendly solvent for clean- ing applications, ranging from dry cleaning clothing to degreasing machine parts to photoresist stripping. A key advantage of CO2 is the ease with which it is separated from “dirt” and detergents. When its temperature and pressure are reduced below the critical temperature and vapor pressure respectively, it vaporizes, leaving dissolved substances behind. For a change in state of CO2 from 70°C and 150 bar to 20°C and 15 bar, estimate the changes in its molar enthalpy and entropy. Solution 6.4 We follow the three-step computational path of Fig. 6.3. Step 1 transforms the real fluid at 70°C and 150 bar into its ideal-gas state at the same conditions. Step 2 changes conditions in the ideal-gas state from the initial to the final conditions of T and P. Step 3 transforms the fluid from its ideal-gas state to the real-gas final state at 20°C and 15 bar. 6.4. Generalized Property Correlations for Gases 233 The residual-property values required for calculating the changes of Steps 1 and 3 depend on the reduced conditions of the initial and final states. With critical properties from Table B.1 of App. B: Tr1= 1.128 Pr1= 2.032 Tr2= 0.964 Pr2= 0.203 A check of Fig. 3.10 indicates that the Lee/Kesler tables are required for the initial state, whereas the second-virial-coefficient correlation should be suitable for the final state. Thus, for Step 1, interpolation in Lee/Kesler tables D.7, D.8, D.11, and D.12 provides the values: (HR)0 _____ (HR)1 (SR)0 (SR)1 = − 2.709, _____ = − 0.921, _____ = − 1.846, _____ = − 0.938 RTc RTc R R Then: ΔH1= − HR(343.15 K, 150 bar) = − (8.314)(304.2)[−2.709 + (0.224)(−0.921)]= 7372 J⋅mol−1                ΔS1= − SR(343.15 K, 150 bar) = − (8.314)[ −1.846 + (0.224)(−0.938)] = 17.09 J⋅mol−1⋅K−1 For Step 2, the enthalpy and entropy changes are calculated by the usual heat-capacity integrals, with polynomial coefficients from Table C.1. The ideal-gas-state entropy change caused by the pressure change must also be included. ΔH2= 8.314 × ICPH(343.15, 293.15; 5.547, 1.045 × 10−3, 0.0, −1.157 × 105) = −1978 J⋅mol−1 Δ                S2 = 8.314 × ICPS(343.15, 293.15; 5.547, 1.045 × 10−3, 0.0, −1.157 × 105) − (8.314)ln (15 / 150) = − 6.067 + 19.144 = 13.08 J ⋅mol−1⋅K−1 Finally, for Step 3, ΔH3= HR(293.15 K, 15 bar) = 8.314 × 304.2 × HRB(0.964, 0.203, 0.224)= − 660 J⋅mol−1              ΔS3= SR(293.15 K, 15 bar) = 8.314 × SRB(0.964, 0.203, 0.224)= −1.59 J⋅mol−1⋅K−1 Sums over the three steps yield overall changes, ΔH = 4734 J⋅mol−1 and ΔS = 28.6 J⋅mol−1⋅K−1. The largest contribution here comes from the residual properties of the initial state, because the reduced pressure is high, and the super- critical fluid is far from its ideal-gas state. Despite the substantial reduction in temperature, the enthalpy actually increases in the overall process. For comparison, the properties given in the NIST fluid-properties database, accessed through the NIST Chemistry Webbook, are: H1= 16,776 J⋅mol−1 S1= 67.66 J⋅mol−1⋅K−1       H2= 21,437 J⋅mol−1 S1= 95.86 J⋅mol−1⋅K−1 234 CHAPTER 6. Thermodynamic Properties of Fluids From these values, considered accurate, overall changes are ΔH = 4661 J⋅mol−1 and ΔS = 28.2 J⋅mol 1⋅K−1. Even though the changes in residual properties make up a substantial part of the total, the prediction from generalized correlations agrees with the NIST data to within 2 percent. Extension to Gas Mixtures Although no fundamental basis exists for extension of generalized correlations to mixtures, reasonable and useful approximate results for mixtures can often be obtained with pseudocrit- ical parameters resulting from simple linear mixing rules according to the definitions: ω ≡ ∑ yiωi (6.78) Tp  c ≡ ∑ y  i Tc   i (6.79) Pp  c ≡ ∑ y  i Pc   i (6.80) i i i The values so obtained are the mixture ω and pseudocritical temperature and pressure, Tpc and Ppc, which replace Tc and Pc to define pseudoreduced parameters: T P = ___ Tpr = ___ (6.81) Ppr (6.82) Tpc Ppc These replace Tr and Pr for reading entries from the tables of App. D, and lead to values of Z by Eq. (3.57), HR/RTpc by Eq. (6.66), and SR/R by Eq. (6.67). Example 6.5 Estimate V, HR, and SR for an equimolar mixture of carbon dioxide(1) and propane(2) at 450 K and 140 bar by the Lee/Kesler correlations. Solution 6.5 The pseudocritical parameters are found by Eqs. (6.78) through (6.80) with criti- cal constants from Table B.1 of App. B: ω = y1ω1+ y2ω2= (0.5)(0.224)+ (0.5)(0.152)= 0.188 = y1Tc 1+ y2Tc 2= (0.5)(304.2)+ (0.5)(369.8)= 337.0 K             Tpc = y1Pc 1+ y2Pc 2= (0.5)(73.83)+ (0.5)(42.48)= 58.15 bar Ppc Then, 450 140 = _____ Tpr = _____ = 1.335 Ppr = 2.41 337.0 58.15 Values of Z0 and Z1 from Tables D.3 and D.4 at these reduced conditions are: Z0= 0.697 and Z1= 0.205 6.5. Two-Phase Systems 235 By Eq. (3.57), Z = Z0+ ω Z1= 0.697 + (0.188)(0.205)= 0.736 Whence, (0.736)(83.14)(450) ZRT _________________ V = ____ =    = 196.7cm3⋅mol−1 P 140 Similarly, from Tables D.7 and D.8, with substitution into Eq. (6.66): ( RTpc ) ( RTpc ) 0 1 HR HR _____       = − 1.730 _____       = − 0.169      HR _____ = − 1.730 + (0.188)(− 0.169)= − 1.762 RTpc and HR= (8.314)(337.0)(−1.762)= − 4937 J⋅mol−1 From Tables D.11 and D.12 and substitution into Eq. (6.67), S R ___ = − 0.967 + (0.188)(− 0.330)= − 1.029 R and SR= (8.314)(− 1.029)= − 8.56 J⋅mol−1⋅K−1 6.5 TWO-PHASE SYSTEMS The curves shown on the PT diagram of Fig. 3.1 represent phase boundaries for a pure sub- stance. A phase transition at constant temperature and pressure occurs whenever one of these curves is crossed, and as a result the molar or specific values of the extensive thermodynamic properties change abruptly. Thus the molar or specific volume of a saturated liquid is very different from the molar or specific volume of saturated vapor at the same T and P. This is true as well for internal energy, enthalpy, and entropy. The exception is the molar or specific Gibbs energy, which for a pure species does not change during a phase transition such as melting, vaporization, or sublimation. Consider a pure liquid in equilibrium with its vapor and existing in a piston/cylinder arrangement that undergoes a differential evaporation at temperature T and corresponding vapor pressure Psat. Equation (6.7) applied to the process reduces to d(nG) = 0. Because the number of moles n is constant, dG = 0, and this requires the molar (or specific) Gibbs energy of the vapor to be identical with that of the liquid. More generally, for two phases α and β of a pure species coexisting at equilibrium, Gα= Gβ (6.83) where Gα and Gβ are the molar or specific Gibbs energies of the individual phases. 236 CHAPTER 6. Thermodynamic Properties of Fluids The Clapeyron equation, introduced as Eq. (4.12), follows from this equality. If the tem- perature of a two-phase system is changed, then the pressure must also change in accord with the relation between vapor pressure and temperature if the two phases continue to coexist in equilibrium. Because Eq. (6.83) applies throughout this change, dGα= dGβ Substituting expressions for dGα and dGβ as given by Eq. (6.11) yields: VαdPsat− SαdT = VβdPsat− SβdT which upon rearrangement becomes: dPsat _______ _____ Sβ− Sα _____ ΔSαβ = β = αβ dT V − V ΔV α The entropy change ΔSαβ and the volume change ΔVαβ are changes that occur when a unit amount of a pure chemical species is transferred from phase α to phase β at the equilibrium T and P. Integration of Eq. (6.9) for this change yields the latent heat of phase transition:

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