Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap06

6 Physical transformations of pure substances Solutions to exercises Discussion questions E6.1(b) Refer to Fig 6.8 The white lines represent the regions of superheating and supercooling The chemical potentials along these lines are higher than the chemical potentials of the stable phases represented by the colored lines Though thermodynamically unstable, these so-called metastable phases may persist for a long time if the system remains undisturbed, but will eventually transform into the thermodynamically stable phase having the lower chemical potential Transformation to the condensed phases usually requires nucleation centers In the absence of such centers, the metastable regions are said to be kinetically stable E6.2(b) At 298 K and 1.0 atm, the sample of carbon dioxide is a gas (a) After heating to 320 K at constant pressure, the system is still gaseous (b) Isothermal compression at 320 K to 100 atm pressure brings the sample into the supercritical region The sample is now not much different in appearance from ordinary carbon dioxide, but some of its properties are (see Box 6.1) (c) After cooling the sample to 210 K at constant pressure, the carbon dioxide sample solidifies (d) Upon reducing the pressure to 1.0 atm at 210 K, the sample vapourizes (sublimes); and finally (e) upon heating to 298 K at 1.0 atm, the system has resumed its initial conditions in the gaseous state Note the lack of a sharp gas to liquid transition in steps (b) and (c) This process illustrates the continuity of the gaseous and liquid states E6.3(b) First-order phase transitions show discontinuities in the first derivative of the Gibbs energy with respect to temperature They are recognized by finite discontinuities in plots of H , U , S, and V against temperature and by an infinite discontinuity in Cp Second-order phase transitions show discontinuities in the second derivatives of the Gibbs energy with respect to temperature, but the first derivatives are continuous The second-order transitions are recognized by kinks in plots of H , U , S, and V against temperature, but most easily by a finite discontinuity in a plot of Cp against temperature A λ-transition shows characteristics of both first and second-order transitions and, hence, is difficult to classify by the Ehrenfest scheme It resembles a first-order transition in a plot of Cp against T , but appears to be a higher-order transition with respect to other properties See the book by H E Stanley listed under Further reading for more details Numerical exercises E6.4(b) Assume vapour is a perfect gas and ln p∗ p =+ 1 = ∗+ T T = vap H R R vap H vap H is independent of temperature 1 − ∗ T T ln p∗ p 8.314 J K−1 mol−1 58.0 + × ln 293.2 K 32.7 × 103 J mol−1 66.0 = 3.378 × 10−3 K −1 T = 3.378 × 10−3 K −1 = 296 K = 23◦ C INSTRUCTOR’S MANUAL 88 E6.5(b) Sm Vm dp = dT fus S = fus S assuming fus S Vm dp dT ≈ p T Vm Vm independent of temperature and = (152.6 cm3 mol−1 − 142.0 cm3 mol−1 ) × = (10.6 cm3 mol−1 ) × m3 106 cm3 (1.2 × 106 Pa) − (1.01 × 105 Pa) 429.26 K − 427.15 K × (5.21 × 105 Pa K−1 ) = 5.52 Pa m3 K −1 mol−1 = 5.5 J K−1 mol−1 fus H = Tf S = (427.15 K) × (5.52 J K−1 mol−1 ) = 2.4 kJ mol−1 E6.6(b) Use vap H RT d ln p = ln p = constant − dT vap H RT dependence must be equal, so T 3036.8 K vap H − =− T /K RT Terms with vap H = (3036.8 K)R = (8.314 J K−1 mol−1 ) × (3036.8 K) = 25.25 kJ mol−1 E6.7(b) (a) log p = constant − vap H RT (2.303) Thus vap H = (1625 K) × (8.314 J K −1 mol−1 ) × (2.303) = 31.11 kJ mol−1 (b) Normal boiling point corresponds to p = 1.000 atm = 760 Torr log(760) = 8.750 − 1625 T /K 1625 = 8.750 − log(760) T /K 1625 = 276.87 T /K = 8.750 − log(760) Tb = 276.9 K PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES E6.8(b) 89 Tf fus V Tf pM × p= × H fus fus H [Tf = −3.65 + 273.15 = 269.50 K] T = T = fus V fus S × p= (269.50 K) × (99.9 MPa)M × 8.68 kJ mol−1 ρ 1 − −3 0.789 g cm 0.801 g cm−3 = (3.1017 × 106 K Pa J−1 mol) × (M) × (+ 01899 cm3 /g) × m3 106 cm3 = (+ 5.889 × 10−2 K Pa m3 J−1 g−1 mol)M = (+ 5.889 × 10−2 K g−1 mol)M T = (46.07 g mol−1 ) × (+ 5.889 × 10−2 K g−1 mol) = + 2.71 K Tf = 269.50 K + 2.71 K = 272 K dn dm = × MH2 O dt dt E6.9(b) q where n = vap H dn dq/dt (0.87 × 103 W m−2 ) × (104 m2 ) = = dt 44.0 × 103 J mol−1 vap H = 197.7 J s−1 J−1 mol = 200 mol s−1 dm = (197.7 mol s−1 ) × (18.02 g mol−1 ) dt = 3.6 kg s−1 E6.10(b) The vapour pressure of ice at −5◦ C is 3.9 × 10−3 atm, or Torr Therefore, the frost will sublime A partial pressure of Torr or more will ensure that the frost remains E6.11(b) (a) According to Trouton’s rule (Section 4.3, eqn 4.16) vap H = (85 J K−1 mol−1 ) × Tb = (85 J K−1 mol−1 ) × (342.2 K) = 29.1 kJ mol−1 Solid Liquid Pressure c b Critical point Start d a Gas Temperature Figure 6.1 INSTRUCTOR’S MANUAL 90 (b) Use the Clausius–Clapeyron equation [Exercise 6.11(a)] p2 p1 ln vap H = R × 1 − T1 T2 At T2 = 342.2 K, p2 = 1.000 atm; thus at 25◦ C 29.1 × 103 J mol−1 8.314 J K−1 mol−1 ln p1 = − × 1 − 298.2 K 342.2 K = −1.509 × 1 − 333.2 K 342.2 K = −0.276 p1 = 0.22 atm = 168 Torr At 60◦ C, 29.1 × 103 J mol−1 8.314 J K−1 mol−1 ln p1 = − p1 = 0.76 atm = 576 Torr E6.12(b) T = Tf (10 MPa) − Tf (0.1 MPa) = Tf pM fus H ρ = 6.01 kJ mol−1 fus H (273.15 K) × (9.9 × 106 Pa) × (18 × 10−3 kg mol−1 ) 6.01 × 103 J mol−1 T = 1 − −3 9.98 × 10 kg m 9.15 × 102 kg m−3 = −0.74 K × Tf (10 MPa) = 273.15 K − 0.74 K = 272.41 K E6.13(b) vap H = vap U + vap (pV ) vap H = 43.5 kJ mol−1 vap (pV ) = p vap V = p(Vgas − Vliq ) = pVgas = RT [per mole, perfect gas] vap (pV ) = (8.314 J K−1 mol−1 ) × (352 K) = 2927 J mol−1 Fraction = vap (pV ) vap H = 2.927 kJ mol−1 43.5 kJ mol−1 = 6.73 × 10−2 = 6.73 per cent E6.14(b) Vm = M 18.02 g mol−1 = 1.803 × 10−5 m3 mol−1 = ρ 999.4 × 103 g m−3 2γ Vm 2(7.275 × 10−2 N m−1 ) × (1.803 × 10−5 m3 mol−1 ) = rRT (20.0 × 10−9 m) × (8.314 J K −1 mol−1 ) × (308.2 K) = 5.119 × 10−2 p = (5.623 kPa)e0.05119 = 5.92 kPa PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES E6.15(b) 91 γ = 21 ρghr = 21 (0.9956 g cm−3 ) × (9.807 m s−2 ) × (9.11 × 10−2 m) × (0.16 × 10−3 m) × 1000 kg m−3 g cm−3 = 7.12 × 10−2 N m−1 E6.16(b) pin − pout = 2γ 2(22.39 × 10−3 N m−1 ) = r (220 × 10−9 m) = 2.04 × 105 N m−2 = 2.04 × 105 Pa Solutions to problems Solutions to numerical problems P6.3 (a) (b) dp = dT vap S = vap H [6.6, Clapeyron equation] Tb vap V 14.4 × 103 J mol−1 = + 5.56 kPa K−1 = (180 K) × (14.5 × 10−3 − 1.15 × 10−4 ) m3 mol−1 vap V dp dp vap H × p 11, with d ln p = = p dT RT −1 (14.4 × 10 J mol ) × (1.013 × 105 Pa) = = + 5.42 kPa K −1 (8.314 J K−1 mol−1 ) × (180 K)2 The percentage error is 2.5 per cent P6.5 (a) ∂µ(l) − ∂p T ∂µ(s) = Vm (l) − Vm (s)[6.13] = M ρ ∂p T 1 −1 − = (18.02 g mol ) × −3 1.000 g cm 0.917 g cm−3 = −1.63 cm3 mol−1 (b) ∂µ(g) − ∂p T ∂µ(l) = Vm (g) − Vm (l) ∂p T = (18.02 g mol−1 ) × 1 − 0.598 g L−1 0.958 × 103 g L−1 = + 30.1 L mol−1 At 1.0 atm and 100◦ C , µ(l) = µ(g); therefore, at 1.2 atm and 100◦ C à(g)à(l) (as in Problem 6.4) P6.7 (30.1 ì 10−3 m3 mol−1 ) × (0.2) × (1.013 × 105 Pa) ≈ + 0.6 kJ mol−1 Since µ(g) > µ(l), the gas tends to condense into a liquid pH2 O V The amount (moles) of water evaporated is ng = RT The heat leaving the water is q = n vap H The temperature change of the water is T = −q , n = amount of liquid water nCp,m Vvap p = INSTRUCTOR’S MANUAL 92 T = Therefore, −pH2 O V vap H RT nCp,m −(23.8 Torr) × (50.0 L) × (44.0 × 103 J mol−1 ) = g (62.364 L Torr K−1 mol−1 ) × (298.15 K) × (75.5 J K −1 mol−1 ) × 18.02250 g mol−1 = −2.7 K The final temperature will be about 22◦ C P6.9 (a) Follow the procedure in Problem 6.8, but note that Tb = 227.5◦ C is obvious from the data (b) Draw up the following table θ/◦ C T /K 1000 K/T ln p/Torr 57.4 330.6 3.02 0.00 100.4 373.6 2.68 2.30 133.0 406.2 2.46 3.69 157.3 430.5 2.32 4.61 203.5 476.7 2.10 5.99 227.5 500.7 2.00 6.63 The points are plotted in Fig 6.2 The slope is −6.4 × 103 K, so implying that = +53 kJ mol−1 vap H − vap H = −6.4 × 103 K, R 2.0 2.2 2.4 2.6 2.8 3.0 Figure 6.2 P6.11 (a) The phase diagram is shown in Fig 6.3 Liquid –2 Solid Liquid–Vapour Solid–Liquid –4 Vapour –6 –8 100 200 300 400 500 600 Figure 6.3 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES 93 (b) The standard melting point is the temperature at which solid and liquid are in equilibrium at bar That temperature can be found by solving the equation of the solid–liquid coexistence curve for the temperature = p3 /bar + 1000(5.60 + 11.727x)x, So 11 727x + 5600x + (4.362 × 10−7 − 1) = The quadratic formula yields 727) −1 ± + 4(11 −5600 ± {(5600)2 − 4(11 727) × (−1)}1/2 56002 x= = 2(11 727) 11727 5600 1/2 The square root is rewritten to make it clear that the square root is of the form {1 + a}1/2 , with a 1; thus the numerator is approximately −1 + (1 + 21 a) = 21 a, and the whole expression reduces to x ≈ 1/5600 = 1.79 × 10−4 Thus, the melting point is T = (1 + x)T3 = (1.000179) × (178.15 K) = 178.18 K (c) The standard boiling point is the temperature at which the liquid and vapour are in equilibrium at bar That temperature can be found by solving the equation of the liquid–vapour coexistence curve for the temperature This equation is too complicated to solve analytically, but not difficult to solve numerically with a spreadsheet The calculated answer is T = 383.6 K (d) The slope of the liquid–vapour coexistence curve is given by dp vap H = dT T vap V −− so vap H −− = T vap V −− dp dT The slope can be obtained by differentiating the equation for the coexistence curve dp d ln p d ln p dy =p =p dT dT dy dT dp = dT 10.418 − 15.996 + 2(14.015)y − 3(5.0120)y − (1.70) × (4.7224) × (1 − y)0.70 y2 p × Tc At the boiling point, y = 0.6458, so dp = 2.851 × 10−2 bar K−1 = 2.851 kPa K−1 dT and P6.12 vap H −− = (383.6 K) × (30.3 − 0.12) L mol−1 × (2.851 kPa K −1 ) = 33.0 kJ mol−1 1000 L m−3 The slope of the solid–vapour coexistence curve is given by −− dp sub H = dT T sub V −− so sub H −− = T sub V −− dp dT The slope can be obtained by differentiating the coexistence curve graphically (Fig 6.4) INSTRUCTOR’S MANUAL 94 60 50 40 30 20 10 144 146 148 150 152 154 156 Figure 6.4 dp = 4.41 Pa K−1 dT according to the exponential best fit of the data The change in volume is the volume of the vapour Vm = RT (8.3145 J K−1 mol−1 ) × (150 K) = = 47.8 m3 p 26.1 Pa So sub H −− = (150 K) × (47.8 m3 ) × (4.41 Pa K −1 ) = 3.16 × 104 J mol−1 = 31.6 kJ mol−1 Solutions to theoretical problems P6.14 P6.16 ∂Gβ ∂ G ∂Gα = − = Vβ − V α ∂p T ∂p T ∂p T Therefore, if Vβ = Vα , G is independent of pressure In general, Vβ = Vα , so that though small, since Vβ − Vα is small pV Amount of gas bubbled through liquid = RT (p = initial pressure of gas and emerging gaseous mixture) m Amount of vapour carried away = M Mole fraction of vapour in gaseous mixture = Partial pressure of vapour = p = m M m M + pV RT m M G is nonzero, m M pV + RT ×p = p PmRT VM mRT PVM +1 = mP A , mA + A= RT PVM For geraniol, M = 154.2 g mol−1 , T = 383 K, V = 5.00 L, p = 1.00 atm, and m = 0.32 g, so A= (8.206 × 10−2 L atm K−1 mol−1 ) × (383 K) = 40.76 kg−1 (1.00 atm) × (5.00 L) × (154.2 × 10−3 kg mol−1 ) Therefore p= (0.32 × 10−3 kg) × (760 Torr) × (40.76 kg−1 ) = 9.8 Torr (0.32 × 10−3 kg) × (40.76 kg−1 ) + PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES P6.17 p = p0 e−Mgh/RT [Box 1.1] vap H × p = p∗ e−χ χ = R 95 1 − ∗ T T [6.12] Let T ∗ = Tb the normal boiling point; then p ∗ = atm Let T = Th , the boiling point at the altitude h Take p0 = atm Boiling occurs when the vapour (p) is equal to the ambient pressure, that is, when p(T ) = p(h), and when this is so, T = Th Therefore, since p0 = p∗ , p(T ) = p(h) implies that e−Mgh/RT = exp − It follows that vap H R × 1 − Th Tb 1 Mgh = + Th Tb T vap H where T is the ambient temperature and M the molar mass of the air For water at 3000 m, using M = 29 g mol−1 1 (29 × 10−3 kg mol−1 ) × (9.81 m s−2 ) × (3.000 × 103 m) = + Th 373 K (293 K) × (40.7 × 103 J mol−1 ) 1 = + 373 K 1.397 × 104 K Hence, Th = 363 K (90◦ C) P6.20 Sm = Sm (T , p) ∂Sm dT + dSm = ∂T p ∂Sm dp ∂p T Cp,m ∂Sm = [Problem 5.7] ∂T p T ∂Vm ∂Sm =− [Maxwell relation] ∂p T ∂T p ∂Vm dp ∂T p ∂q ∂p Hm = Cp,m − T Vm α = Cp,m − αVm × [6.7] CS = ∂T s ∂T s Vm −− C(graphite) C(diamond) = 2.8678 kJ mol−1 at TC rG dqrev = T dSm = Cp,m dT − T P6.22 We want the pressure at which r G = 0; above that pressure the reaction will be spontaneous Equation 5.10 determines the rate of change of r G with p at constant T (1) (2) ∂ rG = r V = (VD − VG )M ∂p T where M is the molar mas of carbon; VD and VG are the specific volumes of diamond and graphite, respectively − C, p) may be expanded in a Taylor series around the pressure p − C = 100 kPa at T r G(T C, r G(T p) = rG + ∂ r G−− (TC, p −− ) (p − p −− ) ∂p T ∂ r G−− (TC, p −− ) (p − p −− )2 + θ(p − p −− )3 ∂p −− C (T, p −− ) + T INSTRUCTOR’S MANUAL 96 We will neglect the third and higher-order terms; the derivative of the first-order term can be calculated with eqn An expression for the derivative of the second-order term can be derived with eqn (3) ∂VG ∂VD M = {VG κT (G) − VD κT (D)}M [3.13] − ∂p T ∂p T T Calculating the derivatives of eqns and at TC and p −− ∂2 rG ∂p = (4) ∂ r G(TC, p −− ) = (0.284 − 0.444) × ∂p T (5) ∂ r G(TC, p −− ) ∂p cm3 g × 12.01 g mol = −1.92 cm3 mol−1 = {0.444(3.04 × 10−8 ) − 0.284(0.187 × 10−8 )} T × cm3 kPa−1 g × 12.01 g mol = 1.56 × 10−7 cm3 (kPa)−1 mol−1 It is convenient to convert the value of r G−− to the units cm3 kPa mol−1 8.315 × 10−2 L bar K−1 mol−1 103 cm3 −− × = 2.8678 kJ mol−1 rG L 8.315 J K−1 mol−1 (6) × 105 Pa bar = 2.8678 × 106 cm3 kPa mol−1 Setting χ = p − p −− , eqns and 3–6 give 2.8678 × 106 cm3 kPa mol−1 − (1.92 cm3 mol−1 )χ + (7.80 × 10−8 cm3 kPa−1 mol−1 )χ = when r G(TC, p) = One real root of this equation is rG −− χ = 1.60 × 106 kPa = p − p −− or p = 1.60 × 106 kPa − 102 kPa = 1.60 × 106 kPa = 1.60 × 104 bar Above this pressure the reaction is spontaneous The other real root is much higher: 2.3×107 kPa Question What interpretation might you give to the other real root? ... amount of liquid water nCp,m Vvap p = INSTRUCTOR S MANUAL 92 T = Therefore, −pH2 O V vap H RT nCp,m −(23.8 Torr) × (50.0 L) × (44.0 × 103 J mol−1 ) = g (62.364 L Torr K−1 mol−1 ) × (298.15 K) × (75.5... curve is given by −− dp sub H = dT T sub V −− so sub H −− = T sub V −− dp dT The slope can be obtained by differentiating the coexistence curve graphically (Fig 6.4) INSTRUCTOR S MANUAL 94 60... −5◦ C is 3.9 × 10−3 atm, or Torr Therefore, the frost will sublime A partial pressure of Torr or more will ensure that the frost remains E6.11(b) (a) According to Trouton’s rule (Section 4.3, eqn