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## Chapter 2. Phase Diagrams of Pure Fluids

The strength of molecular interactions is determined by the mean intermolecular distance and the property that most directly reflects this distance is molar density, or its reciprocal, molar volume. The approximate relationship between molar volume and mean intermolecular distance is given by (see Example 1.1), where NA is Avogadro’s number. The packing density of molecules in given volume reflects the strength of the potential interaction. In gases (large V), distances are large and interactions weak. In liquids, the opposite is true. The volume that is occupied by a fixed number of molecules depends both on temperature and pressure. The relationship between volume, pressure, and temperature is of fundamental importance and its mathematical form is known as equation of state. In this chapter we examine this relationship in graphical and mathematical form. The learning objectives of this chapter are to develop the following skills:

1. Using the PVT graph to identify the phase of a pure fluid.

2. Working with tabulated values of P, V, T (steam tables).

3. Identify the region of applicability of the ideal-gas law and the truncated virial equation.

4. Working with cubic equations of state.

5. Working with generalized correlations for the compressibility factor.

6. Representing processes on the PV graph.

### 2.1 The PVT Behavior of Pure Fluid

The molar volume, V, is the volume occupied by 1 mol of the substance, and is the inverse of the molar density, ρ:1

1. We will use the same general symbol, ρ, for both the molar density and the mass density. We will annotate them differently only if they both appear in the same equation. The specific volume is the volume occupied by 1 kg of substance. The specific volume, the mass density, and the molar densities are related to each other: where ρ′ is the mass density (kg/m3) and Mm is molar mass (kg/mol). The volume occupied by a given amount of matter depends on temperature and pressure: it decreases under compression and (for most substances) increases upon heating. The relationship between P, V, T is characteristic of a substance but the general features of this relationship are common to all pure fluids and will be examined below.

The relationship between volume, pressure, and temperature is represented graphically by a three-dimensional surface whose general shape is shown in Figure 2-1. This graph has been rotated to show pressure in the vertical axis, with the mesh lines on the surface representing lines of constant temperature (left to right), and lines of constant volume (front to back). A point on the surface gives the molar volume at the indicated pressure and temperature. The bell-shaped curve facing the pressure-volume plane is the vapor liquid boundary. To its left is the liquid region (steep part of the surface), to its right the vapor region. At sufficiently low temperatures a system exhibits a phase transition to a solid; this region is not shown on this graph.2

2. The surface in Figure 2-1 was calculated using the Soave-Redlich-Kwong equation of state (see Section 2.6), which is appropriate for liquids and gases but not for solids. Figure 2-1: The PVT surface of a pure fluid.

The three-dimensional representation is useful for the purposes of visualizing the PVT relationship but is impractical for routine use. We work, instead, with projections of the PVT surface on one of the three planes, PV, PT, or VT. A projection is a view of the three-dimensional surface from an angle perpendicular to the projection plane and reduces the graph into a two-dimensional plot. The most commonly used projections are those on the PV and the PT planes.

#### The PV Graph

The PV graph is the projection on the PV plane and is shown in Figure 2-2. The characteristic feature of this graph is the vapor-liquid region, represented by a bell-shaped curve that consists of two branches, saturated liquid (to the left) and saturated vapor (to the right). The two branches meet at the top and this point defines the critical point of the fluid. Temperature is indicated by contours of constant temperature (isotherms). These are lines with the general direction from the upper left corner to the lower right. The isotherm that passes through the critical point corresponds to the critical temperature, TC. This isotherm has an inflection point at the critical point, namely, its first and second derivatives are both zero:  Figure 2-2: The PV graph of pure fluid (the solid phase is not shown).

From there on, it decreases smoothly into the vapor region. The region to the left of the saturated liquid is called subcooled liquid to indicate that its temperature is below the boiling point that corresponds to its pressure. For example, at A the temperature is T1, lower than the saturation temperature (T2) that corresponds to pressure PA. Alternatively, it is called compressed liquid to indicate that pressure is higher than the saturation pressure that corresponds to its temperature (the terms subcooled and compressed liquid are used interchangeably). Vapor to the right of the saturated line is superheated because its temperature is higher than the boiling temperature that corresponds to its pressure. At B, for example, temperature (T4) is higher than the saturation temperature (T2) that corresponds to its pressure.

The organization of information on the PV graph can be better understood by conducting a heating or cooling process and following the path on the graph. Suppose we add heat under constant pressure starting with liquid at state A, which is at pressure PA and temperature TA = T1. The process is depicted by the line AB, drawn at constant pressure PA. Between states A and L, heating causes the volume to increase somewhat but the increase is relatively small because the thermal expansion of liquids is small. At point L the liquid is saturated and at the verge of boiling. Adding heat at this point causes liquid to evaporate and produce more vapor, moving the state along the line LV. During this process both pressure and temperature remain constant (line LV is both an isotherm and an isobar). At point V all the liquid has evaporated and the system is saturated vapor. Adding more heat causes temperature and molar volume to increase and moves the state along the line VB. If we start at state B and perform a constant-pressure cooling process, we will observe the reverse course of events. During BV, the vapor is cool. At point V, the state is saturated vapor at the verge of condensation. Removing heat at this point causes vapor to condense until the steam becomes 100% saturated liquid (state L). Upon further cooling, the system moves further into the subcooled region.

Note

Boiling in Open Air

If the heating/cooling process that is described here is conducted in an open container, for example, by heating water in an open flask at atmospheric pressure, the behavior will be somewhat different than the one described here. In an open container, water forms vapor at any temperature below boiling, not just at the boiling point. The important difference is that in an open container we are dealing with a multicomponent system that contains not only water, but also air. A vapor-liquid mixture with two or more components behaves differently from the pure components. Multicomponent phase equilibrium is treated in the second part of this book and until then, it should be understood that we are dealing with pure fluids. The process described by the path AB may be thought to take place inside a sealed cylinder fitted with a piston and initially filled with liquid containing no air at all.

##### The Critical Point

The critical point is an important state and its pressure and temperature have been tabulated for a large number of pure substances. In approaching the critical point from below, the distance between points L and V decreases, indicating that the molar volume of the saturated liquid and saturated vapor come closer together. At the critical point the two saturated phases coincide: vapor and liquid become indistinguishable and the phase boundary seizes to exist. The region of the phase diagram above Pc and Tc is referred to as supercritical fluid. No isotherms or isobars in this region intersect the vapor-liquid boundary. If point E is heated isobarically to final state F, one will observe a continuous transition from a dense, liquid-like state, to a dilute, gaslike state. In the supercritical region the notions of “liquid” and “vapor” are not helpful. These terms are meaningful when both phases can exist simultaneously and can be identified as distinct from each other. The term supercritical fluid avoids these ambiguities.

Properties near the critical point are quite different compared to states at lower temperatures and pressures. As the difference between vapor and liquid becomes less clear near the critical point, the liquid becomes substantially more compressible than typical liquids. This is indicated on the PV graph by the gentle slope of the isotherm as it approaches the critical point. Isotherms below but near the critical temperature (not shown in Figure 2-2) show similar behavior. The usual approximation that treats liquids as incompressible is acceptable only at temperatures well below the critical. In the supercritical region, the behavior of a fluid is somewhere between that of a liquid and a gas. The gentle slope of the isotherms indicates that the fluid is quite compressible, even at high, liquidlike densities (low molar volumes). Other properties, in particular, the solubility of various nonvolatile solutes, are often found to be quite enhanced compared to the subcritical region. As an example, the enhanced solubility of caffeine in supercritical carbon dioxide (Tc = 304.1 K, Pc = 73.8 bar) makes it possible to use carbon dioxide as a solvent to extract caffeine from coffee, thus avoiding the use of other solvents with potential toxic effects.

##### A Special Limit: The Ideal-Gas State

If the molar volume is increased sufficiently, the effect of molecular interactions decreases, and in the limit that it becomes infinite, it vanishes completely. When this condition is met we say that the system is in the ideal-gas state. The mathematical specification of the ideal-gas state is : or, equivalently, The stipulation of constant temperature is necessary. Without it, it would be possible to maintain the system in the liquid (or even solid) phase, even at very low pressures, thus never reaching a state where intermolecular distances are large. On the PV graph, the ideal-gas state is found near the lower-right corner.

In the ideal-gas state, the PVT relationship is universal for all gases, regardless of chemical composition and this relationship is given by the ideal-gas law: where Vig is molar volume, T is absolute temperature, and R is a universal constant (ideal-gas constant), whose value in the SI system is The superscript “ig” will be used to indicate results that are valid only in the ideal-gas state. The ideal-gas law should be viewed as the limiting form of the equation of state of any real fluid when pressure is reduced under constant temperature. Even though the ideal-gas state represents an idealization (infinite distance between molecules), in practice eq. (2.6) provides satisfactory results if the actual state of a gas is sufficiently close to the ideal-gas state. More often than not, engineering problems require calculations at conditions where the ideal-gas law is not valid. It is important to be aware of the limitations of the ideal-gas law and never use it without proper justification.

Note

Directions to the Ideal-Gas State

The specification in eq. (2.5) offers simple directions to the ideal-gas state from any initial state: move along the current isotherm until the phase is vapor and pressure is sufficiently low (or the volume sufficiently large). The pressure does not have to be absolute vacuum as long as it is “sufficiently low.” What constitutes a sufficiently low pressure will be answered in Section 2.4.

##### Two-Phase Region–The Lever Rule

A point E in the two-phase region represents a two-phase system that contains portions of liquid and vapor. Both phases are at the same temperature and pressure but each has its own molar volume. Therefore, the state at E should be viewed as a mixture of states L (saturated liquid) and V (saturated vapor). Line LV that connects the two pure phases is a tie line. Points along the tie line have the same temperature and pressure but differ with respect to the amount of liquid and vapor they contain. A vapor-liquid mixture is also referred to as wet vapor. The amount of vapor as a fraction of the total mass is the quality of the vapor, xV. It varies from 0% (point L, saturated liquid) to 100% (point V, saturated vapor). The mass fraction of the liquid is xL = 1 − xV.

Consider a two-phase system (state E in Figure 2-3) that contains nL moles of liquid and nV of vapor. The molar fractions of vapor and liquid are3

3. For pure components, mass and molar fraction in a liquid/vapor mixture are the same. Figure 2-3: Setup for the application of the lever rule on a tie line. where nV + nL = n is the total mass. The total volume of the two-phase mixture is the sum of the liquid and vapor portions:

Vtot = nLVL + nVVV.

The molar volume is obtained through division by the total number of moles, ntot = nL + nV: This equation gives the molar volume of the two-phase system, if the fractions of vapor and liquid are known. Solving for the liquid and vapor fractions and using xL + xV = 1, we obtain These equations give the fraction of liquid and vapor, if the volume of the two-phase system is known. The two equations in (2.10) are known as the lever rule: if LV is viewed as a lever with force xL acting on point L, force xV on point V, and the pivot placed at E, the lever would be in mechanical equilibrium (see Example 2.1).

Equations (2.9) and (2.10) are applicable on a molar or specific basis with the understanding that both volume and phase fractions are to be expressed on the same basis (molar or specific).

#### The PT Graph

The PT graph is the projection of the three-dimensional surface (Figure 2-1) on the pressure/temperature plane. This graph is shown in Figure 2-4. The vapor-liquid boundary is shown by the line FC and corresponds to the vapor/liquid dome in Figure 2-1 viewed from its side. Line FS marks the solid/liquid boundary. This line is nearly vertical because the melting temperature is not affected strongly by pressure. Unlike the vapor/liquid boundary (FC), which terminates at the critical point, there is no critical point on the solid/liquid boundary. The intersection between the solid/liquid and the liquid/vapor line is the triple point of the fluid. At this point—and only at this point—in the system, all three phases coexist in equilibrium. Most of our applications will be in the liquid and vapor regions, that is, to the right of line FS. Line FC marks the boundary between the vapor phase (to the right) and the liquid (to the left and above). The dotted lines are lines of constant molar volume, also known as isochores. Molar volume increases from left to right, that is, moving on a line of constant pressure from lower temperature to high we encounter increasingly higher molar volumes. Line AB represents a constant-pressure process and corresponds to the same states as in Figure 2-2. Points L and V coincide on this graph (all tie lines are viewed from their edge in this projection). Line LL′ is a line at constant volume V = VL, equal to the volume of saturated liquid. Line VV′ is a line at constant volume V = VV, (saturated vapor). Although the two lines meet on the saturation line, they correspond to different values of the molar volume, as we recall from the PV graph. Figure 2-4: Pressure-temperature graph of pure fluid showing the solid, liquid, vapor and supercritical regions. Dotted lines are lines of constant molar volume.

In addition to marking the phase boundary, line FC expresses the relationship between saturation pressure and temperature. The saturation pressure generally increases quickly with temperature up to the critical point. There is no vapor-liquid transition above the critical point; therefore, the relationship between saturation pressure and temperature exists only below the critical point. The saturation pressure of pure component is an important physical property and a required parameter in many calculations of phase equilibria. Several equations have been developed to describe the mathematical relationship between saturation pressure and temperature. One of the most widely used is the Antoine equation: where A, B, and C are numerical constants specific to the substance. These constants have been obtained by numerical fitting against experimental data and are available for a large number of pure fluids. The Antoine equation is generally very accurate, but this accuracy is guaranteed only within a limited range of temperatures. This range must be indicated in the literature source used to obtain the Antoine constants. A recommended resource for parameters of the Antoine equation is the Properties of Gases and Liquids, by Poling, Prausnitz and O’Connell (5th ed., 2007), which contains extensive tabulations. A more limited tabulation can be found in the online database maintained by NIST.4 The boiling temperature at 1 atm is the normal boiling point. This is a useful property that is often found in tables.

Note

Units in the Antoine Equation

The numerical values of the Antoine parameters depend on the working units of Psat and T, which can vary from source to source. It is important to establish, by checking with the source of the data, what are the proper units of pressure and temperature to be used with a given set of parameters. Also important is to establish whether the logarithm on the left-hand side should be natural or base 10, since parameters may be given for either form of the equation.

### 2.2 Tabulation of Properties

As a state property, the molar (or specific) volume can be determined once as a function of pressure and temperature, and tabulated for future use. Tabulations have been compiled for a large number of pure fluids. In very common use are the steam tables, which contain tabulations of the properties of water. Steam is a basic utility in chemical plants as a heat transfer fluid for cooling or heating, as well as for power generation (pressurized steam), and its properties are needed in many routine calculations. Thermodynamic tables for water are published by the American Society of Mechanical Engineers (ASME) and are available in various forms, printed and electronic. A copy is included in the appendix. We will use them not only because water is involved in many industrial processes but also as a demonstration of how to work with tabulated values in general.

Note

Interpolations

When working with tabulated values it is necessary to perform interpolations if the desired conditions lie between entries in the table. Suppose that a table contains the values of a function f (x) at x = x1 and x = x2 and we wish to obtain the value of f at an intermediate point x such that x1 < x < x2, we assume a linear relationship between f and x and write where f1, f2 are the values of the function at points x1, x2, respectively. The procedure is shown graphically in Figure 2-5a. If the value of x is outside the interval (x1, x2), the same formula may be used and this calculation is called an extrapolation. Extrapolations should be avoided because they can be subject to large error. They may be used if the desired value is beyond the last tabulated entry, however, one cannot be certain about the accuracy of the result. Figure 2-5: Linear interpolation: (a) simple interpolation; (b) double interpolation.

In thermodynamics we are usually dealing with functions of two variables, for example, f (x, y). If point (x, y) is such that the value y is found in the table but the value of x falls between tabulated values, the above equation may be used to interpolate with respect to x. If both x and y fall between tabulated entries, a double interpolation is necessary. The procedure involves three simple interpolations, and can be outlined as follows (Figure 2-5b): first interpolate between the tabulated values at (x1, y1) and (x1, y2) to obtain the value of f at point C; do the same between points (x2, y1) and (x2, y2) to obtain f at point C′. Finally, interpolate between C and C′ to obtain the value at the desired point B. Alternatively, interpolate to obtain points A and A′ followed by interpolation between A and A′ to obtain B—the result is the same. These steps can be combined into a single equation which takes the form: with Here, the notation fij refers to f(xi, yj). If both a and b are between 0 and 1, this calculation is indeed an interpolation and produces a result that is surrounded by the four tabulated values used in the calculation. This equation can be used for extrapolations outside these four values provided the distance is not large.

### 2.3 Compressibility Factor and the ZP Graph

The compressibility factor, Z, is defined as the ratio where V is the molar volume, P is pressure, and T is absolute temperature. It is a dimensionless quantity and a state function. In the ideal-gas state, Vig = RT/P, and the compressibility factor is unity: More precisely, this is the limiting value of the compressibility factor of any real gas when pressure is reduced to zero under constant temperature: This result states that the compressibility factor along a line of constant temperature goes to 1 as P is reduced to zero. In other words, on a graph of Z versus pressure, all isotherms at zero pressure must meet at Z = 1.

The compressibility factor represents an alternative way of presenting molar volume, since the molar volume can be obtained easily if the compressibility factor is known at a given pressure and temperature: Mathematically, the equation of state can be represented as either a relationship between P, V, and T, or between Z, P, and T. The latter relationship is quite useful in presenting the volumetric behavior of fluids. Its graphical representation is given on the ZP graph, whose general form is shown in Figure 2-6. Here again we have the vapor-liquid boundary in the form of bell-shaped curve that is now seated on the vertical axis. The vapor region is at the top, the liquid at the bottom. Generally, compressibility factors in the liquid phase are smaller than those in the vapor phase because the molar volume of the liquid is small. All isotherms meet at P = 0, Z = 1, as anticipated on the basis of eq. (2.18). This point of convergence represents the ideal-gas state. The ZP graph illustrates the path to the ideal-gas state: it is approached by following an isotherm to zero pressure. Since all isotherms converge to this point, the ideal-gas state can be reached from any initial state. We can now see why there is no such a thing as an ideal-gas state: if such a gas existed, its ZP graph would consist of a single horizontal line at Z = 1, which would represent all isotherms. No substance exists that exhibits such behavior. Figure 2-6: The ZP graph of pure fluid.

Although the mathematical definition places the ideal-gas state at a single point (P = 0, Z = 1), from a practical point of view we will consider a gas to be in the ideal-gas state if the compressibility factor is sufficiently close to 1. For calculations that do not require high accuracy we will assume a gas to be in the ideal-gas state if the compressibility factor is within ±5% of the theoretical value of 1. The pressure range over which this approximation is valid varies with temperature.

With reference to Figure 2-6, the isotherm at T4 remains closer to 1 over a wider interval of pressures, compared to the isotherm at T1, which has a larger negative slope and decreases faster. In general, to determine whether a gas at given pressure and temperature can be treated as ideal we must check with a ZP graph. We will return to this question in the next section.

### 2.4 Corresponding States

It is found experimentally that the ZP graphs of different fluids look very similar to each other as if they are scaled versions of a single, universal, graph. This underlying graph is revealed if pressure and temperature are rescaled by appropriate factors. We introduce a set of reduced (dimensionless) variables by scaling pressure and temperature with the corresponding values at the critical point: Using the reduced coordinates it is possible to combine ZP data for several compounds on the same graph. Figure 2-7 shows the resulting graph for selected molecules. The remarkable feature of this graph is that the compressibility factor of the six compounds shown on this figure agree with each other quite well. Even though the dependence of the compressibility factor on temperature and pressure is different for each fluid, they all seem to arise from the same reduced function. This observation gives rise to the correlation of corresponding states, which is expressed as follows:

At the same reduced temperature and pressure, fluids have approximately the same compressibility factor. Figure 2-7: The compressibility factor of selected molecules as a function of reduced pressure and temperature. Data compiled from E. W. Lemmon, M. O. McLinden and D. G. Friend, “Thermophysical Properties of Fluid Systems.” In NIST Chemistry WebBook, NIST Standard Reference Database No. 69, eds. P. J. Linstrom and W. G. Mallard, National Institute of Standards and Technology, Gaithersburg MD, 20899, http://webbook.nist.gov (retrieved December 11, 2010).

We express this mathematically by writing, The practical implication is important: to the extent that eq. (2.20) is obeyed, the compressibility factor of any fluid can be described by a universal equation that is a function of reduced temperature and reduced pressure. Figure 2-7 is a graphical representation of this equation. Accordingly, the compressibility factor of a fluid can be determined at any pressure and temperature using just two parameters, the critical temperature and critical pressure.

The correlation of corresponding states is not an exact physical law and is not obeyed to the same degree of accuracy by all fluids. Although the agreement in Figure 2-7 between different fluids is impressive, it is not exact. The spread of the data points in Figure 2-7 is not due to experimental error (the values have been calculated from validated models) but reflect systematic deviations. These are more clearly seen near the saturation curve and in the liquid region. The correlation arises from similarities in the intermolecular potential. In general, agreement is very good in the gas phase, where interactions are unimportant. In the liquid region, interactions are important and the individual chemical character of molecules becomes more apparent. Even so, nonpolar molecules that are nearly spherical in shape (e.g., Ar, CH4) agree remarkably well. Polar molecules or molecules with more complex structures (e.g., normal heptane) show the largest deviations because their interactions are more complex and dissimilar. Molecules like water, which is both polar and associates strongly via hydrogen bonding, exhibit even more deviations from this correlation. The principle of corresponding states should be treated as a working hypothesis that can provide useful but not always highly accurate estimates of the compressibility factor.

##### Acentric Factor and the Pitzer Method

Eq. (2.20) is a two-parameter correlation because it requires two physical properties, critical temperature and critical pressure. To improve the predictive power of the principle of corresponding state while retaining its simplicity, a third parameter is introduced, the acentric factor. It is a dimensionless parameter that is defined according to the equation, where is the reduced saturation pressure of the fluid at reduced temperature Tr = 0.7. It is a characteristic property of the fluid and is found in tables, usually along with the critical properties of the fluid. It was introduced by Pitzer as a measure of the sphericity of the molecule. More generally, it should be understood as a combined measure of the shape and polarity. Symmetric nonpolar molecules, such as Ar, have ω = 0. These are called simple fluids and are found to obey the two-parameter correlation of corresponding states quite well. Nonspherical or polar molecules have a larger acentric factor. For most fluids the acentric factor is positive and in the range 0 to 0.4, although small negative values of ω are also possible.

In the Pitzer method, the compressibility factor is expressed in the form: where Z(0) and Z(1) are universal functions that depend on Tr and Pr. Function Z(0) represents the compressibility factor of simple fluids (ω = 0) and Z(1) represents a correction that is proportional to the acentric factor. The two functions Z(0) and Z(1) may be calculated once and tabulated against reduced pressure and temperature for future use. The resulting tables and charts are called generalized because they are not limited to a specific molecule.

Various methodologies have been developed for the calculation of the functions in eq. (2.22), but the most widely used is that of Lee and Kesler.5 The Lee-Kesler result for Z(0) is plotted in Figure 2-8. Recall that this term represents the compressibility factor of a simple fluid, therefore, it has the familiar appearance of the ZP graph. Figure 2-9 shows the same graph in semi-logarithmic scales that cover an expanded range of pressures. The Z(1) is shown in Figure 2-10. The correction factor is zero in the vicinity of the ideal-gas state. It increases in absolute value with increasing pressure and it may take positive or negative values. The sharp changes in subcritical temperatures correspond to a shift from the vapor branch to the liquid branch of the isotherm. The vapor/liquid transitions in Figures 2-8, 2-9, and 2-10 apply only to simple fluids (ω = 0). For nonsimple fluids the phase boundary also depends on the acentric factor ω and is generally shifted to lower reduced pressure compared to simple fluids.6

5. B. I. Lee and M. G. Kesler. A generalized thermodynamic correlation based on three-parameter corresponding states. AIChE J., 21(3):510, 527 1975. doi: 10.1002/aic.690210313.

6. The determination of the precise location of the phase boundary is discussed in Chapter 7. Figure 2-8: Generalized graph of Z(0) based on the Lee-Kesler method. Figure 2-9: Generalized graph of Z(0) based on the Lee-Kesler calculation (extended range of pressures). Figure 2-10: Generalized graph of Z(1) based on the Lee-Kesler calculation.

Note

Simple, Normal, and Polar Fluids

Noble gases such as Ar, Kr, Xe, and other spherical, nonpolar molecules such as CH4, interact through similar potential that is spherically symmetric and which can be described through a combination of van der Waals attraction and hard-core repulsion. These fluids are called simple. Their acentric factor is zero or nearly zero, and they are described accurately by the two-parameter correlation of corresponding states. A notable exception is the group of quantum gases, He, Ne, and H2, whose behavior at low temperatures is dominated by quantum effects.

Nonpolar molecules that deviate from spherical shape and small molecules of only moderate polarity are described quite well by the three-constant correlation that utilizes the acentric factor. The acentric factor, therefore, can be viewed as an empirical parameter that encompasses the effect of molecular shape, and to some extent polarity. Molecules in this category constitute the class of normal fluids, and include small molecules such as O2, N2, as well many lower hydrocarbons.

Molecules that are strongly polar or associate strongly (e.g., via hydrogen bonding) are problematic when it comes to predicting their properties using correlations with a small number of parameters. The difficulty arises from the fact that the intermolecular potential of polar and associating molecules is fairly complex: it depends on the relative orientation of molecules and cannot be represented by a small number of parameters. Corresponding-states correlations are generally not accurate for such molecules except in the gas phase and away from the phase boundary.

As a rule, generalized methods, that is, equations or graphs and tables that depend on critical temperature, critical pressure and acentric factor, should be applied to simple normal fluids.

### 2.5 Virial Equation

All isotherms on the ZP graph begin at P = 0, Z = 1, and grow outward with pressure. This behavior can be described mathematically as power series in P: where b, c, ... are coefficients that are characteristic of the isotherm, in other words, for a given fluid they depend only on temperature. With P = 0, we obtain Z = 1, that is, this equation gives the proper result in the ideal-gas state. In the vicinity of the ideal-gas state, the most significant term in the series is the linear term. In this region isotherms are fairly linear in P. Upon further increasing pressure, first the quadratic and then other higher-order terms become increasingly more important and isotherms are no longer linear. If instead of P and T we select V and T as the state variables, a similar series expansion can be constructed in terms of volume: Here the expansion is done in powers of inverse volume, which guarantees that in the ideal-gas state (V = ∞) the compressibility factor is Z = 1. This series expansion is known as the virial equation and its coefficients as the virial coefficients: B is the second virial coefficient, C is the third, and so on. These are characteristic of the fluid and depend on temperature only.

Eqs. (2.23) and (2.24) are different expansions (one is in P, the other in V) and their coefficients are different. They are, however, related. The relationship for the first two coefficients is given by the following equations  As a matter of nomenclature, only eq. (2.24) is known as “virial,” and its coefficients as “virial coefficients,” whereas eq. (2.23) is referred to as the pressure expansion of the compressibility factor.

Note

Second Virial Coefficient and Intermolecular Potential

The second virial coefficient is related to the intermolecular potential according to the equation, where Φ(r) is the intermolecular potential, k is the Boltzmann factor, T is absolute temperature, and the integral is evaluated by letting the distance r between two molecules range from 0 to ∞. If the intermolecular potential is weak relative to kT, the ratio Φ(r)/kT is small, the exponential term is nearly equal to 1, and the second virial coefficient is nearly zero. Under these conditions, deviations from ideal-gas behavior are small. The magnitude of B increases when the potential is strong relative to temperature, a condition that leads to stronger deviations from ideal-gas behavior.

##### Truncated Virial Equation

If all the coefficients of the expansion were known, the virial equation would provide complete description of the vapor isotherm up to the point it intersects the saturated vapor line. In practice, only the second virial coefficient is widely available and as a result, this equation is most often used in truncated form by dropping terms that include higher coefficients. Either series can be used, but eq. (2.23) is more convenient because it expresses the compressibility factor in terms of pressure and temperature. Dropping the quadratic and higher terms in eq. (2.23), the compressibility factor becomes This equation is a linear approximation of the isotherm and is valid in the region near P = 0 that the actual isotherm is linear in pressure. Eq. (2.27) should be considered as a first-order correction to the ideal-gas law that provides us with the equation of state at pressures that are low but not sufficiently so to assume ideal-gas behavior. The range of pressures that satisfy this condition depends on temperature. Figure 2-8 may be used as a guide: Eq. (2.27) is to be used only within a pressure interval from 0 to pressure P such that the isotherm in this interval is a straight line. This pressure interval is narrow at lower temperatures but expands with increasing temperature.

As we can verify by checking with a ZP graph, there is always a neighborhood of pressures near P = 0 where linearity is observed. We also see that the linear range depends on temperature and the linear interval generally increases with increasing temperature. We refer to the linear range of an isotherm as a region of moderate nonideality because the second virial coefficient suffices to describe deviations from ideality. Writing Z = PV/RT and with some rearrangement, eq. (2.27) becomes The truncated virial equation provides us with a simplified equation of state in the vicinity of the ideal-gas state where the ideal-gas law alone is not sufficiently accurate.

##### Pitzer Method for the Second Virial Coefficient

A useful empirical correlation for the second virial coefficient is based on the Pitzer method. In this method, the dimensionless ratio, BPc/RTc is expressed in the form where  Since the second virial coefficient is a function of temperature only, the coefficients B0 and B1 are independent of Pr and functions of Tr, only. These equations can be used to estimate the second virial coefficient if other data are not available. Figure 2-11: Calculated isotherms of ethylene at 40 °C using the ideal-gas law, the truncated virial equation, and the Pitzer method with the Lee-Kesler values of Z(0) and Z(1) (see Example 2.9).

### 2.6 Cubic Equations of State

For engineering calculations it is important to have equations of state that are accurate over a wide range of pressures and temperatures. The ideal gas law is very simple to use, but its validity is restricted to gases at low pressures. The truncated virial equation is applicable over a somewhat wider range of pressures, but only for gases. If the pressure is high or the phase liquid, neither of these equations can be used. Numerous empirical equations of state have appeared in the literature to overcome these difficulties. Such equations usually have some theoretical basis, but the primary consideration is sufficient accuracy for engineering applications. It is typical for these equations to contain parameters that are fine-tuned to improve accuracy. No single mathematical equation of state can describe all fluids. Nonetheless, it is convenient in having one equation whose mathematical form is the same for many fluids but with parameters that are specific to a particular fluid. Among the most important engineering equations of state is the family of cubic equations, which can be viewed as variants of the van der Waals equation of state.

##### Van der Waals Equation of State

The van der Waals equation of state is given by where a and b are parameters specific to the fluid and are given in terms of the critical pressure, Pc, and critical temperature, Tc, of the fluid:  The equation can be rearranged in the form: and upon taking the limit V → ∞, we obtain This proves that the van der Waals equation has the correct behavior in the ideal-gas limit.

The van der Waals equation can be rearranged to obtain the compressibility factor in terms of pressure and temperature. Starting with eq. (2.35) and using V = ZRT/P to eliminate V, we obtain the following expression for Z: where A′ and B′ are defined for convenience as Rearranging to solve for Z, the previous result takes the form The equation for the compressibility factor is a cubic polynomial in Z. This is the reason that the van der Waals equation is referred to as a cubic equation of state. The parameters A′ and B′ depend on pressure and temperature. If T and P are given, the compressibility factor can be calculated from eq. (2.37). As a cubic polynomial, this equation may have one or three real roots. Multiplicity of roots is a characteristic property of equations of state that are capable of describing phase transitions.

The predictions of the van der Waals equation are generally in good qualitative agreement with experimental data, but the accuracy is not sufficiently high for routine engineering calculations. Moreover, the equation makes use of only two physical parameters, critical pressure and critical temperature, and as we saw in Section 2.4, these are not enough to provide accurate estimates of the compressibility factor. This has motivated the development of variants of the van der Waals equation with increased accuracy. Two of the most commonly used equations in current practice are the Soave-Redlich-Kwong (SRK) and the Peng-Robinson (PR) equations of state.

Note

van der Waals and His Equation

The van der Waals equation was the first equation proposed that is capable of predicting the phase behavior of vapors and liquids. It was developed by van der Waals in his doctoral thesis, in 1873. This equation introduces in an approximate but effective way the effect of molecular repulsion and attraction, an element that is important in describing phase transformation, as discussed in Section 1.1. Repulsion is represented by the term RT /(Vb), which ensures that at high pressures the molar volume approaches a constant value b. This effectively accounts for the strong repulsion at close molecular distances, an interaction that gives molecules the appearance of possessing a solid core. Attraction is represented by the term −a/V2. This term is negative because the effect of attractive forces is to pull molecules together and reduce the pressure of the fluid. The factor V2 in the denominator has its origin in the van der Waals force, which is the typical form of molecular attraction. The simultaneous presence of terms representing attraction and repulsion endows this equation with the ability to describe both the vapor and the liquid region of the phase diagram. In 1910, van der Waals received the Nobel Prize in Physics.

Soave-Redlich-Kwong (SRK) Equation of State The Soave-Redlich-Kwong equation of state is given by where a and b are parameters specific to the fluid and are given in terms of the critical pressure, critical temperature, and acentric factor:   Here, Tr = T/Tc is the reduced temperature. The equation for the compressibility factor is with the parameters A′ and B′ defined in eq. (2.36).

Peng-Robinson (PR) Equation of State The Peng-Robinson equation of state is given by The parameters a and b are related to the critical pressure, critical temperature and acentric factor as follows:   where Tr = T/Tc is the reduced temperature. The equation for Z is with the A′ and B′ given by eq. (2.36).

### 2.7 PVT Behavior of Cubic Equations of State

To generate a pressure-volume isotherm, temperature is fixed and the pressure is calculated for various volumes. We demonstrate the behavior of isotherms in Figure 2-12 using the Soave-Redlich-Kwong equation for ethylene (Tc = 282.35 K, Pc = 50.418 bar, ω = 0.0866). All volumes lie to the right of b, which for ethylene has the value 4.03395 m3/mol. The isotherm at T = 340 K has the general shape of supercritical isotherms and decreases gently as volume is increased. The critical isotherm (Tc = 282.35 K) exhibits an inflection point, which occurs at the critical pressure. At this point the first and second derivative of pressure with respect to volume are both zero. The subcritical isotherm (T = 260 K) has unusual behavior: it decreases steeply in the liquid region, as expected, but then is goes through a minimum and a maximum before it emerges in the vapor region, where it resumes the expected behavior. The “hump,” shown by the dashed line in Figure 2-12, represents metastable and unstable parts of the isotherm and must be removed. To do this, we draw a horizontal line at the saturation pressure at 260 K, which from tables is found to be 30.25 bar. The intersection of this line with the liquid branch of the isotherm defines the saturated liquid (point L) and its intersection with the vapor branch defines the saturated vapor (point V). The corrected isotherm is obtained by drawing the tie line between points L and V and removing the dashed part of the isotherm. Figure 2-12: Isotherms of ethylene calculated by the Soave-Redlich-Kwong equation.

##### Unstable and Metastable Parts of Subcritical Isotherm

Along the portion of the isotherm between points L′ (local minimum of the isotherm) and V′ (local maximum) the isotherm has a positive slope, which implies that volume increases with increasing pressure. This unphysical behavior makes this part of the isotherm mechanically unstable. Suppose that the system is brought to a point between L′ and V′. If we increase the external pressure by a small amount, the volume contracts and the state moves to the left. This causes the pressure of the fluid to decrease and the volume to contract even more, as the external pressure remains unchanged. This unstable behavior continues until the system reaches point L′. The branches LL′ and V′V are mechanically stable but thermodynamically metastable. In LL′ the liquid is supersaturated, meaning that it is in the liquid phase, even though its pressure is lower than the saturation pressure. The stable phase is vapor, but the metastable liquid state will be maintained for some time, if care is taken to avoid contaminants and surface defects on the container walls, which can act as nucleation sites. Eventually, thermal fluctuations will cause a phase change to take place from the supersaturated phase to the one that is most stable. The situation is analogous along the metastable vapor part, VV′.

Note

Choosing Among Multiple Roots of Cubic Equations of State

When working with cubic equations of state it is important to remember that subcritical isotherms obtained by direct calculation of the equation of state include the metastable/unstable loop. This requires special attention when solving for V at temperatures below the critical. Suppose we seek the molar volume at specified pressure P and temperature T. The solution is graphically given on the PV graph by the intersection between the isotherm at T and the horizontal line drawn at pressure P. As we see in Figure 2-12, it is possible to have one, or three points of intersection, each of them corresponding to a molar volume that satisfies the equation of state at the given pressure and temperature. If there is only one intersection point, there is no ambiguity and this point defines the unknown volume. If there are three intersections only one (or two, if the system is saturated) is physically acceptable. To determine the correct volume we must determine the saturation pressure at the given temperature, T. This can be obtained from tables or using the Antoine equation. The proper volume is now selected as follows:

1. If pressure is higher than the saturation pressure, the phase is compressed liquid: select the smallest root.

2. If pressure is lower than the saturation pressure, the phase is superheated vapor: select the largest root.

3. If pressure is equal to the saturation pressure, then the system is saturated: select the smallest root to be that of the saturated liquid, and the largest root to be that of the saturated vapor.

The middle root is never selected. Cubic equations of state may also have spurious roots for the volume to the left of b. These roots are not physical and should be rejected. Acceptable roots must be positive and larger than b. In terms of the compressibility factor, acceptable roots must be positive and larger than Pb/RT.

### 2.8 Working with Cubic Equations

The typical problem involving cubic equations of state asks for the molar volume at given pressure and temperature. This volume can be calculated by solving the equation of state for V. Alternatively, it can be calculated by solving the cubic equation (see eq. [2.44]) for Z. The latter method is recommended for a number of reasons. The equation for Z is dimensionless; this reduces the chances of errors involving units. The values of Z generally range from 0 to about 1; by contrast, molar volume spans several orders of magnitude. Finally, as a cubic equation, its roots can be computed using standard formulas (see below). Alternatively, the roots can be obtained by iterative numerical procedures. Numerical methods require suitable starting guesses. For the gas-phase root, a starting guess is Z = 1; for the liquid-phase root, the starting guess is Z = B = Pb/RT. Nonetheless, there is no guarantee that these guesses will converge to the proper roots, therefore, the results of iterative numerical procedures must always be checked.

Note

Equations for the Roots of Cubic Equation

Here is the analytic procedure for the calculation of the real roots of a cubic polynomial. The equations shown here are taken from the book Numerical Recipes:7 but they can be found in many other books, including Perry’s Chemical Engineers’ Handbook.

7. W. H. Press, S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery. Numerical Recipes in Fortran, 2nd ed. (New York: Cambridge University Press, 1992), p. 179.

1. Given the cubic polynomial with real coefficients

x3 + ax2 + bx + c = 0

first compute Q and R as shown below: 2. If R2 < Q3, the polynomial has three real roots. Calculate them as follows: where The roots as obtained here are not necessarily sorted.

3. If R2 > Q3, the polynomial has only one real root. Calculate the parameters A and B as follows: and where sign(x) is the function that returns the sign of x. The real root is Since the complex roots are of no interest to us, we do not provide the formulas for their calculation.

Notice that the various parameters defined here (for example, a, R, etc.) are not related to thermodynamic quantities with the same symbol defined elsewhere in the book.

### 2.9 Other Equations of State

Cubic equations of state are useful because they are capable of handling both liquids and gases, but they are not the only type of equations that have this feature. The mathematical requirement is that subcritical isotherms must exhibit the characteristic h-shaped portion, with an unstable part between the liquid and vapor branches. This behavior can be reproduced by other equations that are not necessarily cubic in Z. One that is worth mentioning is the Benedict-Webb-Rubin equation (BWR), which has the form, It requires eight parameters (A0, B0, C0, a, b, c, α, and γ) that are specific to the fluid. The Benedict-Webb-Rubin equation is modeled after the virial equation and expresses pressure as a finite sum of powers of 1/V, up to the sixth power. The exponential term on the right is meant to account for the higher terms of the series that have been dropped. A modified form of this equation was used by Lee and Kesler8 in the calculation of Z(0) and Z(1). This equation is not cubic but its subcritical isotherms have the same general behavior as those in Figure 2-12, namely, they exhibit an unstable part where the isotherm has a positive slope.

8. B. I. Lee and M. G. Kesler. A generalized thermodynamic correlation based on three-parameter corresponding states, AIChE J., 21(3):510, 527 1975, http://dx.doi.org/10.1002/aic.690210313. Figure 2-13: Isotherms of the modified BWR equation for the simple and reference fluid at Tr = 0.95 (see Example 2.12).

Table 2-1: Constants of the modified BWR used in the Lee-Kesler correlation (see Example 2.12). Note

Another Look at the Pitzer Correlation

Using eq. 2.51 in the Pitzer equation, the compressibility factor is expressed in the form where ωref = 0.3978. With ω = 0, this gives Z(0); that is, Z(0) is the compressibility factor of a simple fluid, as we have already seen. Setting ω = ωref we obtain

Z = Z ref,

that is, Zref is the compressibility factor of a fluid with acentric factor ω = ωref. We may now interpret the Pitzer correlation as a linear interpolation between the compressibility factor of a simple fluid (ω = 0) and that of the reference fluid (ωref). In the Lee-Kesler method, the reference fluid is octane (ωref = 0.3978).

### 2.10 Thermal Expansion and Isothermal Compression

The effect of temperature and pressure on volume is quantified by two coefficients, the volumetric coefficient of thermal expansion, and the coefficient of isothermal compression. The volumetric coefficient of thermal expansion is defined as and gives the fractional change of volume per unit change of temperature under constant pressure. Its units in the SI system are K−1. The coefficient of isothermal compressibility is defined as9

9. Not to be confused with the compressibility factor, Z. and gives the fractional change of volume under pressure at constant temperature. Since volume decreases with increasing pressure, the negative sign is used to ensure that κ is a positive number. Using these definitions, the equation of state can be expressed in a alternative form. First, we write the differential of V in terms of P and T: Using eqs. (2.52) and (2.53) to eliminate the partial derivatives that appear in eq. (2.54), this equation becomes If the coefficients β and κ are known, this equation can be integrated to calculate changes in V.

These coefficients are themselves functions of pressure and temperature, but for liquids, the effect of pressure is quite weak; we may take them to depend on temperature only. The coefficient of isothermal compressibility expresses the degree to which the volume of a liquid responds to pressure. Since liquids at temperatures well below the critical are nearly incompressible, the coefficient of isothermal compressibility is approximately zero. The coefficient of thermal expansion is usually small, as liquids expand much less than gases, but it is not zero. Perry’s Chemical Engineers’ Handbook provides data on the thermal expansion of selected liquids.

### 2.11 Empirical Equations for Density

#### Rackett Equation

A useful and accurate method for the calculation of liquid molar volumes at saturation is the Rackett equation. In its modified form, it gives the molar volume of saturated liquid as where VL is the molar volume at saturation, Tc is the critical temperature, Pc is the critical pressure, Tr = T/Tc is the reduced temperature and ZR is a parameter specific to the fluid. Perry’s Chemical Engineers’ Handbook  lists the values for some common fluids10. If ZR is not available, it can be replaced by the compressibility factor at the critical point, Zc = PcVc/RTc, which is usually tabulated. If we use the critical compressibility factor in eq. (2.56), the Rackett equation is simplified and takes the form

10. Table 2-396 in Ref. . where Vc is the critical volume. The Rackett equation is empirical but quite accurate, even in the above simplified form, which contains no adjustable parameters.

Note

Molar Volume of Compressed Liquid

At temperatures well below critical, liquids are practically incompressible and we may assume κ ≈ 0. In this approximation, an isotherm on the PV graph is almost perpendicular to the volume axis. Accordingly, the molar volume of a compressed liquid is essentially the same as that of the saturated liquid at the same temperature. The volume of saturated liquid is often tabulated, or it can be calculated from empirical equations such as the Rackett equation. Therefore, the volume of compressed liquid can be estimated quite accurately from the volume of the saturated liquid at the same temperature. This approximation breaks down close to the critical point where the liquid phase becomes quite compressible.

### 2.12 Summary

The equation of state is a fundamental property of fluids that describes the relationship between molar volume, temperature, and pressure. This relationship is expressed mathematically as an equation between V, P, and T, or alternatively, between Z, P, and T. The PV graph is a graphical representation of this relationship and a convenient way to present the phase behavior of a pure component. It is a good idea to draw a qualitative PV graph when solving problems involving heating, cooling, compression, or expansion of pure fluids, as a way of becoming oriented in thermodynamic space. Simple processes on this graph is represented by simple paths and help visualize any phase transformations that may occur.

The accurate prediction of the molar volume, or, equivalently, of the compressibility factor, is a requirement in engineering calculations. Several methodologies were discussed in this chapter:

Tabulated values. Tabulations are available for a large number of pure components. The steam tables is one such example. Usually, however, property tabulations are not as detailed or extensive as the steam tables. Tables are generally the most accurate source for properties. They are not convenient, however, for large-scale calculations.

Generalized graphs. These graphs are correlations based on corresponding states. Using just three physical constants, critical pressure and temperature, and acentric factor, one can estimate the molar volume of a pure fluid over a very wide range of conditions. This is a very important advantage but it comes with certain limitations. The method is approximate and is based on graphs that have been tweaked to provide overall good accuracy for several different fluids. Necessarily, the agreement will be better for some and worse for others. The method should be used only for normal fluids that are not polar to a significant degree. This covers a large number of compounds but excludes many industrially important molecules, chiefly among them, water.

Equations of state. An equation of state in mathematical form has the advantage that it can be used in repetitive calculations and is especially suited for computer-based calculations. Industrial software for chemical process design makes extensive use of such equations. The equations discussed in this chapter (van der Waals, Soave-Redlich-Kwong, and Peng-Robinson) incorporate the principle of corresponding states in their constants, which can be computed for any fluid given the critical pressure, critical temperature, and acentric factor. These equations must be used with the same caution as generalized graphs: they should be applied to nonpolar fluids only. The constants of these equations, specifically the part that involves the acentric factor, have been obtained from fitting data for several different fluids to obtain overall good agreement among them. It is possible to modify the constants of an equation of state to improve accuracy for a particular substance, however, the constants given here are specifically for normal fluids.

Of the many equations that were discussed here, two deserve a special note:

Ideal-gas law. The ideal-gas law is the theoretical limit of any equation of state at low pressure. It is a universal limit but its range of applicability is limited to the region we call ideal-gas state. You should resist the temptation to apply the ideal-gas law without justification.

Virial equation. The virial equation has rigorous basis on theory. However, its truncated form, it should be treated as an extrapolation to somewhat higher pressures that those in the ideal-gas state.

Empirical equations. Empirical equations usually have no basis on theory but give nonetheless very accurate predictions. The Rackett equation is one such example. The limitation is that the range of applicability is typically narrow. The Rackett equation, for example, is only good along the line of saturated liquid.

The availability of several alternative methodologies is a toolbox that facilitates the job of the engineer. It is the responsibility of the engineer to choose the right tool from this toolbox for a given situation.

### 2.13 Problems

Problem 2.1: a) Use the steam tables to determine the phase of water (liquid or vapor) at following conditions: 25 °C, 1 bar; 80 °C, 10 bar; 120 °C; 50 bar.

b) The vapor pressure of bromobenzene at 40 °C is 10 mm Hg. Determine the phase of bromobenzene at 40 °C, 1 atm.

c) The normal boiling point of fluorobenzene is 84.7 °C. What is the phase of fluorobenzene at 25 °C, 1 atm?

Problem 2.2: Determine the temperature and phase of water from the following information. If the phase is a vapor-liquid mixture, report the fraction of vapor and liquid.

a) The specific volume of water is 100 cm3/g and the pressure 40 bar.

b) The specific volume of water is 100 cm3/g and the pressure 6 bar.

Problem 2.3: Use the steam tables to do the following:

a) A drum 3.5 m3 in volume contains steam at 1 bar, 210 °C. Determine the mass of steam in the drum.

b) Wet steam with quality 15% vapor is to be stored under pressure at 20 bar in a thermally insulated vessel. What is the temperature?

c) If the total mass to be stored is 525 kg, what is the required volume of the vessel?

The conditions are the same as in part (b).

Problem 2.4: A closed tank contains a vapor-liquid mixture of steam at 45 bar with liquid content 25% by mass.

a) Heat is added until the pressure becomes 80 bar. What is the temperature?

b) The pressure is changed until the contents become 100% saturated vapor. What is that pressure?

c) With pressure held constant at 45 bar, steam is added to or removed from the tank, as needed, until the contents are 25% liquid by volume. Calculate the mass of steam that must be added or removed and report it as a percentage of the total mass originally in the tank.

Problem 2.5: The following data are available for a gas: To calculate the molar volume at 25 °C, 12 bar, our team came up with two different suggestions: (a) linear interpolation for V between the given pressure, or (b) linear interpolation for the density ρ = 1/V between the two pressures. Which method do you recommend and why? Under what conditions is your recommendation accurate?

Problem 2.6: A 12 m3 tank contains a liquid/vapor mixture of steam at 15 bar. The volume of the liquid in the tank is 0.5 m3.

a) What is the temperature?

b) What is the total mass in the tank?

c) What is quality of the steam?

d) 87% of the mass in the tank is removed while keeping pressure constant at 15 bar. What is the final temperature in the tank?

Problem 2.7: An eight-liter pressure cooker contains a mixture of steam and liquid water at 2 bar. Through a level indicator we can see that the liquid occupies 25% of the volume inside the cooker.

a) What is the temperature inside the cooker?

b) What is the total mass of water (liquid plus vapor) in the cooker?

c) What is the mass fraction of the liquid?

d) The cooker, while it remains sealed, is placed under running water until its temperature cools to 25 °C. What is the pressure in the cooker?

e) What force does it take to unseal the cooker? The cover is circular with a radius of 20 cm.

Problem 2.8: a) A pressure cooker is filled to the brim with water at 80 °Cand the lid is locked. The temperature is then changed until the contents become saturated liquid. What is the temperature and pressure at that point?

b) A closed pressure cooker contains 50% by volume liquid and 50% water vapor at 1 bar. The pressure is then changed until the point where the contents become a single phase. Is that phase saturated liquid or saturated vapor?

c) A closed rigid vessel that contains a pure fluid is cooled until the contents become saturated vapor. Determine whether the initial state is superheated vapor, compressed liquid, or vapor/liquid.

d) A closed rigid vessel that contains a pure fluid is heated until the contents become saturated liquid. Determine whether the initial state is superheated vapor, compressed liquid, or vapor/liquid.

Problem 2.9: A tank whose volume is 12 m3 contains 6.2 kg of water at 1.4 bar.

a) What is the phase (liquid, vapor, liquid/vapor mixture)?

b) What is the temperature?

c) We add more steam to the tank while maintaining its temperature constant at the value calculated in part b. As a result, the pressure in the tank increases. Determine how much water (in kg) of steam must be added to bring the steam in the tank to the point of condensation.

d) Draw a qualitative PV graph and show the path of the process for part c.

Problem 2.10: a) Use data from the steam tables to construct the PV graph of water. Show the saturated liquid, the saturated vapor, the critical point. Include the isotherms at 100 °C, 200 °C, 300 °C and 400 °C. Make two plots, one using linear axes and one in which the pressure axis is linear but the volume axis is logarithmic.

b) Make a ZP plot of water using data from the steam tables showing the same information as the PV graph above.

Problem 2.11: Determine whether the truncated virial equation is valid for ethane at the following states:

a) 10 bar, 25 °C.

b) Saturated vapor at 10 bar.

c) 10 bar, −35 °C.

Additional data: The boiling point of ethane at 10 bar is −29 °C.

Problem 2.12: The R&D division of your company has released the following limited data on proprietary compound X-23: Using this incomplete information estimate as best as you can the following:

a) Phase of X-23 at 12 bar, 25 °C.

b) The molar mass of X-23.

c) The second virial coefficient at 25 °C.

d) The required volume of a tank that is needed to store 20 kg of X-23 at 12 bar, 25 C.

e) State clearly and justify as best as you can all your assumptions and the methods you use.

Problem 2.13: Methane is stored under pressure in a 1 m3 tank. The pressure in the tank is 20 bar and the temperature is 25 °C.

a) Calculate the compressibility factor of methane in the tank from the virial equation truncated after the second term.

b) What is the amount (moles) of methane in the tank?

c) You want to store twice as much methane in the tank at the same temperature. What will be the pressure in the tank?

d) Is it appropriate to use the virial equation for this problem? Explain.

At 25 °C the second virial coefficient of methane is −4.22 × 10−5 m3/mol.

Problem 2.14: Use the truncated virial equation to answer the questions below:

a) A 5 m3 tank contains nitrogen at 110 K, 7 bar. How many kg of nitrogen are in the tank?

b) The tank is cooled until the contents become saturated vapor. What is the pressure and temperature in the tank?

c) Is the use of the truncated virial equation justified in this problem? Additional data: The saturation pressure of nitrogen is given by the following empirical equation: with Psat is in mm Hg and T is in kelvin.

Problem 2.15: a) Calculate the second virial coefficient of water at 200 °C using only data from the steam tables.

b) Use the truncated virial equation along with the second virial coefficient calculated above to estimate the volume of water at 200 °C, 14 bar. How does this value compare with the volume obtained from the steam tables? Discuss this comparison.

Problem 2.16: 1000 kg of methane is to be stored in a tank at 25 °C, 75 bar.

What is the required volume of the tank? (Hint: Use the Pitzer equation with the Lee-Kesler values.)

Problem 2.17: 200 kg of carbon dioxide are stored in a tank at 25 °C and 70 bar.

a) Is carbon dioxide an ideal gas under the conditions in the tank?

b) What is the volume of the tank?

c) How much carbon dioxide must be removed for the pressure of the tank to fall to1bar?

The molecular weight of CO2 is 44 g/mol.

Problem 2.18: A full cylinder of ethylene (C2H4) at 25 °C contains 50 kg of gas at 80 bar.

a) Is ethylene an ideal gas under these conditions? Explain.

b) What is the volume of the cylinder?

c) What is the pressure in the cylinder after 90% of the ethylene has been removed, if temperature is 25 °C?

The molecular weight of ethylene is 28 g/mol.

Problem 2.19: Use the Lee-Kesler method to answer the following: 2000 kg of krypton is to be stored under pressure in a tank at 110 bar, 20 °C. The tank is designed to withstand pressures up to 180 bar.

a) Determine the volume of the tank.

b) Is it safe to store 2500 kg in the tank at 25 °C?

c) Is the Lee-Kesler method appropriate?

Problem 2.20: A tank is divided by a rigid, thermally conducting partition into two equal parts, A and B, each 10 m3 in volume. Part A contains saturated liquid n-butane at 20 °C, 2.07 bar; part B contains saturated n-butane vapor, also at 20 °C, 2.07 bar. Each part is equipped with a safety alarm that will go off if pressure exceeds 40 bar.

a) How many moles of n-butane is in part A of the tank?

b) How many moles of n-butane is in part B of the tank?

c) The tank is heated slowly in such a way that the temperature in both parts rises at the same rate. As soon as the alarm goes off, the heating stops. Which alarm goes off, that of part A or part B?

d) What is the temperature when the alarm sounds?

Additional data: The volume expansivity and the isothermal compressibility of liquid n-butane are given below and may be assumed to be constant:

β = 2.54 × 10−3 K−1, κ = 3.4 × 10−4 bar−1.

Problem 2.21: a) A tank contains 10,000 kg of xenon at 132 °C, 82 bar. The plant supervisor asks you to remove xenon and fill the tank with 10,000 kg of steam at 200 °C. What is the pressure in the tank when it is filled with steam?

b) After the tank has been filled with steam, 5000 kg are withdrawn for use elsewhere in the plant. What is the pressure, if temperature remains at 200 °C?

Problem 2.22: The boiling point of o-xylene at 1 bar is 139 °C.

a) What is the state of o-xylene at 0.1 bar, 200 °C?

b) 100 moles of o-xylene are to be loaded in a tank at 0.1 bar, 200 °C. What is the required volume of the tank?

c) The tank can safely withstand pressures up to 44.9 bar. How much o-xylene can be stored in the tank under maximum pressure at 200 °C?

Problem 2.23: a) Determine the percent change in volume when olive oil is heated at constant pressure from 18 °C to 40 °C.

b) Olive oil is stored in a full container at 18 °C. Determine the pressure that will develop in the container if the temperature in the storage room rises to 40 °C.

Additional data: The volume expansion of olive oil is given by the empirical equation

V = V0(1 + a1t + a2t2 + a3t3)

where t is temperature in °C, V0 is the volume at 0 °C, and the coefficients in the equation are The coefficient of isothermal compression is

κ = 52 × 10−6 bar−1.

(Data from Perry’s Chemical Engineers’ Handbook, 7th ed., Tables 2-147 and 2-188.)

Problem 2.24: A 0.5 m3 tank will be used to store CO2 at 20 °C. Using the SRK equation answer the following:

a) Determine the maximum amount (kg) of CO2 that can be stored safely if the tank can withstand a maximum pressure of 70 bar.

b) Repeat if the maximum pressure is 60 bar.

c) Repeat at 50 bar.

Problem 2.25: Use the SRK equation to answer the following:

a) 5000 kg of isobutane is to be stored in a tank at 60 psi, 70°F. What is the required tank volume?

b) Since the temperature in the summer can get as high as 95°F, determine the pressure that the tank must withstand to avoid rupture.

Problem 2.26: Use the SRK equation to calculate the molar volume of isobutane at the following states:

a) 30 °C, 1 bar.

b) 30 °C, 10 bar.

c) Saturated liquid at 30 °C.

d) Saturated vapor at 30 °C.

For parts c and d, compare with the saturated molar volumes reported in the NIST Web Book.

Additional information: The saturation pressure at 30 °C is 4.05 bar.

Problem 2.27: Use the SRK equation to perform the following calculations for isobutane to make a graph that shows three isotherms, one at 30 °C, one at the critical temperature, and one at 150 °C. Make the axis logarithmic in volume and linear in pressure, and select the range in the two axes so that the graph is not crowded and its important features are seen clearly. Report volume in m3/mol, pressure in bar and temperature in °C. Annotate the graph and label the axes properly.

Problem 2.28: The parameters β and κ of a substance are reported to be functions of pressure and temperature and are given below:

β = 1 /T, κ = 1 /P.

a) Determine the equation of state. Assume that at pressure P0 and temperature T0 the volume is known to be V0. Your final answer then should be in terms of P, V, T, P0, T0 and V0.

b) Is this equation appropriate for liquids?

Problem 2.29: Calculate the coefficient of isothermal compressibility of isobutane as saturated liquid and saturated vapor at 30 °C, Psat = 4.05 bar, using the SRK equation. Report the result in bar−1.

Hint: Recall from calculus that Problem 2.30: Use the steam tables to calculate the value of β of water at the following states:

a) 1 bar, 25 °C.

b) 20 bar, 25 °C.

c) 1 bar, 200 °C.

Problem 2.31: Use the Rackett equation to estimate the volume expansivity of liquid ethanol at 25 °C. State your assumptions clearly.

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