Kim, Eugene – Homework 12 – Due: Dec 4 2007, midnight – Inst: McCord
1
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32
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AMa
WEL 3.502
McZ
WEL 2.246
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.
False
correct
2.
True
Explanation:
002
(part 1 of 1) 10 points
Entropy is a state function.
1.
True
correct
2.
False
Explanation:
003
(part 1 of 1) 10 points
Place the following in order of increasing en
tropy.
1.
solid, gas, and liqiud
2.
solid, liquid, and gas
correct
3.
gas, liqiud, and solid
4.
gas, solid, and liqiud
5.
liqiud, solid, and gas
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.
O
2
(g) at 273 K and 1.00 atm
2.
They are the same
3.
O
2
(g) at 450 K and 1.00 atm
correct
4.
Unable to determine
Explanation:
O
2
(g) at 450 K has increased molecular
randomness (1 mol of O
2
at 1.00 atm pressure
will occupy a larger volume at 450 K than at
273 K), 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.
+115 J
·
K

1
·
mol

1
correct
2.

40.5 kJ
·
K

1
·
mol

1
3.
+40.5 kJ
·
K

1
·
mol

1
4.
+513 J
·
K

1
·
mol

1
5.

115 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 4
.
08 mol of ideal gas atoms
from 102
.
61
◦
C to

27
.
23
◦
C at constant vol
ume?
Correct answer:

21
.
5817 J
/
K.
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Kim, Eugene – Homework 12 – Due: Dec 4 2007, midnight – Inst: McCord
2
Explanation:
T
1
= 102
.
61
◦
C + 273 = 375
.
61 K
T
2
=

27
.
23
◦
C + 273 = 245
.
77 K
n
= 4
.
08 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
= (4
.
08 mol)
3
2
8
.
314
J
K
·
mol
¶
×
ln
245
.
77 K
375
.
61 K
¶
=

21
.
5817 J
/
K
.
007
(part 1 of 1) 10 points
What is the entropy change associated with
the isothermal compression of 5
.
73 mol of
ideal gas atoms from 9
.
51 atm to 14
.
33 atm?
Correct answer:

19
.
5326 J
/
K.
Explanation:
P
1
= 9
.
51 atm
P
2
= 14
.
33 atm
n
= 5
.
73 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
¶
= (5
.
73 mol)
8
.
314
J
K
·
mol
¶
×
ln
9
.
51 atm
14
.
33 atm
¶
=

19
.
5326 J
/
K
.
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 Fall '07
 Fakhreddine/Lyon
 Chemistry, Thermodynamics, Entropy, Correct Answer

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