DGD 6
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Course
2132
Subject
Chemistry
Date
Feb 20, 2024
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CHM2132 DGD Assignment #6 •
One copy of your group’s solutions must be uploaded to Brightspace on your Group page by 8 pm on Wednesday Nov 3
rd
. •
You must show your work to receive full credit. •
Appropriate sig figs and units are required in the final answer. •
Assignments that are not legible cannot be marked. •
After the DGD, every student must record participation marks for each of their group members using the link: https://forms.gle/qDdtZn1C2ZD2reuX7 •
Students who do not complete the form will not receive credit for participation at that DGD. 1)
The enthalpy of vaporization for a liquid is 32.6 kJ mol
-1
at its normal boiling point of 368.4 K. The molar volumes of the liquid and the vapour at the boiling point are are 18.6 mL mol
-1
, and 30.56 L mol
-1
, respectively. a)
Use the Clapeyron equation to estimage the temperature dependence of the vapour pressure (dp/dT) at the normal boiling point. b)
Use the Clausius-Clapeyron equation to estimate dp/dT. c)
Does the Clausius-Clapeyron equation overestimate or underestimate the temperature dependence of the vapour pressure? What is the percent error of this estimation? 2)
Suppose the partial pressure of water is 245 Pa on a fall morning at -8.38 o
C. Use a calculation to predict whether frost will form under these conditions. You may need to use some of the data below, and assume that the enthalpy values are temperature independent. Data for H
2
O Δ
vap
H
m
o
=
44.0 kJ mol
-1
Δ
fus
H
m
o
=
6.01 kJ mol
-1
C
p
,
m
o
l
( )
=
75.3 J K
-1
mol
-1
C
p
,
m
o
g
(
)
=
33.6 J K
-1
mol
-1
T
triple point
=
273.16 K
p
triple point
=
611.2 Pa
M
=
18.02 g mol
-1
I
Buap
Ho
32.6
kt
molt
Trap
368.4K
Vm
l
18.6
10
310
Um
g
30.56
It
ftp.nlgj.my
32600J
mot
368
412
30.5640
18.6
100
1110
73
2.897
0
Pak
b
CLAUSIUS
CLAPEYRON
DUM
Ving
day
Duepit
Trap
Um
g
32600J
mot
368
411
30.5641
110
1
7
2.896
103
Pa
K
l
C
CLAUSIUS
CLAPEYRON
UNDERESTIMATES
It
ERROR
DIFFERENCE
ACTUAL
dog
4100
2.897
103 2.896
103
2
897
103
100
0.06
2
Pino
245
Pa
T
8.388
264.77k
NEED
TO
FIND
PHO
EQUILIBRIUM
THIS
TEMPERATURE
E
degli
t
y
Asb
Amphorae
Ho
44.0
6.01
KI
50.0
II
Pz
p
É
FROM
TABLE
TRIPLE
point
27.3.16
K
PTRIPLE
Point
64.2Pa
P2
611.2Pa
expf
III
É
ata
304Pa
SINCE
ACTUAL
Pao
7245
Pa
LESS
THAN
VAPOUR
PRESSURE
OF
H2O
EQUILIBRIUM
FROST
WILL
NOT
FORM
g
I
S
EQUILIBRIUM WILL
PROCEED
IN
DIRECTION
OF
GAS
SO
IF
ANY
FROST
IS
PRESENT
IT
WILL
SUBLIME
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