Iype, Shelby – Homework 12 – Due: Dec 4 2007, midnight – Inst: Vandenbout
1
This
printout
should
have
32
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
EXAM 4 is Thursday, 12/6 from 79 PM
Go to the right room
AK
WEL 1.316
LO
WEL 2.122
PZ
WEL 1.308
001
(part 1 of 1) 10 points
For a given transfer of energy, a greater change
in disorder occurs when the temperature is
high.
1.
True
2.
False
correct
Explanation:
002
(part 1 of 1) 10 points
Entropy is a state function.
1.
False
2.
True
correct
Explanation:
003
(part 1 of 1) 10 points
Place the following in order of increasing en
tropy.
1.
solid, gas, and liqiud
2.
gas, solid, and liqiud
3.
liqiud, solid, and gas
4.
solid, liquid, and gas
correct
5.
gas, liqiud, and solid
Explanation:
Entropy (
S
) is high for systems with high
degrees of freedom, disorder or randomness
and low for systems with low degrees of free
dom, disorder or randomness.
S
(g)
> S
(
‘
)
> S
(s)
.
004
(part 1 of 1) 10 points
Which substance has the higher molar en
tropy?
1.
Unable to determine
2.
Ne(g) at 298 K and 1.00 atm
3.
They are the same
4.
Kr(g) at 298 K and 1.00 atm
correct
Explanation:
Kr(g) is more massive and has more elemen
tary particles, hence a higher molar entropy.
005
(part 1 of 1) 10 points
Calculate the standard entropy of vaporiza
tion of ethanol at its boiling point 352 K. The
standard molar enthalpy of vaporization of
ethanol at its boiling point is 40.5 kJ
·
mol

1
.
1.

40.5 kJ
·
K

1
·
mol

1
2.
+40.5 kJ
·
K

1
·
mol

1
3.
+115 J
·
K

1
·
mol

1
correct
4.

115 J
·
K

1
·
mol

1
5.
+513 J
·
K

1
·
mol

1
Explanation:
006
(part 1 of 1) 10 points
Assuming that the heat capacity of an ideal
gas is independent of temperature, what is
the entropy change associated with lowering
the temperature of 5
.
25 mol of ideal gas atoms
from 102
.
27
◦
C to

25
.
68
◦
C at constant vol
ume?
Correct answer:

27
.
2997 J
/
K.
Explanation:
T
1
= 102
.
27
◦
C + 273 = 375
.
27 K
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Iype, Shelby – Homework 12 – Due: Dec 4 2007, midnight – Inst: Vandenbout
2
T
2
=

25
.
68
◦
C + 273 = 247
.
32 K
n
= 5
.
25 mol
R
= 8
.
314
J
K
·
mol
At constant volume,
dq
=
n C
V
dT
, so
dS
=
dq
T
=
n C
V
dT
T
Z
dS
=
n C
V
Z
dT
T
Δ
S
=
n C
V
ln
T
2
T
1
¶
For an ideal monatomic gas
C
V
=
3
2
R
, so
Δ
S
= (5
.
25 mol)
3
2
8
.
314
J
K
·
mol
¶
×
ln
247
.
32 K
375
.
27 K
¶
=

27
.
2997 J
/
K
.
007
(part 1 of 1) 10 points
What is the entropy change associated with
the isothermal compression of 7
.
68 mol of
ideal gas atoms from 8
.
88 atm to 12
.
78 atm?
Correct answer:

23
.
2471 J
/
K.
Explanation:
P
1
= 8
.
88 atm
P
2
= 12
.
78 atm
n
= 7
.
68 mol
R
= 8
.
314
J
K
·
mol
Because the process is isothermal, Δ
U
= 0,
so
q
=

w
, where
w
=

P dV
.
Because the process is isothermal and re
versible,
dS
=
dq
T
=
P
dV
T
=
n R T
T V
dV
=
n R
V
dV
Z
dS
=
n R
Z
dV
V
Δ
S
=
n R
ln
V
2
V
1
¶
.
Since
P
1
V
1
=
P
2
V
2
,
V
2
V
1
=
P
1
P
2
and
Δ
S
=
n R
ln
P
1
P
2
¶
= (7
.
68 mol)
8
.
314
J
K
·
mol
¶
×
ln
8
.
88 atm
12
.
78 atm
¶
=

23
.
2471 J
/
K
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Fakhreddine/Lyon
 Chemistry, Thermodynamics, Entropy, Correct Answer, Iype

Click to edit the document details