Solucionario Felder PDF

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CHAPTER TWO 3 wk 7d 24 h 3600 s 1000 ms 2.1 (a) = 18144 . × 10 9 ms 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h 554 m 4 1d 1h 1 kg 108 cm 4 (c) = 3.85 × 10 4 cm 4 / min⋅ g d ⋅ kg 24 h 60 min 1000 g 1 m 4 760 mi 1 m 1 h 2.2 (a) = 340 m / s h 0.00062...

Description

CHAPTER TWO 3 wk

7d

24 h 3600 s 1000 ms

= 18144 . × 10 9 ms 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h

2.1 (a)

(c)

2.2 (a)

554 m 4 1d 1h d ⋅ kg 24 h 60 min

1 kg 108 cm 4 = 3.85 × 10 4 cm 4 / min⋅ g 1000 g 1 m 4

1 m 1 h 760 mi = 340 m / s h 0.0006214 mi 3600 s 1 m3 = 57.5 lb m / ft 3 35.3145 ft 3

(b)

921 kg 2.20462 lb m m3 1 kg

(c)

5.37 × 10 3 kJ 1 min 1000 J min 60 s 1 kJ

1.34 × 10 -3 hp = 119.93 hp ⇒ 120 hp 1 J/s

2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12) 3 in 3 1 ball n balls = . × 10 6 ≈ 5 million balls = 518 ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h 1 yr

3600 s 1.86 × 10 5 mi

1d

1 h

1

s

3.2808 ft 0.0006214 mi

1 step = 7 × 1016 steps 2 ft

2.5 Distance from the earth to the moon = 238857 miles 238857 mi

1

m

0.0006214 mi

1 report 0.001 m

= 4 × 1011 reports

2.6

19 km 1000 m 0.0006214 mi 1000 L = 44.7 mi / gal 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. Total Cost

Total Cost

American

European

= $14,500 +

= $21,700 +

$1.25 1 gal gal 28 mi

x (mi)

= 14,500 + 0.04464 x

$1.25 1 gal x (mi) = 21,700 + 0.02796 x gal 44.7 mi

Equate the two costs ⇒ x = 4.3 × 10 5 miles

2-1

2.7 5320 imp. gal

106 cm3

14 h 365 d

plane ⋅ h

1 d

1 yr

0.965 g

220.83 imp. gal

1

cm

3

1 kg

1 tonne

1000 g

1000 kg

tonne kerosene plane ⋅ yr

= 1.188 × 105

4.02 × 109 tonne crude oil 1 tonne kerosene

plane ⋅ yr

7 tonne crude oil 1.188 × 10 tonne kerosene 5

yr

= 4834 planes ⇒ 5000 planes

2.8 (a) (b) (c)

2.9

2.10 2.11

32.1714 ft / s 2

25.0 lb m

1

lb f

32.1714 lb m ⋅ ft / s 2 25 N

1 kg ⋅ m/s 2

1 9.8066 m/s 2

10 ton

1N

1 lb m 5 × 10

-4

50 × 15 × 2 m 3

500 lb m

= 2.5493 kg ⇒ 2.5 kg

980.66 cm / s 2

1000 g ton

= 25.0 lb f

1 g ⋅ cm / s 2

2.20462 lb m

35.3145 ft 3 1 m3

85.3 lb m 1 ft 3

1

1 m3

kg

2.20462 lb m

1 dyne

11.5 kg

= 9 × 10 9 dynes

32.174 ft 1 lb f = 4.5 × 10 6 lb f 2 2 1 s 32.174 lb m / ft ⋅ s

≈ 5 × 10 2

FG 1 IJ FG 1 IJ ≈ 25 m H 2 K H 10K

3

(a) mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2

ρc

ρfh

(30 cm − 14.1 cm)(100 . g / cm 3 ) ρc = = = 0.53 g / cm 3 H 30 cm ρ H (30 cm)(0.53 g / cm 3 ) (b) ρ f = c = = 171 . g / cm 3 (30 cm - 20.7 cm) h

2.12

Vs =

πR 2 H 3

⇒ Vf =

; Vf =

3

−

πr 2 h 3

;

πh Rh

2

2

f

H H−

h3 H2

3

2

= ρs

H3 = ρs H 3 − h3

h

3

r

2

2

ρf h

R r R = ⇒r = h H h H

FG IJ = πR FG H − h IJ − 3 3 H HK 3 H H K πR F πR H h I ⇒ρ =ρ H− G J 3 H 3 H K

πR H 2

ρ f V f = ρ sVs ⇒ ρ f = ρs

πR 2 H

H

H

2

ρs

s

R

1

1−

FG h IJ H HK

2-2

3

ρf

2.13

Say h( m) = depth of liquid

y y= 1 dA y=y=1––1+h h xx

⇒ A(m 2 )

1− y

dA = dy ⋅

∫

h

x = 1– y 2 y= –1

dA

2

−1+ h

( )=2 ∫

dx = 2 1 − y dy ⇒ A m 2

2

1 − y 2 dy

−1

− 1− y 2

⇓

1m

Table of integrals or trigonometric substitution

π 2 A m 2 = y 1 − y 2 + sin −1 y ⎤⎥ = ( h − 1) 1 − ( h − 1) + sin −1 ( h − 1) + ⎦ −1 2 h −1

( )

b g

W N =

4 m × A( m 2 ) 0.879 g 10 6 cm 2 cm

3

1m

E Substitute for A L W b N g = 3.45 × 10 Mbh − 1g 1 − bh − 1g N 4

2.14

3

1 kg

9.81 N

3

kg N

10 g

= 3.45 × 10 4 A

g g0

2

b g π2 OPQ

+ sin −1 h − 1 +

1 lb f = 1 slug ⋅ ft / s 2 = 32.174 lb m ⋅ ft / s 2 ⇒ 1 slug = 32.174 lb m 1 1 poundal = 1 lb m ⋅ ft / s 2 = lb f 32.174 (a) (i) On the earth: 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 5.63 × 10 3 poundals 2 s 1 lb m ⋅ ft / s 2 (ii) On the moon 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 938 poundals 2 6 s 1 lb m ⋅ ft / s 2 (b) F = ma ⇒ a = F / m =

355 poundals 25.0 slugs

1 lb m ⋅ ft / s 2 1 poundal

= 0.135 m / s 2

2-3

1 slug 32.174 lb m

1m 3.2808 ft

2.15 (a) F = ma ⇒ 1 fern = (1 bung)(32.174 ft / s 2 )

⇒

FG 1IJ = 5.3623 bung ⋅ ft / s H 6K

2

1 fern 5.3623 bung ⋅ ft / s 2

3 bung 32.174 ft 1 fern = 3 fern 2 6 s 5.3623 bung ⋅ ft / s 2 On the earth: W = (3)( 32.174) / 5.3623 = 18 fern

(b) On the moon: W =

2.16 (a) ≈ (3)(9) = 27

(b)

(2.7)(8.632) = 23 (c) ≈ 2 + 125 = 127

(d)

2.365 + 125.2 = 127.5 2.17 R ≈

4.0 × 10−4 ≈ 1× 10−5 40 (3.600 ×10−4 ) / 45 = 8.0 × 10−6 ≈

≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4 4.753 × 10 4 − 9 × 10 2 = 5 × 10 4

(7 ×10−1 )(3 × 105 )(6)(5 × 104 ) ≈ 42 × 102 ≈ 4 × 103 (Any digit in range 2-6 is acceptable) (3)(5 × 106 )

Rexact = 3812.5 ⇒ 3810 ⇒ 3.81× 103 2.18 (a) A: R = 731 . − 72.4 = 0.7 o C X=

. + 72.6 + 72.8 + 73.0 72.4 + 731 = 72.8 o C 5

s=

(72.4 − 72.8) 2 + (731 . − 72.8) 2 + (72.6 − 72.8) 2 + (72.8 − 72.8) 2 + (73.0 − 72.8) 2 5−1

= 0.3o C B: R = 1031 . − 97.3 = 58 . oC X=

97.3 + 1014 . + 98.7 + 1031 . + 100.4 = 100.2 o C 5

s=

(97.3 − 100.2) 2 + (1014 . − 100.2) 2 + (98.7 − 100.2) 2 + (1031 . − 100.2) 2 + (100.4 − 100.2) 2 5−1

= 2.3o C

(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.

2-4

2.19 (a)

12

X=

∑X

12

i

i =1

C min=

= 73.5 s= 12 = X − 2 s = 73.5 − 2(1.2) = 711 .

∑ ( X − 735. )

2

i =1

= 12 .

12 − 1

C max= = X + 2 s = 735 . + 2(12 . ) = 75.9

(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 2.20 (a), (b) 1 2 (a) Run 134 131 X Mean(X) 131.9 Stdev(X) 2.2 127.5 Min 136.4 Max (b) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14

X 128 131 133 130 133 129 133 135 137 133 136 138 135 139

Min 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5

3 129

Mean 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9

4 5 6 7 8 9 10 11 12 13 14 15 133 135 131 134 130 131 136 129 130 133 130 133

Max 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4

140 138 136 134 132 130 128 126 0

5

10

15

(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

(b) Q 'approximate ≈

h

2.20462 lb 3.28082 ft 2 m2

kg

1

h

3600 s

(2 × 10−4 )(2)(9) ≈ 12 × 10( −4−3) ≈ 1.2 × 10−6 lb ⋅ ft 2 / s 3 × 103

Q 'exact =1.56 × 10−6 lb ⋅ ft 2 / s = 0.00000156 lb ⋅ ft 2 / s

2-5

2.22 N Pr = N Pr ≈

Cpμ

=

k

0.583 J / g ⋅ o C

1936 lb m

0.286 W / m ⋅ C −1

1 h 3.2808 ft

ft ⋅ h

o

3600 s

1000 g

m 2.20462 lb m

(6 × 10 )(2 × 10 )(3 × 10 ) 3 × 10 ≈ ≈ 15 . × 10 3 . The calculator solution is 163 . × 10 3 −1 3 2 (3 × 10 )(4 × 10 )(2) 3

3

3

2.23 Re = Re ≈

2.24 (a)

Duρ

μ

=

0.48 ft

1

m

2.067 in

1 m

1 kg 10 6 cm 3

0.805 g

s 3.2808 ft 0.43 × 10 −3 kg / m ⋅ s 39.37 in

cm 3

1000 g

1 m3

(5 × 10 −1 )(2)(8 × 10 −1 )(10 6 ) 5 × 101− ( −3) ≈ ≈ 2 × 10 4 ⇒ the flow is turbulent 3 (3)(4 × 10)(10 3 )(4 × 10 −4 )

kg d p y D

1/ 3

⎛ μ ⎞ = 2.00 + 0.600 ⎜ ⎟ ⎝ ρD ⎠

⎛ d p uρ ⎞ ⎜ ⎟ ⎝ μ ⎠

1/ 2

1/ 3

⎡ ⎤ 1.00 × 10−5 N ⋅ s/m 2 = 2.00 + 0.600 ⎢ ⎥ −5 3 2 ⎣ (1.00 kg/m )(1.00 × 10 m / s) ⎦ k g (0.00500 m)(0.100) = 44.426 ⇒ = 44.426 ⇒ k g 1.00 × 10−5 m 2 / s

1/ 2

⎡ (0.00500 m)(10.0 m/s)(1.00 kg/m3 ) ⎤ ⎢ ⎥ (1.00 × 10−5 N ⋅ s/m 2 ) ⎣ ⎦ = 0.888 m / s

(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) dp (m)

y

0.005 0.010 0.005 0.005 0.005

0.1 0.1 0.1 0.1 0.1

D (m2/s) μ (N-s/m2) ρ (kg/m3) u (m/s) 1.00E-05 1.00E-05 1 10 1.00E-05 1.00E-05 1 10 2.00E-05 1.00E-05 1 10 1.00E-05 2.00E-05 1 10 1.00E-05 1.00E-05 1 20

kg 0.889 0.620 1.427 0.796 1.240

2.25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2

200 crystals 0.050 in 25.4 mm 10 crystals 0.050 2 in 2 − min ⋅ mm in min ⋅ mm 2 238 crystals 1 min = 238 crystals / min ⇒ = 4.0 crystals / s 60 s min

(25.4) 2 mm 2 in 2

(b) r =

b g

(c) D mm =

b g

D ′ in

FG H

IJ K

crystals 60 s 25.4 mm crystals = 25.4 D ′ ; r = r′ = 60r ′ s 1 min min 1 in

b

g b

⇒ 60r ′ = 200 25.4 D ′ − 10 25.4 D ′

g

2

2-6

b g

⇒ r ′ = 84.7 D ′ − 108 D ′

2

2.26 (a) 70.5 lb m / ft 3 ; 8.27 × 10 -7 in 2 / lb f ⎡8.27 × 10−7 in 2 9 × 106 N 14.696 lbf / in 2 ⎤ (b) ρ = (70.5 lb m / ft 3 )exp ⎢ ⎥ lbf m 2 1.01325 × 105 N/m 2 ⎥⎦ ⎢⎣ 70.57 lb m 35.3145 ft 3 1 m3 1000 g = = 1.13 g/cm3 3 3 6 3 ft m 10 cm 2.20462 lbm

(c)

F lb IJ = ρ ′ g ρG H ft K cm F lb IJ = P' N PG H in K m m 3

3

f 2

1 lb m

28,317 cm 3

453.593 g

1 ft 3

0.2248 lb f

12

2

m2

39.37 2 in 2

1N

d

= 62.43ρ ′ = 145 . × 10 −4 P '

id

i

d

i

⇒ 62.43ρ ′ = 70.5 exp 8.27 × 10 −7 1.45 × 10 −4 P ' ⇒ ρ ′ = 113 . exp 120 . × 10 −10 P '

P ' = 9.00 × 10 6 N / m 2 ⇒ ρ ' = 113 . exp[(1.20 × 10 −10 )(9.00 × 10 6 )] = 113 . g / cm 3 cm = 16.39V ' ; t bsg = 3600t ′b hr g d i V ' din i 28,317 1728 in ⇒ 16.39V ' = expb3600t ′ g ⇒ V ' = 0.06102 expb3600t ′ g

2.27 (a) V cm 3 =

3

3

3

(b) The t in the exponent has a coefficient of s-1. 2.28 (a) 3.00 mol / L, 2.00 min -1 (b) t = 0 ⇒ C = 3.00 exp[(-2.00)(0)] = 3.00 mol / L

t = 1 ⇒ C = 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 − 3.00 For t=0.6 min: (0.6 − 0) + 3.00 = 14 . mol / L Cint = 1− 0 Cexact = 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L For C=0.10 mol/L:

t int = t exact

1− 0 (010 . − 3.00) + 0 = 112 . min 0.406 − 3 1 C 1 0.10 =ln = - ln = 1.70 min 2.00 3.00 2 3.00

(c) 3.5 C exact vs. t

3 C (mol/L)

2.5 2

(t=0.6, C=1.4)

1.5 1

(t=1.12, C=0.10)

0.5 0 0

1

2

t (min)

2-7

p* =

2.29 (a)

(b)

60 − 20 (185 − 166.2) + 20 = 42 mm Hg 199.8 − 166.2

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 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END 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 OUTPUT DATA 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 # # 215.5 100.0 105.0 1.8 # # 215.0 98.7

2.30 (b) ln y = ln a + bx ⇒ y = ae bx b = (ln y 2 − ln y1 ) / ( x 2 − x1 ) = (ln 2 − ln 1) / (1 − 2) = −0.693 ln a = ln y − bx = ln 2 + 0.63(1) ⇒ a = 4.00 ⇒ y = 4.00e −0.693 x

(c) ln y = ln a + b ln x ⇒ y = ax b b = (ln y 2 − ln y1 ) / (ln x 2 − ln x1 ) = (ln 2 − ln 1) / (ln 1 − ln 2) = −1

ln a = ln y − b ln x = ln 2 − ( −1) ln(1) ⇒ a = 2 ⇒ y = 2 / x (d) ln( xy ) = ln a + b( y / x) ⇒ xy = aeby / x ⇒ y = (a / x)eby / x [can't get y = f ( x)] b = [ln( xy ) 2 − ln( xy )1 ]/[( y / x) 2 − ( y / x)1 ] = (ln 807.0 − ln 40.2) /(2.0 − 1.0) = 3 ln a = ln( xy ) − b( y / x) = ln 807.0 − 3ln(2.0) ⇒ a = 2 ⇒ xy = 2e3 y / x [can't solve explicitly for y ( x)]

2-8

2.30 (cont’d) (e) ln( y 2 / x ) = ln a + b ln( x − 2) ⇒ y 2 / x = a ( x − 2) b ⇒ y = [ax ( x − 2) b ]1/ 2

b = [ln( y 2 / x ) 2 − ln( y 2 / x ) 1 ] / [ln( x − 2) 2 − ln( x − 2) 1 ] . ) = 4.33 = (ln 807.0 − ln 40.2) / (ln 2.0 − ln 10 ln a = ln( y 2 / x ) − b( x − 2) = ln 807.0 − 4.33 ln(2.0) ⇒ a = 40.2 ⇒ y 2 / x = 40.2( x − 2) 4.33 ⇒ y = 6.34 x 1/ 2 ( x − 2) 2.165 2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope = m, Intcpt = − n (c)

1 1 a 1 = + x ⇒ Plot vs. ln( y − 3) b b ln( y − 3)

x [rect. axes], slope =

a 1 , intercept = b b

(d)

1 1 = a ( x − 3) 3 ⇒ Plot vs. ( x − 3) 3 [rect. axes], slope = a , intercept = 0 2 2 ( y + 1) ( y + 1) OR 2 ln( y + 1) = − ln a − 3 ln( x − 3) Plot ln( y + 1) vs. ln( x − 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a ⇒ slope = − , intercept = − 2 2 (e) ln y = a x + b

Plot ln y vs.

x [rect.] or y vs.

x [semilog ], slope = a, intercept = b

(f) log10 ( xy ) = a ( x 2 + y 2 ) + b Plot log10 ( xy ) vs. ( x 2 + y 2 ) [rect.] ⇒ slope = a, intercept = b

(g)

x b x 1 vs. x 2 [rect.], slope = a , intercept = b = ax + ⇒ = ax 2 + b ⇒ Plot y x y y OR

b 1 1 b 1 1 vs. 2 [rect.] , slope = b, intercept = a = ax + ⇒ = a + 2 ⇒ Plot y x xy xy x x

2-9

2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0.011 ) and ( R = 80 , y = 0169 . ).

0.18 0.16 0.14 0.12 y

0.1 0.08 0.06 0.04 0.02 0 0

20

40

60

80

100

R

y=aR+b

U| V| W

. − 0.011 0169 = 2.11 × 10 −3 80 − 5 ⇒ y = 2.11 × 10 −3 R + 4.50 × 10 −4 −3 −4 b = 0.011 − 2.11 × 10 5 = 4.50 × 10

a=

ib g

d

ib g

d

(b) R = 43 ⇒ y = 2.11 × 10 −3 43 + 4.50 × 10 −4 = 0.092 kg H 2 O kg

b1200 kg hgb0.092 kg H O kgg = 110 kg H O h 2

2

2.33 (a) ln T = ln a + b ln φ ⇒ T = aφ b

b = (ln T2 − ln T1 ) / (ln φ 2 − ln φ 1 ) = (ln 120 − ln 210) / (ln 40 − ln 25) = −119 . ln a = ln T − b ln φ = ln 210 − ( −119 . ) ln(25) ⇒ a = 9677.6 ⇒ T = 9677.6φ −1.19

b

(b) T = 9677.6φ −1.19 ⇒ φ = 9677.6 / T

b

g

T = 85o C ⇒ φ = 9677.6 / 85

0.8403

b g T = 290 C ⇒ φ = b9677.6 / 290g T = 175o C ⇒ φ = 9677.6 / 175 o

g

0.8403

. L/s = 535

0.8403

= 29.1 L / s

0.8403

= 19.0 L / s

(c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.

2-10

ln ((CA-CAe)/(CA0-CAe))

2.34 (a) Yes, because when ln[(C A − C Ae ) / (C A0 − C Ae )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0

50

100

150

200

0 -0.5 -1 -1.5 -2 t (m in)

Slope = -0.0093 ⇒ k = 9.3 × 10-3 min −1

(b) ln[(C A − C Ae ) /(C A0 − C Ae )] = − kt ⇒ C A = (C A0 − C Ae )e − kt + C Ae

C A = (0.1823 − 0.0495)e − (9.3×10 C =m /V ⇒ m =CV =

−3

)(120)

+ 0.0495 = 9.300 × 10-2 g/L

9.300 × 10-2 g 30.5 gal 28.317 L = 10.7 g L 7.4805 gal

2.35 (a) ft 3 and h -2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353 . × 10 −2 ) ; or V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 . × 10−2 (c) V ( m3 ) = 100 . × 10 −3 exp(15 . × 10 −7 t 2 ) 2.36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k lnV 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 = e12.736 = 3.40 × 105 mm Hg ⋅ cm4.719 G − GL 1 G −G G −G = ln K L + m ln C = ⇒ 0 = K L C m ⇒ ln 0 m G0 − G K L C G − GL G − GL ln (G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5

3 ln(G 0-G)/(G-G L )

2.37 (a)

2 1 0 -1 3 .5

4

4 .5

5

ln C

2-11

5 .5

2.37 (cont’d)

m = slope = 2.483 (dimensionless) Intercept = ln K L = −10.045 ⇒ K L = 4.340 × 10 −5 ppm-2.483

G − 180 . × 10 −3 = 4.340 × 10 −5 (475) 2.483 ⇒ G = 1806 × 10 −3 . ...