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SOLUTIONS OF EXERCISES LEVEL 0 001 equilibrium 001:1 vessels with water in thermal contact It is plausible to assume that the heat involved in changing the temperature of a quantity of water by one degree is proportional to its mass and more or less constant between the given temperatures, say C units per kg, so that . 10 (te − 50)·C + 5 (te − 25)·C = 0. Equilibrium temperature te = 41 ⅔ °C 001:2 vessels with water in thermal contact The equilibrium temperature (46.4 °C) differs considerably from 41 ⅔ °C (foregoing exc). Apparently the property C is different for different substances. 10 ·(tC−+−= 50) · 5 ·( t 25 ) · C0 eHOeHSO224 CC= 0.34 HSO24 HO 2 001:3 sulphuric acid poured into water The heat of mixing (of two different substances) comes into action; the final temperature is not the same. 002 variables 002:1 the air’s pressure at the Puy de Dôme 725 Torr and 639 Torr, respectively. 002:2 pressure at top and bottom 101329 Pa. 002:3 Fahrenheit’s temperature scale t' t 9 =×+ 32 −17.8 °C 35.6 °C °°FC5 002:4 two phases and their amounts of two substances added together in phase L in phase L I II n(A) n(C) n(A) n(C) n(A) n(C) 10 10 6 2 4 8 5 5 3 1 2 4 6 4 4.8 1.6 1.2 2.4 2 8 0 0 2 8 320 Solutions 002:5 ethanol and water saturate a space P = (300/760) · 101325 Pa PV· V = 50·10-3 m3 n ==0.722 mol RT T = 333.15 K ethanol mole fraction in vapour - in equilibrium with liquid at 60 °C and 300 Torr - is about 0.57. amount of ethanol = 0.57 · 0.722 mol = 0.41 mol amount of water = 0.43 · 0.722 mol = 0.31 mol 003 the rules of the game 003:1 three variables subjected to two conditions There is one independent variable, because of M − N = 3 − 2 = 1. Taking X as the independent variable, then Y (X) =− 2 X {Z (X) = X 003:2 phase diagram or not? The phase diagram is supposed to be the graphical representation of the mole fractions of pairs of coexisting phases as a function of T (or P). Hence, for every point of the vaporus there has to be a corresponding point on the liquidus. In the case of figure a) this obvious rule is violated. In the case of b) the rule is respected; however, for thermodynamic reasons, as will be seen later on, the two curves are allowed to make contact in an extremum only. ? a) b) 003:3 derivation of lever rule from “law of conservation of substance B” i.e. overall amount of B = amount of B in α + amount of B in β Xo {n(α) + n(β)} = Xα · n(α) + Xβ · n(β) 003:4 a system formulation In addition to the three liquid phases I, II and III there, obviously, is a vapour phase V MM⎡⎤PXXXXXXXX ,I , I , II , II , III , III , V , V = ⎣⎦BCBCB C BC NN⎡⎤I II III V , I II III V , I II III V =⎣⎦ μμμAAA === μμμμ ABBB === μμμμμ BCCC === C f = M − N = 9 – 9 = 0 Solutions 321 003:5 amounts of three phases out of three substances n(α) = 0.846 mol ; n(β) = 1.538 mol ; n(γ) = 0.615 mol 003:6 the experimental advantage of a small vapour phase In liquid + vapour equilibrium experiments with low vapour pressures and carried out such that the vapour phase occupies but a small fraction of the space, the equilibrium composition of the liquid phase is ‘forced’ to be equal to the overall composition, Xo (before the experiment adjusted by the investigator), within experimental uncertainty. As an (extreme) example, if Xo = 0.0500 and Xvap = 0.99 and n(liq) = 105 n(vap), then Xliq = (calculated value) 0.0499906. As a result: the answer is Xliq = 0.20 and P = 287.7 003:7 naphthalene is added little by little to toluene P 0.25 0.5 X P 0.25 0.5 o X 003:8 does an empty place matter? Each new open place makes that the number of variables will be one less, just like the number of conditions; in the end f remains equal to c − p + 2. 004 pure substances 004:1 the position of phase symbols fields: β top; γ right; α bottom left increasing entropy: α, β, γ 322 Solutions 004:2 zero Celsius and zero Celsius TV·Δ Δ=TP =Δ Q apply SI units 273 K·(− 1.63 · 10-6 m 3 ·mol− 1 ) =− (611 101325) Pa 6008J·mol−1 = 0.0075 K the answer is 7 mK 004:3 water’s triple point pressure A priori: the data indicate that the result will be close to 0 °C and (4.579/760)·101325 Pa = 610.5 Pa Calculations: • lin. least sq. of pairs {y = ln (P sol /P liq); x = (273.15 + t) -1} for y = 0, calculate x → 273.15 + t /°C = 273.159 • lin. least sq. of pairs { y = ln (P sol ); x = (273.15 + t) -1} for x = (273.159)-1, calculate y → P = 4.5822 Torr or P liq data → P = 4.5822 Torr result: t = 0.01 °C; P = 4.5822 Torr = 610.9 Pa; rounded: 611 Pa. 004:4 carbon dioxide’s metastable normal boiling point liq → vap -1 Q ≈ 17 kJ·mol Tnbp ≈ 185 K 004:5 the substance water under high pressure liquid → solid at 100°C: at about 2.5 GPa triple point (II + V + l): in vicinity of (−19 °C; 0.32 GPa) see also (Fletcher 1970) 004:6 a rule to be respected by metastable extensions → for spontaneous change the shaded field is • at the α side of (α + β): β → α⎫ ⎫ ⎬ γ → α ⎪ • at the β side of (β + γ) : γ → β ⎭ ⎬ ⊗ ⎪ • at the γ side of (α + γ): α → γ ⎭ ⊗ this is the absurdity! 004:7 a phase diagram acts as a thermobarometer 800 • Triple point at 620 ºC and I 0.55GPa • Equilibrium (I+III): 600 dt/dP = 530 K·GPa-1. II • Rules for equilibrium t/°C III lines (metastable 400 extensions) • III at high-pressure side 0 P/GPa 1 Solutions 323 004:8 superposition of stable and metastable P L ()β α V T 004:9 Antoine’s equation A = 15.7576 B = 2408.66 C = 62.060 with these values the following pressures, expressed in Torr, are calculated 289.15 433.56 759.94 1267.95 2025.98 the mean absolute difference in experimental and calculated pressure is just 0.03 Torr (and partly due to round-off effects)! 004:10 supercritical fluid L S P V T 004:11 iron: the heat effect of magnetic change rounded to an integer in kJ·mol-1: 8; heat needed 35.6 kJ 004:12 boiling water altimeter Combine d(lnP)/dT (from the Clapeyron equation) with d(lnP)/dh (from the barometric formula ←002); heat of vaporization 41kJ.mol-1. 005 binary and ternary systems 005:1 a unary diagram made to look like a binary one triple point → eutectic type of three-phase equilibrium boiling curve → two-phase region like region of demixing in Figure 3 324 Solutions 005:2 the amounts of the phases during an experiment n(α) n(β) n(L) i = 0 0 ii < 0 > iii < > > iv 0 < > v 0 0 = 005:3 phase diagram and cooling curve temp. (interval) phase(s) change in temp. is 600 to 515 °C liq fast 515 to ~ 350 °C liq + sol (LiCl) slow at 350 °C liq + sol (LiCl) + sol (KCl) zero 350 to 200 °C sol (LiCl) + sol (KCl) fast 005:4 a reciprocal system Two variables: X (the fraction of the cations that are K+) Y (the fraction of the anions that are Br-) a composition square, of which the vertexes are occupied by the pure substances: NaCl (X= 0; Y = 0 ); KCl (1; 0); KBr (1; 1); NaBr (0; 1) 005:5 increasing repulsive interaction and the phase diagram The change to liquid is more and more “postponed” to higher temperatures: the A liquidus (except for its initial part) is moving upwards, and the eutectic point is moving up to B’s melting point. From a certain “moment” on the A liquidus will be interrupted by a (liquid + liquid) region of demixing - together involving a monetectic three-phase equilibrium. 005:6 overlapping two-phase regions The two three-phase equilibrium temperatures are given by the intersections of the (β + L) solidus and the upper solvus of the (α + β) field. The (α + L) field is between the two three-phase equilibrium lines; and such that the (metastable parts of) the (β + L) liquidus and the (α + β) lower solvus are inside the field. 005:7 the construction of ternary phase diagrams R = racemate = compound AB; Q = quasiracemate = compound AC section stable solids single-phase fields two-phase fields invariant triangles i) C L C + L R R + L ii) *) (A) L (A + L) (B) (B + L) C C + L C L Q Q Q + L R R + L iii) A 2 times L A + L A L R B B + L B L R Solutions 325 C C + L C L R Q Q + L C Q R R R + L L Q R • temperature is just above or just below m.p. of A and B 005:8 the appearance of an incongruently melting compound The sequences of the (two -) and single-phase fields are, when considered from high to low temperature and from left to right (assuming that B’s melting point is below the peritectic temperature): L (A + L) L (A + AB) (AB + L) L (A + AB) (AB + L) L (L + B) (A + AB) (AB + B) 005:9 ternary compositions having a constant ratio of the mole fractions of two components AR : XB for Q; SB: XA for Q AP : XB for P; PB; XA for P Ratios: AP : PB = RP : PS = (AP − RP) : (PB − PS) = AR : SB QED 005:10 cyclohexane with aniline - mixing and demixing Temperature 19.5 ºC - the milky aspect of tube 1.
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