# Properties of pure substances

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# 7 Thermodynamic Surfaces and Equations of State

## 7.1 Thermodynamic Surfaces of a Single-Component Substance

Thermodynamic surfaces are three-dimensional diagrams that describe every equilibrium point of a pure substance, including the vapor, liquid, and solid phases. Two types of thermodynamic surfaces – pressure-volume-temperature (pvT) surfaces and temperature-entropy-pressure (Tsp) surfaces – are discussed here. Figure 11 shows a typical stable equilibrium pvT surface for a pure substance that contracts upon freezing. The primary point of interest on this surface is the critical point, which describes the equilibrium point of the substance where the saturated-liquid and saturated-vapor states are identical. This point exists at the inflection point of the two-phase hump that delineates the phase change from liquid to vapor. The phase change from solid to liquid (melting) is also described by the narrow surface between the solid and liquid phases. One last interesting feature of the pvT surface is the surface that shows sublimation, solid-to-vapor phase change. Sublimation occurs if the pressure of the substance at a constant temperature is low enough. The solid-vapor surface exists below the narrow solid-liquid region and the liquid-vapor two-phase hump. All of these phenomena are more easily observed on the phase diagram that will be described in Section 2.8.2. It should be noted that the thermodynamic surface in Fig. 11 describes a pure substance that contracts upon freezing. For a substance that expands upon freezing, such as water-ice, the solid-liquid two-phase region would have a negative slope in the pressure-temperature plane, instead of the positive slope shown here.

Figure 11 pvT surface of a pure substance that contracts upon freezing.
Figure 12 Tsp surface of a pure substance.

Very similar to the pvT surface is the Tsp surface shown in Figure 12. As in the pvT surface, the critical point lies at the top of the liquid-vapor two-phase hump. However, the two-phase surfaces show how entropy of the substance changes during phase change, instead of how the volume changes, as in the pvT surface. This diagram also demonstrates that the entropy of the substance increases from the densest phase (solid) to the least dense phase (vapor).

## 7.2 p-T, p-v and T-s Phase Diagrams for a Pure Substance

Phase diagrams are created by taking a slice from a thermodynamic surface; the slice is perpendicular to one of its principal axes. Phase diagrams are usually more convenient to use than thermodynamic surfaces because they show the behavior of a substance more clearly in a two-dimensional diagram, i.e., one thermodynamic property from the thermodynamic surface is assumed constant. The three most common phase diagrams for pure substances are: (1) the pressure-temperature phase diagram from the pvT surface, (2) the pressure-volume phase diagram from the pvT surface, and (3) the temperature-entropy phase diagram from the Tsp surface. Figure 13 shows the pT and pv phase diagrams for a pure substance that contracts upon freezing. The pT diagram is relatively straightforward and shows the three phases separated by saturation curves. Along the saturation curves, the two adjacent phases are at equilibrium. The pT diagram also shows the triple point where all three phases coexist at equilibrium. The critical point denotes the highest pressure and temperature at which the distinction between the liquid and vapor phase can be made. The pvT diagram for a pure substance that expands upon freezing is similar to that shown in Fig. 13(a), except that the solid-liquid saturation curve has a negative slope instead of a positive slope.

(a) p-T diagram. (b) p-v diagram.
Figure 13 pT and pv phase diagrams.
Figure 14 Ts phase diagram.

The more interesting phase diagram shown here is the pv diagram. The liquid-vapor two-phase dome and solid-vapor two-phase regions are clearly shown in this diagram. Also shown are the critical point and the triple line, the latter of which coincides with the triple point in the pT diagram. However, here, the range of volumes over which all three phases are present is shown. Additional phase diagrams can be obtained by combination of two-principle axes in the thermodynamic surfaces discussed above. One example is a Ts phase diagram, shown in Fig. 14, which is used mostly in investigations of power or refrigeration cycles. Similar to the pv diagram, the liquid-vapor two-phase region, solid-vapor two-phase region, critical point, and triple line are clearly shown in Fig. 14.

## 7.3 Equations of State for Pure Substances

Proceeding from the state postulate for simple, compressible, pure substances, any intensive property is solely a function of two other independent, intensive properties. In general, a functional relationship among any three properties could be called an equation of state. However, in common usage, the expression equation of state usually refers only to equilibrium relationships involving the pressure p, temperature T, and specific volume v having the functional form of

$f(p,v,T) = 0\qquad \qquad( )$

(183)

An equation of state serves two useful purposes. Its most obvious use is for predicting pvT behavior over a desired range of values. The equilibrium states of a simple compressible substance can be represented by a surface on pvT Cartesian coordinates. The other major use of pvT data is in the evaluation of thermodynamic property data that are not directly measurable; these include internal energy, enthalpy, entropy, Gibbs function, and Helmholtz function. The theoretical relationships for these properties contain first and second derivatives involving p,v, and T, so it is important that the mathematical format of an equation of state lend itself to the required differentiation and subsequent integration. If the vapor or gas can be approximated as an ideal gas, the equation of state is simply

$pV = n{R_u}T{\rm{ or }}pv = {R_g}T\qquad \qquad( )$

(184)

where Ru = 8.3143kJ / kmol − K is the universal gas constant, n is mole number, and Rg = Ru / M (M is molecular mass in kg/kmol) is the particular gas constant. The equation of state for an ideal gas is only applicable to a situation where the pressure of the gas is very low. At higher pressures, the behavior of a gas or vapor deviates substantially from that of the ideal gas. In addition, the ideal gas law is also invalid in the two-phase region represented by the “hump” on a pv diagram, where liquid and vapor phases coexist. As is shown in the pv diagram, the properties experience a discontinuous change when liquid-vapor phase change takes place. It was known that if the critical isotherm – which passed through the critical point on the pv diagram – was followed, the change from vapor to liquid would be continuous. It is noted that continuous transition isotherms below the critical isotherm must exist in order to bridge the gap between ideal gas and incompressible liquid. This idea was realized in the theoretical van der Waals equation constructed by Johannes Diderik van der Waals in 1873. The theoretical equation of state developed by van der Waals is a cubic polynomial:

$p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{{v^2}}}\qquad \qquad( )$
	 (185)

Figure 15 van der Waals polynomial showing three roots for v at a given p.

where the constant b represents the minimum volume occupied by the pure substance in the limit as p→∞, i.e., the volume occupied by the molecules of the substance. The a / v2 term is the additional pressure term that accounts for mutual attraction between the molecules and is proportional to the density squared. The van der Waals equation bridges the gap between ideal gas behavior and incompressible liquid behavior, which can be shown in two limiting cases. The first is the limit as the specific volume of the substance approaches infinity, i.e., the very dilute gas limit, where the van der Waals equation reduces to the ideal gas equation of state, i.e.,

$pv = {R_g}T\begin{array}{*{20}{c}} {} & {v \to \infty } \\ \end{array}\qquad \qquad( )$

(186)

The second is the limit as the pressure of the system approaches infinity, and the volume of the system approaches the minimum volume that the molecules of the substance can occupy, i.e.,

$\begin{array}{*{20}{c}} {v = b} & {p \to \infty } \\ \end{array}\qquad \qquad( )$

(187)

The van der Waals equation is a cubic polynomial that provides three roots for v at any given pressure. This can be seen in Fig. 15, which shows the shape of the van der Waals isotherms below the critical temperature of the substance. While an isotherm passing the two-phase hump on a pv diagram makes a straight line, the van der Waals isotherm has minimum and maximum points below and above the straight line, respectively. It will be shown in the next section that this wavy line is inherently unstable, and therefore the van der Waals equation of state is not a good approximation in the two-phase region. However, it is a very helpful equation because on either side of the two-phase “hump,” it very accurately represents the ideal gas and incompressible liquid behavior of a given substance. It is necessary to know the constants a and b for different substances in order to use the van der Waals equation to evaluate the pvT relation. Because the critical isotherm passes through a point of inflection at the critical point, and the slope is zero at this point, the van der Waals equation can be differentiated with respect to v at constant temperature:

${\left( {\frac{{\partial p}}{{\partial v}}} \right)_T} = - \frac{{{R_g}T}}{{{{\left( {v - b} \right)}^2}}} + \frac{{2a}}{{{v^3}}} = 0\qquad \qquad( )$
 	   (188)

${\left( {\frac{{{\partial ^2}p}}{{\partial {v^2}}}} \right)_T} = \frac{{2{R_g}T}}{{{{\left( {v - b} \right)}^3}}} - \frac{{6a}}{{{v^4}}} = 0\qquad \qquad( )$

(189)

These two derivatives, along with the van der Waals equation at the critical conditions,

${p_c} = \frac{{{R_g}{T_c}}}{{{v_c} - b}} - \frac{a}{{v_c^2}}\qquad \qquad( )$

(190)

can be solved to find the two constants and the critical specific volume, which are not as amenable to measurement as are the critical temperature and pressure. These values are as follows:

${v_c} = 3b\qquad \qquad( )$

(191)

$a = \frac{{27}}{{64}}\frac{{R_g^2T_c^2}}{{{p_c}}}\qquad \qquad( )$

(192)

Table 6 Critical point properties for selected fluids

 Substance Formula Molecular Mass Critical Temperature (K) Critical Pressure (MPa) Air -- 28.97 133.2 3.77 Ammonia NH3 17.031 405.5 11.35 Carbon dioxide CO2 44.01 304.1 7.38 Carbon monoxide CO 28.01 132.9 3.5 Nitrogen N2 28.013 126.2 3.39 Oxygen O2 31.999 154.6 5.04 Water H2O 18.015 647.3 22.12 Propane C3H8 44.094 369.8 4.25 R-12 CCl2F2 120.914 385.0 4.14 R-134a CF3CH2F 102.03 374.2 4.06
$b = \frac{{{R_g}{T_c}}}{{8{p_c}}}\qquad \qquad( )$

(193)

The critical properties of selected substances are tabulated in Table 6.

Example 4 A 1-m3 vessel is filled with 80 kg of propane at a temperature of 120 C. The gas constant for propane is 0.188 kJ/kg-K. Find the pressure of the propane by using the van der Waals equation. What will the pressure of the propane be if the ideal gas law is used?

Solution: The critical pressure and temperature for propane can be found from Table 6; they are pc = 4.25MPa

and Tc = 369.8K.
The constants a and b can be determined using eqs. (192) and (193):

$a = \frac{{27}}{{64}}\frac{{R_g^2T_c^2}}{{{p_c}}} = \frac{{27}}{{64}}\frac{{{{(0.188 \times {{10}^3})}^2}{{369.8}^2}}}{{4.25 \times {{10}^6}}} = 479.78{\rm{ Pa - }}{{\rm{m}}^{\rm{6}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}$ $b = \frac{{{R_g}{T_c}}}{{8{p_c}}} = \frac{{0.188 \times {{10}^3} \times 369.8}}{{8 \times 4.25 \times {{10}^6}}} = 2.04 \times {10^{ - 3}}{{\rm{m}}^{\rm{3}}}{\rm{/kg}}$

The specific volume of the propane is

$v = \frac{V}{m} = \frac{1}{{80}} = 0.0125{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/kg}}$

The pressure of the propane can be found by using the van der Waals equation, eq. (185), i.e.

$p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{{v^2}}} = \frac{{0.188 \times {{10}^3} \times (120 + 273.15)}}{{0.0125 - 0.00204}} - \frac{{479.78}}{{{{0.0125}^2}}} = 4.00{\rm{ MPa}}$

If the ideal gas law is used, the pressure of propane will be

$p = \frac{{{R_g}T}}{v} = \frac{{0.188 \times {{10}^3} \times (120 + 273.15)}}{{0.0125}} = 5.91{\rm{ MPa}}$

It can be seen that the ideal gas law significantly overpredicts the pressure in this case.

The van der Waals equation is one of the most compact equations of state since it has only two empirical constants. Many other equations of state have been developed in an effort to improve on the accuracy of the van der Waals equation. The Redlich-Kwong equation (1949) is generally considered to be the best among the two-constant equations of state. The Redlich-Kwong equation is

$p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{v(v + b){T^{1/2}}}}\qquad \qquad( )$

(194)

where the two constants a and b are obtained by applying the critical point conditions: ${(\partial p/\partial v)_T} = 0$ and ${({\partial ^2}p/\partial {v^2})_T} = 0.$

$a = 0.4275\frac{{R_g^2T_c^{2.5}}}{{{p_c}}}\qquad \qquad( )$

(195)

$b = 0.08664\frac{{{R_g}{T_c}}}{{{p_c}}}\qquad \qquad( )$

(196)

Also widely used is the Beattie-Bridgeman equation of state

$p = \frac{{{R_g}T}}{{{v^2}}}\left( {1 - \frac{c}{{v{T^3}}}} \right)(v + B) - \frac{A}{{{v^2}}}\qquad \qquad( )$

(197)

Where

$A = {A_0}\left( {1 - \frac{a}{{\bar v}}} \right)\qquad \qquad( )$

(198)

$B = {B_0}\left( {1 - \frac{b}{{\bar v}}} \right)\qquad \qquad( )$

(199)

Table 7 Constants for Beattie-Bridgeman equation of state (Bejan, 1997)

 Gas Rg(J/kg-K) A0(N-m4/kg2) ax103(m3/kg) B0x103(m3/kg) bx103(m3/kg) cx10-3(m3-K3/kg) Air 286.95 157.12 0.66674 1.5919 -0.03801 1.498 Argon 208.14 81.99 0.58290 0.98451 0.0 1.499 Carbon dioxide 188.93 262.07 1.62129 2.3811 1.6444 14.997 Helium 2078.18 136.79 1.49581 3.5004 0 9.955 Hydrogen 4115.47 4904.92 -2.5053 10.376 -1.582 0.24942 Nitrogen 296.69 173.54 0.93394 1.8011 -0.24660 1.498 Oxygen 259.79 147.56 0.80097 1.4452 -0.13148 1.498

where the values of the five constants (A0,a,B0,b, and c) for selected substances are listed in Table 7 [note that a and b in eqs. (198) and (199) are different from a and b in eq. (194)], and $\bar v$ is specific molar volume (m3/kmol). The Beattie-Bridgeman equation of state is valid for densities up to 0.8ρc, where ρc is density at the critical point. An improved equation of state that is valid for densities up to 2.5ρc is the Benedict-Webb-Rubin equation of state:

Failed to parse (PNG conversion failed;

check for correct installation of latex, dvips, gs, and convert): \begin{array}{l} p = \frac{{{R_g}T}}{v} + \left( {{B_0}{R_g}T - {A_0} - \frac{{{C_0}}}{{{T^2}}}} \right)\frac{1}{{{v^2}}} + \left( {b{R_g}T - a} \right)\frac{1}{{{v^3}}}(v + B) \\ {\rm{ }} + \frac{{a\alpha }}{{{v^6}}} + \frac{c}{{{v^3}{T^2}}}\left( {1 + \frac{\gamma }{{{v^2}}}} \right){{\mathop{\rm e}\nolimits} ^{ - \frac{\gamma }{{{v^2}}}}} \\ \end{array}\qquad \qquad( )

(200)

which has eight constants (A0,B0,C0,a,b,c,α,γ) that are listed in Table 8.

Table 8 Constants for Benedict-Webb-Rubin equation of state (Bejan, 1997)

 Gas Formula A0x10-2 (N-m4/kg2) B0x103 (m3/kg) C0x10-7(N-m4K2/kg2) a(N-m7/kg2) bx105(m6/kg2) cx10-5(N-m7K2/kg3) αx109(m9/kg3) γx105(m6/kg2) Methane CH4 7.31195 2.65735 0.889635 1.21466 1.31523 0.62577 30.1853 2.33469 Ethylene C2H4 4.3055 1.98649 1.69071 1.19119 1.09451 0.97139 8.08173 1.17469 Ethane C2H6 4.66269 2.08914 2.01509 1.28892 1.23191 1.22361 8.9722 1.30701 Propylene C3H6 3.50217 2.02308 2.51642 1.05482 1.05806 1.39829 6.13014 1.03453 Propane C3H8 3.58575 2.20855 2.65194 1.12224 1.15892 1.52759 7.09776 1.13317 i-Butane C4H10 3.07308 2.36826 2.55256 1.00195 1.25806 1.47891 5.48279 1.00799 i-Butylene C4H8 2.88571 2.06958 2.98871 0.97316 1.10774 1.58056 5.16963 0.941616 n-Butane C4H10 3.02865 2.14127 2.98168 0.97334 1.18582 1.6361 5.62184 1.00799 i-Pentane C5H12 2.49391 2.22006 3.40357 1.01546 1.28545 1.87887 4.53682 0.890805 n-Pentane C5H12 2.37376 2.17426 4.13424 1.10159 1.28545 2.22807 4.83038 0.913893 n-Hexane C6H14 1.97242 2.06498 4.53487 1.12913 1.47181 2.40013 4.40244 0.899353 n-Heptane C7H16 1.77041 1.98756 4.79543 1.04602 1.51575 2.49275 4.33982 0.897754

A more general form of the equation of state can be expressed in the following series form:

$p = \frac{{{R_g}T}}{v} + \frac{{a(T)}}{{{v^2}}} + \frac{{b(T)}}{{{v^3}}} + \frac{{c(T)}}{{{v^4}}} + \frac{{d(T)}}{{{v^5}}} + \cdots$

(201)

where its accuracy depends on the number of terms used. The coefficients $a(T),{\rm{ }}b(T),{\rm{ }}c(T),{\rm{ }}d(T) \cdots$ are functions of temperature only and can be measured experimentally or derived from statistical mechanics. It should be noted that as $v \to \infty$ or $p \to 0$, eq. (201) is reduced to the equation of state for an ideal gas. The equations of state for most substances are too complex to be expressed by simple equations like those presented above. The alternative is to present the thermodynamic properties in the form of tables. While some thermodynamic properties can be measured directly, the others are not directly measurable and must be calculated using their relationships with the measurable properties. The results of these measurements and calculations are usually presented in the form of thermodynamic properties tables. Appendices B to D present thermophysical properties of gases, solids, and phase change materials (PCMs), as well as liquid and vapor properties at saturation. Also presented in Appendix D are temperature-property relationships of saturated liquid and vapor for various substances.

## 7.4 Phase Diagrams for Multicomponent Systems

We have looked at phase diagrams for pure substances, in which the pT phase diagram distinctly shows the equilibrium lines between the various phases. For a multicomponent system, the unique relationship between saturation pressure and temperature no longer exists, because phase change occurs over a range of temperature and pressure. The concentration of each component in the system will affect the range of temperature and pressure over which phase change occurs. For a binary system containing two components, one component – referred to as the solute – is dissolved into another component – referred to as the solvent. The saturation temperature, or melting point, is a function of both the pressure and the mass fraction of the solute. For practical use, the phase diagram is often presented as a temperature-concentration diagram under isobaric (constant pressure) conditions. Phase diagrams for both solid-liquid phase change and liquid-vapor phase change in a binary system are discussed below.

Figure 16 Phase diagram of copper-nickel isomorphous alloy at atmospheric pressure (Reproduced from Smith, W. (1995), Principles of Materials Science and Engineering, 3rd ed. with permission from McGraw-Hill Professional Book Group).

Solid-liquid phase diagrams of binary alloys are extremely useful for material scientists and mechanical engineers. Phase diagrams for multi-component substances differ considerably from single-component phase diagrams. A mixture of two metals is called a binary alloy and constitutes a two-component system, since each metallic element in an alloy is considered a separate component. The phase diagrams of binary alloy systems are usually presented with the temperature of the system as the ordinate and the chemical composition of the system as the abscissa. Binary alloys can be classified in two groups: (1) isomorphous alloys, and (2) eutectic alloys. The two components in an isomorphous alloy are completely soluble in each other in both liquid and solid states; therefore, the solid alloy can be characterized by a single type of crystal structure for different compositions of the components. In eutectic alloys, on the other hand, the solubility between the two components is limited. Consequently, the solid alloy can have different types of crystal structure depending on the chemical composition of the alloys. Figure 16 shows a phase diagram of an isomorphous alloy (copper-nickel) system, which is a slice of a thermodynamic surface taken at atmospheric pressure. The upper line in the diagram is the liquidus, above which lies a stable two-component liquid phase. The lower line in the diagram is the solidus, below which lies a stable solid phase. Between the liquidus and solidus exists a two-phase region in which the alloy contains both liquid and solid phases of both components. However, the average composition of the solid is not the same as that of the liquid. This can be illustrated by focusing on a point where the composition of the alloy is 53 wt% Ni and 47 wt% Cu, and the temperature is 1300 °C. The alloy at this point contains both liquid and solid phases at 1300 °C but neither phase can have an average composition of 53 wt% Ni and 47 wt% Cu. The composition of the liquid phase corresponds to the composition at the point of intersection between the horizontal tie line and the liquidus, which is 45 wt% Ni and 55 wt% Cu. Similarly, the composition of the solid phase can be found by using the compositions at the point of intersection between the horizontal tie line and the solidus, which is 58 wt% Ni and 42 wt% Cu. It is necessary to point out that the diagram shown in Fig. 10 is obtained by slow cooling or heating of the alloy at equilibrium conditions. For rapidly-cooled or rapidly-heated alloys, the alloy system may experience nonequilibrium; in such cases the diagram of Fig. 16 is not applicable. In a eutectic binary alloy, the two components have limited solid solubility in each other. Although some experimental results regarding solid-liquid phase change of binary metallic alloys appear in the literature, many researchers conduct experiments using transparent phase change materials (PCMs) such as NH4Cl-H2O solution, because their solidification is quite similar to that of alloys and, moreover, it is easy to observe (Beckerman and Viskanta, 1988; Braga and Viskanta, 1990). The equilibrium phase diagram for aqueous ammonium chloride is shown in Fig. 17. At the eutectic point, the temperature and composition (NH4Cl-H2O mass fraction) are Te= –15.4 °C and ωe=19.7% respectively. The eutectic point is also the point of intersection of the two liquidus lines, above

Figure 17 Phase diagram for eutectic binary solution (solid-liquid), aqueous ammonium chloride (NH4Cl-H2O) at constant pressure.

which the binary solution is in the liquid phase. When the temperature is below the eutectic temperature, or the temperature corresponding to the second solidus line, the solid phase is present. There are two mushy zones where solid and liquid coexist. Mushy zone 1 is for a subeutectic concentration of NH4Cl-H2O in water (ω < ωe) and is bounded by the solidus 1, the liquidus 1 and the eutectic line. The solid phase in the mushy zone 1 is pure ice. Mushy zone 2 is for the supereutectic concentration (ω > ωe) and is bounded by liquidus 2, solidus 2, and the eutectic line. The solid phase contained in the mushy zone 2 is solid NH4Cl-H2O. The phase diagram can be used, in conjunction with knowledge of the mixture concentration and temperature, to relate the phase concentrations to the mass fraction of the phase on the basis of local thermodynamic equilibrium.

Example 5: The liquidus 1 in the phase diagram, Fig. 11, can be approximated by a $\omega (T) = 1.678 \times {10^{ - 3}} - 1.602 \times {10^{ - 2}}T - 2.857 \times {10^{ - 4}}{T^2}$ $- 4.491 \times {10^{ - 6}}{T^3}$, where the unit of temperature is °C. The local temperature and concentration of NH4Cl at a point in the mushy zone formed by solidification are Tm= - 10°C and ωm=10%, respectively. What is the local solid fraction, f, i.e., mass fraction of the solid in the mushy zone?

Solution: Since the local concentration of NH4Cl-H2O is less than the eutectic concentration, the mushy zone is in the subeutectic region, i.e., mushy zone 1, which contains pure ice and NH4Cl –H2O solution. The mass fraction of NH4Cl in the mushy zone 1 can be obtained by the concentration at the liquidus 1 corresponding to temperature Tm, i.e.,

<center>$\begin{array}{l} \omega ({T_m}) = 1.678 \times {10^{ - 3}} - 1.602 \times {10^{ - 2}}{T_m} - 2.857 \times {10^{ - 4}}T_m^2 - 4.491 \times {10^{ - 6}}T_m^3 \\ {\rm{ }} = 13.8\% \\ \end{array}$

Since the solid phase in the mushy zone contains pure ice, the mass balance of NH4Cl requires that

ωm = ω(Tm)(1 − f)

i.e.,

$f = \frac{{\omega ({T_m}) - {\omega _m}}}{{\omega ({T_m})}} = \frac{{13.8\% - 10\% }}{{13.8\% }} = 0.275$

Our attention so far has been focused on the solid-liquid phase change of a binary solution. Liquid-vapor phase change is also very important, given its application in processes such as distillation and separation. Figure 18 shows a typical phase diagram of a miscible binary system (i.e., the two components in the binary system can dissolve into each other) undergoing liquid-vapor phase change. The system is in the vapor phase above the dew point line. If the temperature of the binary vapor falls below the dew point line, condensation of one component will take place. There is also a bubble point line, below which the system is in the liquid phase. If the temperature of the binary liquid is increased to a temperature above the bubble point line, vaporization of one component will take place and bubbles will be formed. The region between the dew point line and bubble point line is a two-phase mixture that includes both liquid and vapor phases. The characteristics of liquid-vapor phase change in a binary system can be demonstrated by analyzing evaporation and condensation of a binary system with an initial solute mass fraction of ω2a. If the initial temperature of the binary

Figure 18 Phase diagram for liquid-vapor phase change of binary system at constant pressure.

system is below the bubble point, heating the binary liquid beyond the bubble point temperature (point ${a_\ell }$) will result in evaporation that produces vapor at point b. It should be noted that the temperature at b is the same as the bubble point temperature corresponding to its initial mass fraction. The mass fraction at b, however, is lower than that at point a. Since the vapor produced during the evaporation process contains less solute, the solute mass fraction in the remaining liquid will be increased. The temperature of the remaining liquid must be increased in order to continue the evaporation process. A similar analysis can be applied to condensation of the binary vapor with an initial mass fraction of ω2a. Cooling of the binary vapor below the dew point (point av) will result in condensation that produces liquid at point c, the mass fraction of which is greater than that at point a. However, its temperature equals the dew point temperature corresponding to initial mass fraction ω2a. The remaining vapor can condense only at a lower temperature because its solute mass fraction is lower.