This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. tilIT will sometimes be needed in ccatmjmole 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
— = —— — — — 33}
n P! R (Tl. T1 I:
an n as I I H
= — — = —— — + — 34
lnl” constant it?“ or P P espl: R (T T“ l ]
Effect of Mechanical Pressure on Vapor Pressure
P“ .
RTln F =I Ve'Pmd. [35] Semiempirieal Rules and laws Truman‘s rule. de’T}. = If calfdegmole 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 twothirds 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 lowpressure 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 = — {36} 1ivhere r is the radius of the sphere.
Capillary rise. diP = pglt‘ = flyfr 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.‘ Dropweight method. Hid“. = Eary. 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 Waetul = zﬂr‘rif iii3]
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 babblepressure toothed. 21' P = Pt... + — tasi' 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” tsto} .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 ﬂattiffhrftt In 2.4 2.6 2.3 3.0 3.2 3.4
1o’rT“K FIGURE EI 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. 32} 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 22 5. [13 min} The vapor pressure ol‘iCiCl4 increases by 4% per degree around
25‘1“; Calculate oill, 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 pchloroaniline: 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 straightline plots of in P versus llT are sketched in Fig. 33 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 straightline 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 lifllil)"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: ﬁllif, 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. 04. Show qualitatively how 'r a par pi." essure t. T FIGURE 34 Laotrials rmrt' their simple phrase corrr'lr'hria 13] FIGURE 85 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. 35. 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 136 Understanding physical (harrisrry FIGURE 3H surface of the liquid in question, and slowly bubbling an inert gas throu h
the liquid {see Fig. 8l2]. 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 {lgranted1 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
iiIJ. 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 reefr40 = E [L  i] Which gives 0.93“. Liquids and their simple phase commitm I.'_’;'.I [ct From the law oi" rectilinear diameters, the sum olthe liquid and vapor densities should be invariant. At 114°C it isL‘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. iiii"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: iiil x Fiﬁﬂ. 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 dillt" — 5H]. = L35 lteal. 5. We use the Clansins—Clapeyron equation in the form dPIa'T —— F x dHIEIRT‘i IJB Utrri’ci'sroirrfr'ﬂif II'Jl'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 2T.. With these sub stitutions. in P; = {El T.,.'Rllili".l". — 3_i"l]i. since T1 is to he onethird 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 ClaiisiusCiapeyron 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. 8H}. we now draw it betacert 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 glel 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 iiill. 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 — dlURT. 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...
View
Full Document
 Fall '09
 Goux
 Physical chemistry, pH

Click to edit the document details