Chapter 7 Haloalkanes
Multiple Choice
1. Which of the following structures have the correct IUPAC name? (Sec. 7.2)
Cl
H
C CH CH CH
2
2
3
CH3
CH3
F
R-2-chloropentane
I
R-4-fluoro-4-methylcyclohexene
II
H3CO
Br
H
CH3
Cl
Br
H
S-2-chloro-2-fluorobutane
IV
I,
Chapter 9 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 9.1:
a. The rate will increase. As temperature increases, the fraction of molecules having the
energy needed to react goes up.
b. The rate will incre
Chapter 6 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 6.1:
You should have followed the example in the book and printed a spreadsheet page that looks
like the one illustrated in Figure 6.4.
Chapter 6 Ans
Chapter 7 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 7.1:
Gases are more compressible than liquids because the distance between the gas molecules is
much greater than the distance between molecules in a
14.
1.
At a high AT, where the vapor volume is greater,
vapor from lower tubes blankets the upper tubes
leading to a lower average coefficient for the
tube bundle.
Chapter 14
Ta = 500°C or 773 K Tb = 200°C or 473 x
E. = 0.90 _b = 0.25
(a) Fr0m Eq. (14.38)
Chapter 10
10.1 Basis: 1 m2. Let A refer to fireclay brick, B to
kaolin brick, and C to steel. From Appendixes 10
and 11, thermal conductivities are
_
4
Kaolin brick
Steel _
Heat 1055:
q _ _ Th _ Tc
IC + (x3)air_equiv.
kd k8 kc
8.
1.
only,
Chapter 8
(a) Assume the average gas velocity is 20 m/s.
(See p. 189). The actual volume flow is
_ 10,000xlx293
3600x3x273
Crosssectional area required:
3 .= 0.994/20 = 0.0497m2 (0.535 ftz)
Schedule 40
standard pipe size,
(255mm ID).
q = 0.99
cQapfer 17 }
17.1 (21) Use Eq. (17.19) 13T = 3m
1/1 = y = 0.2 y; :1) = 0.02
Dv=0.144 x 3600 x
niol / m3
EDT = 0.0518 m2! h
= 0.0446 k
d 1
Pm 22.4
5 1/2"
I
I
From Eq. (17.19):
JA 2 (0.0518!22.4)(0.2,~ 0.02)! 3 =1.388x10 " kgi mo! lm2 - h
| E
(17)
Chapter 8 - Section B - Non-Numerical Solutions
8.12 (a) Because Eq. (8.7) for the efciency Diesel includes the expansion ratio, re VB / V A , we relate this quantity to the compression ratio, r VC / VD , and the Diesel cutoff ratio, rc V A / VD . Since V
Chapter 7 - Section B - Non-Numerical Solutions
7.2 (a) Apply the general equation given in the footnote on page 260 to the particular derivative of interest here: T S T = P S S P P T The two partial derivatives on the right are found from Eqs. (6.17) and
Chapter 6 - Section B - Non-Numerical Solutions
H S =T
P
6.1 By Eq. (6.8),
and isobars have positive slope T S
Differentiate the preceding equation: 2 H S2
2 H S2 =
P
=
P
P
Combine with Eq. (6.17):
T CP
and isobars have positive curvature.
6.2 (a) Applica
Chapter 5 - Section A - Mathcad Solutions
5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) = TC Work = 1 QH TH TH := 798.15 K TC QH := 250 kJ s kJ s
TC := 323.15 K Work := Q H 1
TH
or
Work = 148.78
Work = 148.78 kW which is t
Chapter 4 - Section B - Non-Numerical Solutions
4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as: CP = A + B C T1 ( + 1) + T12 ( 2 + + 1) 2 3
where T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperatur
Chapter 3 - Section B - Non-Numerical Solutions
3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T : P T =
T
1 V2
V P V T
T
V T V P
+
P
1 V
2V PT 2V T P
= +
2V PT 2V PT
=
P
1 V2
T
P
1 V
=
Addition of these two equations leads i
Chapter 3 - Section A - Mathcad Solutions
3.1 = 1 d dT = P 1 d dP
T
At constant T, the 2nd equation can be written: d = dP ln ( 1.01) ln
2 = P 1
:= 44.18 10
6
bar
1
2 = 1.01 1 Ans.
P :=
P = 225.2bar
3
P2 = 226.2 bar
3.4 b := 2700 bar
c := 0.125
V
c
Chapter 2 - Section A - Mathcad Solutions
2.1 (a) Mwt := 35 kg Work := Mwt g z (b) Utotal := Work g := 9.8 m s
2
z := 5 m Work = 1.715 kJ Ans. Utotal = 1.715 kJ dU + d ( PV) = CP dT Ans.
(c) By Eqs. (2.14) and (2.21): Since P is constant, this can be writ
Chapter 1 - Section B - Non-Numerical Solutions
1.1 This system of units is the English-system equivalent of SI. Thus, gc = 1(lbm )(ft)(poundal)1 (s)2 1.2 (a) Power is power, electrical included. Thus, Nm kgm2 energy [=] [=] time s s3 (b) Electric current
Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this
equation by setting t(F) = t(C).
t := 0
Guess solution:
Given
t = 1.8t + 32
1.5 By definition:
P=
Find ( t ) = 40
F
A
F = mass g
P
6.1
Re
3500 lb/h, and
3500/20.6 = 170 lb/h-ft: From Eq. (4.50),
51 z
I":
with p = 2.42 lb/ft-h:
3 x 2.42 2170
a = (
)U3
62.32 x 4.17 x 109x 0.7071
= 1.03 x 106 ft.
Total holdup is 1.03 x 104 x 29 = 0.03 ft3/ft3
Actual holdup is about twice that predicte
4.6.
p
Force balance:
Use Eq.
From Eq. (4.50),
P
[._
Here |=
(4.50).
998 kg/m,
ti: = 0.15x998!60 = 2.49
b=nx25=14m
kg/s
,u at 40°C = 0.656 c'p {Appendix 6)
q
U3
6 a (3 ><0.656><10"3 x I .495
998% 9.80665 x 7.354
P = 4.0 x 10 moir 0.