6 Physical transformations of pure substances

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1 6 Physical transformations of pure substances E6.b E6.2b E6.3b E6.4b Solutions to exercises Discussion questions Refer to Fig 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. At 298 K and.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 00 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.. c After cooling the sample to 20 K at constant pressure, the carbon dioxide sample solidifies. d Upon reducing the pressure to.0 atm at 20 K, the sample vapourizes sublimes; and finally e upon heating to 298 K at.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. 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 C p. 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 C p 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 C p 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 Assume vapour is a perfect gas and vap H is independent of temperature ln p p + vaph R T T T T + R p ln vap H p 293.2K J K mol ln J mol 66.0 T K K 296 K 23 C

2 88 INSTRUCTOR S MANUAL E6.5b S m V m fus S V m V m p T assuming fus S and V m independent of temperature. fus S 52.6cm 3 mol 42.0cm 3 mol.2 06 Pa Pa K K 0.6cm 3 mol m cm Pa K 5.52 Pa m 3 K mol 5.5JK mol fus H T f S K 5.52JK mol E6.6b Use 2.4 kj mol vap H dlnp 2 ln p constant vaph Terms with T dependence must be equal, so K T/K vaph vap H KR 8.34 J K mol K kj mol E6.7b a log p constant vaph Thus vap H 625 K 8.34 J K mol kj mol b Normal boiling point corresponds to p.000 atm 760 Torr log T/K log760 T/K 625 T/K log T b 276.9K

3 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES 89 E6.8b E6.9b T fusv fus S p T f fus V fus H [T f K] T K 99.9MPaM 8.68 kj mol p T f pm fus H ρ 0.789gcm gcm KPaJ mol M cm 3 /g m cm KPam 3 J g molm Kg molm T g mol Kg mol +2.7 K T f K K 272 K dm dt dn dt dm dt dn dt M q H 2 O where n vap H dq/dt vap H Wm m J mol 97.7Js J mol 200 mol s 97.7 mol s 8.02 g mol 3.6kgs E6.0b The vapour pressure of ice at 5 Cis atm, or 3 Torr. Therefore, the frost will sublime. A partial pressure of 3 Torr or more will ensure that the frost remains. E6.b a According to Trouton s rule Section 4.3, eqn 4.6 vap H 85 J K mol T b 85 J K mol 342.2K 29. kj mol Solid Liquid c b Pressure Critical point d Start Gas a Temperature Figure 6.

4 90 INSTRUCTOR S MANUAL b Use the Clausius Clapeyron equation [Exercise 6.a] p2 ln vaph p R T T 2 E6.2b At T K,p atm; thus at 25 C J mol ln p 8.34 J K mol 298.2K 342.2K p 0.22 atm 68 Torr At 60 C, J mol ln p 8.34 J K mol p 0.76 atm 576 Torr T T f 0 MPa T f 0.MPa T f pm fus H fus H 6.0 kj mol 333.2K 342.2K ρ { } K Pa kg mol T J mol { } kg m kg m K T f 0 MPa K 0.74 K K E6.3b vap H vap U + vap pv vap H 43.5 kj mol vap pv p vap V pv gas V liq pv gas [per mole, perfect gas] vap pv 8.34 J K mol 352 K 2927 J mol Fraction vappv vap H kj mol 43.5 kj mol per cent E6.4b V m M ρ 8.02 g mol gm m 3 mol 2γV m r Nm m 3 mol m 8.34 J K mol 308.2K p kpae kpa

5 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES 9 E6.5b γ 2 ρghr g cm m s m kg m 3 m gcm Nm E6.6b p in p out 2γ r Nm m Nm Pa P6.3 a Solutions to problems Solutions to numerical problems b vaps vap V vaph T b vap V [6.6, Clapeyron equation] J mol 80 K m kpa K mol [ vaph 2 p, with d ln p ] p J mol Pa 8.34 J K mol 80 K kpa K The percentage error is 2.5 per cent P6.5 a µl µs V m l V m s[6.3] M p T p T 8.02 g mol b.63 cm 3 mol ρ.000 g cm gcm 3 µg µl V m g V m l p T p T 8.02 g mol 0.598gL gl L mol At.0 atm and 00 C,µl µg; therefore, at.2 atm and 00 C µg µl V vap p as in Problem m 3 mol Pa kj mol Since µg >µl, the gas tends to condense into a liquid. P6.7 The amount moles of water evaporated is n g p H 2 OV The heat leaving the water is q n vap H The temperature change of the water is T q nc p,m, n amount of liquid water

6 92 INSTRUCTOR S MANUAL Therefore, T p H 2 OV vap H nc p,m 23.8 Torr 50.0L J mol L Torr K mol K 75.5JK mol 250 g 8.02 g mol 2.7K The final temperature will be about 22 C P6.9 a Follow the procedure in Problem 6.8, but note that T b C is obvious from the data. b Draw up the following table θ/ C T/K K/T ln p/torr The points are plotted in Fig The slope is K, so vaph R implying that vap H +53 kj mol K, Figure 6.2 P6. a The phase diagram is shown in Fig Liquid 2 4 Solid Vapour Liquid Vapour Solid Liquid Figure 6.3

7 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 p 3 /bar xx, So 727x x The quadratic formula yields x 5600 ±{ } / { } /2 ± The square root is rewritten to make it clear that the square root is of the form { + a} /2, with a ; thus the numerator is approximately a 2 a, and the whole expression reduces to x / Thus, the melting point is T + xt K 78.8 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 383.6K d The slope of the liquid vapour coexistence curve is given by vaph T vap V so vap H T vap V The slope can be obtained by differentiating the equation for the coexistence curve. p dlnp 0.48 p dlnp dy dy y y y y 0.70 p T c At the boiling point, y , so bar K 2.85 kpa K and vap H L mol 383.6K 000 L m kpa K 33.0 kj mol P6.2 The slope of the solid vapour coexistence curve is given by subh T sub V so sub H T sub V The slope can be obtained by differentiating the coexistence curve graphically Fig. 6.4.

8 94 INSTRUCTOR S MANUAL Figure Pa K according to the exponential best fit of the data. The change in volume is the volume of the vapour V m p J K mol 50 K 47.8m 3 26.Pa So sub H 50 K 47.8m Pa K J mol 3.6 kj mol Solutions to theoretical problems G Gβ Gα P6.4 V β V α p T p T p T Therefore, if V β V α, Gis independent of pressure. In general, V β V α, so that G is nonzero, though small, since V β V α is small. P6.6 Amount of gas bubbled through liquid pv p initial pressure of gas and emerging gaseous mixture Amount of vapour carried away m M Mole fraction of vapour in gaseous mixture Partial pressure of vapour p m M m M + pv m M m M + pv p p m PVM m PVM + mp A ma +, A PVM For geraniol, M 54.2 g mol,t 383 K,V 5.00 L,p.00 atm, and m 0.32 g, so A L atm K mol 383 K.00 atm 5.00 L kg mol 40.76kg Therefore p kg 760 Torr 40.76kg kg kg Torr

9 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES 95 P6.7 p p 0 e Mgh/ [Box.] p p e χ χ vaph R T T [6.2] Let T T b the normal boiling point; then p atm. Let T T h, the boiling point at the altitude h. Take p 0 atm. Boiling occurs when the vapour p is equal to the ambient pressure, that is, when pt ph, and when this is so, T T h. Therefore, since p 0 p, pt ph implies that { e Mgh/ exp vaph R It follows that + Mgh T h T b T vap H } T h T b where T is the ambient temperature and M the molar mass of the air. For water at 3000 m, using M 29 g mol T h 373 K kg mol 9.8 m s m 293 K J mol 373 K K Hence, T h 363 K 90 C P6.20 S m S m T, p Sm Sm ds m + T p p T Sm C p,m Sm Vm [Problem 5.7] [Maxwell relation] T p T p T T p Vm dq rev T ds m C p,m T T p q p C S C p,m TV m α C p,m αv m H m [6.7] T s T s V m P6.22 Cgraphite Cdiamond r G kj mol at TC. We want the pressure at which r G 0; above that pressure the reaction will be spontaneous. Equation 5.0 determines the rate of change of r G with p at constant T. r G r V V D V G M p T where M is the molar mas of carbon; V D and V G are the specific volumes of diamond and graphite, respectively. r GTC, pmay be expanded in a Taylor series around the pressure p 00 kpa at TC. 2 r GTC, p r G TC, p r G TC, p + p p p T + 2 r G TC, p 2 p 2 p p 2 + θp p 3 T

10 96 INSTRUCTOR S MANUAL 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. 2 { } r G VD VG p 2 M {V G κ T G V D κ T D}M [3.3] p T T p T Calculating the derivatives of eqns and 2 at TC and p r GTC, p cm g cm 3 mol p T g mol 2 r GTC, p p 2 T { } cm 3 kpa 2.0 g g mol cm 3 kpa mol It is convenient to convert the value of r G to the units cm 3 kpa mol r G kj mol L bar K mol 0 3 cm Pa 8.35 J K mol L bar 6 r G cm 3 kpa mol Setting χ p p, eqns 2 and 3 6 give cm 3 kpa mol.92 cm 3 mol χ cm 3 kpa mol χ 2 0 when r GTC, p 0. One real root of this equation is χ kpa p p or p kpa 0 2 kpa kpa bar Above this pressure the reaction is spontaneous. The other real root is much higher: kpa. Question. What interpretation might you give to the other real root?

Physical transformations of pure substances Boiling, freezing, and the conversion of graphite to diamond examples of phase transitions changes of

Physical transformations of pure substances Boiling, freezing, and the conversion of graphite to diamond examples of phase transitions changes of phase without change of chemical composition. In this chapter