CHBE 2100 Exam 2 Fall 2019
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Georgia Institute Of Technology *
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Course
2100
Subject
Chemistry
Date
Dec 6, 2023
Type
Pages
10
Uploaded by EarlCaterpillarMaster878
CHBE
2100
Exam
#2,
Fall
2019
page
1/1
ChBE
2100
Chemical
Process
Principles
Fall
2019
Exam
02
There
are
4
problems
on
this
exam.
There
are
some
tables
at
the
end
of
the
exam
that
you
may
find
useful.
Please
read
the
questions
carefully.
This
exam
is
closed-book.
However,
you
are
allowed
one
8.5"
x
11"
piece
of
paper
with
any
information
you
want
on
it.
These
notes
must
be
your
own
and
may
not
be
copied
from
another
person.
All
other
relevant
information
is
provided.
The
use
of
wireless
devices
(e.g.
cell
phones,
IR
transmitters/receivers)
is
not
permitted.
The
use
of
programmable
calculators
is
only
allowed
if
all
relevant
content
has
been
erased
from
the
calculator
memory
and
the
programmable
features
are
not
used.
To
receive
full
credit
on
each
problem,
it
is
advised
to
start
with
the
appropriate
full
form
of
the
balance
equation(s)
needed
to
solve
the
problem.
Label
all
variables
and
equations.
Include
a
brief
word
description
to
explain
each
step
in
your
problem
if
appropriate.
State
all
your
assumptions
clearly.
Present
your
solution
clearly.
Numerical
answers
without
units
or
explanations
will
not
receive
credit.
Name:
Section:
A(9am)
B(10am)
C(2pm)
(Please
write
your
name
on
the
back
of
the
last
page
of
the
exam
too)
The
work
presented
here
is
solely
my
own.
|
did
not
receive
any
assistance
nor
did
|
assist
other
students
during
the
exam.
|
pledge
that
|
have
abided
by
the
above
rules
and
the
Georgia
Tech
Honor
Code.
Signature:
Problem
1
/35
Problem
2
/25
Problem
3
/15
Problem
4
125
Total
/100
CHBE
2100
Exam
#2,
Fall
2019
page
2/10
Problem
1
(35
points)
a)
(2
points)
The
density
of
air
increases
as
it
flows
through
a
tube.
Under
steady-state
conditions,
the
volumetric
flow
rate
of
air
at
the
outlet
is
less
than
/
greater
than
/
equal
to
the
volumetric
flow
rate
at
the
inlet.
Explain
your
answer
to
receive
credit.
b)
(3
points)
On
this
phase
diagram,
identify
which
phase
is
present
in
each
region.
Solid:
region
A
region
B
region
C
v
3
Liquid:
region
A
region
B
region
C
¢
B
£
.
Vapor:
region
A
region
B
region
C
A
C
Temperature
c)
(15
points)
Please
answer
the
following
questions
with
a
brief
answer
of
text
and/or
equations
(i)
(3
points)
If
a
liquid
is
boiling,
what
is
it
vapor
pressure
equal
to?
(i)
(3
points)
If
a
single-component
liquid
is
in
equilibrium
with
a
multicomponent
vapor
phase
(that
contains
vaporized
liquid),
what
is
its
vapor
pressure
equal
to?
(ii1)
(3
points)
A
mixture
of
helium
and
benzene
are
in
a
sealed
container.
Initially
the
mixture
is
at
its
dew
point.
Then,
the
pressure
is
changed
while
the
temperature
is
held
constant.
This
causes
the
mixture
to
become
superheated.
What
happened
to
the
pressure?
Why?
pressure
went
up
pressure
went
down
pressure
stayed
the
same
Why:
WP
(W)
0k
A-d
D
'8N
q
Y
59|
T
CHBE
2100
Exam
#2,
Fall
2019
page
3/3
(iv)
(3
points)
The
partition
coefficient,
K,
of
a
butyl
acetate/oil/water
mixture
is
0.28
(mass
fraction
of
butyl
acetate
in
oil/mass
fraction
of
butyl
acetate
in
water).
Is
butyl
acetate
more
or
less
(circle
one)
soluble
in
oil
than
water?
Explain
your
answer.
(v)
(3
points)
The
mole
fraction
of
benzene
is
0.2
in
a
two
component
benzene
and
toluene
system.
What
is
the
bubble
point
temperature
for
this
mixture?
At
this
temperature,
what
is
the
vapor
mole
fraction
of
toluene?
115
110
105
Temperature
(°C)
0
0.2
0.4
0.6
0.8
1.0
Mole
fraction
benzene
P=1atm
Touwoble
=
___
°C
Ytoluene
bubble
=
mol
toluene/mol
e)
(8
points)
Assess
the
following
statements
by
writing
T
(true)
or
F
(false):
-
The
elevation
of
the
boiling
point
of
a
solution
depends
on
the
identity
of
the
solute.
(
-
The
elevation
of
the
boiling
point
of
a
solution
depends
on the
identity
of
the
solvent.
(
-
In
a
solution,
the
higher
the
concentration
of
the
solute,
the
higher
the
freezing
point.
(
-
At
higher
altitudes,
the
boiling
point
of
a
solution
is
larger.
(
-
The
vapor
pressure
of
1L
of
water
at
25°C
is
equal
to
the
vapor
pressure
of
3L
of
water
at
25°C.
(
-
Water
has
only
1
triple
point.
()
-
If
two
compounds
have
very
different
vapor
pressures,
distillation
can
be
used
to
separate
them.
(__)
-
Sea
water
has
a
lower
vapor
pressure
than
pure
water.
That's
why
sea
water
has
a
higher
boiling
point.
()
113233142
09'0~"2t0)
(N2
592
(9>
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CHBE
2100
Exam
#2,
Fall
2019
page
4/4
f)
(7
points)
During
your
undergraduate
research
work,
you
discovered
that
a
newly-made
solutioq
was
not
properly
labelled
since
it
only
shows
the
solute
(glucose)
and
its
mole
fraction
(Ygucose
=
0.05)
without
naming
the
solvent
and
date
of
the
preparation.
There
is
another
bottle
containing
the
pure
solvent,
but
the
label
is
illegible.
As
a
smart
Yellow
Jacket,
you
decide
to
apply
your
knowledge
in
colligative
properties
to
identify
the
solvent.
You
measure
the
boiling
points
of
the
pure
solvent
and
the
glucose
solution,
and
find
that
the
change
in
boiling
point
of
the
solution
is
2.61
°C.
Therefore:
(i)
(5
points)
Which
of
the
following
solvents
could
have
been
used?
Justify
your
answer.
-
Acetic
Acid
-
Water
(ii)
(2
points)
What
is
the
boiling
point
of
the
solution?
°C
R
I)\e)
\r\fle
1\1‘1
N
CHBE
2100
Exam
#2,
Fall
2019
page
5/5
Problem
2
(25
points)
Contaminated
soil
can
be
remediated
by
injection
of
hot
air
and
steam
(i.e.,
water
vapor)
into
the
soil.
To
carry
out
this
process,
30
m®
of
humid
air
at
100°C
and
98.6
kPa
with
a
dew
point
of
30°C
is
introduced
into
the
contaminated
soil.
Once
in
the
soil,
the
humid
air
cools
to
14°C
at
a
pressure
of
109.1
kPa.
Assume
that
all
components
of
the
soil
remain
in
the
solid
state.
What
fraction
of
the
water
vapor
originally
introduced
into
the
soil
will
condense
after
it
is
introduced
into
the
soil,
assuming
no
gas
escapes
from
the
soil.
%
0
%)
CHBE
2100
Exam
#2,
Fall
2019
page
6/6
Problem
3
(15
points)
A
batch
system
contains
25
moles
of
diisopropyl
ether
and
75
moles
of
acetic
acid
at
1
atm
and
25
°C.
Upon
adding
water,
you
find
that that
the
liquid
separates
into
two
phases:
an
organic
phase
(i.e.,
the
phase
with
more
diisopropyl
ether)
and
an
aqueous
phase
(i.e.,
the
phase
with
more
water).
In
order
to
dispose
of
this
mixture,
the
mole
fraction
of
diisopropy!
ether
should
be
no
greater
than
0.05
in
the
aqueous
phase.
Therefore,
you
decide
to
add
just
enough
water
to
the
mixture
to
achieve
the
target
concentration
in
the
aqueous
phase.
a)
(5
points)
Determine
the
mole
fraction
of
water
in
the
aqueous
phase
___mol
H,O/mol
b)
(6
points)
Determine
the
number
moles
in
the
aqueous
phase.
____moles
)
(4
points)
What
is
the
amount
(in
kg)
of
water
added?
kg
100%
acetic
acid
.
e
&N
e
s
.
.
.
R
Of;\}.
‘7000
005
0.10
0.15
0.20
0.25
0.30
035
0.40
0.45
050
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
QQ
(\9
O
—
f\,
[
@(3
'o’bo
Mole
frac
H20
<
S
S\
wh'geo
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CHBE
2100
Exam
#2,
Fall
2019
page
7/7
Problem
4
(25
points)
Toluene
hydrodealkylation
converts
toluene
(T)
to
benzene
(B).
In
this
process,
toluene
and
hydrogen
are
converted
to
benzene
and
methane
as
shown
in
the
formula
below:
(M
(B)
1
CsHsCH3
+
1
H2
g
1
CsHs
+
1
CH4
A
reaction
vessel
for
a
toluene
hydrodealkylation
has two
input
streams,
one
containing
H,
gas
(stream
1),
and
a
second
containing
a
mixture
of
benzene
and
toluene
(stream
2).
H,
gas
is
flowing
into
the
reaction
vessel
at
a
rate
of
0.50
mol/min
(stream
1),
and
a
liquid
mixture
of
benzene
and
toluene
is
flowing
into
the
reaction
vessel
at
0.40
mol/min
(stream
2).
Inside
the
reaction
vessel,
gas
(stream
3)
and
liquid
(stream
4)
phases
are
in
equilibrium.
The
mole
fraction
of
benzene
in
the
liquid
phase
(stream
4)is
0.20.
The
system
is
initially
at
a
temperature
of
70°C
and
a
pressure
of
4
atm;
at
these
conditions,
the
fractional
conversion
of
toluene
is
negligible
(i.e.,
there
is
no
chemical
reaction).
a.)
(15
points)
Assume
that
no
reaction
occurs.
Calculate
the
mole
fractions
of
all
components
in
the
gas
phase.
Y3
toluene
=
Y3
hydrogen
=
y3,benzene
=
What
are
the
molar
flow rates
of
the
gas and
liquid
streams
leaving
the
vessel?
What
is
the
molar
flow
rate
of
toluene
entering
the
vessel?
N3
gases
=
mol/min
Na
jiquids
=
mol/min
N2
toluene
=
mol/min
b.)
(10
points)
Increasing
reaction
temperature
typically
helps
reactions
occur.
The
temperature
of
the
reaction
vessel
is
increased
to
80°C
while
keeping
pressure
constant
at
4
atm
so
that
50%
of
toluene
undergoes
the
hydrodealkylation
reaction.
The
flow
rate
and
composition
of
streams
1
and
2
are
the
same
as
in
part
(a).
The
reaction
vessel
is
designed
such
that
equilibrium
can
be
assured
between
exiting
liquid
and
gas
streams
(after
the
reaction has
occurred),
and
the
molar
fractions
in
the
liquid
stream
are
measured
to
be
X4
penzene
=
0.555,
and
X4
wiuene
=
0.445.
Determine
the
molar
flow
rate
of
gases
leaving
the
reaction
vessel,
and
give
two
reasons
why
this
value
is
different
than
the
value
calculated
in
part
(a).
Also
determine
the
partial
pressure
of
CH,
leaving
the
reactor.
N3
gases
=
mol/min
Pj3cns
=
atm
Reason
#1:
Reason
#2:
€0t
'\90
1°0"1590
‘bhgo
LOW0
'
9tco0
(Y9800
CHBE
2100
Exam
#2,
Fall
2019
page
8/8
TABLE
B.4
Antoine
Equation
Constants”
B
Tr+C
Example:
The
vapor
pressure
of
acetaldehyde
at
25°C
is
determined
as
follows:
1600017
e
-
orvy
<
—_
108,
Pyi,0(25°C)
=
800552
=T
591E00
2.9551
log,,p"
=
A
p
inmmHbg
7T
in°C
=
pi,ol25°C)
=
1077
=
902
mm
Hg
Compound
Formula
Range
(°C)
A
B
<
Acctaldehyde
C.H
O
0210344
8.00552
1600017
291809
Acetic
acid
GHO:
298101265
7.38782
1533313
222309
Acetic
acid*
CH;0,
01036
718807
1416.7
225
Acctic
anhydnde
CyH.O5
628101394
714948
1444718
199817
Acctone
31,0
12910553
711714
1210595
229.664
Acrvlic
aaid
C:Hi
0,
20.0
10
70.0
5.65204
648.629
154.683
Ammonia*
NH;
83
10
60
7.55466
1002711
247885
Aniline
C;N
102.6
1o
185.2
7.32010
1731.515
206.049
Benzene
CeHe
14.5
10
80.9
6.89272
1203531
219888
Toluene
g
35301115
695805
1346773
219.693
L.1.1-Trichlorocthane
CHLCl
S4t016Y9
8.6434
2136.621
302.769
1.1.2-Trichlorocthane
CHCl
SO0t
113.7
6.95185
1314.410
209.197
Trichlorocthylene
CHCl5
17.8
10
86.5
6.51827
1018.603
192.731
Vinyl
acctate
C:H,.O,
21.810
720
7.21010
1296.130
226.655
Water*
H.0O
010
60
8.10765
1750.286
35.000
Water*
H,0
60
1o
150
7.96681
1668.210
228.000
m-Xylene
m-Cellyy
539.210
1400
700646
1460.183
214.827
o-Xylene
o-Cilly,
63.510
1454
7.00154
1476.393
213872
p-Xylene
p-Cslly
583101393
6.98320
1451.792
215111
CHBE
2100
Exam
#2,
Fall
2019
Table
B.S
Properties
of
Saturated
Steam:
Temperature
Table®
page
9/9
V(m’/kg)
U(kJikg)
H{kJ/kg)
T(°C)
P(bar)
Water
Steam
Water
Steam
Water
Evaporation
Steam
0.01
0.00611
0.001000
206.2
zero
23756
+0.0
2501.6
2501.6
2
0.00705
0.001000
1799
8.4
23783
8.4
2496.8
25052
4
0.00813
0.001000
157.3
168
23811
16.8
2492.1
2508.9
6
0.00935
0.001000
137.8
252
23838
252
2487
4
25126
8
0.01072
0.001000
121.0
336
23866
336
2482.6
2516.2
10
0.01227
0.001000
106.4
420
123893
420
2477.9
2519.9
12
0.01401
0.001000
938
504
23921
504
24732
2523.6
14
0.01597
0.001001
829
588
23948
58.8
2468.5
2527.2
16
0.01817
0.001001
734
67.1
23976
67.1
2463.8
25309
18
0.02062
0.001001
65.1
75.5
24003
755
2459.0
2534.5
20
0.0234
0.001002
578
83.9
24030
839
2454.3
25382
22
0.0264
0.001002
515
922
24058
922
2449.6
2541.8
24
00298
0.001003
45.9
100.6
24085
100.6
2444.9
2545.5
25
0.0317
0.001003
434
104.8
24099
1048
24425
25473
26
0.0336
0.001003
41.0
1089
24112
1089
2440.2
2549.1
28
0.0378
0.001004
36.7
117.3
24140
1173
24354
25529
30
0.0424
0.001004
329
1257
24167
1257
2430.7
2556.4
32
00475
0.001005
29.6
1340
24194
1340
2425.9
2560.0
34
0.0532
0.001006
26.6
1424
24221
1424
24212
2563.6
36
0.0594
0.001006
240
150.7
24248
1507
2416.4
25672
38
0.0662
0.001007
21.6
159.1
24275
1591
24117
2570.8
‘From
R.
W.
Haywood,
Thermodynamic
Tables
in
SI
(Metric)
Units,
Cambridge
University
Press,
London,
1968.V
=
specific
volume,
U
=
specific
internal
energy,and
#
=
specific
enthalpy.
Note:
KJ/kg
%
0.4303
=
Buu/lb,.
Table
B.1
Selected
Physical
Property
Data
Compound
Formula
Mol.
Wt.
SG
T
(°C)
BH,(T,)
To(°C)
AH,(T,)
Tc(K)
Pc(atm)
]
-
)
ki/mol
ki/mol
Ac.
Acid
C:H:0;
60.05
1.049
16.6
12.09
118.2
24.39
594.8
57.1
Water
H;0
18.016
1.00¥
0.00
6.0095
100.00
40.656
647.4
218.3
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CHBE
2100
Exam
#2,
Fall
2019
FACTORS
FOR
UNIT
CONVERSIONS
Quantity
Eyuivalent
Values
Mass
Length
Volume
Force
Pressure
11t
i
1056.68
qt
28317
em’
i
ATOMIC
WEIGHTS
ANDNUMBERS
Tkg
=
1000g
=
0.001
metric
ton
=
2.20462
b,
=
35.27392
oz
1bg
=
160z
=
510
“ton
=
453593
g
=
0.453593
kg
Im
=
100cm
=
1000
mm
=
10°
microns
(um)
=
10'”
angstroms
(A)
39.37in.
=
32808
ft
=
1.0936
yd
=
0.0006214
mile
12in.
=
1/3vd
=
03048
m
=
3048
cm
tm’
=
1000L
=
10°ecm®
=
10°
mL
=
353145
ft"
=
220.83
imperial
gallons
=
264.17
gal
Lit
=
1728in
=
7.4805
gal
=
0.028317
m’®
=
28317
L
IN
=
Tkgm's”
=
107
dynes
=
10°
g-emis®
=
0.22481
Iby
Liby
=
32174
Iby
fUs”
=
44482
N
=
4.4482
%
10°
dynes
latm
=
1.01325
%
10°
N/m?
(Pa)
=
1.01325
%
10"
dynes/cm’
760
mm
Hg
at
(°C
(torr)
=
Atomic
weights
apply
1o
naturally
sccurring
isotopic
compositions
and
are
based
on
an
atomic
mass
of
B¢
12
Atomic
Atomic
Atomic
Atomic
Element
Smbol
Nambor
W
cight
Clement
Svbol
Number
Weight
Actimum
Ac
84
-
Iridium
Ir
7
1922
Aluminum
Al
13
26915
Iron
e
26
SS847
Amerikium
Am
s
Krypton
Kr
36
K380
Antimony
o
1
12178
Lanthanum
fa
57
13891
Argon
Ar
=
30948
Lawrencium
Lr
103
—
Arsenic
As
33
7406
Lead
Pt
82
20719
Astatine
AL
83
Lithium
Li
3
£03G
Barnum
Ba
5n
13734
Lutctivm
Lu
71
17497
Berkelium
Bk
7
—
Magnesium
Mg
12
2312
Rervibum
e
4
GUI12
Manganes
Mn
25
S4.9380
Hismuth
%]
K3
20
O
Mendclevium
Md
1
-
Baoron
n
s
1081
Mercury
Hg
R
20059
Bromine
Br
35
79.%4
Molybdenum
Mo
42
9304
Cadmium
«d
a5
11240
Neadymium
Nd
o0
14424
Calcrar
Ca
20
40.08
Neon
Ne
10
20083
Californium
[8]
%
-
Neplunium
Np
a3
-
Carbon
8
6
1201115
Nickel
Ni
P2
871
Cerum
Ce
Rt
]
140012
Niobium
Nh
41
92906
Cosium
s
55
132U
Nitrogen
N
7
140067
Chlonng
1
17
35353
Nobzlium
No
102
Chromium
Cr
p2
|
$1.%96
Osmium
n
75
1902
€
obait
(o
pa)
SN
0332
Uxypen
(8]
1
189993
Copper
Cu
bl
63540
Palladium
ra
46
106.4
Curium
Cm
9%
Phosphorus
4
1s
U573
Dysprosium
Dy
0
162.50
Platinum
[
™
195.00
Einsteinium
Es
(2]
-
Plutomium
Py
94
-
Erbium
Fr
(2]
16726
Polonium
o
54
Furopium
Fu
[l
151
9%
Potassium
K
19
VR
(1]
Formuum
Fm
1w
~
Prascodymium
Pr
4
JEAT
i
Flucring
I
9
IR9RE
Promethium
Pm
6l
Francium
Ir
87
Protsctinium
"
9l
Gadodinium
G
(2]
15728
Radium
Ra
XK
Coathum
o
i
w7
Radon
Rn
L
-
Guormanium
Ge
32
728
Rhenium
Re
Al
1862
Crald
Au
9
196,967
Rhodium
Rh
44
1.
Hatnwm
H
n
178
40
Rubidium
Rb
37
BLS7
Helwm
He
1002
Rathcnmum
Ru
14
107
Holmwen
Ho
67
1624.9%
Samarium
Sm
a2
15035
Hydrogen
H
i
LOO797
Scambium
S
2
4495
=
101.325
kPa
=
1.01325
bar
10.333
m
H2O
at
4°C
THE
GAS
CONSTANT
8314
m*-Pa’(mol-K)
0.08314
L-bar/(mol-K)
0.08206
L-atm/(mol-K)
62.36
L-mm
He/(mol-K}
0.7302
'
-atm/{Ib-mole
-°R)
10.73
ft*-
psia/(1b-mole
“R
)
8.314
Jimol-K)
LOX7
cal/{mol-K}
L987
Bu/(Ib-mole-*R)
page
10/10
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