Assignment #8 - Solutions

Transcription

1 Assignment #8 - Solutions Individual Exercises:. Consider the G ( ) and G () plots shown in M&S Figs. 9. and 9.. What restrictions, if any, apply to the slopes and curvatures of such plots? he slopes on these plots are the first derivatives of G, G = S G = + V (hese inequalities are required y the positivity of entropy, ensured y the 3 rd Law, and of volume, which is positive y definition.) he curvatures are the second derivatives: G S = C = G V = + = V κ he requirement that oth of these nd derivatives e negative derives from the conditions of thermal ( C ) and mechanical staility ( κ ).. Use the steam chart from the opic 9 handout to estimate the following: (a) the mole fraction of liquid water in a 5 L drum containing 5 kg of saturated steam at 5 C. () the work done and the heat released when steam is compressed isothermally and reversily at C etween pressures of. ar and ar. (a) he specific volume corresponding to these conditions is. m 3 kg -. Measuring the distance etween the coexistence lines as illustrated in red on the chart shown elow, I used the lever rule to estimate the liquid and or mole fractions to e: L 5. x L = =.75 and x V = xl. 5 L + L () he process descried corresponds to the path # # shown in green on the chart elow. Since the process is reversile and isothermal the heat involved can e calculated from

2 ˆ q rev = dsˆ = ( Sˆ Sˆ ) he caret on these symols denotes the fact that I m doing the calculation for kg of water. (I forgot to state the quantity on the prolem, so I ll report the results on this asis.) Reading the values of entropy from the chart, I estimate: qˆ = (473.5 K)( kj K - kg - ) = -3. MJ/kg (Since q is negative the system gives off heat, as expected in an isothermal compression.) Determining the work required requires a little more work (pun intended): wˆ rev = Uˆ qˆ rev = Hˆ ( Vˆ) qˆ rev All of the pieces can e read from the diagram as follows: Ĥ = MJ/kg = -. MJ/kg ( Vˆ) = Vˆ Vˆ = ( a m 3 kg - ) = -.9 MJ/kg so ŵ rev = +.79 MJ/kg he work is positive as it must e for a compression. (Note the relatively small (~%) difference that the V term makes compared to the thermal terms i.e. the relatively small difference etween Ĥ and Uˆ.) # + L L #

3 3. (a) What is the relationship etween chemical potential and fugacity? () M&S 9-38: Show that Eqs. 9.3 and 9.3 for µ (, ) for a monatomic ideal gas are 5 equivalent to using the relation G = H S with H = R and S given y Eq (a) Fugacity can e defined: f G exp G R IG Since the chemical potential is the same as the molar Gis energy for a pure sustance the relationship is simply f µ µ exp R IG () I will ignore the route suggested y the prolem and demonstrate the relationship etween S, H, and µ in Eq. 9.9 in a different (and in my opinion more logical) way. Eq derives from the general relation: S ln Q = k B V, N + k B ln Q Similarly, the relationship etween the internal energy and Q is: U ln Q = k B V, N Using these two expressions, G can e written in terms of Q via: G = H S = U S + V where the use of V = U S + Nk = k ln Q + Nk B B B = Nk specializes these expressions for ideal gas prolems. B N For a partition function of the ideal gas form, Q( N, V, ) = q( V, ) / N!, G can e rewritten as 3

4 G = ( Nk B ln q k B ln N!) + Nk B = ( Nk B ln q Nk B ln N + Nk B ) + Nk B q = Nk B ln N (he nd of these equations makes use of Stirling s approximation ln N! N ln N N.) Considering mole of gas, N N A, G G = µ, and N A k B = R, so that this expression transforms into: q( V, ) q( V, ) V q( V, ) V µ = R ln = R ln = R ln N. A V N A V N 4. (a) M&S 9-4: he or pressures of solid and liquid chlorine are given y S ln( / torr) = K / L ln( / torr) = K / where is the asolute temperature. Calculate the temperature and pressure at the triple point of chlorine. () M&S 9-6: he slope of the melting curve of methane is given y: d / d = (.8446ar K ) from the triple point to aritrary temperatures. Using the fact that the temperature and pressure at the triple point are 9.68 K and.74 ar, calculate the melting pressure of methane at 3 K. Equilirium etween the liquid and solid requires equal or pressures (i.e. the pressures of all three phases must e equal). Equating the two expressions for ln() and solving for temperature one finds: K / = K / tp / K = = he pressure is given y inserting this temperature into either equation: tp 3777 / torr = exp4.3 = α () For a relationship of the form ( d / d ) = a the pressure at any temperature can e determined given data at some reference (,) point via: 4

5 ( ) = ( ) + α a a d = ( ) + α + α + [ ] + α I actually performed the evaluation of this integral in Mathcad using the triple point data for reference with the result: ar := 5 a tp := 9.68K tp :=.74 ar a :=.8446ar K.85 ( ) := tp + tp ax.85 dx ( 3 K) ar = Nearly all oiling and melting points in the literature are reported as the normal ( atm) values. he difference etween the normal and standard melting points of a sustance is completely negligile (Why?). However, the difference etween the normal and standard oiling points, although small, may not e negligile for some purposes. (a) Using the same approximations that led to Eq. 9. in the text, derive the expression: R S () Use this expression and routon s rule to otain an approximate numerical estimate of / ( n s) / n, the relative difference in the normal and standard oiling points, applicale to all sustances (at least to within the approximations specified). (c) Estimate the standard oiling points of H O, CCl 4, and enzene using the data taulated in the opic 9 handout. Compare your results to the accurate values of 37.77, , and K determined for these liquids. (a) he (exact) Clapeyron equation is: d d = V S Neglecting the volume of the liquid compared to the volume of the or, Assuming the or to e an ideal gas, V R /, this equation ecomes V V V. d d R S or d R S d (Although not explicitly suscripted, and refer to values under liquid or coexistence conditions.) For pressures of atm or ar the orization transition is referred 5

6 to as oiling so. For a difference as small as that etween atm and ar, so that one finally otains the desired result: d R S () routon s rule states that the entropy of orization at the (normal) oiling point is approximately.5 R. hus, n n s R S n n s.5 atm (atm ar) 3 =.4 Within the approximations made aove the fractional difference etween the normal and standard oiling points is a constant. (c) he standard oiling points can e estimated from = s n with / either estimated using the universal value of.4-3 ( # in the tale elow) or the more accurate expression / ( R / S ). 38, which does not employ routon s rule ( # elow). I used an Excel spreadsheet to carry out oth calculations with the results taulated elow: Sustance S /R n # # s # s # s Exp. Err # /% Err # /% HO %.% CCl % -3.% C6H % -.7% he errors listed in this tale are the % errors in the difference rather than in the temperatures themselves. (he % error in the latter values would, of course e much smaller.) Only in the case of water is there a sustantial error in the simple (method #) estimate. In all cases the method # estimate provides to within a few percent. 5. M&S 9- with modifications. he or pressure of mercury from 4 C to from 3 C can e expressed y: 76.7 K ln( / torr) = 7.85 he density of the or at its normal oiling point is 3.8 g L - and that of the liquid is.7 g ml -. Estimate the molar enthalpy of orization of mercury at its normal oiling point. 6

7 (a) First solve this prolem using the exact relation etween ( d / d ) and H. () Solve this prolem again using the more approximate Clausius-Clapeyron relation. (c) Calculate the % error in the estimate of part () and identify the largest source of error in this approximation. he exact relationship etween the slope of the coexistence curve and H is: d d H = V or H d = d V he or pressure equation can e expressed in the form / torr = exp( a + / ) which is easily solved for () and for the slope of the coexistence curve: d ( ) = / torr = exp( a + / ) ln( / torr) a d (I ve kept the units on as a reminder that they are not SI units and will need to e converted for use in the Clapeyron equation.) Finally V is given y: V = M V L ρ ρ Using these equations in the Mathcad worklsheet shown on the following page provides the result: H = kj/mol. () he Clausius-Clapeyron equation can e written: d ln d H R or H R d ln d For the form of the or pressure curve provided, simple result: ( d ln / d ) = / and one has the H R = 58.7 kj/mol (c) he error in use of the approximate Clausius-Clapeyron equation is.6%. Given the relative simplicity of the use of this approximate equation, the error is small. here are only two possile sources for this error: i) neglect of the volume of the liquid or ii) the assumption 7

8 that the or ehaves like an ideal gas. As shown y the calculation at the end of the mathcad spreadsheet, nearly all of the error in this case comes from the assumption of ideality. torr = 33.3a L:= 3 m 3 R 8.34 J K := mol := 76 torr a := 7.85 := 76.7 K MW :=.59g mol d v := 3.8g L d l :=.7 3 gl ( ) := ln a torr dd( ) := torr exp a + n := ( ) n = 69.48K dd( n ) a K =.85 3 V MW := d v d l V =.5 m3 mol H a := dd n ( ) ( n V) H a 3 = J mol H := R H 3 = J mol %err H H a := H a %err =.68 V MW := d v d l V app := MW d v %err V app V := V %err =.3 V := MW d v V app := R n %err V app V := V %err = M&S 9-3: he pressures at the solid-liquid coexistence oundary of propane are given y.83 the empirical equation = where is in ar and in Kelvin. Given that = K and H = 3.53 kj mol -, calculate V at K. Solving the Clapeyron equation for V provides the relation: Carrying out this calculation in Mathcad (worksheet elow) yields ar = 5 a := 85.46K H := J mol V = H ( d / d ) V = 38.3 cm 3 mol -. ( ) ar H := + V := V m 3 = K d mol ( ) d d ( ).78 6 kg = d ms K 8

9 Group rolems: 7. (a) M&S 9-: lot the following data for the densities of liquid and gaseous ethane in equilirium with each other as a function of temperature, and determine the critical temperature of ethane. L G () M&S 9-: Use the data in the preceding prolem to plot ( ρ ρ ) / against - c, with c = 35.4 K. he resulting straight line is an empirical law called the law of rectilinear diameters. If this curve is plotted on the same figure as in the preceeding prolem, the intersection of the two curves gives the critical density, ρ c. (c) A variety of scaling laws descrie how the properties of a fluid vary as a function of the distance from the critical point. One such law descries the vanishing difference etween β the coexisting liquid and or densities as ( ρ L ρ V ) ( c ). β in this expression is termed a critical exponent. Determine the value of β for ethane using the data provided in M&S 9- and assuming that c = 35.4 K. (a), () he plot of coexisting densities is shown elow. Using an expanded version of this plot I estimated the critical temperature and density to e those indicated on the plot. he average density versus temperature is indeed a straight line, as expected from the law of rectilinear diameters. (As far as I know this law is purely empirical.) 5 Regressions: all points: β=.35±.4 last 5 pts: β=.376±.6 Density ρ/(mol dm -3 ) 5 5 c = 35.6 ±. K ρ c = 6.84 ±.7 mol dm -3 (ρ L ρ V )/(mol dm -3 ) emperature ( c -)/ c (c) Rather than plot - c itself, I chose to plot (- c )/ c so as to display the normalized displacement from the critical point. In the log-log representation, these data are close to linear, which means that the power law suggested is a good representation here. I estimated values of β y making linear fits to the data in this representation. Since the scaling ehavior that gives rise to this power-law dependence is only expected to e valid close to the critical point, I performed 9

10 two fits one using all of the data and one using only the five points closest to c. he results differ slightly. Based on the two fits, I would estimate β =.36±.. 8. Assuming that the formation of liquid water is a requirement for ice skating, and that the pressure-dependence of the freezing point of water is the only mechanism for producing the requisite liquid, estimate the lowest temperature (in C) ice-skating should e possile. Clearly discuss any assumptions and/or approximations you employ. his prolem requires a knowledge of the change of the melting temperature of ice with pressure, which can e otained in two ways. he simplest and most accurate method is to simply use the graph provided in M&S Fig I read points from this figure in order to otain the analytic representation: / K = ( / kar).45( / kar) he data and fit are plotted at the right. (Note that this expression for () predicts a value of 7.7 K at ar pressure, which is.4 K smaller than the correct value of 73.5 K. he discrepancy is simply due to the inaccuracy in reading data from the figure.) Melting oint / K ( / kar) 55 / K = ( / kar).45( / kar) ressure /kar Aside: As an instructive exercise, one can also estimate the required pressure dependence using the Clapeyron equation and the transition properties at atm pressure. he Clapeyron equation can e written in the form: d V = H d Making the (poor) approximation that the volume and enthalpy change of ion (or at least their ratio) is independent of, this expression integrates to:

11 In the case of water, V ( ) = ( )exp{ α ( )} with α. H = -.63 cm 3 /mol, H =6. kj/mol (from M&S Ex. 9-3) at = V atm and ( )=73.5 K so that α = a - = -.7 kar -. Given the small size of this exponent, for the pressures of interest, the exponential can e expanded and only the first order term kept, which provides the relation: / K = ( / kar) his result is shown for comparison as the dashed line in the figure aove. As would e anticipated the result derived via the Clapeyron equation provides the correct ehavior (apart from the displacement caused y inaccurate reading of the M&S graph) at low pressures. However, the assumption that V and H are pressure independent for these large pressures leads to inaccurate predictions at kar pressures. End Aside We must now consider how much pressure is likely to e generated y a skater. For skater mass m and skate area A, the pressure is given y = mg / A where g is the acceleration of gravity (9.86 m s - ). We can take as a typical ody mass m = 75 kg. he only difficult thing to estimate is the skate area. he lowest pressure will e otained y assuming that the entire area of the skate lade contacts the ice. An estimate for the area in this case might e A = cm 3 mm = 6-4 m. But does the entire skate lade really contact the ice at any instant? roaly not. If one assumes that only a fraction (say /3) of the length of a curved lade and only the width of an edge of the skate lade contacts the surface, the pressures generated would e much higher. Estimating an edge width to e ~µm, the area would now e ~6-6 m. Using this rather road spread of estimated skate areas, I computed the melting temperatures taulated elow. Area /m 3 ressure /kar f f / C f / F It would appear that except for the smallest estimates of skate area, the pressure-induced change in temperature is modest. If the lowest area estimate is reasonale, a 75 kg person should e ale to skate at temperatures as low as C. Of course, the aove judgment rests on skating eing an equilirium processes. his analysis has ignored the effects of friction, which are likely to sustantially lower the temperature at which skating is possile. (I say this mainly ased on the elief that I ve skated at temperatures lower than - C.

12 9. M&S 9-4: Use the or pressure data given in rolem 9-7 and the density data given in rolem 9-3 to calculate for methanol from the triple point (75.6 K) to the critical H point (5.6 K). lot your results. M&S 9-5: Use the results of the previous prolem to plot S of methanol from the triple point to the critical point. M&S 9-6: Use the or pressure data for methanol given in rolem 9-7 to plot ln against /. Using your calculations from prolem 9-4, over what temperature range do you think the Clausius-Clapeyron equation will e valid? (a) & () Noting that equation can e solved for V d H = d = ρ ρ, where the ρ I are molar densities, the Clapeyron V L H and written in the form: ρv ρl S is given y S = H /. Given the expressions for ( ), ρ V (), and ρ L () provided in M&S prolems 9-7 and 9-3 I calculated H ( ) and S ( ) and ln() using the Mathcad worksheet shown elow. In addition, given the remarkale linearity of the ln vs /, it was of interest to calculate the prediction for H ( ) that would e made assuming the validity of the Clausius-Clapeyron equation. Denoting this estimate HCC, the relevant equation is: d H CC ( ) = R. d his quantity was also calculated in the Mathcad worksheet shown elow. he plots of H ( ) and S ( ) and ln(/) follow. p := dl :=.579 dv := c := 5.6K x ( ) ar exp p := + p + p 3 x + p 4 x + p 5 x 3 + p 6 ( x).7 x ρ l () x := ρ c + dl ( x).35 + dl ( x) + dl 3 x + dl 4 x 3 ( x) ρ v () x := ρ c exp dv + dv ( x).35 + dv 3 ( x) + dv 4 ( x) x H () x := p := dl := dv := ρ c := 8.4mol L ρ v () x ρ l () x p 3 := dl 3 := tp := 75.6K d x dx x () dv 3 := p 4 := dl 4 :=.3577 x ( ) ( ) := c dv 4 := ( ) H () x S () x := c x p 5 := p 6 := H cc () x R c x d := x () dx x ()

13 S / (J K - mol - ) H / (kj/mol) H S CC emperature ln(/ar) kj/mol 4 kj/mol / /( -3 K - ) he plots of H ( ) and S ( ) display the ehavior expected: oth functions approach zero as c. More remarkale is the fact that ln is very nearly a linear function of / over the entire range of temperatures considered: all the way from the triple point to the critical point of the fluid. Over this temperature range the pressure varies y a factor of 8! Such approximately linearity is not confined to methanol, ut is oserved in many other systems, i.e. in the examples of Ar, ethane, CCl 4, and enzene I showed in class. he Clausius-Clapeyron equation predicts ln to e linear in / when H is temperature independent. But in the present case, and in general, H is far from constant, especially when approaches c. hus, the Clausius-Clapeyron equation does not provide correct predictions over this whole range of temperatures. o explore the validity of the Clausius-Clapeyron equation further, I calculated H CC ( ) as descried aove and plotted the result (dashed curve) together with the exact values of H ( ). From this comparison one finds that the estimates of H ( ) provided y the Clausius-Clapeyron are accurate to aout the 5% level for temperatures etween tp and ~.6 c. However, at temperatures higher than.8 c the estimates are in error y more than 5%. It is interesting to conjecture aout the origins of the approximate linearity of ln versus /. One can rewrite the Clausius equation in the form: 3

14 d d = H V d or d ln d(/ ) = R H Z where Z = ( V / R ). From this last equation, it would seem that the near linearity oserved here must stem from the near constancy of the ratio ( H / Z) throught the entire temperature range. (Why this ratio should e constant is another question ) 4

15 . his prolem examines the quality of corresponding states correlations under su-critical conditions. (a) M&S -9: Look up the oiling points of the gases listed in ale.5 and plot these values verus the critical temperatures c. Is there a correlation? ropose a reason to justify your conclusion from the plot. () You should find a good correlation etween and c in part (a), ut you should ask yourself whether this is really corresponding states ehavior. (What s eing done with the pressure variale?) Use the perspectives provided y Ch. 9 to rationalize why the correlation is as good as it is. (c) Compare f and c in the same manner as in part (a). Note that f provides a good approximation to the triple-point temperature, tp in most cases. Explain why this is true and comment on the quality of the corresponding states ehavior you oserve. he data I used for this comparison are taulated elow. I have omitted the He point, since it is not expected to follow corresponding states correlations (due to its quantal nature), and I have also added a few more sustances to the list in ale.5 of M&S. Sustance c c /ar f -CS Neon Argon Hydrogen Nitrogen Oxygen Caron monoxide Chlorine Water Ammonia Methane Ethane ropane Butane Methylpropane entane Benzene Caron tetrachloride rimethylpantane Bromine Hydrogen fluoride Hydrogen chloride Ethanol Diethyl ether lots of the correlations etween, f, and c are provided elow (left-hand panel). 5

16 5 4 3 Boiling oint Boiling oint f 4 3 Melting oint C.S. oint =.4 c Critical emperature c 4 6 Critical emperature c One finds that the correlation etween and c is fairly good, ut far from perfect. A listing of some characteristics of this and other correlations is provided elow. f -CS Fit: R Fit: slope Fit: std. err Ratio: average Ratio: std. dev he st three rows laeled Fit refer to linear fits (with a constraint of zero intercept) of and f, versus c. he final two rows refer to ratios / c and f / c -- i.e. to the reduced oiling and melting points. From these data one finds that the oiling point of a sustance is located at (.6±.3) c. Is this ehavior an example of the law of corresponding states? Yes, ut the law of corresponding states has not een rigorously applied here. he oiling point refers to a pressure of atm, and this pressure represents different reduced conditions (i.e. r =/ c ) for the different sustances. A correct corresponding states analysis entails comparison of sustances under the same ( r, r ) conditions, not the same ( r, ) conditions. What should e done here is to compare the temperatures (or reduced temperatures) for which the coexisting liquid-or pressure reaches a particular reduced value. For the range of sustances considered here, atm corresponds to reduced pressures. r. 4, with an average value of r =.4. It is possile to use the Clausius-Clapeyron equation and the enthalpies of orization of these liquids to estimate true corresponding states temperatures y computing the temperature at which =.4 c for all of the sustances. he last two 6

17 columns laeled and -CS show the temperature change and corrected temperature for a true corresponding states comparison. he right-hand figure elow shows the effect of making this more proper corresponding states comparison. he proper and improper comparisons do not differ apprecialy. From the taulated values of, one finds that, for most simple sustances, the differences in the normal oiling temperatures and the more proper corresponding states temperatures is small. Large corrections are found primarily for sustances with strong hydrogen onding interactions. Since these latter sustances are not expected to fit into the same corresponding states correlations as simple non-hydrogen onding sustances, the quality of the more proper corresponding states analysis is little changed (see the statistics tale). he normal melting point and triple point of most sustances should not differ y more than a degree. Examples culled from the textook are [sustance ( tp / f )]: O (54.3 / K), I (386 / 387 K), enzene (78.7 / 78.7 K), methane (9.68 / 9.7 K), and water (73.6 / 73.5). Why should these temperatures e so similar? he difference etween the triple point and the normal melting point of a sustance can e estimated from f tp tp V H tp where = ( atm tp ). Consider enzene as a typical case (since data is availale from M&S p. 36). V =.3 cm 3 mol - and H = 9.95 kj mol - so / =.3-4 /ar -4 (since tp << ar). his result is typical in the sense that the or pressures of solids are usually less than ar, and the ratio of ( V / H ) is not expected to e greatly different from the value -4. hus, in general one should expect tp = f to within % or etter. he triple point is a unique point in the phase diagram of any pure sustance. Having confirmed that the normal melting point provides a good measure of the temperature at the triple point, if corresponding states correlations were to apply to oth the fluid and solid states, one would expect there to e a good correlation etween f and c. he comparisons made here show that there is some degree of correlation, ut it is much poorer than is the case with v. he ion or triple point temperature of a sustance can e estimated to e (.4±.) c, ut the scatter aout this correlation is considerale. he reason for the scatter is that the details of molecular shape and interaction type are more important for determining the properties of ordered solid phases than they are for determining the properties of disordered fluid phases. hus, the underlying idea upon which corresponding states ehavior rests, that molecular details are unimportant apart from a generic size and interaction strength parameter, is a poorer approximation for solid phases. 7