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P2in PI = RT1T2 RT1T2 (P2 ) I (101.325 K Pa = 341.5 K

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P2in PI = RT1T2 RT1T2 (P2 ) I (101.325 K Pa = 341.5 K The boiling point of hexane at 1 atm is 68.7 °C. What is the boiling point at 1 bar? 6.1 Given: The vapor pressure of hexane at 49.6 °C is 53.32 k Pa. SOLUTION In PIP2 = ~vapH(T2RT1T2 -TI) t.vapH = (T2-:-rI)RT1T2 In (P2PI ) (8.3145)(322.8)(3419) I (101.325k Pa ) = 30850 J mnl-l = 19.1n 53.32kPa , ( 101.325 ) -~ (J:-.- -~ In- -8.3145 Tl 341.9 ) 100 1 TI = = 341.5 K ( 101.325 ) ~ 8.314~30850 1n + 341.9 100 Thus the boiling point is reduced 0.4 °C to 68.3 °C. 6.2 What is the boiling point of water 2 miles above sea level? Assume the atmosphere follows the barometric formula (equation 1.54) with M = 0.0289 kg mol-l and T = 300 K. Assume the enthalpy of vaporization of water is 44.0 kJ mol-l independent of temperature. SOLUTION h = (2 miles)(5280 ft/mile)(12 in/ft)(2.54 cm/in)(0.01 m/cm) = 3219 m p = Po exp(- gMh/R7) = (1.01325 b r) [ -(9.8 m s-2)(0.0289 kg mol-l)(3219 ml- a exp (8.314 J K-l mol-l)(300 K) = 0.703 bar In (PlP2 ) = Llvap1l(T2 -Tl)/RT2T2 1.01325 -(44 000)(373 -Tl) In ( ) 0.703 -(8.314)(373)Tl 96 Chapter 6/Phase Equilibrium TI = 364 K = 90.6 °C 6.3 The barometric equation 1.44 and the Clausius-Clapeyron equation 6.8 can be used to estimate the boiling point of a liquid at a higher altitude. Use these equations to derive a single equation to make this calculation. Use this equation to solve problem 6.2. SOLUTION The two equations are In(PJP) = gMhlRTatm In(PJP) = L\vapH(To -T)/RTTo where p ° is the atmospheric pressure at sea level and p is the atmospheric pressure at a higher altitude. T° is the boiling point at sea level, and T is the boiling point at the higher altitude. Setting the right-hand sides 'of these equations equal to each other yields !-1-+ gMh T -T ° T atm~vapll Substituting the values from problem 6.2 yields ! --1- (9.8 m s-2)(0.0289 kg mol-l)(3.219 x 103 m) T- 373.15 + (298.15 K)(44,000 J mol-l) T = 363.58 K = 90.43 °C 6.4 Liquid mercury has a density of 13.690 9 cm-3, and solid mercury has a density of 14.193 9 cm-3, both being measuredat the melting point, -38.87 °C, at 1 bar pressure. The heat of fusion is 9.75 J g-l. Calculate the mel ting points of mercury under a pressureof (a) 10 bar and (b) 3540 bar. The observed melting point under 3540 bar is -19.9 °C. SOLUTION ( 1 1 ~T T(\I] -~ ) (234.3 K) 13.690 -14.193 ) cm 3 9 -1 -= 1 s= = [~ 1 ] (10-2mcm-l)3 L\P ~fusH 9.75 J g- = 6.22 x 10-8 K Pa-l (a) ~T= (6.22 x 10-8 K Pa-l)(9 x 100,000 Pa) = 0.056 K t = -38.87 + 0.06 = -38.81 °C (b) ~T= (6.22 x 10-8 K Pa-l)(3539 x 100,000 Pa) = 22.0 K t = -38.87 + 22.0 = -16.9 °C 6.5 From the L\tG° of Bf2(g) at 25 °C, calculate the vapor pressureof Br2(1). The pure liquid at 1 bar and 25 °C is taken as the standard state. Chapter 6/Phase Equilibrium 97 illL UTlOO PBr2 Br2(1) = Br2(g) K=-p;- L\GO= -RT In ( ~ ) 3110 I J mol-l ] PBr2 - -p;-= exp [ -(8.314 J K-l mol-l)(298.15 K) nol-l)(298.15 K) ] PEr = 0.285 bar 2 Calculate l1Go for the vaporization of water at O °C using data in Appendix C.2 and 6.6 assuming that MlO for the vaporization is independent of temperature. Use l1Go to calculate the vapor pressure of water at O °C. SQLUTION H20(1) = H20(g) ~GO(298.15 K) = -228.572 + 237.129 = 8.577 kJ mol-l MlO(298.15 K) = -241.818 + 285.830 = 44.012 kJ mol-l Using the equation in problem 4.10(a) l1G1T2 ( T l1G2=T!+AH 1- i ) l1G(O °C) = 8.577(~) + 44.012( 1 -~) = 7.858 + 3.690 = 11.548 kJ mol-l = -RTln (~) -11.548 p = Po exp(S.3145 x 10-3 x 273.15) = 6.190 x 10-3 bar 6.7 The change in Gibbs energy for the conversion of aragonite to calcite at 25 °C is -1046 J mol-l. The density of aragonite is 2.93 9 cm-3 at 15 °C and the density of calcite is 2.71 9 cm-3. At what pressureat 25 °C would these two forms of CaCO3 be in equilibrium? SOLUTION 98 Chapter 6/Phase Equilibrium 1102 -1101 = 11V(P -1) 0 + 1046 J mol-l = (2.77 x 10-6 m3 mol-l)(p -1) p = 3780 bar *6.8 n-Propyl alcohol has the following vapor pressures: t/OC 40 60 80 100 P/kPa 6.69 19.6 50.1 112.3 Plot these data so as to obtain a nearly straight line, and calculate (a) the enthalpy of vaporization, (b) the boiling point at 1 bar, and (c ) the boiling point at 1 atm. SOLUTION: In(PJkPs.) 5 4 3 2 1 1000JT 2.6 2.8 3 3.2 3.4 Ll£..1 vapllvapll: slope = -2.34 x 103 KK= = I" '1{\'1\ 'Q '11A T V-l n1-1\ (a) (2.303) (8.314 J K-l mol-l) ~vapll = 44.8 kJ mol-l In ~ = ~ (! -1-- (b) 100 8.3145 T- 373.15 ) T- -8.3145 112.31 1 = 370.18 K = 97.03 °C 44 800 In 100 + 373.15 T- 1 (,..\ -~ 112.3 1 = 370.51 K = 97.36 °C 44 800 In 101.325 + 3nI5 6.9 For uranium hex afluoride the vapor pressures (in Pa) for the solid and liquid are given by In Ps = 29.411 -5893.5/T In PI = 22.254 -3479.9/T Calculate the temper:ature and pressure of the triple point. Chapter 6/Phase Equilibrium 99 SOLUTION The heats of vaporization and of fusion of water are 2490 J g-l and 33.5 J g-l at 0 °C. The vapor pressureof water at 0 °C is 611 Pa. Calculate the sublimation pressureof ice at -15 °C, assuming that the enthalpy changesare independent of temperature. SOLUTION /1subH = /1fusH + /1vapH L\fusH liquid f solid = 33.5 + 2490 = 2824 J g-l .J, 6V.~U~ vapor 8subH (T2 -TI) - In~ -1.PI RTIT2£ - P2 = PI exp[~S~~[ J1.subll(T2 -TI) PI exp RTIT2 RTIT2 ] [ (2824 x 181 J mol-l)(-15 K) ] = , (611 Pa) exp (Sl(8.314 ~1LI. JT K-l1(-1 mol-l)(273.15mn1 K)(258.15 K) = 166 Pa 6.11 The sublimation pressures of solid CI2 are 352 Pa at -112 °C and 35 Pa at -126.5 °C. The vapor pressures of liquid Cl2 are 1590 Pa at -100 °C and 7830 Pa at -80 °C. Calculate (a) ~subll, (b) ~vapll, (c) ~fusH, and (d) the triple point. , SOLUTION !1 .u -!!J:lI1.. In f1 (a) su~~ -T2- TI PI (8.314 J K-I moI-I)(161.15 K)(146.65 K) I 352 = .14:5 K n35 = 31.4 kJ mol-l ~ H -(8.314 J K-l mol-l)(173.15 K)(193.15 K) l ~ (b) yap -20 K n 1590 = 22.2 kJ mol-l (c) ~fusH = ~subH -~vapH = 31.4- 22.2 = 9.2 kJ mol-l ~ 100 Chapter 6/Phase Equilibrium (d) For the solid 3777 r- For the liquid In p = In 1590 + K3I49200 ( r7m1 -f 1) = 13.762 - 1107 r- At the triple point 2670 29.300- ~= 13.762- J~7 T- -15.538~ = 172 K T= 172 K- 273.15 K = -101.2 °C The vapor pressureof solid benzene,C6fl6, is 299 Pa at -30 °C and 3270 Pa at 0 °C, 6.12 and the vapor pressure of liquid C6H6 is 6170 Pa at 10 °C and 15,800 Pa at 30 °C. From these data, calculate (a) the triple point of C6fl6, and (b) the enthalpy of fusion of C6H6. SOLUTION Calculate the enthalpy of sublimation of C6ll6. 1 ~ tisu~..A .u -30-(8.314) (253.15)(223.15) n 299 = 44,030 J mol-l Express sublimation pressures as a function of T. ~subll ~subS lnPsub=-~+ -y 44 030 ~subS At 273.15 K, In 3270 = -RT + ~ 44 030 + 27.481 In Psub = - RT Calculate the enthalpy of vaporization of C6lI6 1 ~ Uvap'A-A -J/ -~8.314)(283.15)(303.15)20 n 6170 = 33,550 J mol-l Express vapor pressures as a function of T. ilvapH ilvapS InPvap=- RT+ -y 33 550 ~vapS At 303.15 K, In 15,800 = -8.314(303.15) + R 33 550 In Pvap = -8.314 T + 22.979 At the triple point, In Psub = In Pvap (a) -~+.27.481 = -33 5~ + 22.979 RT RT Chapter 6/Phase Equilibrium 101 T= 279.99 K = 6.85 °C ( 44 030 p = exp -8.314 x 279.99 + 27.481 ) = 5249 Pa ~fusH = ~sub1l- ~vapH = 44.03 -33.55 = 10.48 kJ mol-l 6.13 The surface tension of toluene at 20 °C is 0.0284 N m-l, and its density at this temperatureis 0.866 9 cm-3.
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