Thermodynamics prausnitz manual solution.pdf

Additional order info. Pearson offers affordable and accessible purchase options to meet the needs of your students. Connect with us to learn more. We're sorry! We don't recognize your username or password. Please try again. The work is protected by local and international copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Thus, M1 y1 x1 M2 0. Then, 61 Solutions Manual k f m2 fd where k is a constant.

However, the di-isopropyl ether will offer some steric hindrance. The cross-coefficient, B12 , is a measure of association. Let D be the fraction of molecules that are dimerized at equilibrium. For this pressure, yA Using the McGlashan and Potter equation [Eq. Consequently, if we start out with 1 mol of H2, the amount of ethane that must be added to have a zero P H is 0.

The fugacity coefficient from the volume-explicit virial equation of state is given by Eq. Solving Eq. Because the gas phase is primarily methane, the amount of water that must be removed per mol methane is mol water removed mol methane Because we are given the initial and final state and enthalpy is a state function, 'H is fixed in this problem.

Consequently, minimum Ws corresponds to maximum Q. Maximum Q occurs when the process is reversible, or equivalently Q T'S. Hence, Eq. Expressions for enthalpy and entropy are given by Eqs. The second virial coefficient for a square-well potential is given by Eq. In this case, the relation between the second virial coefficient and the dimerization constant is given by Eq.

Combining Eqs. T K K Substitution of the square-well potential [Eq. Substitution of Sutherland potential [Eq. We now have to replace the approximate result [Eq. The equation of equilibrium for helium is f1L f1R where superscripts L and R stand for left and right compartments, respectively. Because the equilibrium constant is independent of pressure, the more probable reaction is the one that satisfies this condition. Next we need to check for the pressure independence of K b.

Substitution of Eqs. From the plot P-x-y we can see the unusual behavior of this system: 1. There is a double azeotrope 2. Liquid and vapor curves are very close to each other. To find the constant, use the boundary condition J 1 1 when x1 Hence J 1 x1 1.

Hence, the curve P-x goes through a maximum at x1 trivial one. As y1P 5. But J1 x1 is the activity, and the activity of component 1 cannot reach unity for any x1 less than one because the solution will split into two phases of lower activity. See Fig. We have to solve the equilibrium equations: M i yi P J i xi Pis 1 Because total pressure is low below atmospheric we assume vapor phase as ideal: M i 1. The activity coefficients are obtained from the equation for GE given in the data.

Using Eq. These can be easily obtained from the vaporpressure equations. To obtain the T-x1-y1 diagram we assign values for the liquid mole fraction x1. The sequence of calculations is now repeated for this new temperature we assume here that activity coefficients are independent of temperature , yielding: P2s We can now calculate the vapor phase mole fraction from y1 x1J 1P1s P 0.

The following figure shows the computed T-x1-y1 diagram for this system at 30 kPa. Vapor Mole Fraction Cyclohexanone 1 0 0.

Assume g E Ax1 x 2 , where A is a function of temperature. In Eq. This can be done by using the Gibbs-Duhem equation. Slight positive deviations merely mean that the physical interaction between SO2 and C4H8 makes a larger contribution to the excess Gibbs energy than does the chemical interaction.

Steric hindrance in isobutene is larger than in n-butene The suggested procedure is to integrate numerically a suitable form of the Gibbs-Duhem equation. This method is described by Boissanas, quoted in Prigogine and Defay, Chemical Thermodynamics, page It gives good agreement with experimental partial-pressure data for this particular system.

Choose x1 or x2 2. Calculate y1 or y2 from Eq. Calculate P from Eq. The solution procedure would be: 1. Find P1s and P2s at each T. Repeat; assuming new values. Keep repeating until Pcalc is very close to 0. Reid, J. Prausnitz, B. E Poling 4th.

The following figures show the calculated results in the form of P-x1-y1 and y1-x1 diagrams. However, the procedure is long and tedious. It is easier to graph ' mix g over the composition range and to look for inflection points. For the data given, phase separation occurs at qC. At 50qC, Then x A 0. At 70qC, Then x A b 0. The pure component boiling points are: tbA 67qC tbB 61qC In the range 61qC c The enthalpy of mixing equation cannot be totally consistent since the expression for g E is quadratic in mole fraction and the expression for ' mix h is cubic.

However, they may be close. Let B refer to toluene and C refer to CS2. Therefore, we take the smaller value. The partial molar quantities h E and s E are the contributions to these excess properties per differential amount added to the solution. This will require energy h1E! As x1 gets larger, some dimer will begin to exist in the solution, so these effects will diminish.

Therefore, at small x1, the curves should look something like this: Solutions Manual s1 h1 0 3. Ether and pentachloroethane hydrogen bond with each other but not with themselves. Thus, random mixing predicts a value higher than that given by quasichemical theory.

The observed consolute temperature is likely to be lower than both. To solve for them, we use two equilibrium equations and one mass balance. Then, J1 exp 0. To solve Eqs. This gives, 0. Substituting Eq. The equation of equilibrium is x 3 1 J 31 x 3 2 J 32 1 where subscript 3 denotes cyclohexane and superscripts 1 and 2 denote, respectively, carbon disulfide phase and perfluoro-n-heptane phase. Rearrangement of Eq. From the regular-solution theory [Eq. In dilute solution in iso-octante , methanol is a monomer.

For an order-of-magnitude estimate, we can assume that, to make a monomer, approximately one hydrogen bond must be broken. The molar specific heat is roughly J K-1 mol Strongly hydrogen-bonded liquids e.

Let HA be the acid. Pure methanol is highly hydrogen bonded. However, in dilute solutions of CH2Cl2, methanol exists primarily as monomer. Prausnitz Rdiger N. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.

All product names mentioned herein are the trademarks of their respective owners. From problem statement, we want to find. This entropy calculation corresponds to a series of steps as follows: s T P saturated liq.

P T v 10 8. In the step 1 2 the gas is infinitely rarified, and hence exhibits ideal gas behavior. Then, the second integral in Eq. To make this calculation, we first transform the derivative involved in the integral to one expressed in terms of volumetric proper- ties.

For , and. J g J mol Thus, ln. Substitution in Eq. Therefore falls with rising temperature. This is true for most cases but not always.

It does not assume the Lewis fugacity rule. Pure-component saturation pressures show that water is relatively nonvolatile at 25C. Under these conditions the mole fraction of ethane in the vapor phase y E is close to unity. Alterna- tively, ge e Poynting factor to correct to bar. At 1 btain f at bar and C, use the Steam Ta t f at saturation 3. Isothermal expansion to the ideal-gas state. Isobaric cooling of the ideal gas. A three-ste III. Compression to the final pressure.

Consider mixing as a three-step process: I Expand isothermally to ideal-gas state. II Mix ideal gas. III Compress mixture isothermally.

The attractive forces are London forces and induced dipola forces; we neglect small quadrupo- lar forces. There are no dipole-dipole forces since N 2 is nonpolar. Let 1 stand for N 2 and 2 stand for NH 3. Infrared spectrum will show strong absorption at u. Argon has only translational degrees of freedom while CO has, in addition, rotational and vibrational degrees of freedom. Therefore, the specific heat of CO is larger than that of argon. Electron affinity is the energy released when an electron is added to a neutral atom or mole- cule.

Ionization potential is the energy required to remove an electron from a neutral atom or molecule. A a good Lewis acid. SO 2 is a better a Lewis acid than ammonia. Solutions Manual 35 9. Phenol is more soluble in w. Distribution experiments show that phenol is strongly as Solutions Manual 36 nonpolar v d spectra show absorption at a frequency corresponding to the OH H hydrogen b sol ents like CCl 4.

Infrare " ond. Chlorobenzene is the next best solvent due to its high p it is a Lewis acid while PPD is a Lewis base. Although it can hydrogen- bond with PPD, this requires breaking the H-bonding ne n-butanol molecules. Steric hindrance prevents it from forming H- bonding networks; therefore, it readily exchanges one H-bond for an- other when mixed with PPD.

The lower boiling point of t-butanol sup- ydrogen bonding with itself than does n-butanol. For maximum solu- bility, we want one solvent that can hook up with the ONO 2 group e. At 20C, the dielectric constants are 4 8 CCl 2. At C and 25 bar: is above 1 is well below 1 z H 2 z amine z 1 H 2 HCl 0 1 y amine z HCl is slightly below 1 a A mixture of amine and H is expected to exhibit positive deviations from Amagats la 2 w.

Addition of argon to HCl greatly reduces the attractive forces experienced by the HCl molecules, and the mixture rapidly approaches ideality with addition of argon. Addition l to Ar causes induced dipole attractive forces to arise in argon, but these forces are much smaller than the dipole-dipole forces lost upon add upwards.

Thus the curve is convex Acetylene can therefore com- plex with DMF, explaining its higher solubility. No complexing occurs with octane. There are few interactions be- ene and methane or hydrogen. Therefore, benzene feels equally comfortable in both gases. However, at 40 bar there are many more interactions betwee drogen in the gas phase. Now benzene does care about the nature of its surroundings.

Because meth han hydrogen, benzene feels more comfortable with meth- ane than with hydrogen. At C, for ethane , for helium, At lower values of near unity the molecules have an average thermal kinetic energy on the order of c because is on the order of Acetylene b At the tween benz n benzene and methane or hy- ane has a larger polarizability t Under the same conditions, CO 2 experiences stronger attra polarizability.

At higher , the molecules have such high thermal energies that they are not significantly affected by the attractive part. The mole- cules look like hard spheres to one another, and only the repulsive part of the potential is impor- tant.

However, at 0C and at ideal, becoming more nonideal as temperature falls. As the temperature falls, the solubility of heptane in high-pressure ethane and propane rises due to increased attraction between heptane and e. Classical rather than quantum sta 4. But if we fix the core size to be a fixed fraction of the collision diameter, then Kihara potential is a 2-parameter o, c potential that satisfies corre- sponding states.

Assumption 1 is probably pretty good for H 2 ; assumption 4 is violated slightly. All substances violate assumption 3, but H 2 isnt very polarizable so it might be closer than the average substance to pairwise addi- tivity. Values of for isolated molecules cannot be computed by these methods, because rotation, vibration, translational kinetic energy appear in Q int and the 1. Q can be factored so that Q is indepen tistical mechanics is applicable. Le Schematically we have for the initial state and for the final equilibrium state: t o represent the phase inside the droplet and the surrounding phase.

The osmotic pressure is thus given by [cf. Substitution of values in Eq. Because only water can diffuse across the membrane, we apply directly Eq. Since the aqueous solution in part o is dilute in the sense of Raoults law,.

This re- duces Eq. Substituting these mole fractions in Eq. Therefore the actual volume of the spherical particle is xcluded volume 10 4 2 95 10 24 25 3 1.

For the specific volume: 2 95 10 10 2 52 10 2 52 25 19 6 3 1 3 1. The charged protein particles require counterions so electro e counterions form an ion atmosphere around a central protein particle and therefore this particles and its surrounding ion atmosphere have a larger excluded volum e. Th e than the uncharged particl is the difference between olecule gives the c 4 2 2 2 1 z M m MX In this equation, we take the solu on mass density. It can be shown see, e. Hiemenz, R.

The above equation gives V ex that is 4 times the actual volume of the particles we rical. Calculation gives for the aggregates volume, 5. Because CO 0. To find out how much, assume solid is pure CO 2. Let n be the number of moles of CO 2 left in the gas phase. From the mass balance and, as basis, 1 m le of mixture, o 0 0 Joule- Thomson throttling is an isenthalpic process that may be analyzed for 1 mole of gas through a 3- step I, III: Isothermal pres- sure changes.

II: Isobaric temperature process: change. From Eqs. Second, the fugacities of liquid and vapor phases may be calculated. Using Lee-Kesler charts see, e.

Using the Pitzer-Tsonopoulos equation see Sec. As first guess, assume Then we calculate, and We may estimate feed value. Using the virial equation, the as- sumption is close enough. This calculation is important to determine solvent losses in natural gas absorbers using methanol as solvent. Then, the activity of the monomer. However, the di-isopropyl B , is a measure of association. Let o be the fraction of molecules that are dimerized at equilibrium.

F H G I K 3 3 5 15 b Using an equation that gives fugacities from volumetric data, we obtain J 3 2 2 2 1 1 1 1 2 2 2 1 2 1 1 1. Using the McGlashan and Potter equation [Eq. Consequently, if we start out with 1 mol of H 2 , the amount of ethane that must be added to have a zero is 0.

L N M O P v e j where subscript 2 denotes water. Solving Eq. Because we are given the initial and final state and enthalpy is a state function, W s AH is fixed in this problem.

Consequently, minimum corresponds to maximum Maximum occurs when the process is reversible, or equivalently W s Q. Hence, Eq. Expressions for enthalpy and entropy are given by Eqs. Substituting Eq. The second virial coefficient for a square-well potential is given by Eq.

In this case, the relation between the second virial coefficient and the dimerization constant is given by Eq. Combining Eqs.

Results are Ah 0 1 4. Because o. We substitute now this result in Eq. Substitution of the square-well potential [Eq. Su titution of Sutherland potential [Eq. Th 1 where superscripts L and R stand for left and right compartments, respectively.

Substituting Eqs. Because the equilibrium constant is independent of pressure, the more probable reaction is satisfies this condition. Next we need to check for the pressure independence of 2 Assuming reaction b is more probab K a K b. Substitution of Eqs. O P 1 6 5 1 7 Q Corresponding values of K b at 1. Thus, if Solutions Manual 83 3.