In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the examle, are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic stes. Numbers without underlines (including final answers) are shown with the roer number of sig figs. 1 Exercise 4.1b g 153 Question How many hases are resent at each of the oints marked in Fig. 4.23b? a. 1 - oint is entirely inside a single hase region (not on any boundaries). b. 3 - oint on a boundary where 3 hases meet. c. 3 - (same exlanation as b) d. 2 - oint occurs on boundary between two hases (on a single line). 1
2 Exercise 4.5b g 153 Given Iron is heated from T i = 25 C to T f = 100 C. Over this temerature range S m = 53 J K 1 1. In terms of given variables, this is written: T i = 25 C T f = 100 C S m = 53 J K 1 1 Find By how much does its chemical otential change? Strategy First, temeratures are converted to Kelvin. T i = 100 + 273 C = 373 K T f = 1000 + 273 C = 1273 K We can use text book Equation 4.2 (g 143) to relate temerature changes to changes in chemical otential. which gives the differential ( ) µ = Sm dµ = S m Integrating dµ over the temerature range gives the change in chemical otential. Tf µ = Sm T i = Sm (T f T i ) J = 53 (1273 K 373 K) K = 47700 J = 47.7 kj where the integral has been easily solved, as we ve assumed S m is constant over this temerature range. 2
µ = 50 kj 3
3 Exercise 4.11b g 153 Given The vaour ressure of a liquid between 15 C and 35 C fits the exression log(/ torr) = 8.750 1625/(T/ K) Find Calculate... (a) the enthaly of vaorization (b) the normal boiling oint of the liquid Strategy We ll start with Equation 4.11 (g 148). Through rearrangement we solve for va H. d ln = vah RT 2 va H = RT 2 d ln To find d ln, the given exression for log(/ torr) can be converted to an exression for ln(/ torr) using the change of base formula. and this gives the equation log b x = log k x log k b Differentiating this exression by T gives ln(/ torr) = ln(10) (7.960 1625/ (T/ K)) ln(d) = ln(10) 1625 K T 2 (The Torr units were discarded as the units of ressure would only lead to constant shift in the exression for ln() and this constant is lost on differentiation.) Lastly we can substitute this exression for ln(d) into our earlier exression for va H. 4
va H = R T 2 ln(10) 1625 K T 2 = R ln(10) 1625 K = J 8.314 ln(10) 1625 K K = 31108.5 J = 31.1085 kj The normal boiling oint can be found by solving the given exression (T ) for the temerature at which (T ) = atmoshere where the atmosheric ressure atmoshere = 1.00 atm = 760 torr. log(760 torr/ torr) = 8.750 1625/(T/ K) T = 1625 K 8.750 log 760 = 276.9 K (a) va H = 31.11 kj (b) T = 280 K 5
4 Exercise 4.14b g 153 Given On a cold, dry morning after a frost, the temerature was T = 5 C and the artial ressure of water in the atmoshere fell to H2 O = 0.30 kpa. In terms of given variables, this is written: T = 5 C = 0.30 kpa Find (a) Will the frost sublime? (b) What artial ressure of water H2O would ensure that the frost remained? Strategy In this exercise, the second question answers the first question, in that once we know the artial ressure of water H2 O needed to ensure the frost remains, we know any H2 O below this will lead to frost sublimation. The artial ressure needed to revent sublimation is found by determing the solid-vaor ressure for water at this temerature; this is the ressure at this temerature on the coexistance curve for ice and water vaor. We know that if the atmoshere has a lower artial ressure of water than the solid-vaor ressure, then the water vaour will be favored over the solid and the ice will sublime. At higher artial ressures the solid hase is favored. We can find the solid-vaor ressure using Equation 4.12 (g 149) from our text book = e χ χ = subh R ( 1 T 1 ) T where we ve relaced va H in the original exression with sub H as we re concerned with the sublimation coexistence oint instead of the vaorization oint. The sublimation enthaly va H is found from the vaorization enthaly and the fusion enthaly, va H and fus H resectively. sub H = fus H + va H = 6.008 kj 1 + 44.016 kj 1 = 50.024 kj 1 = 5.0024 10 4 J 1 Using Equation 4.12 we can calculate the ressure at temerature T when we know the reference ressure at the reference temerature T. As we re solving for a (T ) on the solid/gas coexistence curve, we ll need the reference to also fall on this curve. Therefore we ll use the the trile oint of water as our reference giving the following values: T = 273.16 K 6
= 0.61173 kpa This allows us to find the solid-vaor ressure on the solid/gas coexistence curve for the temerature T = 5 C = 268 K. ( 1 T 1 ) T χ = subh R = 5.0024 104 J 1 8.314 J K 1 1 = 0.42409 ( ) 1 268 K 1 273.16 K = e χ = 0.61173 kpa e 0.42409 = 0.40029 kpa Now that we know the solid-vaor ressure = 0.40029 kpa at the given temerature, we know the ice will sublime into any gas system with a artial ressure of water H2O < 0.40029 kpa. This gives us the answer to art b as H2 O = 0.40029 kpa. Additionally, can determine that the ice will sublime into the atmoshere as the artial ressure of water is H2O = 0.30 kpa. (a) Yes. (b) H2 O = 0.40029 kpa 7
5 Exercise 4.17b g 154 Question What fraction of the enthaly of vaorization va H of ethanol is sent on exanding its vaour? Strategy We can use the definition of enthaly, H = U + V (Equation 2.18 g 56) to decomose the enthaly of vaorization va H into an internal energy comonent va U and an exansion work comonent va (V ). va H = va U + va (V ) Additionally, we can assume the ressure is constant and the liquid volume V liq is negligible relative to the gas volume such that va (V ) = (V gas V liq ) V gas Next, we can use the ideal gas law V = nrt to relate V gas to RT for a ar quantity of gas by assuming the ethanol vaor is ideal. This gives va (V ) = RT And thereby the ratio of exansion work va (V ) to exansion enthaly va H is va (V ) va H = RT va H Using text book table 2.3 we find that ethanol vaorizes at T = 352 K and its exansion enthaly is 43.5 kj 1. Substituting these values gives the ratio value va (V ) va H RT = va H = 8.3145 J K 1 1 352 K 43.5 kj 1 1000 J 1 kj = 0.0673 = 6.73% va (V ) va H = 6.73% 8
6 Problem 4.4 g 154 Question Calculate the difference in sloe of the chemical otential against temerature on either side of (a) the normal freezing oint of water and and (b) the normal boiling oint of water. (c) By how much does the chemical otential of water suer cooled to 5.0 C exceed that of ice at that temerature. Strategy To solve arts (a) and (b) we ll use Equation 4.13 (art 2 g 150) that relates how the temerature derivative (sloe) of chemical otential, µ, changes across a coexistence curve. ( ) µ (β) where α and β denote the two hases. ( ) µ (α) = S m (β) + S m (α) = trs S = trsh T trs For art (a) we re interested in the difference between the liquid l and solid s hases, the fusion transition fus H, which gives the following exression. ( ) µ (s) = S m (l) + S m (s) = fus S = fush T f Substituting in the enthaly and temerature of water s fusion transition, fus H = 6.008 kj 1 and T f = 273.15 K resectively as found in Table 2.3 of our text book, gives ( ) µ (s) = fush T f 6.008 kj 1 = 273.15 K = 0.021995 kj 1 K 1 = 21.995 J 1 K 1 Likewise for art (b) we re interested in the difference between the gas g and the liquid l hases which is the vaorization transition va H. ( ) µ (g) = S m (g) + S m (l) = va S = vah T v Substituting in va H = 40.656 kj 1 and T v = 373.15 K from Table 2.3 gives ( ) µ (g) = vah T v 40.656 kj 1 = 373.15 K = 0.108953 kj 1 K 1 = 108.953 J 1 K 1 9
For art (c) we re interested in calculating the difference in chemical otential of liquid water at 5 C, µ(l, 5 C) and the chemical otential of solid water at the same temerature µ(s, 5 C) and we ll call this quantity x. x = µ(l, 5 C) µ(s, 5 C) To calculate x we ll take advantage of the normal freezing oint of water, which imlies that at T = 0 C solid water and liquid water have the same chemical otential: µ(l, 0 C) = µ(s, 0 C) Therefore we can subtract µ(l, 0 C) µ(s, 0 C) (as this quantity is 0) from the exression we re working to solve. x = µ(l, 5 C) µ(s, 5 C) = µ(l, 5 C) µ(s, 5 C) [µ (l, 0 C) µ (s, 0 C)] = [µ (l, 5 C) µ (l, 0 C)] [µ (s, 5 C) µ (s, 0 C)] = µ(l) µ(s) In the third equation above we ve rearranged the righthand side so as to lace the difference in chemical otentials at the two temeratures in brackets for each hase. We call this quantity µ(α) where µ(α) = µ (α, 5 C)µ (α, 0 C) and α secifies either the liquid (l) or solid (s) hase. We can calculate µ(l) and µ(s) using Equation 4.2 (g 143) which gives the change in chemical otential with temerature. ( ) µ = S m Hence the change in chemical otential for the α hase can be calculated as where S m is assumed temerature indeendent. (α) = = Tf T i Tf T i ( ) µ S m = S m (T f T i ) Using T f T i = T = 5 K and the standard entroies of liquid water S m (l) and solid water S m (s) we get the following exressions for chemical otential changes (l) = S m (l) T (s) = S m (s) T Substituting these exression into our equation for x gives x = µ(l) µ(s) = [S m (l) S m (s)] T The above term in brackets, the difference in entroy of liquid and solid water, is just the oosite of the entroy of fusion fus S which we know from art (a). 10
x = [S m (l) S m (s)] T = fus S T ( = +21.995 J 1 K 1 ) 5 K = 109 J 1 We are reminded that x is just the difference in chemical otentials that we were solving. µ(l, 5 C) µ(s, 5 C) = x = 109 J 1 This ositive difference in chemical otential between liquid and solid hases imlies higher free energy for the liquid hase relative to the solid hase (ice) and exlains why ice is favorable at this temerature. (a) (b) (c) ( ) µ (g) ( ) µ (s) = 22.00 J 1 K 1 = 108.95 J 1 K 1 µ(l, 5 C) µ(s, 5 C) = x = 100 J 1 11