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é 2'1 d/ éaé Section 2 Solutions and Comments on Problems Solution The requested plot is given at the end of the problem. The line is a linear least
squares fit to the data. The slope is shown in the figure. Therefore, s11 : R[slope) = —(8.314J mol" K' 1)(*3,067.7 K) mp : 2.550 x irrjmol“. (1) Substituting of Eq. 1 into the Clausius—Clapeyron equation yields PWT) ' W 1 M 2.550 >< 104M
1n = ln _—_—
L227 .1 P(205 K) R m. T‘
A_2.550><104li_i] (2
T R LT” 205 )
Thus,
PWT) = 227 exp{ r — — ﬂ , (3) which is the desired expression. Normal boiling occurs at the point where
P“q = 760 tort = 101.325 kPa. Solving Eq. 3 for 1.3T”, we obtain 1 1 R Emmi Tel ' 2% V 2.550 >< 104 ml. 227 J" (4)
Substituting of P9510") = 101.325 kPa gives
T199 = 205 2.5:03:41041n(1022:25) z 0'005141’ (5)
so that
91 : —1— = 194.5 K, (6)
0.005141 which is the predicted normal boiling point. The experimental result is
T? = 194.65 K. CO: Vapor Pressures vs. T 1 lr1 (Pal) : 2013808 — (3ll67.7 K):"T llll.._.___‘ll 5.“ In P“ (P in kPa) 5.5:E
E IlJIIIJLAJ—l 4.6 4.11 5.0 5.2 x 103
1.711(4) . . ﬁk‘vﬂ east— (1) (2) {3) there (4) (5) (6) .lli is 6.23 Chapter 6 The partial molar enthalpy of fusion of C02(s) is 7,950] mol—l at the melting
point. At this same temperature, the partial molar enthalpy of sublimation is
about 25,505 I mol’l, At 220 K, the equilibrium vapor pressure over liquid
CO2 is 590.6 kPa. Use the Clausius—Clapevron equation to obtain an analytic
expression for the vapor pressure over liquid C02, assuming the enthalpy of
vaporization to be independent of temperature. Solution The enthalpy of vaporization is given by ﬁn“, : Aim 7 mm = 2.550 ><104 ~ 7,9501moi1 = 1.756 x 1041m01‘. {1) 6.24 If we assume this value to be constant over a narrow temperature range
around the melting point, the equilibrium vapor pressure will be given by
Eq. 6.12: . (2} Pp 0 R qu _ To ln[p§q] _ _ AEWP[ 1 1
Taking Tﬁ'q = 220 K and Piq = 590.6 kPa and dropping the subscripts on P3“
and Tﬁq, we obtain 2,111
T“;  1.75 X 1 4 1 1
in P“! = ln(590.6) e #i i : 15.98 — 8.314 1”q 5 E
This gives
2,111 Tm (4) P : exp[15.98 — which is the desired expression. Using the results obtained in Problems 6.22 and 6.23, determine the tempera
ture at which the equilibrium vapor pressure over C02(s) and C020) are equal.
What is the value of the equilibrium vapor pressure at this temperature? What
are the predicted temperature and pressure at the CO2 triple point? Solution The vapor pressure over solid CO; was obtained in Problem 6.22. The result is 2.550 X 104 1 1
peq __ 7 __— _ _ .
(Sf 22 exp{ I R [TN 205] } kPa (1) The equilibrium vapor pressure over liquid C02 was derived in Problem 6.23.
The result is (2) a 2,111
P“? = P = exp[15.98 * J kPa. Te‘i If we are to have PE: = Pail, then . 1 4'
227exp{—255&i[i — = P = exp[15.98  2,111]. (3) R Teq 205 Tact To find the solution of Eq. 4, we conduct a grid search for the root of 2.550 X 104 1 1 2,111
HT) = 227exp{_—R——[ﬁ — e exp[15,98 — Teq ]: 0. (4) Two repetitions of the grid search yields a root at T“1 = 217.0 K. Substituting of
this temperature into Eq. 2 gives Phase Equilibrium 229 230 Section 2 Solutions and Comments on Problems 6.25 6.26 Peq = 5.19 bar + When Peq is the same over solid and liquid C02, all three phases—solid, liquid,
and vapor—are in simultaneous equilibrium. By definition, this is the triple
point. Therefore, the predicted value is Tfjpiwmm : 217.0 K and Pfﬁmpmm :
5.19 bar. The experimental values areTfﬁPmPomt = 216.8 K and Pfﬁpwmm =
5.11 bar. So the foregoing analysis is reasonably accurate. The densities of solid and liquid C0; are 1.5(3 kg L' l and 1.101 kg 1.”, respec
tively. The triple point was predicted in Problem 6.24 to occur at T = 216.8 K, with
a pressure of 5.19 bar. The partial molar enthalpy of fusion of C02(s] is
7,950] mol" = 79.504 L bar mol". Use these data to obtain an analytical equa_
tion showing the dependence of the equilibrium temperature (melting point) of
solid COz(s) on pressure. At what pressure will the CO;(s) melting point be 222 K? Solution
Equation 6.20 gives the dependence of the equilibrium temperature upon
pressure:
[139:‘ — 97‘4" = “2“ ln[TSq]. (1)
‘ ” av TT“ Let us take 1°?“ = 5.19 bar and T?“ : 217.0 K, since we know that liquid and
solid C02 are in equilibrium at this point. The molar volume of the solid is — mass of 1 mole 0.04401 kg mol "1 V : W   :0028211. 11. 2
(S) density 1.56 kg L ' mo ( )
For the liquid,
_ _ 0.04401kgmol '
V0) = ——,— : 0039971,.mor1. (3;
1,101 kg L
Therefore, AV = W1) — W5) = 0.03997 — 0.02821 L moi 1 = 0.01176 L mol' l. (4)
After dropping the subscript substituting into Eq. 1 gives Fq
T J bar 79.504 L bar mol'] [ T“1
—— in 0.01176 L mol 1 217.0 I ‘5’ = 5.19 + 6,760.5l
l “[2170 which is the required expression. The pressure required to make the equilibrium temperature (melting
point) be 222 K is 222 _
13"“ = 5.19 + 6,760.51n(2170) bar : 159.2 bar = 1.592 X lll’kl’a. (6) Use the results of Problems 6.22 through (2.25 to make a careful plot of the CO2
phase diagram. Solution We wish to plot the loci of all points (P, T) at which each pair of phases.
liquid—vapor, solid—vapor and liquid—solid, are in equilibrium. The analytic
equations for these curves were obtained in Problems 6.22 (solidevapor),
6.23 (liquid—vapor), and 6.25 (liquid~solid). The equation giving the loci of nninh; 2f urhirl‘i 1119 :nlir‘l and \rnnrw am in onnililuinna i. id, liquid,
the triple
q qple poin: : "iplr pm‘m ‘ I, respec_
8 K, with
:Oﬂs) is
:al equa_
point) of
1e 222 K? ['0 Upon (1) .lld and
d is (2) (3) ases.
lytic
ior).
:i of Chapter 6
For liquid—vapor equilibrium, we have
P : exp[15.98 — 2;;1] kPa. (2)
The solid—liquid curve is
Peq = 5.19 + 6,760.5 ln[21:0] bar = 519 + 6.7605 x 1051n[21:0] kPa. (3) The plot follows. The measured critical temperature is 304.2 K. Our assumptions
that AH W is constant and that the partial molar volume of the vapor is large rel
ative to that of the liquid break down completely as we approach this tempera
ture. Therefore, we show the phase diagram only up to a temperature of 260 K. C02 Phase Diagram Solid~liquid
equilibrium 2000 — Solid Liquid Pressure (kl’a)
Q
5
i Liquidvapor
equilibrium 100i}
Solid—vapor equilibrium T (Triple point)
(217.0 K. 519.0 kPa) 190 200 210 220 230 240 250 260 Temperature (K) 6.27 10 moles of 1120(1) at 330 K are inserted into an evacuated container that is insu lated so as to prevent heat transfer to or from the surroundings. The volume of
vacuum above the water is 20 L. As the H200) vaporizes to establish equilib—
rium with HZO( g), the temperature drops because the heat of vaporization must
be provided by the internal energy present in the H200), since there is no heat
transfer from the surroundings. If we ignore the change in the volume, mass,
and heat capacity of the liquid, what is the final temperature of the system when
equilibrium is established between H200) and H20(g)? Assume that am“, for H200) is constant and equal to 44,010 J mol’1 and that the total heat capacity of
the H200) is 752.9 J K". Solution Let dn moles of H20( 1) vaporize into the gas phase at temperature T. The total
loss of internal energy by the water is Energy loss by water = —Aﬁvapdn. (1) This energy loss will cause a decrease in the water’s temperature given by
Ca‘T, where C is the total heat capacity of the water and d1" is the differential temperature change caused by the vaporization of dn moles of water. Therefore,
we have Phase Equilibrium 231 ...
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 Spring '08
 OHRN
 Kinetics

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