Chapter 7. Engineering Equations of State for PVT Properties

I am more than ever an admirer of van der Waals.

Lord Rayleigh (1891)

From Chapter 6, it is obvious that we can calculate changes in U, S, H, A, and G by knowing changes in any two variables from the set {P-V-T} plus CP or CV. This chapter introduces the various ways available for quantitative prediction of the P-V-T properties we desire in a general case. The method of calculation of thermodynamic properties like U, H, and so on. is facilitated by the use of departure functions, which will be the topic of the next chapter. The development of the departure functions is a relatively straightforward application of derivative manipulations. What is less straightforward is the logical development of a connection between P, V, and T. We introduced the concept in Chapter 1 that the pressure, temperature, and density (i.e., V–1) are connected through intermolecular interactions. We must now apply that concept to derive quantitative relationships that are applicable to any fluid at any conditions, not simply to ideal gases. You will see that making the connection between P, V, and T hinges on the transition from the molecular-scale forces and potential energy to the macroscopic pressure and internal energy. Understanding the approximations inherent in a particular equation of state is important because effectively all of the approximations in a thermodynamic model can be traced to the assumed equation of state. Whenever deficiencies are found in a process model, the first place to look for improvement is in revisiting the assumptions of the equation of state.

Understanding the transition from the molecular scale to the macroscopic is a major contribution in our conceptual puzzle of calculating energy, entropy, and equilibrium. We made qualitative connections between the microscopic and macroscopic scales for entropy during our introduction to entropy. For energy, however, we have left a gap that you may not have noticed. We discussed the molecular energy in Chapter 1, but we did not quantify the macroscopic implications. We discussed the macroscopic implications of energy in Chapter 2, but we did not discuss the molecular basis. It is time to fill that gap, and in doing so, link the conceptual framework of the entire text.

From one perspective, the purpose of the examples in Chapter 6 was to explain the need for making the transition from the molecular scale to the macroscopic scale. The purpose of the material following this chapter is to demonstrate the reduction to practice of this conceptual framework in several different contexts. So in many ways, this chapter represents the conceptual kernel for all molecular thermodynamics.

Chapter Objectives: You Should Be Able to...

1. Explain and apply two- and three- parameter corresponding states.

2. Apply an equation of state to solve for the density given T and P, including liquid and vapor roots.

3. Evaluate partial derivatives like those in Chapter 6 using an equation of state for PVT properties.

4. Identify the repulsive and attractive contributions to an equation of state and critically evaluate their accuracy relative to molecular simulations and experimental data.

7.1. Experimental Measurements

The preferred method of obtaining P-V-T properties is from experimental measurements of the desired fluid or fluid mixture. We spend most of the text discussing theories, but you should never forget the precious value of experimental data. Experimental measurements beat theories every time. The problem with experimental measurements is that they are expensive, especially relative to pushing a few buttons on a computer.

To illustrate the difficulty of measuring all properties experimentally, consider the following case. One method to determine the P-V-T properties is to control the temperature of a container of fluid, change the volume of the container in carefully controlled increments, and carefully measure the pressure. The required derivatives are then calculated by numerical differentiation of the data obtained in this manner. It is also possible to make separate measurements of the heat capacity by carefully adding measured quantities of heat and determining changes in P, V, and T. These measurements can be cross-referenced for consistency with the estimated changes as determined by applying Maxwell’s relations to the P-V-T measurements. Imagine what a daunting task this approach would represent when considering all fluids and mixtures of interest. It should be understandable that detailed measurements of this type have been made for relatively few compounds. Water is the most completely studied fluid, and the steam tables are a result of this study. Ammonia, carbon dioxide, refrigerants, and light hydrocarbons have also been quite thoroughly studied. The charts which have been used in earlier chapters are results of these careful measurements. Equations of state permit correlation and extrapolation of experimental data that can be much more convenient and more broadly applicable than the available charts.


Image The basic procedure for calculating properties involves using derivatives of P-V-T data.


An experimental approach is naturally impractical for all substances due to the large number of fluids needing to be characterized. The development of equations of state is the engineering approach to describing fluid behavior for prediction, interpolation, and extrapolation of data using the fewest number of adjustable parameters possible for the desired accuracy. Typically, when data are analyzed today, they are fitted with elaborate equations (embellishments of the equations of state discussed in this chapter) before determination of interpolated values or derivatives. The charts are generated from the fitted results of the equation of state.

As a summary of the experimental approach to equations of state, a brief review of the historical development of P-V-T measurements may be beneficial. First, it should be recalled that early measurements of P-V-T relations laid the foundation for modern physical chemistry. Knowing the densities of gases in bell jars led to the early characterizations of molecular weights, molecular formulas, and even the primary evidence for the existence of molecules themselves. At first, it seemed that gases like nitrogen, hydrogen, and oxygen were non-condensable and something quite different from liquids like water or wood alcohol (methanol). As technology advanced, however, experiments were performed at higher temperatures and pressures. Carbon dioxide was a very common compound in the early days (known as “carbonic acid” to van der Waals), and it soon became apparent that it showed a high degree of compressibility. Experimental data were carefully measured in 1871 for carbon dioxide ranging to 110 bars, and these data were referenced extensively by van der Waals. Carbon dioxide is especially interesting because it has some very “peculiar” properties that are exhibited near room temperature and at high pressure. At 31°C and about 70 bars, a very small change in pressure can convert the fluid from a gas-like density to a liquid density. Van der Waals showed that the cause of this behavior is the balance between the attractive forces from the intermolecular potential being accentuated at this density range and the repulsive forces being accentuated by the high-velocity collisions at this temperature. This “peculiar” range of conditions is known as the critical region. The precise temperature, pressure, and density where the vapor and the liquid become indistinguishable is called the critical point. Above the critical point, there is no longer an abrupt change in the density with respect to pressure while holding temperature constant. Instead, the balance between forces leads to a single-phase region spanning vapor-like densities and liquid-like densities. With the work of van der Waals, researchers began to recognize that the behavior was not “peculiar,” and that all substances have critical points.1


Image Fortunately, P,V,T behavior of fluids follows the same trends for all fluids. All fluids have a critical point.


7.2. Three-Parameter Corresponding States

If we plot P versus ρ for several different fluids, we find some remarkably similar trends. As shown in Fig. 7.1 below, both methane and pentane show the saturated vapor density approaching the saturated liquid density as the temperature increases. Compare these figures to Fig. 1.4 on page 23, and note that the P versus ρ figure is qualitatively a mirror image of the P versus V figure. The isotherms are shown in terms of the reduced temperature, TrT/Tc. Saturation densities are the values obtained by intersection of the phase envelope with horizontal lines drawn at the saturation pressures. The isothermal compressibility Image is infinite, and its reciprocal is zero, at the critical point (e.g., 191 K and 4.6 MPa for methane). It is also worth noting that the critical temperature isotherm exhibits an inflection point at the critical point. This means that (∂2P/∂ρ2)T = 0 at the critical point as well as (∂P/∂ρ)T = 0. The principle of corresponding states asserts that all fluid properties are similar if expressed properly in reduced variables.

Image

Figure 7.1. Comparison of the PρT behavior of methane (left) and pentane (right) demonstrating the qualitative similarity which led to corresponding states’ treatment of fluids. The lines are calculated with the Peng-Robinson equation to be discussed later. The phase envelope is an approximation sketched through the points available in the plots. The smoothed experimental data are from Brown, G.G., Sounders Jr., M., and Smith, R.L., 1932. Ind. Eng. Chem., 24:513. Although not shown, the Peng-Robinson equation is not particularly accurate for modeling liquid densities.


Image The isothermal compressibility is infinite at the critical point.


Although the behaviors in Fig. 7.1 are globally similar, when researchers superposed the P-V-T behaviors based on only critical temperature, Tc and critical pressure, Pc, they found the superposition was not sufficiently accurate. For example, one way of comparing the behavior of fluids is to plot the compressibility factor Z. The compressibility factor is defined as

Image

Image The compressibility factor.


Note: The compressibility factor is not the same as the isothermal compressibility. The similarity in names can frequently result in confusion as you first learn the concepts.

The compressibility factor has a value of one when a fluid behaves as an ideal gas, but will be non-unity when the pressure increases. By plotting the data and calculations from Fig. 7.1 as a function of reduced temperature Tr = T/Tc, and reduced pressure, Pr = P/Pc, the plot of Fig. 7.2 results. Clearly, another parameter is needed to accurately correlate the data. Note that the vapor pressure for methane and pentane differs on the compressibility factor chart as indicated by the vertical lines on the subcritical isotherms. The same behavior is followed by other fluids. For example, the vapor pressures for six compounds are shown in Fig. 7.3, and although they are all nearly linear, the slopes are different. In fact, we may characterize this slope with a third empirical parameter, known as the acentric factor, ω. The acentric factor is a parameter which helps to specify the vapor pressure curve which, in turn, correlates the rest of the thermodynamic variables.2

Image

Figure 7.2. The Peng-Robinson lines from Fig. 7.1 plotted in terms of the reduced pressure at Tr = 0.8, 0.9, 1.0, 1.1, and 1.3, demonstrating that critical temperature and pressure alone are insufficient to accurately represent the P-V-T behavior. Dashed lines are for methane, solid lines for pentane. The figure is intended to make an illustrative point. Accurate calculations should use the compressibility factor charts developed in the next section.

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Figure 7.3. Reduced vapor pressures plotted as a function of reduced temperature for six fluids demonstrating that the shape of the curve is not highly dependent on structure, but that the primary difference is the slope as given by the acentric factor.

Image

Image Critical temperature and pressure are insufficient characteristic parameters by themselves. The acentric factor serves as a third important parameter.


Note: The specification of Tc, Pc, and ω provides two points on the vapor pressure curve. Tc and Pc specify the terminal point of the vapor pressure curve. ω specifies a vapor pressure at a reduced temperature of 0.7. The acentric factor was first introduced by Pitzer et al.3 Its definition is arbitrary in that, for example, another reduced temperature could have been chosen for the definition. The definition above gives values of ω ~ 0 for spherical molecules like argon, xenon, neon, krypton, and methane. Deviations from zero usually derive from deviations in spherical symmetry. Nonspherical molecules are “not centrally symmetric,” so they are “acentric.” In general, there is no direct theoretical connection between the acentric factor and the shape of the intermolecular potential. Rather, the acentric factor provides a convenient experimental vapor pressure which can be correlated with the shape of the intermolecular potential in an ad hoc manner. It is convenient in the sense that its value has been experimentally determined for a large number of compounds and that knowing its value permits a significant improvement in the accuracy of our engineering equations of state.


Image The acentric factor is a measure of the slope of the vapor pressure curve plotted as ln Psat versus 1/T.


7.3. Generalized Compressibility Factor Charts

P-V-T behavior can be generalized in terms of Tc, Pc, and ω. The original correlation was presented by Pitzer, and is given in the form

Image

Image Pitzer correlation.


where tables or charts summarized the values of Z0 and Z1 at reduced temperature and pressure. The broad availability of computers and programmable calculators is making this approach somewhat obsolete, but it is worthwhile to visualize the trends. Fig. 7.44 may be applied for most hydrocarbons. The plot of Z0 represents the behavior of a fluid that would have an acentric factor of 0, and the plot of Z1 represents the quantity Image, which is the correction factor for a hypothetical fluid with an acentric factor of 1. By perusing the table on the back flap of this book, you will notice that most fluids fall between these ranges so that the charts may be used for interpolation.

Image

Figure 7.4. Generalized charts for estimating the compressibility factor. (Z0) applies the Lee-Kesler equation using ω = 0.0, and (Z1) is the correction factor for a hypothetical compound with ω = 1.0. Note the semilog scale.

Eqn. 7.3 can be applied to any fluid once Tr, Pr, and ω are known. It should be noted, however, that this graphical approach is rarely used in current practice since computer programs are more conveniently written in terms of the equations of state as demonstrated in Section 7.5 and the homework.


Example 7.1. Application of the generalized charts

Estimate the specific volume in cm3/g for carbon dioxide at 310 K and (a) 8 bar (b) 75 bar by the compressibility factor charts and compare to the experimental values2 of 70.58 and 3.90, respectively.

Solution

ω = 0.228 and Tr = 310/304.2 = 1.02 for both cases (a) and (b), so,

a. Pr = 8/73.82 = 0.108; from the charts, Z0 = 0.96 and Z1 = 0, so Z = 0.96.

V = ZRT/(P·MW) = (0.96·83.14·310)/(8·44) = 70.29, within 0.4% of the experimental value.

b. Pr = 75/73.82 = 1.016 ≈ 1.02; Note that the compressibility factor is extremely sensitive to temperature in the critical region. To obtain a reasonable degree of accuracy in reading the charts, we must interpolate between the reduced temperatures of 1.0 and 1.05 which we can read with more confidence.

At Tr = 1.0, Z0 = 0.22 and Z1 = –0.08 so Z = 0.22 + 0.228·(–0.08) = 0.202

At Tr = 1.05, Z0 = 0.58 and Z1 = 0.03, so Z = 0.58 + 0.228·(0.03) = 0.587

Interpolating, Z = 0.202 + (0.587 – 0.202)·2/5 = 0.356

V = ZRT/(P·MW) = (0.356·8.314·310)/(7.5·44) = 2.78, giving 29% error relative to the experimental value.


It should be noted that the relative error encountered in this example is somewhat exaggerated relative to most conditions because the Z-charts are highly non-linear in the critical region used in this problem. Since the compressibility factor charts essentially provide a “linear interpolation” between Z values for ω = 0 and ω = 1, the error is large in the critical region. If the reduced temperature had been slightly higher, (e.g., Tr = 1.1), then the relative error would have been roughly 1%, as demonstrated in the homework problems. It would be a simple matter to specify conditions that would make the chart look much more reliable, but then students might tend to err liberally rather than conservatively. For better reliability, computer methods provide proper alternatives, and these are easily applied on any modern engineering calculator. Example 7.5 on page 268 will demonstrate in detail the validity of this perspective.

7.4. The Virial Equation of State

At low reduced pressure, deviations from ideal gas behavior are sufficiently small that we can write our equation of state as explicit in a power series with respect to density. That is,

Image

where B, C, and D are the second, third, and fourth virial coefficients. This can be considered an expansion in powers of ρ. Coefficients C and D are rarely applied because this power series is not very accurate over a broad range of conditions. The most common engineering application of the virial equation of state is to truncate it after the second virial coefficient and to limit the range of application appropriately. It provides a simple equation which still has a reasonable number of viable applications. It has become common usage to refer to the equation truncated after the second virial coefficient as the virial equation, even though we know that it is really a specialized form. We, too, will follow this common usage. Furthermore, the truncated form may alternatively be expressed as Z = Z(P,T). Hence, we often refer to the virial equation as:

Image

where B is a function of T. Note that Eqn. 7.5 indicates that Z varies linearly with pressure along an isotherm. Look back at Fig. 7.4 and notice that the region in which linear behavior occurs is limited, but in general, the approximation can be used at higher reduced pressures when the reduced temperature is higher. The virial equation can be generalized in reduced coordinates as given by Eqns. 7.67.9.5 Eqn. 7.10 checks for restriction of the calculation to the linear Z region.


Image Virial equation. B is known as the second virial coefficient, and it is a measure of the slope of the Z-chart isotherms in the linear region.


Image

The temperature dependence of the slope of the Z lines is not sufficiently represented by 1/T, so the temperature dependence of B in Eqns. 7.8 and 7.9 is required. The virial equation is limited in its range of applicability, but it has the advantage of simplicity. Its simplicity is especially advantageous when illustrating derivations of real-fluid behavior for the first time and extending thermodynamic relations to vapor mixtures. Unfortunately, the virial equation does not apply to liquids, and many interesting results in thermodynamics appear in the study of liquids. To develop a global perspective applicable to gases and liquids, we must consider the physics of fluids in a more sophisticated manner. The simplest form which still permits this level of sophistication is the cubic equation, discussed in the following section.


Example 7.2. Application of the virial equation

Estimate the specific volume in cm3/g for carbon dioxide at 310 K and (a) 8 bar (b) 75 bar by the virial equation and compare to the experimental values of 70.58 and 3.90, respectively.

Solution

ω = 0.228 and Tr = 310/304.2 = 1.02 for both cases (a) and (b), so,

B0 = 0.083 – 0.422/1.021.6 = –0.326

B1 = 0.139 – 0.172/1.024.2 = –0.0193

B(T)Pc/RTc = B0 + ωB1) = (–0.326 + 0.228·(–0.0193)) = –0.3304

a. Pr = 8/73.82 = 0.108; so Z = 1 + (B0 + ωB1)Pr/Tr = 1 – 0.3304·0.108/1.02 = 0.965 V = ZRT/(P·MW) = (0.965·83.14·310)/(8·44) = 70.66, within 0.1% of the experimental value.

b. Pr = 75/73.82 = 1.016; applying Eqn. 7.10, 0.686 + 0.439·1.016 = 1.13 > Tr = 1.02. Therefore, the virial equation may be inaccurate using only the second virial coefficient.


There is an adaptation of the form of the virial series which should be mentioned before concluding this discussion. It should not seem surprising that the inclusion of extra adjustable parameters in the form of the virial series is an extremely straightforward task—just add higher order terms to the series. In many cases, exponential terms are also included as in Eqn. 7.11. In this way, it is possible to fit the P-V-T behavior of the liquid as well as the vapor to a reasonable degree of accuracy. It turns out that the theoretical foundation for the series expansion in this way is tenuous, however. Reading “the fine print” in discussions of series expansions like the Taylor series shows that such an approach is only applicable to “analytic” functions. At present, there is a general acceptance that the behavior of real fluids is “non-analytic” at the critical point. This means that application of such a series expansion above the critical density and below the critical temperature is without a rigorous mathematical basis. Nevertheless, engineers occasionally invoke the motto that “we can fit the shape of an elephant with enough adjustable parameters.” It is in this spirit that empirical equations like the Benedict-Webb-Rubin equation are best appreciated. One particular modification of the Benedict-Webb-Rubin form is given below. It is the form that Lee and Kesler6 developed to render the Pitzer correlation in terms of computer-friendly equations. The Lee-Kesler equation was used to generate Fig. 7.4.

Image

Twelve parameters are used to specify the temperature dependence of B, C, D, E0, E1, and E2 for each compound. Readers are directed to the original article for the exact values of the parameters as part of the homework.


Image The acronym EOS will be used to mean equation of state.


7.5. Cubic Equations of State

To apply the relationships that we can develop for relating changes in properties to CP, CV, P, T, V, and their derivatives, we need really general relationships between P, V, and T. These relationships are dictated by the equation of state (EOS). Constructing an equation of state with a firm physical and mathematical foundation requires considering how the intermolecular forces affect the energy and pressure in a fluid. In a dense fluid, we know that the molecules are close together on the average, and such closeness gives rise to an attractive potential energy. A common practical manifestation of this attractive energy is the heat of vaporization of a boiling liquid. But how can we make a quantitative connection between molecular forces and macroscopic properties? A firm understanding of this physical and mathematical foundation is helpful to understand the extensions to multicomponent mixtures and multiphase equilibria. A proper derivation would provide a mathematical connection between the microscopic potential and the macroscopic properties. We will lay the groundwork for such a rigorous derivation later in the chapter. For introductory purposes, however, we would like to see how some typical equations look and how to use them in conjunction with the theorem of corresponding states.

The van der Waals Equation of State

One of the most influential equations of state has been the van der Waals (1873) equation. Even the most successful engineering equations currently used are only minor variations on the theme originated by van der Waals. The beauty of his model is that detailed knowledge of the molecular interactions is not necessary. Simply by noting that there are two characteristic molecular quantities (ε and σ) and two characteristic macroscopic quantities (Tc and Pc), he was able to infer a simple equation that captured the key features of each fluid through the principle of corresponding states. His final equation expressed the attractive energy in terms of a parameter which he referred to as a, and the size parameter b, but the choice of symbols was arbitrary. The key feature to recognize at this stage is that there are at least two parameters in all the equations and that these can be determined by matching experimental data.


Johannes Diderik van der Waals (1837–1923) was a Dutch physicist. He was awarded the 1910 Nobel Prize in physics.


The resulting equation of state is:

Image

Image Van der Waals EOS.


where ρ = molar density = n/V.

Note: Common engineering practice is to use ρ to denote intensive density. We follow that convention here, using ρ as molar density. Advanced chemistry and physics books and research publications frequently use ρ as number density N/V = nNA/V, so the definitions must be carefully determined.


Image ρ will be used to denote molar density.


The exact manner of determining the values for the parameters a and b is discussed in Section 7.8 on page 270.

Image

We may write the equation of state as Z = 1 + Zrep + Zatt, where 1 denotes the ideal gas behavior, Zrep represents the deviations from the ideal gas law due to repulsive interactions, Zatt represents the deviations due to attractive interactions. For the van der Waals equation,

Image

Image The van der Waals equation written in the form Z = 1 + Zrep + Zatt.


where the second and third terms on the right-hand side are Zrep and Zatt, respectively. Eqn. 7.12 is compact, but Eqn. 7.14 more clearly represents the origin of the contributions to Z. In many later applications we will need to use the departure of Z from ideal gas behavior, Z–1, and Eqn. 7.14 will fulfill this need. There are two key features of the van der Waals equation. First, note that the repulsive term accounts for the asymptotic divergence of the compressibility factor as the packing factor bρ increases. The divergence occurs because rapidly increasing large pressures are required to increase the density as close packing is approached. Second, note that the attractive term increases in magnitude as the temperature decreases, and contributes to smaller values of Z at low T. The contributions of the attractive forces increase at low temperature because the kinetic energy can no longer overwhelm the potential attractions at low temperature. As we have discussed, this eventually leads to condensation. The discussion in Section 7.11 provides a better understanding of the molecular basis of the van der Waals equation. Note that the van der Waals EOS does not incorporate the acentric factor and is incapable of representing different vapor pressure slopes. It is thus primarily a pedagogical tool to introduce the forms of cubic EOSs, not a practical tool.

The Peng-Robinson Equation of State

Since the time of van der Waals (1873), many approximate equations of state have been proposed. For the most part, these have been semi-empirical corrections to van der Waals’ characterization of “a = constant,” and most have taken the form a = a(T,ω). One of the most successful examples of this approach is that of Peng and Robinson (1976). We refer to this equation many times throughout this text and use it to demonstrate many central themes in thermodynamics theory as well as useful applications.


Image The Peng-Robinson EOS. Note that a is a temperature-dependent parameter, not a constant. Note the dependence on the acentric factor.


The Peng-Robinson equation of state (EOS) is given by:

Image

where ρ = molar density = n/V, b is a constant, and a depends on temperature and acentric factor,7

Image
Image

Note that α is not the isobaric coefficient of thermal expansion and κ is not the isothermal compressibility – they are simply variables introduced by Peng and Robinson for notational convenience. The temperature derivative of a is useful when temperature derivatives of Z are needed:

Image

Tc, Pc, and ω are reducing constants according to the principle of corresponding states. Expressing the contributions to Z in the manner we followed for the van der Waals equation,

Image

Comparison of the Peng-Robinson equation to the van der Waals equation shows one obvious similarity; the repulsive term is the same. There are some differences: the temperature dependence of the attractive parameter a is incorporated; dependence of a on the acentric factor is introduced; and the density dependence of Zatt is altered by the denominator of the attractive term. The manner in which these extra details were added was almost entirely empirical; different equations were tried until one was found which seemed to fit the data most accurately while retaining cubic behavior. (Many equations can be tried in 103 years.) There is not much to say about this empirical approach beyond the importance of including the acentric factor. The main reason for the success of the Peng-Robinson equation is that it is primarily applied to vapor-liquid equilibria and that the representation of vapor-liquid equilibria is strongly influenced by the more accurate representation of vapor pressure implicit in the inclusion of the acentric factor. Since the critical point and the acentric factor characterize the vapor pressure fairly accurately, it should not be surprising that the Peng-Robinson equation accurately represents vapor pressure.


Image There are a lot of EOSs. We focus on the Peng-Robinson to illustrate the concepts.


One caution is given. The differences in accuracy between various equations of state are subtle enough that one equation may be most accurate for one narrow range of applications, while another equation of state is most accurate over another range. This may require a practicing engineer to adopt an equation of state other than the Peng-Robinson equation for his specific application. Nevertheless, the treatment of the Peng-Robinson equation presented here is entirely analogous to the treatment required for any equation of state. If you understand this treatment, you should have no problem adapting. A brief review of several thermodynamic models commonly encountered in chemical process simulations is given in Appendix D.

Note: The variables a and b are used throughout equation of state literature (and this text) to denote equation of state parameters. The formulas or values of these parameters for a given equation of state cannot be used directly with any other equation of state.


Image The variables a and b are commonly used in EOSs. Do not interchange the formulas.


7.6. Solving the Cubic Equation of State for Z

In most applications we are given a pressure and temperature and asked to determine the density and other properties of the fluid. This becomes slightly difficult because the equation involves terms of density that are to the third power (“cubic”) even when simplified as much as possible. Standard methods for solutions to cubic equations can be applied. The equation can be made dimensionless prior to application of the solution method. Note:

Image

Image Dimensional analysis is an important engineering tool. Here we make the EOS dimensionless so that it can be solved in a generalized way.


Defining dimensionless forms of the parameters

Image
Image

results in the lumped variables

Image

The Peng-Robinson equation of state becomes

Image

Image Do not confuse the EOS parameters A and B with other uses of the variables. The intended use is almost always clear.


Note: The variable A is also used elsewhere in the text to denote Helmholtz energy. The variable B is used elsewhere in the text to represent the second virial coefficient and availability. The context of the variable usage should make the meaning of the variable clear. We choose to use A and B as reduced equation of state parameters for consistency with equation of state discussions in the literature.

Rearranging the dimensionless Peng-Robinson equation yields a cubic function in Z that must be solved for vapor, liquid, or fluid roots:

Image

Isotherm Shape and the Real Roots

Fig. 7.5 shows several P-V isotherms for CO2 as generated with Eqn. 7.15. Comparing to Fig. 1.4, note that the cubic EOS predicts “humps” when T < Tc. The humps are larger at lower temperature, and the pressure can be negative as shown by the 275 K isotherm. The cubic equation always has three roots, but at some conditions two are imaginary roots. For engineering, we are interested in the real roots. For the isotherm at 290 K, three real roots exist at pressures between approximately 39 bar and 58 bar. We will refer to this pressure region as the three-root region meaning that there are three real roots in this region. Above 58 bar, only a liquid root will result, and below 39 bar only a vapor root will result. These two regions are called one-root regions to communicate that only one real root exists in this region. The three-root region size depends on temperature. At 275 K the three-root region extends down to P = 0, but the region is very narrow near 300 K.

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Figure 7.5. Illustration of the prediction of isotherms by the Peng-Robinson equation of state for CO2 (Tc = 304.2 K) at 275 K, 290 K, 300 K, 310 K, 320 K, and 350 K. Higher temperatures result in a high pressure for a given volume. The “humps” are explained in the text. The calculated vapor pressures are 36.42 bar at 275 K, 53.2 bar at 290 K, and 67.21 bar at 300 K.

The cubic equation for Z in Eqn. 7.25 also has similar behavior. Naming this function F(Z), we can plot F(Z) versus Z to gain some understanding about its roots as shown in Fig. 7.6. Considering the case when P = Psat, we see that three real roots exist; the larger root of F(Z) will be the vapor root and will be the value of Z for saturated vapor. The smallest root will be the liquid root and will be the value of Z for saturated liquid. At all other pressures at this temperature, where three real roots are found, one of the roots is always more stable. Below the critical temperature, when P > Psat, the fluid will be a compressed liquid, and the liquid root is more stable. Below the critical temperature, when P < Psat, the fluid will be a superheated vapor and the vapor root is more stable. When T > Tc, we have a supercritical fluid which can only have a single root but it may vary continuously between a “vapor-like” or “liquid-like” densities and compressibility factors. We will discuss stability and how to choose the stable root without generating an entire isotherm in an upcoming section.

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Figure 7.6. Comparison of behavior of cubic in Z for the Peng-Robinson equation of state at several conditions. The labels Znew and Zold in the upper left are described in the iterative description in Appendix B.

Methods of Solving the Cubic Equation

Engineering applications typically specify P and T, and require information about V. Solution of the equation of state in terms of Z is preferred over solution for V, and we can subsequently find V using

Image

The value of Z often falls between 0 and 1. (See Fig. 7.4 on page 257.) V often varies from 50–100 cm3/mole for liquids to near infinity for gases as P approaches zero. It is much easier to solve for roots over the smaller variable range using the compressibility factor Z. There are two basic approaches to solving cubic equations of state. First, we may use an iterative method. One such method is the Newton-Raphson method. Another method is to solve analytically. A computer or calculator program is helpful in either solution method.

Iterative Method

The Newton-Raphson method is described in Appendix B. The Newton-Raphson method uses an initial guess along with the derivative value to rapidly converge on the solution.

Analytical Solution

The other choice we have for solution of the cubic is to analytically obtain the roots as detailed in Appendix B. The method varies depending on whether one or three roots exist at the pressure of interest. Solutions are implemented in a spreadsheet (Preos.xlsx) or a MATLAB script (Preos.m). MATLAB includes a polynomial root finder, so the statement Zvals=roots([1 a2 a1 a0]) results in both real and imaginary roots. The argument in the “roots” function is the vector of coefficients for the polynomial in Z. In MATLAB, the indexes of the real roots can by found with index=find(imag(Zvals)==0) followed by selecting the real parts of the roots using Zreal=real(Zvals(index)).


Image Preos.xlsx uses the procedures from Appendix B and shows the intermediate calculations.



Example 7.3. Peng-Robinson solution by hand calculation

Perform a hand calculation of the real roots for argon at 105.6 K and 0.498 MPa.

Solution

This example is available online and provides an example of hand calculation at the same conditions as the next example.



Image Preos.xlsx, Preos.m.



Example 7.4. The Peng-Robinson equation for molar volume

Find the molar volume predicted by the Peng-Robinson equation of state for argon at 105.6 K and 4.96 bar.

Solution

The critical data are entered from the table on the back flap of the text.

Preos.xlsx output is shown below. The state is in the three-root region, because the cells for the one-root region are labeled #NUM! by Excel. Many of the intermediate calculations are also shown. The volumes are 27.8, 134, and 1581 cm3/mole. The lower value corresponds to the liquid volume and the upper value corresponds to the vapor. Note that Z is close to zero for the liquid and close to one for the vapor.

The output from the Preos.m MATLAB script is also shown below. Though the default output does not include intermediate values, they may be obtained by removing the “;” at the end of any code line and rerunning the script.

Output from Preos.m:

argon  Tc(K)= 150.9 Pc(MPa)= 4.898 w = -0.004
T(K)= 105.600000 P(MPa)= 0.496000
Zvals =
    0.8971
    0.0759
    0.0157
Z= 0.897123  0.015681
V(cm^3/mol)= 1588.066740  27.758560
fugacity (MPa)= 0.449384  0.449903
Hdep (J/mol)= -222.933032  -6002.507074

Image


Example 7.5. Application of the Peng-Robinson equation

Estimate the specific volume in cm3/g for carbon dioxide at 310 K and (a) 8 bars (b) 75 bars by the Peng-Robinson equation and compare to the experimental values of 70.58 and 3.90, respectively.1

Solution

ω = 0.228, Tc = 304.2, Pc = 73.82, MW = 44 g/gmol,

a. Z = 0.961
V = ZRT/(P · MW) = (0.961 · 83.14 · 310)/(8 · 44) = 70.37, within 0.3% of the experimental value.

b. Z = 0.492
V = ZRT/(P · MW) = (0.492 · 83.14 · 310)/(75 · 44) = 3.84 giving 1.5% error relative to the experimental value.



Image Preos.xlsx, Preos.m.


Example 7.5 shows the equation of state is much more reliable than reading the compressibility factor charts in Example 7.1 on page 258.

Determining Stable Roots

When three real roots are found, which is most stable can be quickly determined. As mentioned in Example 7.4, the smallest root usually corresponds to a liquid state, and the largest root usually corresponds to the vapor state.8,9 When three real roots are found over a range of pressures for a given temperature, these roots do not indicate vapor + liquid coexistence at all these conditions. Vapor and liquid phases coexist only at the vapor pressure—above the vapor pressure, the liquid root is most stable—below the vapor pressure, the vapor root is most stable. The most stable root represents the phase that will exist at equilibrium. When three roots are found, the most stable root has the lower Gibbs energy or fugacity. At phase equilibrium, the Gibbs energy and fugacity of the roots will be equal. Fugacity is closely related to the Gibbs energy and will be described in Chapter 9, but we will begin to use the calculated values before we explain the calculation procedures completely. When three roots exist, the center root is thermodynamically unstable because the derivative of pressure with respect to volume is positive, which violates our common sense, and is shown to be thermodynamically unstable in advanced thermodynamics texts. Physically, we can understand the meaning of the unstable root as follows. Imagine placing Lennard-Jones spheres all in a row such that the force pulling on each molecule from one direction is exactly balanced by the force in the other direction. Next, imagine placing similar rows perpendicular to that one until you obtain the desired density at the desired temperature. Physically and mathematically, this configuration is conceivable, but what will happen when one of these atoms moves? The perfect balance will be destroyed, and a large number of atoms will cluster together to form a liquid. The atoms that do not form a liquid will remain in the form of a low-density vapor. Although this discussion has discussed only the energy of interactions, the entropy is also important. It is actually a balance of enthalpy and entropy that results in phase equilibrium, as we will discuss in Chapter 9.


Image The stable root represents the phase that will exist at equilibrium. The stable root has the lower Gibbs energy or fugacity.


A simple way to visualize the conditions that lead to vapor-liquid equilibrium along an isotherm is to consider the P-V diagram illustrated in Fig. 7.5. We will show in Chapter 9 that the condition for equilibrium between vapor and liquid roots occurs when the horizontal line on the P-V diagram is positioned such that the area enclosed above the line is exactly equal to the area enclosed below the line. Even though the enclosed areas have different shapes, imagine moving this line up and down until it looks like the areas are equal. The dots in the figure are the predictions of the saturated liquid and vapor volumes, and form the phase envelope. The parts of the isotherms that are between the saturated vapor and saturated liquid roots are either metastable or unstable. To determine whether a given point is metastable or unstable, look for the point where the isotherm reaches a maximum or a minimum. Points between the liquid root and the minimum are considered metastable liquids; points between the maximum and the vapor root are considered metastable vapors. The metastable state can be experimentally obtained in careful experiments. Under clean conditions, it is possible to experimentally heat a liquid above its boiling point to obtain superheated liquid. Likewise, under clean conditions, it is possible (though challenging) to experimentally obtain subcooled vapor. However, a metastable state is easily disrupted by vibrations or nucleation sites, e.g., provided by a boiling chip or dust, and once disrupted, the state decays rapidly to the equilibrium state. The boundary between the metastable and unstable states is known as the spinodal condition, predicted by the EOS by the maximum and minimum in the humps in sub-critical isotherms. We will discuss more details about characterizing proper fluid roots when we treat phase equilibrium in a pure fluid.

7.7. Implications of Real Fluid Behavior

There is one implication of non-ideal fluid behavior that should be clear from the equations presented above: Real fluids behave differently from ideal gases. How differently? An example provides the most straightforward answer to that question. Here we adapt some of the derivatives from Chapter 6.


Example 7.6. Derivatives of the Peng-Robinson equation

Determine Image, Image, and Image for the Peng-Robinson equation.

Solution

The derivatives (∂U/∂V)T and (∂CV/∂V)T have been written in terms of measurable properties in Examples 6.6 and 6.9, respectively, and have been evaluated for an ideal gas. The analysis with the Peng-Robinson model provides more realistic representation of the properties of real substances. Beginning with the same analytical expressions set forth in the referenced examples, a key derivative is obtained for the Peng-Robinson equation,

Image

which approaches the ideal gas limit: Image. The volume dependence of CV is obtained by the second derivative:

Image

which approaches the ideal gas limit of zero at low density,

Image

which also approaches the ideal gas limit of zero at low density. We have thus shown that CV depends on volume. To calculate a value of CV, first we determine Image, where Image is the heat capacity tabulated in Appendix E. Then, at a given {P,T}, the equation of state is solved for ρ. The resultant density is used as the limit in the following integrals, noting as V → ∞, ρ → 0, and dV = –dρ/ρ2: This method is used for departures from ideal gas properties in Chapter 8.

Image

where Image


7.8. Matching the Critical Point

The capability of a relatively simple equation to represent the complex physical phenomena illustrated in Figs. 7.57.6, and as shown later in Figs. 7.7 and 7.9, is a tribute to the genius of van der Waals. His method for characterizing the difference between subcritical and supercritical fluids was equally clever. He recognized that, at the critical point,

Image
Image

Figure 7.7. Compressed liquid argon. Experimental data from NIST WebBook. Dashed lines characterize the van der Waals model and solid lines correspond to molecular simulation of the square-well model with λ = 1.7. The manner of fitting the molecular parameters (a and b or ε and σ) is described in Example 7.9.

You can convince yourself that this is true by looking at the P versus ρ plots of Fig. 7.1 on page 254. From this observation, we obtain two equations that characterize the equation of state parameters a and b in terms of the critical constants Tc and Pc. In principle, this is all we need to say about this problem. In practice, however, it is much simpler to obtain results by recognizing another key feature of the critical point: The vapor and liquid roots are exactly equal at the critical point (and the spurious middle root is also equal). We can apply this latter insight by specifying that Image (Appendix B). Equating the coefficients of these polynomials gives three equations in three unknowns: Zc, Ac, and Bc.


Example 7.7. Critical parameters for the van der Waals equation

Apply the above method to determine the values of Zc, Ac, and Bc for the van der Waals equation.

Solution

Rearranging the equation in terms of Ac and Bc we have: Image

By comparing coefficients of Image.

Substituting Ac into the last equation, we have: Image.

Cancelling the Image and solving we have Bc = 1/8 = 0.125. The other equations then give Zc = 0.375 and Ac = 27/64.

The solution is especially simple for the van der Waals equation, but the following procedure can be adapted for any cubic equation of state:

1. Rearrange the equation of state into its cubic form: Z3a2Z2 + a1Za0.

2. Guess a value of Zc (e.g., Zc ~ 1/3).

3. Solve the equivalent of expression (1) for Bc.

4. Solve the equivalent of expression (2) for Ac.

5. Solve the equivalent of expression (3) for Zc.

6. If Zc = guess, then stop. Otherwise, repeat.


7.9. The Molecular Basis of Equations of State: Concepts and Notation

In the previous sections we alluded to equations of state as empirical equations that may have appeared by magic. In this section and the next two, we attempt to de-mystify the origins behind equations of state by systematically describing the current outlook on equation of state development. It may seem like overkill to develop so much theory to justify such simple equations. As empirical equations go, equations of state are not much more difficult to accept than, say, Newton’s laws of motion. Nevertheless, our general purpose is for readers to learn to develop their own engineering model equations and to refute models that are not sensible. By establishing the connection between the nanoscopic potential function and macroscopic properties, molecular modeling lays the foundation for design at the nanoscale.

It is feasible to develop equations of state based solely on fitting experimental data. If the fit is insufficiently precise for a given application, simply add more parameters. We see evidence of this approach in the Peng-Robinson equation, where temperature and density dependencies are added to the parameter “a” in order to fit vapor pressure and density better. A more extensive example of this approach is evident in the 32 parameter Benedict-Webb-Rubin equation that forms the basis of the Lee-Kesler model. The IAPWS model of H2O is representative of the current state of this approach. It is the basis of the steam tables in Appendix E.

The shortcoming of this approach is that we lose the connection between the parameters in the equation of state and their physical meaning. For example, the Peng-Robinson “a” parameter must be related to attractive interactions, like the square-well parameter ε. But ε cannot be a function of temperature and density, so what part of the Peng-Robinson model is due to ε and what part is due to something else? If we could recover that physical connection, then all our efforts to fit data would result in systematically refined characterizations of the molecular interactions. With reliable characterizations of the molecular interactions, we could design molecules to assemble into a myriad of nanostructures: membranes for water purification, nanocomposites, polymer wrappers that block oxygen, artificial kidneys small enough to implant. The possibilities are infinite.

Since about 1960, computers have made it feasible to simulate macroscopic properties based on a specified intermolecular potential. With this tool, the procedure is clear: (1) Specify a potential model for a given molecule, (2) simulate the macroscopic properties, (3) evaluate the deviations between the simulated and experimental properties, (4) repeat until the deviations are minimized. This procedure is straightforward but tedious. Each simulation of Z(T,ρ) can take an hour or so.

Corresponding States in Molecular Dimensions

As engineers, we would like to get results faster. One idea is to leverage the principle of corresponding states. We know that ε has dimensions of J/molecule, so NAε has dimensions of J/mol. Therefore, RT/(NAε) would be dimensionless and serve in similar fashion to the usual reduced temperature, T/Tc. Similarly, the molecular volume, vmol, has dimensions of cm3/molecule and NAvmol has dimensions of cm3/mol. Therefore, NAvmolρ would be dimensionless and serve in very similar fashion to the usual reduced density, ρc. Another idea would be to tabulate the dimensionless properties from the simulation at many state points, then interpolate, similar to the steam tables. The interpolating equations might even resemble traditional equations of state in form and speed. The difference would be that they retain the connection between the nanoscopic potential model and macroscopic properties. In other words, we can engineer our equations of state to be consistent with specific potential models by expressing our “reduced” temperature and density using molecular dimensions. Then the principle of corresponding states can be applied to match the ε and σ for a particular molecule in the same way that we match a and b parameters in the van der Waals model.

Perhaps the most difficult part of understanding the molecular perspective is making the transformations from the macroscopic scale to the nanoscopic. For example, the “b” parameter has dimensions of cm3/mol. What does that imply about the diameter of the molecule in nm? As another example, the “a” parameter has dimensions of J-cm3/mol2. How does that relate to the molecular properties? Answering these questions leads to the introduction of a few “conversion shortcuts” to facilitate the scale transformations. One valuable conversion shortcut is to note that the transformation from cm3 to nm3 involves a factor of (107)3 or 1021. This transformation from cm3 to nm3 usually goes hand-in-hand with a transformation from moles to molecules, involving a factor of NA. If we write NA = 602(1021) instead of 6.02(1023), then factors of 1021 cancel conveniently. This convenience motivates us to work in cm3 at the macroscopic level. Another shortcut is to write the volume of a sphere in terms of diameter instead of radius. Finally, we note that the molecular volume on a molar basis is equivalent to the “b” parameter in cm3/mol. Altogether,

Image

Example 7.8. Estimating molecular size

Example 7.4 shows that b = 19.9cm3/mol for argon. Estimate the diameter (nm) of argon according to the Peng-Robinson model.

Solution

NAπσ3/6 = 19.9 cm3/mol; σ3 = 6(19.9cm3/mol)(1mol/602(1021)molecules)(1021 nm3/cm3)/π Thus, σ3 = 6(19.9)/(602π) = 0.06313 nm3; σ = (0.06313) = 0.398 nm.


Speaking of the “b” parameter, it is useful to note that the combination of bρ appears in the equations quite often. This combined variable is very important. In addition to being dimensionless, and a convenient reduced density, its meaning is quite significant. It represents the volume occupied by molecules divided by the total volume. It makes sense intuitively that the density cannot be higher than when the total volume is completely filled. So this explains why the van der Waals equation includes (1–bρ) in the denominator, forcing divergence as this limit is approached. The prevalence of this combined variable suggests that we give it a special symbol and name, ηP = bρ = b/V, the packing efficiency (aka. packing fraction).10

Finally, we should consider the square-well energy parameter, ε, and the van der Waals parameter, a. Applying Eqn. 7.13 indicates that the dimensions of the “a” parameter are J-cm3/mol2. We can rewrite the van der Waals equation as Z = 1/(1–ηP) – (a/bRTP. In this format, it is clear that the combination of variables “a/b” represents an attractive energy in J/mol. In other words, a/b ~ NAε. Another shortcut for quickly transforming from the macro scale to the nano scale is to recognize that ε/k = NAε/R and both have dimensions of absolute temperature, K. In this context, the combination of variables ε/kT = βε is an especially convenient characterization of dimensionless reciprocal temperature, where β=1/kT.

Distinguishing Repulsive and Attractive Effects

One of the advantages of molecular modeling is that the potential model can be dissected into various parts: the repulsive core, attractive wells, dipole moments, hydrogen bonding, and so forth. The total potential function is the sum of all of these interactions, but simulations can be done separately with one, two, or all interactions. Then we can understand which parts of the equation of state come from each part of the potential model. Fig. 7.7 illustrates Z versus reciprocal temperature. This shows a specific y-intercept at infinite temperature. Analyzing the van der Waals equation shows that this y-intercept corresponds to Z0 = 1/(1–), where the subscript “0” designates the point where reciprocal temperature reaches zero. This contribution represents positive deviations from ideality, and therefore we can call it repulsive. The temperature-dependent part of the van der Waals equation is negative and represents attractive contributions. The reason that the attractive contribution becomes negligible at high temperature is that such a large molecular kinetic energy overwhelms the relatively small “stickiness” of the molecular attractions. Only the repulsive interaction is large enough to contribute at high temperature.

If the y-intercept is so important then what does the x-intercept mean? We have seen Z ~ 0 before, in Example 7.4, where Z = 0.016 for the liquid root. On the scale of Fig. 7.7, Z = 0.016 is practically zero. The pressure of the saturated liquid is low despite having a high density (and high repulsion) because the attractions are comparable to the repulsions when the temperature is low enough. Slow-moving molecules show a greater tendency to “stick together.”

Ultimately, it is necessary to characterize the parameters that relate the intermolecular potential to experimental data. With sufficient data for the compressed liquid density, the problem of characterizing ε and σ becomes a simple matter of matching the slope and intercept of a plot like Fig. 7.7. The procedure is illustrated in Example 7.9(b). This is the most straightforward approach because it relates experimental PVT data directly to PVT data from a molecular simulation.

Fig. 7.7 also compares to experimental compressed liquid data for argon. These data transition quickly to supercritical temperatures and pressures, but the trend is smooth (almost linear) because the density is constant (i.e., isochoric). We can fit the van der Waals equation at one density by tuning the a and b parameters, as shown in Example 7.9(a). Deviations are large, however, when we apply the van der Waals model to a different density using the same a and b. This reflects deficiencies in the physics of the van der Waals model.

We can improve the characterization of argon by using the square-well potential with λ = 1.7. Once again, the parameters (ε and σ this time) are tuned to the Z versus 1/T data at the high density as shown in Example 7.9(b). The predictions (using the same ε and σ) are much better, as shown by the solid lines in Fig. 7.7, reflecting the improved physics underlying the square-well model coupled with molecular simulation. Systematically studying the square-well model, and dissecting the repulsive and attractive contributions, leads to a better understanding of the molecular interactions, and this leads to better predictions.

Other approaches exist to infer potential parameters from experimental data, but they are too complicated for our introductory treatment. One alternative is to apply experimental data for saturated vapor pressure and density. This approach accounts better for vapor pressure being a very important property in chemical engineering, and more data are available. As another alternative, you may be wondering about fitting the critical point, as done by van der Waals. Unfortunately, the behavior at the critical point does not conform to the rules of normal calculus and even simulations are challenging. If you read the fine print on the theorems of calculus, you find the stipulation that functions must be analytic for the theorems to apply. Phenomena in the critical region are non-analytic. The non-analytic behavior is universal for all compounds, so the principle of corresponding states is still valid. On the other hand, fitting a simple analytic function to data outside the critical region leads to inconsistencies inside the critical region, and vice versa. Cubic equations exhibit this inconsistency by predicting a PV phase envelope that is not flat enough on the top. Dealing further with these inconsistencies is a topic of current research. Methods that avoid the critical region are gaining favor at present.

We have barely scratched the surface of what is necessary to characterize the intermolecular forces between all the molecules that we can imagine. For example, we have only considered the square-well potential, but the Lennard-Jones model would be more realistic, and those are just two of the possibilities. As another example, the presentation here is limited to spherical molecules. The molecular perspective is not extended to non-spherical molecules until Chapter 19, and then only briefly.11 You might say that the Peng-Robinson equation can be applied to non-spherical molecules, but only because of clever fitting. The physics behind the Peng-Robinson is simply the same as that of van der Waals: spherical. Simple physics and educated empirical fitting are cornerstones of engineering models. The Peng-Robinson model is a prime example of what can be accomplished with that approach. But recognize that we are always learning more about physics and those new insights are the cornerstones of new technology. Accurately characterizing intermolecular forces involves characterizing many small molecules that share common fragments. When those fragments are characterized, they can be assembled to predict the properties of large molecules. Ultimately, we can imagine a time when nanostructures can be designed and constructed the way civil engineers build bridges today. These structures occur naturally in everything from sea shells to proteins. Learning how to do it is a basis for modern research.


Example 7.9. Characterizing molecular interactions

Based on Fig. 7.7, trend lines indicate y-intercept values of, roughly, 5.7 and 4.7 when fit to the isochoric PVT data for argon at 1.38g/cm3 and 1.25 g/cm3, respectively. Similarly, the x-intercepts are roughly 11.2 and 9.5, respectively. Use these values to estimate the EOS parameters.

a. Estimate the values of a and b at 1.38 g/cm3 according to the van der Waals model.

b. Predict the values of x- and y- intercepts at 1.25 g/cm3 using the a and b from part (a).

c. Suppose the square-well simulation data can be represented by:

Z = 1+4 ηP/(1–1.9 ηP)–15.7 ηP βε/(1–0.16 ηP)

Estimate the values of σ and ε/k at 1.38g/cm3 and predict the x- and y-intercepts at 1.25 g/cm3.

Solution

a. At 1.38g/cm3, y-intercept, Z0 = 1/(1 – ηP) = 5.7 => ηP = 1 – 1/5.7 = 0.825 = .
b = 0.825·39.9(g/mol)/1.38(g/cm3) = 23.9 cm3/mol.

At the x-intercept, 0 = 5.7 – (a/bRT)·ηP = 5.7 – (a/bRT)·0.825 => a/bRT = 5.7/0.825 = 6.91. Using the x-intercept to determine temperature, 1000/T = 11.2 => T = 1000/11.2 = 89.3K => a = 23.9(8.314)89.3(6.91) = 123 kJ-cm3/mol2.

b. At 1.25 g/cm3, ηP = 23.9(1.25)/39.9 = 0.7487 => Z0 = 1/(1 – ηP) = 4.0 = y-intercept.
At the x-intercept, 0 = 4.0 – 123000/(23.9RT)·0.7487 = 4.0 – 463/T => T = 463/4 = 116.
Therefore, the x-intercept is 1000/T = 1000/116 = 8.6. These x- and y- intercepts form the basis for the dashed line in Fig. 7.7 at 1.25 g/cm3. The prediction of the van der Waals model is poor.

c. The procedure for finding σ and ε/k is similar. At the 1.38g/cm3,
Z0 = 1 + 4 ηP/(1 – 1.9 ηP) = 5.7 => ηP (4 + 4.7·1.9) = 4.7 => ηP = 0.363 =
b = 0.363·39.9/1.38 = 10.5 cm3/mol = NAπσ3/6 => σ = 0.322 nm

At the x-intercept, 0 = 5.7 – 15.7(0.363)βε/(1 – 0.16·0.363) => βε = 0.942;
1000/T = 11.2 => T = 1000/11.2 = 89.3K => ε/k = (0.942)89.3= 84.1 K

At 1.25 g/cm3, following the same procedure:

ηP = 10.5(1.25)/39.9 = 0.329 => Z0 = 1 + 4ηP/(1 – ηP) = 4.5 = y-intercept.
At the x-intercept, 0 = 4.5 – 15.7(0.329)βε/(1 – 0.16·0.329) = 4.5 – (5.452)βε => βε = 0.827 => T = 84.1/0.827 = 102. Therefore, the x-intercept is 1000/T = 1000/102 = 9.8. These x- and y-intercepts form the basis for the solid line in Fig. 7.7 at 1.25 g/cm3, and the prediction is quite good.


To put the significance of this analysis in perspective, imagine you were designing a material with pores just the right size to capture argon from air using the van der Waals model. The diameters would indicate the appropriate pore size. For the van der Waals model σ = (6·23.9/602π)1/3 =0.423nm compared to 0.322nm by the square-well estimate. This means that your pores could be over sized by more than 30%. Improved physical insight can suggest more successful experiments.

7.10. The Molecular Basis of Equations of State: Molecular Simulation

In Chapter 1 we developed an ultrasimplified kinetic theory based on ideal gas interactions. Constructing a sophisticated kinetic theory simply means accounting for molecular interactions like attractions and collisions. Unfortunately, accounting for these detailed interactions requires a sequential calculation of collisions that does not lend itself to explicit solution. Fortunately, modern computers make it easy to characterize these collisions over time.

Computation involving finite systems of molecules colliding with one another and with walls is called molecular dynamics simulation. In this section, we apply a simple computation from elementary physics: the collision of two particles, as illustrated in Fig. 7.8. If we can compute a single collision, a computer can be programmed to compute the next trillion collisions. Thus, we see how the average properties over all collisions can be computed from the intermolecular potential. Comparing the computer model to experimental data leads to a connection between the molecular properties (e.g., ε, σ) and the macroscopic properties (e.g., a, b). Seeing the molecular motions and relating the molecular properties to changes in density, temperature, energy, and pressure sheds light on both the nanoscopic and macroscopic levels.

Image

Figure 7.8. Molecular collision in 2D. The dashed disk is a disk image that will be discussed in the text.

Elastic Collision of Two Particles in Two Dimensions

Tools for molecular simulation are freely available, highly visual, and conveniently interactive.12 On the other hand, using a simulator as a mysterious “black box” is not satisfying. We should feel confident that a molecular dynamics simulation is nothing more than application of Newton’s laws. In this way, we can leverage our confidence in Newton’s laws to infer the behavior of large systems of molecules. Then the connection between elementary physics and molecular thermodynamics should not seem mysterious at all.

For simplicity consider two disks moving in two dimensions (2D). Typical physics courses describe 2D particle collision.13 Consider the special case of a disk1 with an x-component of velocity and a stationary disk2 of identical size as shown in Fig. 7.8. Conservation of kinetic energy and momenta provides:

Image
Image
Image

This provides three equations and four unknowns (v1, θ1, v2, θ2). For hard disks, the fourth equation derives from noting that the force of impact changes the momentum of each particle. The collision force acts on disk2 in the direction of the vector between the two centers at contact and the force on disk1 is equal and opposite. The impact force changes the momentum in the corresponding direction. Referring to Fig. 7.8. disk2’s angle of recoil θ2 can be found by the shaded right triangle. For the purpose of this presentation, we consider the case of identical disk diameters, σ. Note that the shaded triangle hypotenuse is σ and the vertical side is y2cy1c where the “c” denotes collision location. The angle is given by

Image

For purposes of this presentation, we solve the special case where both masses are equal, in which case Eqn. 7.29 and the Pythagorean theorem show that

Image

The simultaneous solution of the remaining three equations provides:

Image

Image An online supplement provides a generalized derivation for different sizes and masses.


Details of the simultaneous solution are provided as an online supplement. The solution in 2D is sufficient in principle because other orientations can be rotated into this reference frame.14

The final element of the computation is to calculate the collision time. The strategy is to calculate the collision time for all possible collisions, then execute the one that occurs first. Collisions with walls are the simplest. If the molecule has a positive x-velocity, then you know it collides with the east wall before it collides with the west wall. Similar reasoning applies for the y-velocity and north or south walls. A wall collision occurs when the molecule travels from its current position until its outer edge touches the wall. To be specific, we can define the “simulation box” to have its southwest corner at the origin and walls of length “L.” For example, the collision time with the east wall (E) would be:

Image

The collision time between two molecules follows a similar procedure. In this case, they touch when their distance is (σ/2 + σ/2) = σ, but only if they approach at the proper angle. Approach angle is not a concern for a wall collision because the wall extends forever, but a molecule is finite in size. The strategy is to translate the molecules to the reference frame of Fig. 7.8, compute the time until crossing the x- position of molecule 2, then check if the molecules are close enough at that point to collide. The first step of the translation simply involves subtracting the position of the first molecule to put it at the origin, and subtracting the velocity of molecule 2 to make it stationary.

Image

The position of the second molecule must be rotated in accordance with the velocity of the first molecule being on the x-axis, such that Image. In polar coordinates,

Image

Transforming the coordinates by rotation,15

Image
Image

A collision occurs if |y2| < σ. If there is a collision,

Image
Image

Immediately after the collision, the positions become

Image

The velocities can be reported in the original reference frame by reversing the φ1 rotation,

Image

For the second molecule, it is most convenient to apply Eqns. 7.30 and 7.31 in their resolved forms:

Image

Example 7.10. Computing molecular collisions in 2D

Let the diameters of two disks, σ, be 0.4 nm, the masses be 16 g/mole, and the length of the square box, L, be 5nm. Start the disks at [1.67 1.67], [3.33 3.33] and initial velocities (nm/ps): [0.167 0.222], [–0.167 –0.222] where 1 nm = 10-9 m and 1ps = 10-12 s. Note that the gas constant 8.314 J/mol-K = 8.314 kg-nm2/(ns2-mol-K) = 8.314(10-6) kg-nm2/(ps2-mol-K).

a. Compute the temperature (K).

b. Compute the collision times with the walls.

c. Compute the collision times with the disks. Which event occurs first?

d. Compute the velocity vectors (m/s) after the first collision event.

Solution

a. T2D = Mw<v2>/(2R); <v2> = (0.1672 + 0.2222 + 0.1672 + 0.2222)/2 = 0.07717

T2D = (0.016kg/mol)(0.07717 nm2/ps2)/(2·8.314(10–6) kg-nm2/(ps2-mol-K)) = 89K.

b. The collision time with the walls depends on the wall being approached. Note that the molecular coordinate will be within 0.5σ = 0.2 nm of the wall coordinate when a wall collision occurs. Disk1 is approaching the north wall and east wall (using superscripts to denote geographic directions), the collision times are t1N = (4.8 – y1o)/v1,y = (4.8 – 1.67)/0.333 = 9.40ps, t1E = (4.8 – x1o)/v1,x = (4.8 – 1.67)/0.222 = 14.1ps. Similarly, t2S = (0.2 – y2o)/v2,y = (0.2 – 1.67)/(–0.222) = 6.62ps; t2W = (0.2 – x2o)/v2,x = (0.2 – 3.33)/(–0.111) = 28.2ps. Molecule 2 collides with the south wall first among wall collisions.

c. Translating by Eqn. 7.36, x2′ = y2′ = 3.33 – 1.67 = 1.66. Translating the velocities to make molecule 2 stationary: v1,x′ = 0.167 – (–0.167) = 0.334. v1,y′ = 0.444. Using Eqn. 7.371 = tan–1(v1,y′/v1,x′) = tan–1(0.444/0.334) = 53.13o2 = tan–1(1.66/1.66) = 45o. r2= 1.66(2)½ = 2.35nm. x2= 2.35 cos(45 – 53.13) = 2.326; y2= 2.35 sin(45 – 53.13) = –0.332. Since |y2″| < σ, these molecules do collide. By Eqn. 7.40, θ2 = sin–1(–0.332/0.4) = –56.10o. Then, x1c” = 2.326 – 0.4 cos(–56.10) = 2.103; noting v1,x″ = (0.3342+0.4442)½ = 0.5556. t12c = 2.103/0.556 = 3.78 ps. So the intermolecular collision occurs first.

d. Computing the velocities after collision requires Eqn. 7.34, noting by Eqn. 7.33 that θ1 = 90 – 56.10 = 33.9. v2 = v1″cos θ2 = 0.5556 cos(–56.10) = 0.3099, v1 = (0.55562– 0.30992)½ = 0.4612. Also note that Eqn. 7.34 gives only the magnitude of the velocity and we are still in the reference frame of Fig. 7.8. Rotating to the original reference frame: v1,xf = v1cos(φ1 + θ1) = 0.4607 cos(33.9 + 53.13) = 0.0239. v1,yf = 0.4607 sin(33.9 + 53.13) = 0.4605; v2,xf = v2,xo + v1,xov1,xf = –0.167 + 0.167 – 0.0239 = –0.0239. v2,yf = –0.222 + 0.222 – 0.4605 = –0.4605. Finally, we update all the positions to the time of the collision. x1f = [1.67 + 0.167·3.78 1.67 + 0.222·3.78] = [2.301 2.514]; x2f = [3.33 – 0.167·3.78 3.33 – 0.222·3.78] = [2.695 2.486]. From this point, the procedure for the next collision is exactly the same.

In retrospect, a major oversimplification of this problem deserves comment. By restricting the system to two particles, it is necessary that the components of velocity be equal and opposite in sign. Otherwise, the system itself would have a net velocity. You should not mistake this equality as a general result. If there were three particles, for example, the velocities would sum to zero, but the individual magnitudes could vary quite substantially.


Analyzing MD Results

For our purposes, we can assume that you have sufficiently grasped the principles of molecular simulation if you can compute a single collision. A second collision is much like the first and computers are made for these kinds of repetitive calculations. At that point, the challenge becomes analyzing the results of the simulations. We can illustrate this kind of analysis with simulations of the hard-sphere fluid to infer the repulsive contribution of the square-well fluid’s equation of state. As shown in Fig. 7.9(a), the hard-sphere (HS) potential can be considered as a special case of the square-well potential when the depth of the well approaches zero. Thus, there are two ways that βε can approach zero: the temperature can approach infinity, or the well depth, ε, can approach zero. Both results lead to the hard-sphere repulsive term.

Image

Figure 7.9. (a) The hard-sphere potential as a special case of the square-well model; (b) results of DMD simulations for the hard-sphere potential compared to simulation data of Erpenbeck and Wood cited in the text.

The results of hard-sphere simulations by Erpenbeck and Wood16 are presented in Fig. 7.9(b). Three equations of state are compared to the simulation results: the van der Waals model, the Carnahan-Starling model, and the ESD model. These models are listed below, along with another called the Scott model.

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It is immediately apparent that the van der Waals model is quite inaccurate while the Carnahan-Starling model is practically quantitative. The ESD model is imprecise when the packing fraction exceeds ηP > 0.40, but it does preserve the prospect of forming the basis for a cubic equation of state. The Scott equation is not shown, but it is slightly less precise than the ESD model and slightly simpler. The precision of the Carnahan-Starling model makes it a popular choice for many of the equations of state discussed in the text. Nevertheless, it is feasible to mix and match various characterizations of the repulsive and attractive contributions to construct an equation of state that is applicable to any particular situation. Constructing your own equation of state is the best way to appreciate the advantages and disadvantages underlying models like the Peng-Robinson equation. We should probably warn you that it is hard to stop once you start down this path of “observe, predict, test, and evaluate.” It is a very good sign, however, if you feel yourself being drawn that way.

Molecular dynamics simulation was first accomplished in 1959.17 Until that time, it was impossible to resolve arguments about whose characterization of the hard-sphere reference system was best. In the final analysis, only a molecular simulation can resolve this debate conclusively. Today, several such programs can be accessed online and some are open source. In particular, the discontinuous molecular dynamics (DMD) module at Etomica.org has been designed to simplify visual and interactive exploration of the relations between temperature, pressure, density, internal energy, and the choice of potential model.


Image Online exercises may help your understanding of molecular simulations.



Example 7.11. Equations of state from trends in molecular simulations

Use the 3D DMD module at Etomica.org to characterize the trends of the attractive contributions for argon with λ = 1.7 at densities of 1.25 and 1.38 g/cm3 assuming a diameter of 0.323 nm and ε/k = 87 K. Use the results to obtain a cubic equation of state.

Solution

It is straightforward to set a diameter of 0.323nm, NAε = 87·8.314 = 723 J/mol, MW = 40, and λ =1.7. For purposes of this problem, we assume the ESD form suffices over the density range of interest with the objective of obtaining a cubic equation.

The next step is to simulate the full potential and solve for the attractive contribution by subtraction. Fig. 7.7 suggests that a linear function in βε should suffice, and we know that the attractive contribution increases with density. These observations suggest an equation of state of the form

Z = 1 + 4 ηP/(1 – 1.9 ηP) – z11 ηP βε,

where z11 designates a constant corresponding to first order in both ηP and βε. By regressing the slope of the attractive contribution at the two given densities, we can characterize z11 as a function of density. We can also infer the zero density limit of z11 from the second virial coefficient as z11(0) = 4(λ3 – 1) = 15.7. The results of these characterizations give z11 = 16.3 at ηP = 0.333 and 17.0 at ηP = 0.367. In order to obtain a cubic equation, we must restrict our attention to equations of the form,

z11 = z11(0)/(1 – z12 ηP).

Plotting z11(0)/z11 and fitting a trendline gives z12 = 0.16 and the final model is,

Z = 1 + 4 ηP/(1 – 1.9 ηP) – 15.7 ηP βε/(1 – 0.16 ηP)

This fit of the attractive trend is crude, but it would be difficult to improve given the constraints imposed by the cubic form. This leaves the door open to future improvements beyond the cubic form. The approach would be similar, however.


7.11. The Molecular Basis of Equations of State: Analytical Theories

Molecular simulation provides a numerical connection between the intermolecular potential model and the macroscopic properties, but it does so one state point at a time. For an equation of state, we need an equation that makes this connection over all state points. The key to making this kind of connection is to consider the average number of neighbors for each molecule within range of the potential model. We alluded to this in Example 1.1(e), and simply called it “four,” but this number must vary with density and strength of attraction and with the precise distance between molecules. Therefore, we must define a quantity representing the average number of molecules at each distance from the center of an average molecule, and study its dependence on density and temperature. To get the configurational internal energy,18 multiply this average number of molecules by the amount of potential energy at that distance and integrate over all distances. To get the pressure, multiply this average number of molecules by the amount of force per unit area at that distance and integrate over all distances. The average number of molecules at a particular distance from an average molecule is characterized by the “radial distribution function,” which is discussed in detail below. If you have ever seen a parking lot, you already know more about radial distribution functions than you may realize.

The Energy Equation

The ideal gas continues to be an important concept, because it is a convenient reference fluid. To calculate the internal energy of a real gas, we simply need to compute the departure from the ideal gas. In this way, the kinetic energy of the gas is included in the ideal gas internal energy, and we calculate the contribution to internal energy due to the intermolecular potentials of the real gas,

Image

where u is the pair potential and g(r) ≡ the radial distribution function defined by Eqn. 7.55. This is often called the configurational energy to denote that it relates to summed potential energy of the configuration. This equation can be written in dimensionless form as

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The Pressure Equation

We also may choose to solve for the pressure of our real fluid. Once again it is convenient to use the ideal gas as our reference fluid and calculate the pressure of the real fluid relative to the ideal gas law. Since intermolecular force is the derivative of the intermolecular potential, we note the derivative of the intermolecular potential in the following equation.

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This equation is typically derived by determining the product PV, but we have multiplied by density to show the pressure.19 This equation can also be written in dimensionless form, recalling the definition of the compressibility factor:

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Note in both the energy equation and the pressure equation, that our integral extends from 0 to infinity. Naturally, we never have a container of infinite size. How can we represent a real fluid this way? Look again at the form of the intermolecular potentials in Chapter 1. At long molecular distances, the pair potential and the derivative of the pair potential both go to zero. Long distances on the molecular scale are 4 to 5 molecular diameters (on the order of nanometers), and the integrand is practically zero outside this distance. Therefore, we may replace the infinity with dimensions of our container, and obtain the same numerical result in most situations. This substitution makes a single equation valid for all containers of any size greater than a few molecular diameters.20

An Introduction to the Radial Distribution Function

As a prelude to a general description of atomic distributions, it may be helpful to review the structure of crystal lattices like those in body-centered cubic (bcc) metals, as shown in Fig. 7.10. Such a lattice possesses long-range order due to repetitive arrangements of the unit cell in three dimensions. This close-packed arrangement of atoms gives a single value for the density, and the density correlates with many of the macroscopic properties of the material (e.g., strength, ductility). These are some of the key considerations fundamental to materials science, and more details are given in common texts on the subject. One goal of introducing the radial distribution function is to generalize the concept of atomic arrangements so that non-lattice fluids can be included.

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Figure 7.10. The body centered cubic unit cell.

The distribution of atoms in a bcc crystal is fairly easy to understand, but how can we address the distribution of atoms in a fluid? For a fluid, the positions of the atoms around a central atom are less well defined than in a crystal. To get started, think about the simplest fluid, an ideal gas.

The Fluid Structure of an Ideal Gas

Consider a fluid of point particles surrounding a central particle. What is the number of particles in a given volume element surrounding the central particle? Since they are point particles, they do not influence one another. This means that the number of particles is simply related to the density,

Image

where dNV is the number of particles in the volume element, N is the total number of particles in the total volume, V is the total volume, dV is the size of the volume element, and dNV = NAρ dV

If we would like to know the number of particles within some spherical neighborhood of our central particle, then,

dV = 4π r2 dr

where r is the radial distance from our central particle,

Image

where R0 defines the range of our spherical neighborhood, Nc is the number of particles in the neighborhood (coordination number).

The Fluid Structure of a Low-Density Hard-Sphere Fluid

Now consider the case of atoms which have a finite size. In this case, the number of particles within a given neighborhood is strongly influenced by the range of the neighborhood. If the range of the neighborhood is less than two atomic radii, or one atomic diameter, then the number of particles in the neighborhood is zero (not counting the central particle). Outside the range of one atomic diameter, the exact variation in the number of particles is difficult to anticipate a priori. You can anticipate it, however, if you think about the way cars pack themselves into a parking lot. We can express these insights mathematically by defining a “weighting factor” which is a function of the radial distance. The weighting factor takes on a value of zero for ranges less than two atomic radii, and for larger ranges, we can consider its behavior undetermined as yet.


Image The hard-sphere fluid has been studied extensively to represent spherical fluids.


Then we may write

Image

where g(r) is our average “weighting function,” called the radial distribution function. The radial distribution function is the number of atomic centers located in a spherical shell from r to r + dr from one another, divided by the volume of the shell and the bulk number density.

This is a lot like algebra. It helps us to organize what we do know and what we do not know. The next task is to develop some insights about the behavior of this weighting factor so that we can make some engineering approximations.

As a first approximation, we might assume that atoms outside the range of two atomic radii do not influence one another. Then the number of particles in a given volume element goes back to being proportional to the size of the volume element, and the radial distribution function has a value of one for all r greater than one diameter. The approximation that atoms outside the atomic diameter do not influence one another is reasonable at low density. An analogy can be drawn between the problem of molecular distributions and the problem of parking cars. When the parking lot is empty, cars can be parked randomly at any position, as long as they are not parked on top of one another. Recalling the relation between a random distribution and a flat radial distribution function, Fig. 7.11 should seem fairly obvious at this point.

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Figure 7.11. The radial distribution function for the low-density hard-sphere fluid.

The Structure of a bcc Lattice

Far from the low-density limit, the system is close-packed. The ultimate in close-packing is a crystal lattice. Let’s clarify what is meant by the radial distribution function of a lattice. The radial distribution function of a bcc lattice can be deduced from knowledge of Nc and the defining relation for g(r).

Image

If we assume that the atoms in a crystal are located in specific sites, and no atoms are out of their sites, then g(r) must be zero everywhere except at a site. For a body-centered cubic crystal these sites are at r = {σ, 1.15σ, 1.6σ,...} g(r) looks like a series of spikes. In the parking lot analogy, the best way of parking the most cars is to assign specific regular spaces with regular space between, as shown in Fig. 7.12.

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Figure 7.12. The radial distribution function for the bcc hard-sphere fluid.

The Fluid Structure of High-Density Hard-Sphere Fluid

The distributions of atoms in a fluid are most conveniently referred to as the fluid’s structure. The structures of these simple cases clarify what is meant by structure in the context that we will be using, but the behavior of a dense liquid illustrates why this concept of structure is necessary. Dense-liquid behavior is something of a hybrid between the low-density fluid and the solid lattice. At large distances, atoms are too far away to influence one another and the radial distribution function approaches unity because the increase in neighbors becomes proportional to the size of the neighborhood. Near the atomic diameter, however, the central atom influences its neighbors to surround it in “layers” in an effort to approach the close packing of a lattice. Thus, the value of the radial distribution function is large, very close to one atomic diameter. Because liquids lack the long-range order of crystals, the influence of the central atom on its neighbors is not as well defined as in a crystal, and we get smeared peaks and valleys instead of spikes. Returning to the parking lot analogy once again, the picture of liquid structure is considerably more realistic than the assumption of a regular lattice structure. There are no “lines” marking the proper “parking spaces” in a real fluid. If a few individuals park out of line, the regularity of the lattice structure is disrupted, and it becomes impossible to say what the precise structure is at 10 or 20 molecular diameters. It is true, however, that the average parking around any particular object will be fairly regular for a somewhat shorter range, and the fluid structure in Fig. 7.13 reflects this by showing sharp peaks and valleys at short range and an approach to a random distribution at long range.

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Figure 7.13. The radial distribution function for the hard-sphere fluid at a packing fraction of bρ = 0.4.

The Structure of Fluids in the Presence of Attractions and Repulsions

As a final case, consider the influence of a square-well potential (presented in Section 1.2) on its neighbors. The range r < σ is off-limits, and the value of the radial distribution function there is still zero. But what about the radial distribution function at low density for the range where the attractive potential is influential? We would expect some favoritism for atoms inside the attractive range, σ < r < λσ, since that would release energy. How much favoritism? It turns out to be simply related to the energy inherent in the potential function.

Image

This exponential function, known as a Boltzmann distribution, accounts for the off-limits range and the attractive range as well as the no-influence (r > λσ) range. Referring to the parking lot analogy again, imagine the distribution around a coffee and doughnut vending truck early in the morning when the parking lot is nearly empty. Many drivers would be attracted by such a prospect and would naturally park nearby, if the density was low enough to permit it.


Image The low-density limit of the radial distribution function is related to the pair potential.


As for the radial distribution function at high density, we expect packing effects to dominate and attractive effects to subordinate because attaining a high density is primarily affected by efficient packing. At intermediate densities, the radial distribution function will be some hybrid of the high and low density limits, as shown in Fig. 7.14.

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Figure 7.14. The square-well fluid (R = 1.5) at zero density and at a packing fraction of bρ = 0.4. The variable β ≡ 1/kT.

A mathematical formalization of these intuitive concepts is presented in several texts, but the difficulty of such a rigorous treatment is beyond the scope of our introductory presentation. For our purposes, we would simply like to understand that the number of particles around a central particle has some character to it that depends on the temperature and density, and that representing this temperature and density dependence in some way will be necessary in analyzing the energetics of how molecules interact. In other words, we would like to appreciate that something called “fluid structure” exists, and that it is described in detail by the “radial distribution function.” This appreciation will be of use again when we extend these considerations to the energetics of mixing. Then we will develop expressions that can be used to predict partitioning of components between various phases (e.g., vapor-liquid equilibria).

The Virial Equation

The second virial coefficient can be easily derived using the concepts presented in this section, together with a little more mathematics. Advanced chemistry and physics texts customarily derive the virial equation as an expansion in density:

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The result of the advanced derivation is that each virial coefficient can be expressed exactly as an integral over the intermolecular interactions characterized by the potential function. Even at the introductory level we can illustrate this approach for the second virial coefficient. Comparing the virial equation at low density, Z = 1 + βρ, with Eqn. 7.52, we can see that the second virial coefficient is related to the radial distribution function at low density. Inserting the low-density form of the radial distribution function as given by Eqn. 7.57, and subsequently integrating by parts (the topic of homework problem 7.29), we find

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This relationship is particularly valuable, because experimental virial coefficient data may be used to obtain parameter values for pair potentials.


Example 7.12. Deriving your own equation of state

Appendix B shows how the following equation can be derived to relate the macroscopic equation of state to the microscopic properties in terms of the square-well potential for λ = 1.5.

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Apply this result to develop your own equation of state with a radial distribution function of the form:

Image

where x = r/σ, b = πNAσ3/6, and S is the “Student” parameter. You pick a number for S, and this will be your equation of state. Evaluate your equation of state at ε/kT = 0.5 and bρ = 0.4.

Solution

At first glance, this problem may look outrageously complicated, but it is actually quite simple. We only need to evaluate the radial distribution function at x = 1 and x = 1.5 and insert these two results into Eqn. 7.60.

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Supposing S = 3, Z(0.5,0.4) = 1 + 4·0.4·1.649/(0.2·2.4) – 13.5·0.4·0.648/(0.789·1.211) = 2.83.

Congratulations! You have just developed your own equation of state. Have fun with it and feel free to experiment with different approximations for the radial distribution function. Hansen and McDonalda describe several systematic approaches to developing such approximations if you would like to know more.


a. Hansen, J.P., McDonald, I.R. 1986. Theory of Simple Liquids. New York:Academic Press.

We conclude these theoretical developments with the comment that a similar analysis appears in the treatment of mixtures. At that time, it should become apparent that the extension to mixtures is primarily one of accounting; the conceptual framework is identical. The sooner you master the concepts of separate contributions from repulsive forces and attractive forces, the sooner you will master your understanding of fluid behavior from the molecular scale to the macro scale.

7.12. Summary

The simple physical observations and succinct mathematical models set forth in this chapter provide powerful tools for current chemical applications and excellent examples of model development that we would all do well to emulate. This chapter has illustrated applications of physical reasoning, dimensional analysis, asymptotic analysis, and parameter estimation that have set the standard for many modern engineering developments.

Furthermore, the final connection has been drawn between the molecular level and the macroscopic scale. In retrospect, the microscopic definition of entropy is relatively simple. It follows naturally from the elementary statistics of the binomial distribution. The qualitative description of molecular interaction energy is also simple; it was discussed in the introductory chapter. Last, but not least, the macroscopic description of energy is easy to understand; it gives the macroscopic energy balance. What is not so simple is the connection of the qualitative description of molecular energies with the macroscopic energy balance. This is the significant development of this chapter. Having complete descriptions of the molecular and macroscopic energy and entropy, all the “pieces to the puzzle” are now in our hands. What remains is to put the pieces together. This final step requires a fair amount of mathematics, but it is largely a straightforward application of tools that are readily available from elementary courses in calculus and the background of Chapter 6.

Important Equations

Several equations stand out in this chapter because we apply them repeatedly going forward. These are Eqns. 7.2, 7.12, and 7.157.19. Eqn. 7.2 is the definition of acentric factor (ω), which provides a convenient standard vapor pressure (cf. back flap) and a crude characterization of the molecular shape. Eqn. 7.12 is the van der Waals equation of state, one of the greatest model equations of all time. Eqns. 7.157.19 describe the Peng-Robinson equation of state, a remarkably small evolution from the van der Waals model considering the 100 years of intervening research. We refer to these often as the basis for further derivations and applications to energy and entropy balances for chemicals other than steam and refrigerants.

Finally, Eqns. 7.51 and 7.52 convey the foundation for understanding the connection between the molecular scale and the macroscopic scale. There is no simpler way to see the connection between u and U than Eqn. 7.51. Understanding the molecular interactions becomes essential when we consider why some mixtures behave ideally while others do not. This becomes apparent when we extend Eqn. 7.51 to mixtures.

7.13. Practice Problems

P7.1. For Tr < 1 and PrPrsat, the Peng-Robinson equation of state has three roots corresponding to compressibility factors between zero and 10. The smallest root is the compressibility factor of the liquid. The largest root is the compressibility factor of the vapor and the middle root has no physical significance. This gives us a general method for finding the compressibility factor of any fluid obeying the Peng-Robinson equation. For the iterative method, use an initial guess of Z = 0 to find the liquid roots and Z = 1 to find the vapor roots of methane at the following conditions:

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Compare to experimental data from N.B. Vargaftik. 1975. Handbook of Physical Properties of Liquids and Gases, 2nd ed., New York: Hemisphere.

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ANS. The liquid roots are very close. The vapor roots are accurate for Tr < 0.9.

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P7.2.

a. Estimate the value of the compressibility factor, Z, for neon at Pr = 30 and Tr = 15.

b. Estimate the density of neon at Pr = 30 and Tr = 15. (ANS. 1.14, 0.25 g/cm3)

P7.3. Above the critical point or far from the saturation curve,21 only one real root to the cubic equation exists. If we are using Newton’s method, we can check how many phases exist by trying the two different initial guesses and seeing if they both converge to the same root. If they do, then we can assume that only one real root exists. Find the compressibility factors for methane at the following conditions, and identify whether they are vapor, liquid, or supercritical fluid roots. Complete the table. Compare your results to Z-charts.

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When Newton’s method is applied with an initial guess of zero, erratic results are obtained at these conditions. Explain what is happening, and why, by plotting F(Z) versus Z for each iteration.

P7.4. A rigid vessel is filled to one-half its volume with liquid methane at its normal boiling point (111 K). The vessel is then closed and allowed to warm to 77°F. Calculate the final pressure using the Peng-Robinson equation. (ANS. 33.8 MPa)

P7.5. 4 m3 of methane at 20°C and 1 bar is roughly equivalent to 1 gal of gasoline in an automotive engine of ordinary design. If methane were compressed to 200 bar and 20°C, what would be the required volume of a vessel to hold the equivalent of 10 gal of gasoline? (ANS. 16 L)

P7.6. A carbon dioxide cylinder has a volume of 0.15 m3 and is filled to 100 bar at 38°C. The cylinder cools to 0°C. What is the final pressure in the cylinder and how much more CO2 can be added before the pressure exceeds 100 bar? If you add that much CO2 to the cylinder at 0°C, what will the pressure be in the cylinder on a hot, 38°C day? What will happen if the cylinder can stand only 200 bar? [Hint: log (Prsat) ≈ (7(1 + ω)/3) (1 – 1/Tr)] (ANS. 3.5 MPa, 38 MPa, boom!)

7.14. Homework Problems

7.1. The compressibility factor chart provides a quick way to assess when the ideal gas law is valid. For the following fluids, what is the minimum temperature in K where the fluid has a gas phase compressibility factor greater than 0.95 at 30 bar?

a. Nitrogen

b. Carbon dioxide

c. Ethanol

7.2. A container having a volume of 40 L contains one of the following fluids at the given initial conditions. After a leak, the temperature and pressure are remeasured. For each option, determine the kilograms of fluid lost due to the leak, using:

a. Compressibility factor charts

b. The Peng-Robinson equation

Options:

i. Methane Ti = 300 K, Pi = 100 bar, Tf = 300 K, Pf = 50 bar

ii. Propane Ti = 300 K, Pi = 50 bar, Tf = 300 K, Pf = 0.9 bar

iii. n-butane Ti = 300 K, Pi = 50 bar, Tf = 300 K, Pf = 10 bar

7.3. Estimate the liquid density (g/cm3) of propane at 298 K and 10 bar. Compare the price per kilogram of propane to the price per kilogram of regular gasoline assuming the cost of 5 gal of propane for typical gas grills is roughly $20. The density of regular gasoline can be estimated by treating it as pure isooctane (2,2,4-trimethylpentane ρ = 0.692 g/cm3) at 298 K and 1 bar.

7.4. From experimental data it is known that at moderate pressures the volumetric equation of state may be written as PV = RT + B · P, where the second virial coefficient B is a function of temperature only. Data for methane are given by Dymond and Smith (1969) as,22

Image

a. Identify the Boyle temperature (the temperature at which B = 0) and the inversion temperature (the temperature at which (∂T/∂P)H = 0) for gaseous methane. [Hint: Plot B versus T–1 and regress a trendline, then differentiate analytically.]

b. Plot these data versus T–1 and compare to the curve generated from Eqn. 7.7. Use points without lines for the experimental data and lines without points for the theoretical curve.

7.5. Data for hydrogen are given by Dymond and Smith (1969) as,

Image

a. Plot these data versus T–1and compare to the results from the generalized virial equation (Eqn. 7.7). Suggest a reason that this specific compound does not fit the generalized equation very accurately. Use points without lines for the experimental data and lines without points for the theoretical curve.

b. Use the generalized virial equation to speculate whether a small leak in an H2 line at 300 bar and 298 K might raise the temperature of H2 high enough to cause it to spontaneously ignite.

7.6. N.B. Vargaftik (1975)23 lists the following experimental values for the specific volume of isobutane at 175°C. Compute theoretical values and their percent deviations from experiment by the following:

a. The generalized charts

b. The Peng-Robinson equation

Image

7.7. Evaluate (∂P/∂V)T for the equation of state where b is a constant:

P = RT/(Vb)

7.8. Evaluate (∂P/T)V for the equation of state where a and b are constants:

P = RT/(V = b) + a/T3/2

7.9. Evaluate Image for the Redlich-Kwong equation of state Image, where a and b are temperature-independent parameters.

7.10.

a. The derivative (∂V/T)P is tedious to calculate by implicit differentiation of an equation of state such as the Peng-Robinson equation. Show that calculus permits us to find the derivative in terms of derivatives of pressure, which are easy to find, and provide the formula for this equation of state.

b. Using the Peng-Robinson equation, calculate the isothermal compressibility of ethylene for saturated vapor and liquid at the following conditions: {Tr = 0.7, P = 0.414 MPa}; {Tr = 0.8, P = 1.16 MPa}; {Tr = 0.9, P = 2.60 MPa}.

7.11. When cubic equations of state give three real roots for Z, usually the smallest root is the liquid root and the largest is the vapor root. However, the Peng-Robinson equation can give real roots at high pressure that differ from this pattern. To study this behavior, tabulate all the roots found for the specified gas and pressures. As the highest pressures are approached at this temperature, is the fluid a liquid or gas? Which real root (smallest, middle, or largest) represents this phase at the highest pressure, and what are the Z values at the specified pressures?

a. Ethylene at 250 K and 1, 3, 10, 100, 150, 170, 175, and 200 MPa

b. n-Hexane at 400 K and 0.2, 0.5, 1, 10, 100, 130 and 150 MPa

c. Argon at 420 K and 0.1, 1, 5, and 10 MPa

7.12. Plot Pr versus ρr for the Peng-Robinson equation with Tr = [0.7,0.9,1.0], showing both vapor and liquid roots in the two-phase region. Assume ω = 0.040 as for N2. Include the entire curve for each isotherm, as illustrated in Fig. 7.1 on page 254. Also show the horizontal line that connects the vapor and liquid densities at the saturation pressure. Use lines without points for the theoretical curves. Estimate Trsat by log(Prsat) = 2.333(1 + ω) (1 – 1/Trsat).

7.13. Within the two-phase envelope, one can draw another envelope representing the limits of supercooling of the vapor and superheating of liquid that can be observed in the laboratory; along each isotherm these are the points for which (∂P/∂ρ)T = 0. Obtain this envelope for the Peng-Robinson equation, and plot it on the same figure as generated in problem 7.12. This is the spinodal curve. The region between the saturation curve and the curve just obtained is called the metastable region of the fluid. Inside the spinodal curve, the fluid is unconditionally unstable. The saturation curve is called the binodal curve. Outside, the fluid is entirely stable. It is possible to enter the metastable region with hot water by heating at atmospheric pressure in a very clean flask. Sooner or later, the superheated liquid becomes unstable, however. Describe what would happen to your flask of hot water under these conditions and a simple precaution that you might take to avoid these consequences.

7.14. Develop a spreadsheet that computes the values of the compressibility factor as a function of reduced pressure for several isotherms of reduced temperature using the Lee-Kesler (1975) equation of state (AIChE J., 21:510). A tedious but straightforward way to do this is to tabulate reduced densities from 0.01 to 10 in the top row and reduced temperatures in the first column. Then, enter the Lee-Kesler equation for the compressibility factor of the simple fluid in one of the central cells and copy the contents of that cell to all other cells in the table. Next, copy that entire table to a location several rows lower. Replace the contents of the new cells by the relation Pr = Z·ρr·Tr. You now have a set of reduced pressures corresponding to a set of compressibility factors for each isotherm, and these can be plotted to reproduce the chart in the chapter, if you like. Copy this spreadsheet to a new one, and change the values of the B, C, D, and E parameters to correspond to the reference fluid. Finally, copy the simple fluid worksheet to a new worksheet, and replace the contents of the compressibility factor cells by the formula: Z = Z0 + ω(ZrefZ0)/ωref, where the Zref and Z0 refer to numbers in the cells of the other worksheets.

7.15. The Soave-Redlich-Kwong equation24 is given by:

Image

where ρ = molar density = n/V

Image
Image

Tc, Pc, and ω are reducing constants according to the principle of corresponding states. Solve for the parameters at the critical point for this equation of state (ac, bc, and Zc) and list the next five significant figures in the sequence 0.08664.......

7.16. Show that Bc = bPc/RTc = 0.07780 for the Peng-Robinson equation by setting up the cubic equation for Bc analogous to the van der Waals equation and solving analytically as described in Appendix B.

7.17. Determine the values of ε/kTc, Zc, and bc in terms of Tc and Pc for the equation of state given by

Image

where F = exp(ε/kT) – 1. The first term on the right-hand side is known as the Scott equation for the hard-sphere compressibility factor.

7.18. Consider the equation of state

Image

where ηP = b/V. The first term on the right-hand side is known as the Carnahan-Starling equation for the hard-sphere compressibility factor.

a. Determine the relationships between a, b, and Tc, Pc, Zc.

b. What practical restrictions are there on the values of Zc that can be modeled with this equation?

7.19. The ESD equation of state25 is given by

Image

ηP = bρ, c is a “shape parameter” which represents the effect of non-sphericity on the repulsive term, and q = 1 + 1.90476(c – 1). A value of c = 1 corresponds to a spherical molecule. Y is a temperature-dependent function whose role is similar to the temperature dependence of the a parameter in the Peng-Robinson equation. Use the methods of Example 7.7 to fit b and Y to the critical point for ethylene using c = 1.3.

7.20. A molecular simulation sounds like an advanced subject, but it is really quite simple for hard spheres.26 Furthermore, modern software is readily available to facilitate performing simulations, after an understanding of the basis for the simulations has been demonstrated. This problem provides an opportunity to demonstrate that understanding. Suppose that four hard disks are bouncing in two dimensions around a square box. Let the diameters of the disks, σ, be 0.4 nm, masses be 40 g/mole, and length of the square box, L, be 5 nm. Start the four disks at (0.25L, 0.25L), (0.75L, 0.25L), (0.25L, 0.75L), (0.75L, 0.75L) and with initial velocities of (v, v/(1 + 2½)), (–v, v), (v/2½, – v/2½), (–v/2½, – v/2½), where v designates an arbitrary velocity. (Hint: you may find useful information in the DMD module at Etomica.org.)

a. Compute v initially assuming a temperature of 298 K.

b. Sum the velocities of all four particles (x and y separately). Explain the significance of these sums.

c. Sketch the disks using arrows to show their directions. Make the sizes of the arrows proportional to the magnitudes of their velocities.

d. Solve for the time of the first collision. Is it with a wall or between particles? Compute the velocities of all disks after the first collision.

7.21. Suppose you had a program to simulate the motions of four molecules moving in 2D slowly enough that you could clearly see the velocities of all disks. (Hint: The Piston-Cylinder applet in the DMD module at Etomica.org is an example of such a program when kept in “adiabatic” mode.)

a. Let the disk interactions be characterized by the ideal gas potential. Describe how the disks would move about. Note that the slow particles would always stay slow, and the fast particles stay fast. Why is that?

b. Change the potential to “repulsion only” as modeled by a hard disk model. Compare the motions of the “repulsion only” particles to the ideal gas particles. Explain the differences. Which seems more realistic?

c. Set the potential to “repulsion and attraction,” as modeled by the square-well model with λ=2.0. Compare the motions of these disks to the “repulsion only” particles and ideal gas particles. Explain the differences.

7.22. Suppose you had a program to simulate the motions of N molecules moving in 2D. (Hint: The 2D applet in the DMD module at Etomica.org is an example of such a program when kept in “adiabatic” mode.)

a. Simulate the motions of the disks using each potential model (ideal gas, hard disk, square well) for 1000 ps (1 picosecond=10–12 second) at a density of 2.86E-6mol/m2 with an initial temperature of 300K. Which would have the higher pressure, ideal gas or hard disks? Explain. Which would have the higher pressure, ideal gas or square well disks? Explain.

b. Simulate the motions of the disks using each potential model for 1000 ps each at a density of 2.86E-6mol/m2 with an initial temperature of 300 K. Sketch the temperature versus time in each case. Explain your observations.

c. Suppose you simulated the motions of the disks using each potential model for 1000 ps each at a density of 2.86E-6mol/m2 with an initial temperature of 300 K. How would the internal energy compare in each case? Explain.

7.23. Sphere and disk collisions can be expressed more compactly and computed more conveniently in vector notation. Primarily, this involves converting the procedures of Example 7.10 to use the dot product of the relative position and relative velocity. (Hint: You may find useful information in the DMD module at Etomica.org.)

a. Write a vector formula for computing the center to center distance between two disks given their velocities, u, and their positions, r0, at a given time, t0.

b. Write a vector formula for computing the distance of each disk from each wall. (Hint: Use unit vectors x=(1, 0) and y=(0, 1) to isolate vector components.)

c. Noting that energy and momentum must be conserved during a collision, write a vector formula for the changes in velocity of two disks after collision.
Hints: (1) ab=abcosθ. (2) A unit vector with direction of a is: a/a.

d. Write a vector formula for the change in velocity of a disk colliding with a wall.

7.24. Molecular simulation can be used to explore the accuracy and significance of individual contributions to an equation of state. Here we explore how the σ parameter relates to experimental data.

a. Erpenbeck and Wood have reported precise simulation results for hard spheres as listed below. Plot these data and compare the ESD and Carnahan-Starling (CS) equations for hard spheres.

Image

b. According to the CS equation, what value do you obtain for ZHS at ηP=0.392?

c. What value of b corresponds to ηP = 0.392 for Xenon at 22.14 mol/L? What value of σ corresponds to that value of b?

d. The simulation results below have been tabulated at ηP = 0.392. Plot Z versus βε for these data. Estimate the value of βε that corresponds to the saturation temperature.

e. Referring to Xenon on the NIST WebBook, estimate the saturation temperature at 22.14mol/L. Referring to part (d) for the value of βε, estimate the value of ε(J/mol).

f. Plot Z versus 1000/T for the simulation data using your best ε and σ at ηP=0.392. Referring to the “fluid properties” link, plot the isochoric data for Xenon from the NIST WebBook at 22.14mol/L on the same axes.

g. What values of a and b of the vdW EOS match the simulation data of this plot? Compute Zvdw versus 1000/T and show the vdW results as a dashed line on the plot.

h. Using the values of ε and σ from parts (c) and (e), simulate the system “isothermally” at 225 K and 20.0 mol/L for ~400 ps (got pizza?). Use the CS equation to estimate the y-intercept for Z. Plot these points including a trendline with equation. Plot the NIST data for this isochore on the same axes. This represents a prediction of the data at 20.0 mol/L since the parameters were determined at other conditions.

i. Using the values of a and b from part (g), plot the vdW results at 20.0 mol/L as a dashed line on the plot. This represents the vdW prediction.

j. Which model (SW or vdW) matches the experimental trend best? Why?

k. Neither prediction is perfect. Suggest ways that we may proceed to improve the predictions further.

SW results at ηP = 0.392, λ = 2.0.

Image

7.25. Molecular simulation can be used to explore the accuracy and significance of individual contributions to an equation of state. Use the DMD module at Etomica.org to tune Xe’s ε and σ parameters.

a. According to the Carnahan-Starling (CS) model, what value do you obtain for ZHS at ηP=0.375?

b. What value of σ corresponds to ηP = 0.375 for Xe at 22.14 mol/L?

c. The simulation results below have been tabulated at ηP = 0.375, λ = 1.7. Plot Z versus βε for these data. Referring to the NIST WebBook for Xe, estimate the saturation T and Z at 22.14 mol/L. Estimate the value of βε that corresponds to the saturation Z. Estimate the value of ε(J/mol).

d. Plot Z versus 1000/T, using your best ε and σ at ηP = 0.375 and showing the fluid properties (isochoric) data from WebBook.nist.gov at 22.14 mol/L on the same axes.

e. What values of a and b of the vdW EOS will match the simulation data of this plot? Show the vdW results as a dashed line on the plot.

f. Using the values of ε and σ from parts (b) and (c), simulate the system at 225 K and 20.0 mol/L for ~400 ps (got pizza?). Use the CS equation to estimate the y-intercept value for Z and connect the dots on a new plot with a straight line extrapolating through the x-axis. Plot the NIST data for this isochore on the same axes. This represents a prediction of the data at 20.0 mol/L since the parameters were determined at other conditions.

g. Using the values of a and b from part (e), plot the vdW results at 20.0 mol/L as a dashed line on the plot. This represents the vdW prediction. Comment critically.

h. Compare to Problem 7.24. Summarize your observations.

SW results at ηP = 0.375, λ = 1.7.

Image

7.26. The discussion in the chapter focuses on the square-well fluid, but the same reasoning is equally applicable for any model potential function. Illustrate your grasp of this reasoning with some sketches analogous to those in the chapter.

a. Sketch the radial distribution function versus radial distance for a low-density Lennard-Jones (LJ) fluid. Describe in words why it looks like that.

b. Repeat the exercise for the high-density LJ fluid. Also sketch on the same plot the radial distribution function of a hard-sphere fluid at the same density. Compare and contrast the hard-sphere fluid to the LJ fluid at high density.

7.27. Suppose that a reasonable approximation to the radial distribution function is

Image

where x = r/σ, F = exp(ε/kT) – 1 and b = πNAσ3/6. Derive an expression for the equation of state of the square-well fluid based on this approximation. Evaluate the equation of state at bρ =0.6 and ε/kT = 1.

7.28. Suppose that a reasonable approximation for the radial distribution function is g(r) = 0 for r < σ, and

Image

for r≥ σ, where u is the square-well potential and b = πNAσ3/6. Derive an equation of state for the square-well fluid based on this approximation.

7.29. The truncated virial equation (density form) is Z = 1 + Bρ. According to Eqn. 7.52, the virial coefficient is given by

Image

where the low pressure limit of g(r) given by Eqn. 7.57 is to be used. Another commonly cited equation for the virial coefficient is Eqn. 7.59. Show that the two equations are equivalent by the following steps:

a. Beginning with Image, insert the low-pressure limit for g(r), and simplify as much as possible.

b. Integrate by parts to obtain

Image

c. Show that the left-hand side of the answer to part (b) may be written as Image for a physically realistic pair potential. Then combine integrals to complete the derivation of Eqn. 7.59.

7.30. The virial coefficient can be related to the pair potential by Eqn. 7.59.

a. Derive the integrated expression for the second virial coefficient in terms of the square-well potential parameters ε/k, σ, and R.

b. Fit the parameters to the experimental data for argon.27

Image

c. Fit the parameters to the experimental data for propane.1

Image

7.31. One suggestion for a simple pair potential is the triangular potential

Image

Derive the second virial coefficient and fit the parameters σ, ε, and R to the virial coefficient data for argon tabulated in problem 7.30.

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