{"id":4713,"date":"2024-10-05T17:53:42","date_gmt":"2024-10-05T17:53:42","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=4713"},"modified":"2024-10-05T17:53:43","modified_gmt":"2024-10-05T17:53:43","slug":"equations-of-state","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/10\/05\/equations-of-state\/","title":{"rendered":"\u00a0Equations of State"},"content":{"rendered":"\n<p>The simplest example of what is called an equation of state is the ideal gas law itself. Equations of state for nonideal gases can be just empirical relations selected to fit a data set, or they can be based on theory, or a combination of\u00a0the two.\u00a0Figure 7.4\u00a0shows the measurements by Andrews in 1863 of the pressure versus the specific volume for CO<sub>2<\/sub>\u00a0at various constant temperatures. Note that point C at 31\u00b0C is the highest temperature at which liquid and gaseous CO<sub>2<\/sub>\u00a0can coexist in equilibrium. Above 31\u00b0C only\u00a0<strong>critical<\/strong>\u00a0fluid exists, so that what is called the\u00a0<strong>critical temperature<\/strong>\u00a0for CO<sub>2<\/sub>\u00a0is 31\u00b0C (304 K). The corresponding gas (fluid)\u00a0<strong>pressure<\/strong>\u00a0is 72.9 atm (7385 kPa). Also note that at higher temperatures, such as 50\u00b0C, the data can be represented by the ideal gas law because\u00a0<em>pV<\/em>\u00a0is constant, a hyperbola. You can find experimental values of the critical temperature (<em>T<\/em><sub><em>c<\/em><\/sub>) and the critical pressure (<em>p<\/em><sub><em>c<\/em><\/sub>) for various compounds on the CD that accompanies this book. If you cannot find a desired critical value in this text or in a handbook, you can consult Reid et al. (refer to the references at the end of this chapter), who describe and evaluate methods of estimating critical constants for various compounds.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/fig07-04.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Figure 7.4. Experimental measurements of carbon dioxide by Andrews (+). The solid lines represent smoothed data. C is the highest temperature at which any liquid exists. At the big solid dots liquid and vapor start to coexist. Note the nonlinear scale on the horizontal axis.<\/p>\n\n\n\n<p>How can you predict the\u00a0<em>p<\/em>,\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">, and\u00a0<em>T<\/em>\u00a0properties of a gas between the region in which the ideal gas law is valid and the region in which the gas condenses into liquid? One way is to use one or more equations of state. Where substantial changes in curvature occur, perhaps several different equations must be used to cover a region accurately.\u00a0Table 7.2\u00a0lists some of the well-known single equations of state.<\/p>\n\n\n\n<p><a><\/a>Table 7.2. Examples of Equations of State (for 1 g mol)<a href=\"https:\/\/learning.oreilly.com\/library\/view\/basic-principles-and\/9780132885478\/ch07.xhtml#ch07tn01\">*<\/a><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07tab02.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07tab02a.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p><a href=\"https:\/\/learning.oreilly.com\/library\/view\/basic-principles-and\/9780132885478\/ch07.xhtml#ch07tn01a\">*<\/a>&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&nbsp;is the specific volume,&nbsp;<em>T<sub>c<\/sub><\/em>&nbsp;and&nbsp;<em>p<sub>c<\/sub><\/em>&nbsp;are explained in the text, and \u03c9 is the acentric factor, also explained in the text.<\/p>\n\n\n\n<p>The units used in calculating the coefficients in the equations are determined by the units selected for R.<\/p>\n\n\n\n<p>Some of the classical equations of state are formulated as a power series (called the\u00a0<strong>virial<\/strong>\u00a0form) with\u00a0<em>p<\/em>\u00a0being a function of 1\/<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">\u00a0or\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">\u00a0being a function of\u00a0<em>p<\/em>\u00a0with three to six terms. You should note that the coefficients in the\u00a0van der Waals, Peng-Robinson (PR),\u00a0Soave-Redlich-Kwong (SRK), and Redlich-Kwong (RK) equations can be calculated from certain physical properties (discussed below), whereas in the virial equations the coefficients are strictly determined from experimental measurement. Because of its accuracy, the databases in many commercial process simulators make extensive use of the SRK equation of state. These equations in general will not predict\u00a0<em>p<\/em>&#8211;<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">&#8211;<em>T<\/em>\u00a0values across a phase change from gas to liquid very well. Keep in mind that under conditions such that the gas starts to liquefy, the\u00a0<em>gas laws apply only to the vapor phase portion<\/em>\u00a0of the system for this book.<\/p>\n\n\n\n<p>How accurate are equations of state? Cubic equations of state such as Redlich-Kwong, Soave-Redlich-Kwong, and Peng-Robinson listed in\u00a0Table 7.2\u00a0can exhibit an accuracy of 1%\u20132% over a large range of conditions for many compounds. Equations of state in databases may have as many as 30 or 40 coefficients to achieve high accuracy (see, for example, the AIChE DIPPR reports that can be located on the AIChE Web site). Keep in mind that you must know the region of validity of any equation of state and not extrapolate outside that region, particularly not into the liquid region, by ignoring the possibility of condensation for gases such as CO<sub>2<\/sub>, NH<sub>3<\/sub>, and low-molecular-weight organic compounds such as acetone, ethyl alcohol, and so on. If you plan to use a specific equation of state such as one of those listed in\u00a0Table 7.2, you have numerous choices, no one of which will consistently give the best results.<\/p>\n\n\n\n<p><em>Although this may seem a paradox, all exact science is dominated by the idea of approximation.<\/em><\/p>\n\n\n\n<p>Bertrand Russell<\/p>\n\n\n\n<p>Other than the use of equations of state to make predictions of values of&nbsp;<em>p<\/em>,&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">, and&nbsp;<em>T<\/em>, what good are they?<\/p>\n\n\n\n<p><strong>1.<\/strong>&nbsp;They permit a concise summary of a large mass of experimental data and also permit accurate interpolation between experimental data points.<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;They provide a continuous function to facilitate calculation of physical properties based on differentiation and integration of&nbsp;<em>p<\/em>&#8211;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&#8211;<em>T<\/em>&nbsp;relationships.<\/p>\n\n\n\n<p><strong>3.<\/strong>&nbsp;They provide a point of departure for the treatment of the properties of mixtures.<\/p>\n\n\n\n<p>In addition, some of the advantages and disadvantages of using equations of state versus other methods to make predictions are the following:<\/p>\n\n\n\n<p>Advantages:<\/p>\n\n\n\n<p><strong>1.<\/strong>&nbsp;Values of&nbsp;<em>p<\/em>&#8211;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&#8211;<em>T<\/em>&nbsp;can be predicted with reasonable error in regions where no data exist.<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;Only a few values of coefficients are needed in the equation to be able to predict gas properties, versus collecting large amounts of data by experiment for tables and graphs.<\/p>\n\n\n\n<p><strong>3.<\/strong>&nbsp;The equations can be manipulated on a computer whereas graphics methods of prediction cannot.<\/p>\n\n\n\n<p>Disadvantages:<\/p>\n\n\n\n<p><strong>1.<\/strong>&nbsp;The form of an equation is hard to change to fit new or better data.<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;Inconsistencies may exist between equations for&nbsp;<em>p<\/em>&#8211;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&#8211;<em>T<\/em>&nbsp;and equations for other physical properties.<\/p>\n\n\n\n<p><strong>3.<\/strong>&nbsp;Usually the equation is quite complicated and may not be easy to solve for&nbsp;<em>p<\/em>,&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">, or&nbsp;<em>T<\/em>&nbsp;because of its nonlinearity.<\/p>\n\n\n\n<p><em>I consider that I understand an equation when I can predict the properties of its solutions, without actually solving it.<\/em><\/p>\n\n\n\n<p>Paul Dirac<\/p>\n\n\n\n<p>Disadvantage 3 prevented the widespread use of equations of state until computers and computer programs for solving nonlinear algebraic equations came into the picture. You can see that it is easy to solve the SRK equation in\u00a0Table 7.2\u00a0for\u00a0<em>p<\/em>\u00a0given values for\u00a0<em>T<\/em>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">, or for\u00a0<em>T<\/em>\u00a0given values of\u00a0<em>p<\/em>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">, but quite difficult and tedious to solve for\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">\u00a0given values for\u00a0<em>T<\/em>\u00a0and\u00a0<em>p<\/em>\u00a0without the aid of computers. Similar remarks apply to virial equations.<\/p>\n\n\n\n<p>For example, look at the Redlich-Kwong (RK) equation. Given&nbsp;<em>p<\/em>&nbsp;and&nbsp;<em>T<\/em>, is the RK equation cubic in&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">? Yes. Given&nbsp;<em>p<\/em>,&nbsp;<em>T<\/em>, and&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">, is the RK equation cubic in&nbsp;<em>n<\/em>? Yes. Given&nbsp;<em>p<\/em>&nbsp;and&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">, is it cubic in&nbsp;<em>T<\/em>? No. We will not focus in this book on how to solve cubic and more complex equations for&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&nbsp;but instead use an equation solver such as Polymath on the CD in the back of the book.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><a><\/a>Example 7.6. Use of the RK Equation to Calculate&nbsp;<em>p<\/em>&nbsp;or&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\"><\/p>\n\n\n\n<p>Determine the pressure (in atmospheres) of 1 g mol of C<sub>2<\/sub>H<sub>4<\/sub>&nbsp;at 300 K with&nbsp;<em>V<\/em>&nbsp;= 0.674 L using the RK equation. From the CD that accompanies the book, for C<sub>2<\/sub>H<sub>4<\/sub>:&nbsp;<em>T<\/em><sub><em>c<\/em><\/sub>&nbsp;= 282.8 K and&nbsp;<em>p<\/em><sub><em>c<\/em><\/sub>&nbsp;= 50.44 atm.<\/p>\n\n\n\n<p>Solution<\/p>\n\n\n\n<p>The RK equation is (from\u00a0Table 7.2)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/371equ01.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>What should you do first? Take a basis of the given values of&nbsp;<em>V<\/em>,&nbsp;<em>n<\/em>, and&nbsp;<em>T<\/em>. Then calculate a, b, and&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\"><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/371equ02.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Next, insert the known values of the variables and coefficients:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/371equ03.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>If you know\u00a0<em>p<\/em>\u00a0and\u00a0<em>V<\/em>\u00a0instead of\u00a0<em>p<\/em>\u00a0and\u00a0<em>T<\/em>, you can solve explicitly for\u00a0<em>T<\/em>\u00a0by rearranging the equation or by using an equation solver.\u00a0Figure E7.6\u00a0compares the prediction of\u00a0<em>p<\/em>\u00a0by the ideal gas law and the RK equation.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/ex07-06.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Figure E7.6<\/p>\n\n\n\n<p>Next, let\u2019s determine the specific volume of C<sub>2<\/sub>H<sub>4<\/sub>&nbsp;at 300 K and 30 atm using the RK equation. Because with the RK equation you cannot explicitly solve for&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">, what should you do now? You can solve the RK equation in the usual format after introducing the known values and then use Polymath to determine&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&nbsp;= 0.6748 L\/g mol.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"ch07lev2sec9\">7.2.2. The Critical State and Compressibility<\/h4>\n\n\n\n<p>We mentioned the critical pressure\u00a0<em>p<\/em><sub><em>c<\/em><\/sub>\u00a0and critical temperature\u00a0<em>T<\/em><sub>c<\/sub>\u00a0in connection with\u00a0Figure 7.4. The\u00a0<strong>critical state\u00a0(point)<\/strong>\u00a0for the gas-liquid transition is the set of physical conditions at which the density and other properties of the liquid and vapor become identical. In\u00a0Figure 7.5\u00a0the points on the constant temperature lines at which the gas (vapor) starts to condense have been connected by a dashed line (<strong>&#8212;<\/strong>). On the opposite side of the figure, the dot-dashed curve (___ . ___) shows the locus of points at the respective temperatures at which completion of the condensation occurs; that is, the vapor becomes all liquid. Between the two bounds a mixture of vapor and liquid exists.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/fig07-05.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Figure 7.5. The critical point is located where the lengths of the (solid) lines are zero. The solid lines connect the points at which condensation starts at various temperatures to the corresponding points for that temperature at which condensation is complete.<\/p>\n\n\n\n<p>The intersection of the two bounds is denoted as the&nbsp;<strong>critical point<\/strong>, and it occurs at the highest temperature and pressure possible (<em>T<\/em><sub><em>r<\/em><\/sub>&nbsp;= 1,&nbsp;<em>p<\/em><sub><em>r<\/em><\/sub>&nbsp;= 1) at which gas and liquid can coexist.<\/p>\n\n\n\n<p>A\u00a0<strong>supercritical fluid<\/strong>\u00a0is a compound in a state above its critical point. Supercritical fluids are used to replace solvents such as trichloroethylene and methylene chloride, the emissions from which, and the contact with which, have been severely limited. For example, coffee decaffeination, the removal of cholesterol from egg yolk with CO<sub>2<\/sub>, the production of vanilla extract, and the destruction of undesirable organic compounds all can take place using supercritical water. Supercritical water has been shown to destroy 99.99999% of all of the major types of toxins in these organic compounds.<\/p>\n\n\n\n<p>Other terms with which you should become familiar are the\u00a0<strong>reduced variables<\/strong>. These are conditions of temperature, pressure, or specific volume\u00a0<em>normalized<\/em>\u00a0(divided) by their respective critical conditions, as follows:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/373equ01.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>In theory, the\u00a0<strong>law of corresponding states<\/strong>\u00a0indicates that any compound should have the same reduced volume at the same reduced temperature and reduced pressure so that a universal gas law might be<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07equ07.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Unfortunately Equation (7.7) does not universally make accurate predictions. You can check this conclusion by selecting a compound such as water, applying Equation (7.7) at some low temperature and high pressure to calculate\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">, and comparing your results with the value obtained for\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">\u00a0with the corresponding conditions from the tables for water vapor that are in the folder in the back of this book.<\/p>\n\n\n\n<p>The concept of reduced variables nevertheless has been applied to prediction of real gas properties. One common way is to modify the ideal gas law by inserting an adjustable coefficient\u00a0<em>z<\/em>, the\u00a0<strong>compressibility factor<\/strong>, a factor that compensates for the nonideality of the gas, and can be looked at as a measure of nonideality. Thus, the ideal gas law is turned into a real gas law called a\u00a0<strong>generalized equation of state<\/strong>:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07equ08.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>or<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07equ08a.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>One way to look at\u00a0<em>z<\/em>\u00a0is to consider it to be a factor that makes Equation (7.8) an equality. Note that\u00a0<em>z<\/em>\u00a0= 1 is for an ideal gas. Although we treat only gases in this chapter, Equation (7.8) has been applied to liquids.<\/p>\n\n\n\n<p>If you plan on using Equation (7.8), where can you find the values of\u00a0<em>z<\/em>\u00a0to use in it? Equations exist in the literature for specific compounds and classes of compounds, such as those found in petroleum refining. Theoretical calculations based on molecular structure sometimes prove to be useful. Refer to the references at the end of this chapter. Usually you will find graphs or tables of\u00a0<em>z<\/em>\u00a0to be quite convenient sources for engineering purposes. If the compressibility factor derived from experiment is plotted for a given temperature against the pressure for different gases, figures such as\u00a0Figure 7.6a\u00a0result. However, if the compressibility factor is plotted against the reduced pressure as a function of the reduced temperature, then for like gases the compressibility values at the same reduced temperature and reduced pressure fall at about the same point, as illustrated in\u00a0Figure 7.6b.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/fig07-06.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Figure 7.6. (a) Compressibility at 100\u00b0C for several gases as a function of pressure; (b) compressibility factor for several gases as a function of reduced temperature and reduced pressure<\/p>\n\n\n\n<p>You can use the charts described in the next section,\u00a0Section 7.3, to get approximate values of\u00a0<em>z<\/em>\u00a0as a function of the reduced temperature and pressure. Or, you can use one of the numerous methods that have appeared in the literature and in process simulation codes to calculate\u00a0<em>z<\/em>\u00a0via an equation in order to obtain more accurate values of\u00a0<em>z<\/em>\u00a0than can be obtained from charts. Equation (7.9) employs the\u00a0Pitzer acentric factor, \u03c9:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07equ09.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Tables in\u00a0Appendix C\u00a0list values of\u00a0<em>z<\/em><sup>0<\/sup>\u00a0and\u00a0<em>z<\/em><sup>1<\/sup>\u00a0as a function of\u00a0<em>T<\/em><sub><em>r<\/em><\/sub>\u00a0and\u00a0<em>p<\/em><sub><em>r<\/em><\/sub>; \u03c9 is unique for each compound, and you can find values for it on the CD that accompanies this book.\u00a0Table 7.3\u00a0is an abbreviated table of the acentric factors from Pitzer.<\/p>\n\n\n\n<p><a><\/a>Table 7.3. Selected Values of the Pitzer<a href=\"https:\/\/learning.oreilly.com\/library\/view\/basic-principles-and\/9780132885478\/ch07.xhtml#ch07tn02\">*<\/a>&nbsp;Acentric Factor<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/07tab03.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p><a href=\"https:\/\/learning.oreilly.com\/library\/view\/basic-principles-and\/9780132885478\/ch07.xhtml#ch07tn02a\">*<\/a>K. S. Pitzer,&nbsp;<em>J. Am. Chem. Soc.,<\/em>&nbsp;<strong>77<\/strong>, 3427 (1955).<\/p>\n\n\n\n<p>The acentric factor \u03c9 indicates the degree of acentricity or nonsphericity of a molecule. For helium and argon, \u03c9 is equal to zero. For higher-molecular-weight hydrocarbons and for molecules with increased polarity, the value of \u03c9 increases.<\/p>\n\n\n\n<p>As an example of calculating&nbsp;<em>z<\/em>&nbsp;via the Pitzer correlation, calculate the compressibility factor&nbsp;<em>z<\/em>&nbsp;for ethylene (C<sub>2<\/sub>H<sub>4<\/sub>) at 300 K and 3000 kPa. First get&nbsp;<em>T<\/em><sub><em>c<\/em><\/sub>&nbsp;(283.1 K) and&nbsp;<em>p<\/em><sub><em>c<\/em><\/sub>&nbsp;(50.5 atm) from the CD, and then calculate&nbsp;<em>T<\/em><sub><em>r<\/em><\/sub>&nbsp;and&nbsp;<em>p<\/em><sub><em>r<\/em><\/sub>:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/375equ01.jpg\" alt=\"Image\"\/><\/figure>\n\n\n\n<p>Next, look up (using interpolation)\u00a0<em>z<\/em><sup>0<\/sup>\u00a0(0.812) and\u00a0<em>z<\/em><sup>1<\/sup>(\u20130.01) as well as \u03c9 (0.089) from\u00a0Appendix C\u00a0or\u00a0Table 7.3.<\/p>\n\n\n\n<p><em>z<\/em>&nbsp;= 0.812 + (\u20130.01)(0.089) = 0.812<\/p>\n\n\n\n<p>The calculation of&nbsp;<em>z<\/em>&nbsp;is easy if you know the values of&nbsp;<em>p<\/em>&nbsp;and&nbsp;<em>T<\/em>&nbsp;for the gas and just want to calculate&nbsp;<em>z<\/em>&nbsp;for those two conditions. But if you know&nbsp;<em>p<\/em>&nbsp;and&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&nbsp;or&nbsp;<em>T<\/em>&nbsp;and&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">, you have to employ a trial-and-error solution to get&nbsp;<em>z<\/em>. What you do is assume a sequence of values of&nbsp;<em>p<\/em>, calculate the related sequence of values of&nbsp;<em>p<\/em><sub>r<\/sub>, and next calculate the associated sequence of values of&nbsp;<em>z<\/em>. Then you calculate values of&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&nbsp;from&nbsp;<em>p<\/em><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">&nbsp;=&nbsp;<em>z<\/em>R<em>T<\/em>. When you find the value of the specific volume that matches the value specified in the problem, you have an appropriate&nbsp;<em>z<\/em>&nbsp;(and&nbsp;<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\" width=\"14\" height=\"18\">).<\/p>\n\n\n\n<p>A different way of predicting\u00a0<em>p<\/em>,\u00a0<img loading=\"lazy\" decoding=\"async\" width=\"14\" height=\"18\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9780132885478\/files\/graphics\/v-cap.jpg\" alt=\"Image\">, and\u00a0<em>T<\/em>\u00a0properties is the\u00a0<strong>group contribution method<\/strong>\u00a0which has been successful in estimating properties of pure components. This method is based on combining the contribution of each functional group of a compound. The key assumption is that a group such as \u2013CH<sub>3<\/sub>\u00a0or \u2013OH behaves identically irrespective of the molecule in which it appears. This assumption is not quite true, so any group contribution method yields approximate values for gas properties. Probably the most widely used group contribution method is\u00a0UNIFAC,<sup>1<\/sup>\u00a0which forms a part of many computer databases.\u00a0UNIQUAC\u00a0is a variant of UNIFAC and is widely used in the chemical industry in the modeling of nonideal systems (systems with strong interaction between the molecules).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"ch07lev1sec4\">Self-Assessment Test<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"ch07lev2sec10\">Questions<\/h4>\n\n\n\n<p><strong>1.<\/strong>\u00a0Explain why the van der Waals and Peng-Robinson equations (of state) are easy to solve for\u00a0<em>p<\/em>\u00a0and hard to solve for\u00a0<em>V<\/em>.<\/p>\n\n\n\n<p><strong>2.<\/strong>\u00a0Under what conditions will an equation of state be the most accurate?<\/p>\n\n\n\n<p><strong>3.<\/strong>\u00a0What are the units of\u00a0<em>a<\/em>\u00a0and\u00a0<em>b<\/em>\u00a0in the SI system for the Redlich-Kwong equation?<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"ch07lev2sec11\">Problems<\/h4>\n\n\n\n<p><strong>1.<\/strong>\u00a0Convert the virial (power series) equations of\u00a0Kammerlingh-Onnes\u00a0and\u00a0Holborn\u00a0(in\u00a0Table 7.2) to a form that yields an expression for\u00a0<em>z<\/em>.<\/p>\n\n\n\n<p><strong>2.<\/strong>\u00a0Calculate the temperature of 2 g mol of a gas using van der Waals\u2019 equation with\u00a0<em>a<\/em>\u00a0= 1.35 \u00d7 10<sup>\u20136<\/sup>\u00a0m<sup>6<\/sup>\u00a0(atm)(g mol<sup>\u20132<\/sup>),\u00a0<em>b<\/em>\u00a0= 0.0322 \u00d7 10<sup>\u20133<\/sup>\u00a0(m<sup>3<\/sup>)(g mol<sup>\u20131<\/sup>) if the pressure is 100 kPa and the volume is 0.0515 m<sup>3<\/sup>.<\/p>\n\n\n\n<p><strong>3.<\/strong>\u00a0Calculate the pressure of 10 kg mol of ethane in a 4.86 m<sup>3<\/sup>\u00a0vessel at 300 K using two equations of state: (a) ideal gas and (b) Soave-Redlich-Kwong. Compare your answer with the experimentally observed value of 34.0 atm.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"ch07lev2sec12\">Thought Problems<\/h4>\n\n\n\n<p><strong>1.<\/strong>&nbsp;Data pertaining to the atmosphere on Venus (which has a gravitational field only 0.81 that of the Earth) shows that the temperature of the atmosphere at the surface is 474 \u00b1 20\u00b0C and the pressure is 90 \u00b1 15 atm. What do you think is the reason(s) for the difference between these figures and those at the Earth\u2019s surface?<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;A method of making protein nanoparticles (0.5 to 5.0 \u03bc in size) has been patented by the Aphio Corp. Anti-cancer drugs that small can be used in novel drug delivery systems. In the process described in the patent, protein is mixed with a gas such as carbon dioxide or nitrogen at ambient temperature and 20,000 kPa pressure. When the pressure is released, the proteins break up into fine particles.<\/p>\n\n\n\n<p>What are some of the advantages of such a process versus making powders by standard methods?<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"ch07lev2sec13\">Discussion Question<\/h4>\n\n\n\n<p>Fossil fuels provide most of our power, and the carbon dioxide produced is usually discharged to the atmosphere. The Norwegian company Statoil separates carbon dioxide from its North Sea gas production and, since 1996, has been pumping it at the rate of 1 million tons per year into a layer of sandstone 1 km below the seabed. The sandstone traps the gas in a gigantic bubble that in 2001 contained 4 million tons of carbon dioxide.<\/p>\n\n\n\n<p>Will the bubble of carbon dioxide remain in place? What problems exist with regard to the continuous addition of carbon dioxide in future years? Is the carbon dioxide actually a gas?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The simplest example of what is called an equation of state is the ideal gas law itself. Equations of state for nonideal gases can be just empirical relations selected to fit a data set, or they can be based on theory, or a combination of\u00a0the two.\u00a0Figure 7.4\u00a0shows the measurements by Andrews in 1863 of the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":4604,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[570],"tags":[],"class_list":["post-4713","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ideal-and-real-gases"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/09\/gases.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4713","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=4713"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4713\/revisions"}],"predecessor-version":[{"id":4714,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4713\/revisions\/4714"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/4604"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=4713"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=4713"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=4713"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}