2- 1 CHAPTER TWO 2.1 (a) 3 24 3600 1 18144 10 9 wk 7 d h s 1000 ms 1 wk 1 d h 1 s ms= ×. (b) 38 3600 2598 26 0 . . . 1 ft / s 0.0006214 mi s 3.2808 ft 1 h mi / h mi/ h= ⇒ (c) 554 1 1 1000 g 3 m 1 d h kg 10 cm d kg 24 h 60 min 1 m 85 10 cm g 4 8 4 4 4 4 ⋅ = × ⋅. / min 2.2 (a) 760 mi 3600 340 1 m 1 h h 0.0006214 mi s m / s= (b) 921 kg 35.3145 ft 57 5 2.20462 lb 1 m m 1 kg lb / ft m 3 3 3 m 3 = . (c) 537 10 1000 J 1 1 11993 120 3 . . × × = ⇒ kJ 1 min .34 10 hp min 60 s 1 kJ J / s hp hp -3 2.3 Assume that a golf ball occupies the space equivalent to a 2 2 2 in in in × × cube. For a classroom with dimensions 40 40 15 ft ft ft × × : n balls 3 3 6 ft (12) in 1 ball ft in 10 5 million balls= × × = × ≈ 40 40 15 2 518 3 3 3 3 . The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4 3 24 3600 s 1 0 0006214 . . light yr 365 d h 1.86 10 mi 3.2808 ft 1 step 1 yr 1 d h 1 s mi 2 ft 7 10 steps 5 16 × = × 2.5 Distance from the earth to the moon = 238857 miles 238857 mi 1 4 10 11 1 m report 0.0006214 mi 0.001 m reports= × 2.6 19 00006214 1000 264 17 447 500 25 1 14 500 0 04464 700 25 1 21700 002796 km 1000 m mi L 1 L 1 km 1 m gal mi/ gal Calculate the total cost to travel miles. Total Cost gal (mi) gal 28 mi Total Cost gal (mi) gal 44.7 mi Equate the two costs 4.3 10 miles American European 5 . . . $14, $1. , . $21, $1. , . = = + = + = + = + ⇒ = × x x x x x x 2- 2 2.7 63 3 5 5320 imp. gal14 h365 d10 cm0.965 g 1 kg1 tonne planeh1 d1 yr 220.83 imp. gal 1 cm1000 g1000 kg tonne kerosene 1.18810 planeyr ⋅ =× ⋅ 9 5 4.0210 tonne crude oil1 tonne kerosene planeyr yr7 tonne crude oil1.188 10 tonne kerosene 4834 planes5000 planes ×⋅ × =⇒ 2.8 (a) 250 250 . . lb 32.1714 ft / s 1 lb 32.1714 lb ft / s lb m 2 f m 2 f ⋅ = (b) 25 2 55 2 6 N 1 1 kg m / s 9.8066 m / s 1 N kg kg 2 2 ⋅ = ⇒. . (c) 10 1000 g 980.66 cm 1 9 10 9 ton 1 lb / s dyne 5 10 ton 2.20462 lb 1 g cm / s dynes m 2 -4 m 2 × ⋅ = × 2.9 50 15 2 853 32 174 1 45 10 6 × × ⋅ = × m 35.3145 ft lb ft 1 lb 1 m 1 ft s 32.174 lb ft s lb 3 3 m f 3 3 2 m 2 f . . / . 2.10 500 lb 5 10 1 2 1 10 25 2m 3 m 3 1 kg 1 m 2.20462 lb 11.5 kg m≈ × F H G I K J F H G I K J ≈ 2.11 (a) m m V V h r H r h H f f c c f c c f displaced fluid cylinder 3 3 cm cm g / cm 30 cm g / cm = ⇒ = ⇒ = = = − = ρ ρ ρ π ρ π ρ ρ 2 2 30 14 1 100 053 ( . )( . ) . (b) ρ ρ f c H h = = = ( )( . ) . 30 053 171 cm g / cm (30 cm- 20.7 cm) g / cm 3 3 H h ρ f ρ c 2- 3 2.12 V R H V R H r h R H r h r R H h V R H h Rh H R H h H V V R H h H R H H H h H H H h h H s f f f f s s f s f s s s = = − = ⇒ = ⇒ = − F H G I K J = − F H G I K J = ⇒ − F H G I K J = ⇒ = − = − = − F H G I K J π π π π π π ρ ρ ρ π ρ π ρ ρ ρ ρ 2 2 2 2 2 2 3 2 2 3 2 2 3 2 3 3 3 3 3 3 3 3 3 3 3 3 1 1 ; ; ρ f ρ s R r h H 2.13 Say h m ( ) = depth of liquid A ( ) m 2 h 1 m ⇒ y x y = 1 y = 1 – h x = 1 – y 2 d A dA dy dx y dy A m y dy A y y y h h h y y h h = ⋅ = − ⇒ = − = − + = − − − + − + − − − − − + − − − − z z E 1 1 2 2 2 1 1 2 1 1 1 2 1 2 2 2 1 2 1 1 1 1 1 1 2 d i d i b g b g b g Table of integrals or trigonometric substitution m 2 sin sin π W N A A A g g b g= × = × E 4 0879 1 1 10 345 10 2 3 4 0 m m g 10 cm kg 9.81 N cm m g kg Substitute for 2 6 3 3 ( ) . . { W h h hN b g b g b g b g = × − − − + − + L N M O Q P − 345 10 1 1 1 1 2 4 2 1 . sin π 2.14 1 1 32 174 1 1 1 32174 lb slug ft / s lb ft / s slug= 32.174 lb poundal=1 lb ft / s lb f 2 m 2 m m 2 f = ⋅ = ⋅ ⇒ ⋅ = . . y= – 1 y= – 1+h dA 2- 4 2.14(cont’d) (a) (i) On the earth: M W = = = ⋅ = × 175 lb 1 544 175 1 1 m m m 2 m 2 3 slug 32.174 lb slugs lb 32.174 ft poundal s lb ft / s 5.63 10 poundals . (ii) On the moon M W = = = ⋅ = 175 lb 1 5 44 175 1 1 m m m 2 m 2 slug 32.174 lb slugs lb 32.174 ft poundal 6 s lb ft / s 938 poundals . ( ) /b F ma a F m= ⇒ = = ⋅ = 355 pound 1 1als lb ft / s 1 slug m 25.0 slugs 1 poundal 32.174 lb 3.2808 ft 0.135 m/ s m 2 m 2 2.15 (a) F ma= ⇒ F H G I K J = ⋅ ⇒ ⋅ 1 1 6 53623 1 fern = (1 bung)(32.174 ft / s bung ft / s fern 5.3623 bung ft / s 2 2 2 ) . (b) On the moon: 3 bung 32.174 ft 1 fern 6 s 5.3623 bung ft / s fern On the earth: =18 fern 2 2 W W = ⋅ = = 3 3 32174 5 3623( )( . ) / . 2.16 (a) ≈ = = ( )( ) ( . )( . ) 3 9 27 2 7 8 632 23 (b) 4 5 46 4.010 110 40 (3.60010)/458.010 − − −− × ≈≈× ×=× (c) ≈ + = + = 2 125 127 2 365 1252 127 5. . . (d) ≈ × − × ≈ × ≈ × × − × = × 50 10 1 10 49 10 5 10 4 753 10 9 10 5 10 3 3 3 4 4 2 4 . 2.17 154 23 6 3 (710)(310)(6)(510) 4210410 (3)(510) 3812.538103.8110 exact R R − ××× ≈≈×≈× × =⇒⇒× (Any digit in range 2-6 is acceptable) 2- 5 2.18 (a) A: C C C o o o R X s = − = = + + + + = = − + − + − + − + − − = 731 724 0 7 724 731 726 728 73 0 5 72 8 72 4 728 731 72 8 726 728 72 8 728 730 72 8 5 1 0 3 2 2 2 2 2 . . . . . . . . . ( . . ) ( . . ) ( . . ) ( . . ) ( . . ) . B: C C C o o o R X s = − = = + + + + = = − + − + − + − + − − = 1031 97 3 58 97 3 1014 987 1031 100 4 5 100 2 973 1002 1014 1002 987 1002 1031 1002 1004 1002 5 1 2 3 2 2 2 2 2 . . . . . . . . . ( . . ) ( . . ) ( . . ) ( . . ) ( . . ) . (b) Thermocouple B exhibits a higher degree of scatter and is also more accurate. 2.19 (a) X X s X X s X s i i i = = = − − = = − = − = = + = + = = = ∑ ∑ 1 12 2 1 12 12 735 735 12 1 12 2 735 2 12 711 2 735 2 12 759 . ( . ) . . ( . ) . . ( . ) . C C min= max= (b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness 2- 6 2.20 (a),(b) (c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12. 2.21 (a) Q' . = × ⋅ − 2 36 10 4 kg m 2.10462 lb 3.2808 ft 1 h h kg m 3600 s 2 2 2 2 (b) Q Q ' ( )( )( ) . / ' . / / ( approximate 2 exact 2 2 lb ft s = lb ft s 0.00000148 lb ft s ≈ × × ≈ × ≈ × ⋅ × ⋅ = ⋅ − − − − − 2 10 2 9 3 10 12 10 12 10 148 10 4 3 4 3) 6 6 2.22 N C k C C N p o o Pr Pr . . . ( )( )( ) ( )( )( ) . . . = = ⋅ ⋅ ⋅ ≈ × × × × × ≈ × ≈ × × − − µ 0583 1936 3 2808 0 286 6 10 2 10 3 10 3 10 4 10 2 3 10 2 15 10 163 10 1 3 3 1 3 3 3 3 J / g lb 1 h ft 1000 g W / m ft h 3600 s m 2.20462 lb The calculator solution is m m 2.23 Re . . . . Re ( )( )( )( ) ( )( )( )( ) ( = = × ⋅ ≈ × × × × ≈ × ≈ × ⇒ − − − − − − Duρ µ 048 2 067 0 805 0 43 10 5 10 2 8 10 10 3 4 10 10 4 10 5 10 3 2 10 3 1 1 6 3 4 1 3) 4 ft 1 m in 1 m g 1 kg 10 cm s 3.2808 ft kg / m s 39.37 in cm 1000 g 1 m the flow is turbulent 6 3 3 3 (a) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X 134 131 129 133 135 131 134 130 131 136 129 130 133 130 133 Mean(X) 131.9 Stdev(X) 2.2 Min 127.5 Max 136.4 (b) Run X Min Mean Max 1 128 127.5 131.9 136.4 2 131 127.5 131.9 136.4 3 133 127.5 131.9 136.4 4 130 127.5 131.9 136.4 5 133 127.5 131.9 136.4 6 129 127.5 131.9 136.4 7 133 127.5 131.9 136.4 8 135 127.5 131.9 136.4 9 137 127.5 131.9 136.4 10 133 127.5 131.9 136.4 11 136 127.5 131.9 136.4 12 138 127.5 131.9 136.4 13 135 127.5 131.9 136.4 14 139 127.5 131.9 136.4 126 128 130 132 134 136 138 140 0 5 10 15 2- 7 2.24 (a) k d y D D d u k k g p p g g = + F H G I K J F H G I K J = + × ⋅ × L N M O Q P × ⋅ L N M O Q P = ⇒ × = ⇒ = − − − − 200 0600 200 0600 100 10 100 100 10 000500 100 100 100 10 44426 000500 0100 100 10 44 426 0 1 3 1 2 5 5 1 3 5 1 2 5 . . . . . ( . )( . / ) ( . )( . )( . ) ( . ) . ( . ( . ) . / . / / / / / µ ρ ρ µ N s / m kg / m m s m m/ s kg / m N s/ m m) m s .888 m s 2 3 2 3 2 2 (b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) d p (m) y D (m 2 /s) µ (N-s/m 2 ) ρ (kg/m 3 ) u (m/s) k g 0.005 0.1 1.00E-05 1.00E-05 1 10 0.889 0.010 0.1 1.00E-05 1.00E-05 1 10 0.620 0.005 0.1 2.00E-05 1.00E-05 1 10 1.427 0.005 0.1 1.00E-05 2.00E-05 1 10 0.796 0.005 0.1 1.00E-05 1.00E-05 1 20 1.240 2.25 (a) 200 crystals/ min mm; 10 crystals / min mm 2 ⋅ ⋅ (b) r = ⋅ − ⋅ = ⇒ = 200 10 4 0 crystals 0.050 in 25.4 mm min mm in crystals 0.050 in (25.4) mm min mm in 238 crystals / min 238 crystals 1 min 60 s crystals/ s 2 2 2 2 2 2 min . (c) D D Dmm in mm 1 in b g b g = ′ = ′ 254 254 . . ; r r r crystals min crystals 60 s s 1 min F H G I K J = ′ = ′60 ⇒ ′ = ′ − ′ ⇒ ′ = ′ − ′60 200 254 10 254 84 7 108 2 2 r D D r D D. . . b g b g b g 2.26 (a) 705. / ; lb ft 8.27 10 in / lb m 3 -7 2 f × (b) 7262 f 3 m252 f 33 m 3363 m 8.2710 in910 N14.696 lb/in (70.5 lb/ft)exp lb m1.01325 10 N/m 70.57 lb35.3145 ft 1 m1000 g 1.13 g ft m10 cm2.20462 lb ρ −  ×× =  ×   == 3 /cm (c) ρ ρ ρ lb ft g lb cm cm g 1 ft m 3 m 3 3 3 F H G I K J = ′ = ′ 1 28317 453593 6243 , . . P P P lb in N .2248 lb m m N 39.37 in f 2 f 2 2 2 2 F H G I K J = = × − ' . ' 0 1 1 145 10 2 4 ⇒ ′ = × × ⇒ ′ = × − − − 62 43 705 8 27 10 145 10 113 120 10 7 4 10 . . exp . . ' . exp . 'ρ ρ d i d i d i P P P' . ' . exp[( . )( . )] .= × ⇒ = × × = − 9 00 10 113 120 10 9 00 10 113 6 10 6 N / m g / cm 2 3 ρ 2- 8 2.27 (a) V V Vcm in 28,317 cm in 3 3 3 3 d i d i = = ' . ' 1728 16 39 ; t ts hr b g b g = ′3600 ⇒ = ′ ⇒ = ′16 39 3600 0 06102 3600. ' exp ' . expV t V t b g b g (b) The t in the exponent has a coefficient of s -1 . 2.28 (a) 300. mol/ L, 2.00 min -1 (b) t C C = ⇒ = ⇒ = 0 3 00 300 . . exp[(-2.00)(0)]= 3.00 mol / L t = 1 exp[(-2.00)(1)]= 0.406 mol / L For t=0.6 min: C C int . . ( . ) . . . = − − − + = = 0406 300 1 0 06 0 300 14 300 mol / L exp[(-2.00)(0.6)]= 0.9 mol / L exact For C=0.10 mol/L: t t int exact min = - 1 2.00 ln C 3.00 = - 1 2 ln 0.10 3.00 = 1.70 min = − − − + = 1 0 0406 3 010 300 0 112 . ( . . ) . (c) 0 0.5 1 1.5 2 2.5 3 3.5 0 1 2 t (min) C (mol/L) (t=0.6, C=1.4) (t=1.12, C=0.10) C exact vs. t 2.29 (a) p* . . ( . )= − − − + = 60 20 199 8 166 2 185 166 2 20 42 mm Hg (b) c MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, * ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 2 FORMAT (10X, F5.1, 10X, F5.1) CONTINUE END 2- 9 2.29 (cont’d) SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I = 1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I = I + 1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END DATA OUTPUT 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) M M 100.0 1.2 215.5 100.0 105.0 1.8 M M 215.0 98.7 2.30 (b) ln ln (ln ln ) / ( ) (ln ln ) / ( ) . ln ln ln . ( ) . . y a bx y ae b y y x x a y bx a y e bx x = + ⇒ = = − − = − − = − = − = + ⇒ = ⇒ = − 2 1 2 1 0.693 2 1 1 2 0 693 2 0 63 1 4 00 4 00 (c) ln ln ln (ln ln ) / (ln ln ) (ln ln ) / (ln ln ) ln ln ln ln ( )ln( ) / y a b x y ax b y y x x a y b x a y x b = + ⇒ = = − − = − − = − = − = − − ⇒ = ⇒ = 2 1 2 1 2 1 1 2 1 2 1 1 2 2 (d) ln( ) ln ( / ) ( / ) ( )] [ln( ) ln( ) ] / [( / ) ( / ) ] (ln . ln . ) / ( . . ) ln ln( ) ( / ) ln . ln( . ) ( / ) / / / / xy a b y x xy ae y a x e y f x b xy xy y x y x a xy b y x a xy e y x e by x by x y x y x = + ⇒ = ⇒ = = = − − = − − = = − = − ⇒ = ⇒ = ⇒ = [can' t get 2 1 2 1 3 3 8070 40 2 2 0 10 3 8070 3 2 0 2 2 2 (e) ln( / ) ln ln( ) / ( ) [ ( ) ] [ln( / ) ln( / ) ]/ [ln( ) ln( ) ] (ln . ln . ) / (ln . ln . ) . ln ln( / ) ( ) ln . . ln( . ) . / . ( ) . ( ) / .33 / . y x a b x y x a x y ax x b y x y x x x a y x b x a y x x y x x b b2 2 1 2 2 2 2 1 2 1 2 2 4 1 2 2 165 2 2 2 2 2 8070 40 2 2 0 10 4 33 2 807 0 4 33 2 0 402 402 2 6 34 2 = + − ⇒ = − ⇒ = − = − − − − = − − = = − − = − ⇒ = ⇒ = − ⇒ = − 2.31 (b) Plot vs. on rectangular axes. Slope Intcpt 2 3 y x m n= = −, (c) 111a1 Plot vs. [rect. axes], slope = , inter cept = ln(3)ln(3)bb a xx ybby =+⇒ −− (d) 1 1 3 1 1 3 2 3 2 3 ( ) ( ) ( ) ( ) , , y a x y x a + = − ⇒ + − Plot vs. [rect. axes] slope = intercept = 0 OR 2- 10 2.31 (cont’d) 2 1 3 3 1 3 2 ln( ) ln ln( ) ln( ) ln( ) ln y a x y x a + = − − − + − ⇒ − − Plot vs. [rect.] or (y +1) vs. (x -3) [log] slope= 3 2 , intercept = (e) ln ln y a x b y x y x = + Plot vs. [rect.] or vs. [semilog ], slope = a, intercept = b (f) Plot vs. [rect.] slope = a, intercept = b log ( ) ( ) log ( ) ( ) 10 2 2 10 2 2 xy a x y b xy x y = + + + ⇒ (g) Plot vs. [rect.] slope = , intercept= OR b Plot 1 vs. 1 [rect.] , slope= intercept = 1 1 1 2 2 2 2 y ax b x x y ax b x y x a b y ax b x xy a x xy x b a = + ⇒ = + ⇒ = + ⇒ = + ⇒ , , 2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0 011. ) and ( R = 80 , y = 0169. ). 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0 20 40 60 80 100 R y y a R b a b y R = + = − − = × = − × = × U V | W | ⇒ = × + × − − − − − 0169 0011 80 5 211 10 0011 211 10 5 4 50 10 2 11 10 4 50 10 3 3 4 3 4 . . . . . . . . d i b g (b) R y= ⇒ = × + × = − − 43 211 10 43 4 50 10 0 092 3 4 . . . d i b g kg H O kg 2 1200 0 092 110 kg kg h kg H O kg H O h 2 2 b g b g . = [...]... 447.7 471.2 D 1.34E-06 2.50E-06 4.55E-06 8.52E-06 1.41E-05 2.00E-05 1/T 2.88E-03 2.67E-03 2.52E-03 2.38E-03 2.23E-03 2.12E-03 Sx Sy Syx Sxx -E/R ln D0 lnD (1/T)*(lnD) -1 3.5 -0 .03897 -1 2.9 -0 .03447 -1 2.3 -0 .03105 -1 1.7 -0 .02775 -1 1.2 -0 .02495 -1 0.8 -0 .02296 2.47E-03 -1 2.1 -3 .00E-02 6.16E-06 -3 666 -3 .0151 D0 7284 E 0.05 2- 17 (1/T)**2 8.31E-06 7.14E-06 6.37E-06 5.65E-06 4.99E-06 4.50E-06 CHAPTER THREE... 0 800 ln(1-Cp/Cao) 400 ln(1-Cp/Cao) 0 200 0 -1 -2 -3 -4 ln(1-Cp/Cao) = -0 .0062t 100 400 500 t Lab 1 600 400 600 -4 -6 ln(1-Cp/Cao) = -0 .0111t t Lab 2 k = 0.0111 s-1 800 0 0 ln(1-Cp/Cao) ln(1-Cp/Cao) 200 300 -2 k = 0.0062 s-1 0 200 0 -2 -4 200 400 600 800 0 -2 -4 -6 ln(1-Cp/Cao)= -0 .0064t -6 ln(1-Cp/Cao) = -0 .0063t t Lab 3 k = 0.0063 s-1 t Lab 4 k = 0.0064 s-1 (c) Disregarding the value of k that is... = 6.536, b = −4 206 2- 16 =0 2.45 (a) E(cal/mol), D0 (cm2 /s) (b) ln D vs 1/T, Slope=-E/R, intercept=ln D0 (c) Intercept = ln D0 = -3 .0151 ⇒ D0 = 0.05 cm 2 / s 3.0E-03 2.9E-03 2.8E-03 2.7E-03 2.6E-03 2.5E-03 2.4E-03 2.3E-03 2.2E-03 2.1E-03 2.0E-03 Slope = − E / R = -3 666 K ⇒ E = (3666 K)(1.987 cal / mol ⋅ K) = 7284 cal / mol -1 0.0 ln D -1 1.0 -1 2.0 -1 3.0 -1 4.0 ln D = -3 666(1/T) - 3.0151 1/T (d) Spreadsheet... mL lL 1 mol 3-1 3 ln(CA) 3.31 (a) kt is dimensionless ⇒ k (min-1 ) (b) A semilog plot of CA vs t is a straight line ⇒ ln CA = ln CAO − kt 1 0 -1 -2 -3 -4 -5 y = -0 .4137x + 0.2512 R2 = 0.9996 0.0 5.0 t (min) 10.0 k = 0.414 min −1 ln CAO = 02512 ⇒ CAO = 1286 lb - moles ft 3 FG 1b - molesIJ = C ′ mol 28.317 liter H ft K liter 1 ft t ′b sg 1 min t b ming = = t ′ 60 60 s (c) C A A 3 2.26462 lb- moles = 0.06243C′... coordinates, the plot is a straight line 0 50 100 150 200 0 -0 .5 -1 -1 .5 -2 t (min) Slope = -0 .0093 ⇒ k = 9.3 × 10 -3 min −1 (b) ln[(CA − C Ae ) / (C A0 − C Ae )] = −kt ⇒ C A = (C A 0 − CAe )e − kt + CAe −3 CA = (0.1823 − 0.0495)e− (9.3×1 0 )(120) + 0.0495 =9.300 ×1 0-2 g/L 9.300 ×1 0-2 g 30.5 gal 28.317 L C =m/ V ⇒ m=CV = = 10.7 g L 7.4805 gal 2.35 (a) ft 3 and h-2 , respectively (b) ln(V) vs t2 in rectangular... 8.5 lnP 8 7.5 7 6.5 6 2.5 3 lnP = -1 .573(lnV ) + 12.736 3.5 4 lnV k = − slope = −( −1573) = 1573 (dimensionless) Intercept = ln C = 12.736 ⇒ C = e 12.736 = 3.40 × 10 5 mm Hg ⋅ cm 4.719 G − GL 1 G −G G −G = ⇒ 0 = K L Cm ⇒ ln 0 = ln K L + mln C m G0 − G K L C G − GL G − GL ln(G 0 -G)/(G-G L )= 2.4835lnC - 10.045 3 ln(G 0-G)/(G-G L) 2.37 (a) 2 1 0 -1 3.5 4 4.5 5 lnC 2- 12 5.5 2.37 (cont’d) m = slope... converted; the decomposition products might not be harmless 2- 14 2.41 (a) and (c) y 10 1 0.1 1 10 100 x (b) y = ax b ⇒ ln y = ln a + b ln x; Slope = b, Intercept = ln a ln y = 0.1684ln x + 1.1258 2 ln y 1.5 1 0.5 b = slope = 0168 0 -1 0 1 2 ln x 3 4 5 Intercept = ln a = 11258 ⇒ a = 3.08 2.42 (a) ln(1-Cp /CA0 ) vs t in rectangular coordinates Slope=-k, intercept=0 (b) 600 0 800 ln(1-Cp/Cao) 400 ln(1-Cp/Cao)... b f nf 3- 5 Conc (g Ile/100 g H2O) 3.12 (a) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.987 y = 545.5x - 539.03 R2 = 0.9992 0.989 0.991 0.993 0.995 0.997 Density (g/cm3) From the plot above, r = 5455ρ − 539.03 (b) For ρ = 0.9940 g / cm 3 , & m Ile = 150 L 0.994 g h 3 cm r = 3.197 g Ile / 100g H 2O 1000 cm 3 3.197 g Ile L 1 kg 103.197 g sol 1000 g = 4.6 kg Ile / h (c) The density of H2 O increases as T decreases,... SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X 'PROBLEM 2-3 9'/) WRITE (6, 4) A, B 4FORMAT (1H0, 'SLOPEb bAb =', F6.3, 3X 'INTERCEPTb b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X 'Xb... & V ( mL s ) 62 87 107 123 138 151 5 10 15 20 25 30 h ( mm ) ∆P ( mm Hg ) 0.114 0.227 0.341 0.455 0.568 0.682 b g & (b) lnV = n ln ∆P + ln K 6 ln(V) 5.5 y = 0.4979x + 5.2068 5 4.5 4 -2 .5 -2 -1 .5 -1 -0 .5 0 ln( P) 3-1 9 760 mm H g 2 1.01325×106 dyne/cm2 h . cal / mol− ⇒ ⋅E R E/ ln D = -3 666(1/T) - 3.0151 -1 4.0 -1 3.0 -1 2.0 -1 1.0 -1 0.0 2.0E-03 2.1E-03 2.2E-03 2.3E-03 2.4E-03 2.5E-03 2.6E-03 2.7E-03 2.8E-03 2.9E-03 3.0E-03 1/T ln D (d) Spreadsheet. 1.34E-06 2.88E-03 -1 3.5 -0 .03897 8.31E-06 374.2 2.50E-06 2.67E-03 -1 2.9 -0 .03447 7.14E-06 396.2 4.55E-06 2.52E-03 -1 2.3 -0 .03105 6.37E-06 420.7 8.52E-06 2.38E-03 -1 1.7 . -0 .02775 5.65E-06 447.7 1.41E-05 2.23E-03 -1 1.2 -0 .02495 4.99E-06 471.2 2.00E-05 2.12E-03 -1 0.8 -0 .02296 4.50E-06 Sx 2.47E-03 Sy -1 2.1 Syx -3 .00E-02 Sxx 6.16E-06