Adamson Phase Equilibria - 8 LIQUIDS AND THEIR SIMPLE PHASE...

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Unformatted text preview: 8 LIQUIDS AND THEIR SIMPLE PHASE EQUILIBRIA CflMM ENTS We now consider the application of the combined first and second laws of thermodynamics to simple phase equilibria. The general cross~dilferentiation equation [dfl-“rll‘ip = {ii-$.35 VJ}. becomes the Clapeyron equation dPM T = dltr'TdV when applied to the equilibrium between two phases, since :15 can now. be written as di'i'y’T. and it is not necessary to retain the partial differential form. ffone ofthe two phases is an ideal gas. then further approx- Imilllfll‘lh' lead to the very useful Clausius—Clapeyron equation. These two relationships. plus some semiempirical rules such as Trouton‘s rule. are presented to you in this chapter in quite a variety ofdisguises. The present chapter is an appropriate one in which to introduce surface tension. ils manifestations. and its determination. The Laplace equation is fundamental to this topic ofcapillarity. and its use will he required over and over again in the problems that follow. 1Il'llateh for variations such as the case of maximum bubble pressure. and the rise ofa liquid between parallel plates. Be careful in your choice of units. til-IT will sometimes be needed in cc-atmjmole units and pressure differences in capillarity situations will generally he In dynetern”. not atm. souartows awn concsrrs Clapeyron Equation as _ an in" _ r a V {3— I] 114 Liquids ri'nrir their simple phase equilibn'o 125 Clausius—Clapeyron Equation F dH _ = _ 3—2 dln alT RT: l l P1 dH 1 l — = —— — — — 3-3} n P! R (Tl. T1 I: an n as I I H = — — = —— — + — 3-4 lnl” constant it?“ or P P espl: R (T T“ l ] Effect of Mechanical Pressure on Vapor Pressure P“ . RTln F =I Ve'Pmd. [3-5] Semiempirieal Rules and laws Truman‘s rule. de’T}. = If calfdeg-mole where 1']. denotes the normal boiling point and AH... the beat of vaporization per mole. Another observau tion is that the normal boiling point of a liquid is often about two-thirds of its critical temperature. Law of Reetilinear Diameters The sum of the densities of a liquid and its equilibrium vapor is a constant. independent of temperature. A more realistic version is that the sum varies linearly with temperature. Capillarity Laplace equation. .t‘tP = yllli'll1 + ly’RE] where flP is the pressure dif— ference across a curved surface. '.I is the surface tension. and R] and R1 are the two radii of curvature. The signs are such that the surface is convex toward the low-pressure side: thus the pressure inside a spherical drop or _ soap bubble is greater inside than outside. For surfaces that are sections of a sphere. the Laplace equation becomes as = — {3-6} 1ivhere r is the radius of the sphere. Capillary rise. diP = pglt‘ = fly-fr where r is the radius of the capillary and ii. the height of rise. For a nonzero contact angle. the equation becomes 1v cost} r {Hill lift Uh o'ers ro riding physr'eaf Ntfmfa'itj.‘ Drop-weight method. Hid“. = Ear-y. This is known as Tate‘s law. and actual drop weights will difler from the ideal weight by a substantial correc- tion factor I; so that Waetu-l = zflr‘rif iii-3] This correction factor can be expressed as a graphical function of rfIr"-"3 where Via the drop voturne. Note that in the ease ofa liquid that wets the tube. r is the outside radius. Moxittrttm babble-pressure toothed. 21' P = Pt... + — ta-si' r P...” is the hydrostatic pressure as determined by the depth of immersion of the tube out of which bubbles are formed. Pendant drop method. For a pendant drop. the basic equation of Laplace can be maneuvered into the form 2 = ”if” ts-to} .l where ff“ is the equatorial or maximum diameter of the drop. and H is a function of a shape factor 3'. S is defined as d,.-’d,. where a. is the diameter of the drop measured a distance at. up from the bottom. The Following is an abbreviated UH versus 3 table: 5 flfffl [US [LSD {1.35 {11.91} [:95 I110 Ir'H (13ft flab? 0.5T L148 {1.41 0.36 [Lil PRDHIEMS 1. t2! mintThe normal boiling point ofpyridine is ] 14°C. At thistempera- lure. its vapor density is 2.5 gjliter. and the density ofthe liquid is DECK!) grcc. At a certain higher temperature T'. the liquid has expanded to where its density is it‘iflflfl gree. Calculate. or estimate with explanation. ta] the heat of vaporization of pyridine. {b} the boiling point of pyridine on a mountain top where the atmospheric pressure is "HUI mm t—tg instead of roe mm Hg. and tel the vapor density of pyridine at the temperature T'. 2. [l4 mint The semilog plots for the vapor pressures of liquids A. and E are shown in Fig. B—I. tail Calculate arr. for liquid A... lb} Liquids A and B Liquids and their st'mlm't' phase flat-tiffhrftt In 2.4 2.6 2.3 3.0 3.2 3.4 1o’rT“K FIGURE E-I cannot both have the same Trouton constant. Espiain how_thils can be concluded from the data. Give a brief argument as to which liquid sltould he eonsidered the more associated. 3. [4.5 min} The law oi rectilinear diameters [in its simplest form as illus- trated itt Fig. 3-2} is not compatibie with the van der Waals statement that K. = 3b. Show why this statement is true. A [IR min}The vapor pressure ofliquid at is St) rmp Hg at 445°C.- this vapor pressure is t].5[t mrn greater than that of solid a at the same temperature. At 45°C. the vapor pressure of the liquid is Liltt mm Hg greater than that of the solid. no. is 9.0 kcal. tal Estimate the melting point of a. fb]: Calculate the heat of fusion of A. and its heat of sublimation. 123 Understanding physical difflffl'ull'y liq vap p FIGURE 2-2 5. [13 min} The vapor pressure ol‘iCiCl4 increases by 4% per degree around 25‘1“; Calculate oil-l, and the normal boiling point ol‘ CCl4. List all the assumptions and approximations involved in the derivation oFthe equation used to obtain dH,.. ME. [I 2 min] A particular liquid obeys both the llCllausius—Clapeyron equation and Trouton's rule. With this information calculate tho vapor pressure of the liquid, in atmospheres, at a temperature equal to one—third ol‘ the normal boiling point {expressed in ”Kl 'l. {15 min} The heat of vaporization of water that For liquid A. is Till kcalfmole. The vapor pressure of water and of a are the same at 150°C. [aJ Plot P versus UT on a semilog plot, for water and for A. [The plots may he approximate, but it must be clear how they were obtained.) lb] Determine from the plot the normal boiling point of liquid a. to} Which olthe two liquids more nearly obeys Trouton's rule? [On a problem such as this, the appropriate graph paper would be available.) is 9.? kealy’mole, whereas 8. {31} min] The following data are given For p-chloroaniline: mol. wt: i2? vp at the mp: 5.0 mm Hg normal mp: 'lfl‘C vp at 100T : Elli} mm Hg arr, = 4,?[lfl callmole densities at the mp .' solid 1.45 glee liquid 1.15 glee {a1 Calculate the heat ol‘ vaporiration. {b} State the assumptions and approximations involved in the equationls} used in [a]. {e} Estimate the normal boiling point and the heat of sublimation. State whether the mp under 100 atm prmsuie would be greater or less than 'lfl‘t': [explain]. {d} Air at I aim and lflt’i‘C is bubbled through the liquid {also at tlitl°Cl at the rate of 3.3 moleyh. Assuming the ell-'luent air to be saturated with vapor, how long will it talre for I1? g of the liquid to be evaporated? [Effluent gas also at 1 atm.} Liquids and their simple phase aquililirr'o 119 9. {3 min} The normal boiling point of a liquid obeying Trouton‘s rule is INT. Calculate the vapor pressure of the liquid at l2I°C (or, better, the increase in vapor pressure from ETC to 121°C}. III]. [12 min} The vapor pressure of acetonitrile is changing at the rate of 0.030 atm,r'deg in the vicinity of its normal boiling point, which is EDT. Calculate the heat of vaporization. 1]. {l2 min} The straight-line plots of in P versus llT are sketched in Fig. 3-3 for various liquids obeying the simple Clausius—Clapeyron equation. These lines all meet at I ,l Tequaf to zero ; show that this behavior is required by Trouton‘s rule, 12. ' [12 min] The straight-line plot ol'ln P versus ly'T for the vapor pressure ol‘ a certain liquid whose normal boiling point is ETC extrapolates to P = lliS atn1 at infinite temperature. Using the Clausius—Clapeyron equa‘ tion, calculate from these data the heat of vaporization olthe liquid. M13. [I8 min} A. certain liquid of molecular weight tit] has a critical tempera— ture of 400°C. [ts melting point is liflli-l)"C as normally measured and is HHSDDC at its triple point {where the system is subjected only to its own low vapor pressure}. The solid and liquid densities are {135 and {181} glee, respectively. Calculate, using empirical or semicrnpirical relationships where necessary, arrf: fill-if, filler, and the vapor pressure at the triple point. 14. {9 min} Explain whether the melting point of a solid substance a will be raised or lowered by pressure, given that solid A does not float on liquid a. Illustrate your explanation by suitable equations. l5. {l3 min} The melting point of glacial acetic acid is rare at latm preosu re. Calculate the melting point under its own vapor pressure {ease utiall y % MT FIGURE 3—3 l3" E-‘trrlei'srrurdr'ug plrt'rt'c'rrl' chemistry zero pressure]. The heat ol‘ fusion is ll‘00 calfrnolc, densities For liquid and solid acetic acid are l.05 and l.l0 glee, respectively, and the molecttlar weight is 60. Set up the appropriate equation and insert the proper data so as to obtain an equation whose only unknown is the desired quantity. The normal boiling point ol'acetic acid is l 10°C. Show how to estimate the heat of sublimation of solid acetic acid from this and the other data given {obtain a numerical answer]. 16. {ii mitt] It is desired to calculate the vapor pressure of an equilibrium misture of solid and liquid benzene which is under l00atin of inert gas pressure. The information available includes Mfr and oH”, the molar heats of Fusion and vaporization at the melting point {under normal pressure], which is T, ; the densities of the solid and liquid p. and p2 : the molecular weight M; the normal boiling point oithe liquid T1 .- and general constants such as R. Set up equations For calculating this vapor pressure. Make clear what equations you would ttse, the sequence oltheir use, and the actual data needed. 11". [24 mitt] For a certain substance the change in entropy on melting is 3 caly’deg per cc of solid which melts. The melting point under I atm pressure is 0°C. and the densities of the solid and liquid are 0.90 and 0.35 glee, re— spectively. Calculate the melting point under l04 atiu pressure. Also, the vapor pressure versus temperature curve for the solid and the liquid. at l atm pressure, are sketched in Fig. 0-4. Show qualitatively how 'r a par pi." essure t. T FIGURE 3-4 Lao-trials rmrt' their simple phrase corrr'lr'hria 13] FIGURE 8-5 these two curves should look for the solid and liquid under 100 atm pressure [the pressurization is done by means oi" inert gas pressure, so these vapor pressure curves can be obtained experimentally]. [The curves should be of the correct shape, relative position, etc] 18. [24 min] A substance A ol‘ mol. wt 44 melts at 20°C under its own vapor pressure of 0.02 atm. The low pressure region of the phase dtagram ti shown in Fig. 3-5. In addition the melting point changes by 2%. {using K] per 100 atm applied pressure. The densities of the solid and bound forms are about US and 1.] glee, respectively. Calculate, or ii' semiempirical rtiles arc applicable, estimate {a} the heat ol‘ fusion, {b} the heats of vaporization and sublimation, and {c} the value of 5G for the process A is, 10°C, latm] = A{|,10°C,tatm]. J/IEI. to min] at}! is 0.0? keal for the transition S [rhotnbic] = S lrnonoclinic]. I Monoclinic sulfur is in equilibrium with rhombic at latm pressure and 115°C, and at [00 atm pressure the two are in equilibrium at 120°C. Show which of the two forms is the more dense. 20. [15 min] Water, which wets glass, rises in a given capillary to a height it. ll, as shown in Fig. 3—6, the capillary is broken oft", so that its length above the surface is only lit-“l will the water then flow over the edge? Explain your 13-6 Understanding physical (harrisrry FIGURE 3-H surface of the liquid in question, and slowly bubbling an inert gas throu h the liquid {see Fig. 8-l2]. a manometer connected to the tube permits the 3:22;:2:tpingzlegfimnclfihin pgessure bet-.veen the gas in the tube and ‘ .. . _ ow at t e gas pressure is a maximum :dius R of the bubble is equal to the radius r of the tube. in bribli‘hlhrgiitiii‘i Asgflmg? :1 {lg-ranted1 based on the laws governing surface tension effects] |iqhid fight proposltton to be correct, calculate the surface tension of the maximum pressure difference is use mm Hg. given that the temperature is 25%? the density of the li uid ' . . is I5 ' 13.6 giee, and the radius of the tube is Drill} em. Ema "“11 Of ““3?ch '5 30. [I2 min} The profile oian enlarged benzene suriaee tension 3t] dyneiem and ii-IJ. Calculate the magnification factor. photograph of a pendant drop of density 0.9 glee1 is shown in Fig. ANSWERS I. {a} Using Trouton’s rule. off" 2 2] x JEFK = 8J3 lteal mtlgitTFrom the Clatgsiusthlapeyron equation1 d in PMT = c-.l'-iri'l{’l“2 or :1 Rid?" = iii; P2211216? an approximation. take P to be 'iStl mm Hg so y =. , _ x . x ssr = ass I'nm Hgi'deg. The tern t '1 which P is "£40 mm Hg is then reduced by 2tl_i'2tl.5 or by Lil“. lvloii:rejrali:rti:yat log reef-r40 = E [L - i] Which gives 0.93“. Liquids and their simple phase commit-m I.'_’;'.|I [ct From the law oi" rectilinear diameters, the sum olthe liquid and vapor densities should be invariant. At 114°C it is-L‘Eflfl + 2.5 giliter. The vapor density at 'l'" is then 392.5 — Tilt] or 12.5 glitch-"‘- 1. {a} From the Clausius—Clapeyron equation, ii In PidfliT] = —iltH,-'R, st:- thc slope of the line for liquid A can be used directly. It is [log t — log tllli [2.9 — 3.41 X Ill":1| or about —2,tlflt’l. iii-i"n is then 2.0m x 2.3 x 1.93 = 9.2 lteal. [b]: The simplest way oishotving this is by writing the integrated form of the Clausius—Clapeyron equation: in P = const — hHiRT. By Trouton‘s rule, when P = 1 atm. dHf T = 21. so the constant of integration must be 213R. It is thus the same for all liquids or in other words all liquids should show the same intercept at IIT = 0. This is clearly not the case here, so Trouton‘s rule cannot be valid in this instance. to] Liquid B has the higher heat of vaporization [steeper slope} and hence is probably the more associated. 3. In its simplest form, the law of rectilinear diameters states that the sum of liquid and equilibrium vapor densities is a constant. For a liquid around its boiling point, the vapor density is so small that this sum is essentially just the liquid density. Then1 according to the LED... at the critical tempera— ture, the liquid and vapor densities are equal, and each must then be half the density oi'the liquid around its boiling point. in other words. the critical volume is predicted to be twice {not three times} that ol‘ the liquid and hence only twice that olthe volume of I mole of molecules [to which it is supposed to correspond]. 4. la} At the melting point. liquid and solid must have the same vapor pressure. lithe difference is 1 mm at 45“C and [LS mm at 46°C, it should be close to zero at 4T“C. [bl The most efficient way to get at the heat of fusion is probably From the Clausius—Clapeyron equation: rlPirlT: iii-l x Fifi-fl. For the vapor at 445°C. then [dFidT].. = sans 3-: Shilflfl x 3!? = 2.2 mmideg. Since the vapor pressure of the solid is increasing 0.5 mrnfdcg faster than is that of the liquid. thMTL = 2.? niml'deg. The heat of sublimation oi the solid is then 2:: x 1.93 5: see1 an =_—= mask I " 51.? t“ The heat of fusion is then dill-t" — 5-H]. = L35 lteal. 5. We use the Clansins—Clapeyron equation in the form dPIa'T -—— F x dHIEIRT‘i IJB Utrri’ci'sroirrfr'flif II'J-l'ti's'r't'al :‘llerrir'sii- |-' If P increases by 4% per degree then a'P-“P it?" 's {104- ' - rats at 2932 = 1.04 ltcal. { . I; I I M M" _ mm x The normal boiling point can be estimated from Trouton‘s rule.- i}. = leaflets] = 339K. Thelassuniptioris and approximations involved in the derivation of the (Thames Clapcyi'on equation are as Follow. [l] First and second laws of thermodynamics: [2] equilibrium between liquid and vapor: [3i neglect molar volume olliquid in comparison with that olvapor: l4las- some ideal behavior for the vapor - 6. Using the integrated Form. In PiiF. = toHrRttii'T. — trig}. let P. — latm. and TI the normal boiling point: all is then 2|T.. With these sub- stitutions. in P; = {El T.,.'Rllili".l". — 3_i"l]i. since T1 is to he one-third of T]. Simplification then gives in P3 = — 42m. from which P2 = 6.4 K II] ' '“atm 'i'. tat. We know P... is ] atm a1333°K .' bythe Claiisius--Ciapeyron equation log FLIP. = th_."2_3thlfT. - UTE}. the vapor pressure at 150°C will be log P: _._ [9.3013513 x lfifiltlbfit — 2.53} x ill":1 2 MD or P2 2 Satin {bl Since tho plot will be a straight line [see Fig. 8-H}. we now draw it beta-cert the two points. For liquid a. the point at l5ti°C is the same. and at ttlt'i"C its vapor pressure is given by P. lflflll "I = -—- — x [1.33 x ill—5 = __ 9.? r 2.3 x iss " “5" = “'5 _,.-F{].t‘i.. asst (LI 2.2 2.4 2.5 2.5. .‘Lfl ltJafT FIGURE g-lel Liquids rirrd' t'Iir'i'i' si'iiqlnle' plrnsr' Mtrflihrlo 139 or F. = [132 x 5 = Lu atm The line for liquid A can then be located. Extension oftlie line gives a normal boiling point ol' lflJ.-"T = 2.86 or T = 35TH. tel diH_.t'l], is 9.7flfll333 = so for water and 'l.t}tltlf35tl = 20 for liquid r't. so liquid at is the closer to obeying the Trouton rule value of 2|. ll. talThe lieatoi' vaporization is given by the Clausius Clapeyron equation: log Ett-‘S = [AH-"2.3 x l.98}tlr‘343 — 15333] or all = tldfi x 2.3 x l.93lfl.24 x It'll—'1' = 11.4 lical. Hill The assumptions and approximations involved in the derivation of the Clausius—Clapeyren equation are ill first and second laws of thermov dynamics. {2} molar volume of liquid negligible oompared with that of the vapor. [.3] vapor behaves ideally. {4i arr does not vary with temperature. iind [5} liquid and vapor are in equilibrium. [c] Front Trouton‘s rule, the normal boiling point would be I"; = tlfiflflle 543°K.The heat ofsubliination is the sum 6rd”! and iii-ll. or 4.1" + ll.4 = Ital kea]. From the Clapeyron equation. dedT = dHflToV. Since the solid is more dense. its molar volume is less than that for the liqtiid and M" is positive for the process; solid .T—‘liquid. Since an is also positive for this process. til’laiT must be positive. and the melting point under 1013 atm pressure will be greater than TQDC. {til The mole fraction of vapor in the effluent gas will be 2|]le or 0.0163. or the ratio of moles of air to moles of vapor will be 314. New ll'lg ol‘ liquid corresponds to [it] mole: hence 3.14 moles olair are needed. and air must be bubbled through for 1 It. 9. We write din PM?" = dil'i'i'li‘l'"2 or ill-",r'dT = PonRTI. At the boiling point. P = iatm, and dHiT = 2|. so ilFi‘rlT: BURT: [3.011 The vapor pressure itt 121°C is then LDIT atm. It]. The Clausitts Clapeyron equation is best used here in the form lP-‘F‘l tT— 33"— “ -' l‘ *siv where thfPlli'dT is given a [LISP-[l [P is I aim at the normal boiling point}. then all! = fl.fl3tl x 1.93 x 3532 = 7.4 liealli'mele. ll. The integrated form of the Clausius—Clapeyren equation may be wrilten lnP = eonst — dHli’RT. According to Trouton‘s rule. when P = Iaim. Mill" = 2] : hence the constant equals Eii'R. that is. in P = 2|,“R r—_._._ 149 Understanding physical chemistry — dl-URT. Thus all liquids are supposed to have a common UT 2 41 intercept at 12. The. integrated Iforrn ol the Clausius—Clapeyron equation is In P = A —- dHrRT. and. evidently. A = In IDS. so the equation becomes log lflllS = —dHy’R x are x 2.3...
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