triesault (njt298) – H13: Thermo  2nd Law – McCord – (53740)
1
This printout should have 32 questions.
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beFore answering.
001
10.0 points
±or a given transFer oF energy, a greater change
in disorder occurs when the temperature is
high.
1.
True
2.
±alse
correct
Explanation:
±rom Δ
S
=
q
T
since
T
is in the denomina
tor, Δ
S
will be
larger
(more positive) when
ever
T
is
smaller
.
002
10.0 points
Entropy is a state Function.
1.
±alse
2.
correct
Explanation:
State Functions are denoted by capital
ized letters. They are
P
(ressure),
V
(olume),
T
(emperature),
S
(entropy),
G
(ibb’s ±ree en
ergy),
H
(enthalpy). The change in the value
oF a state Function is independent oF the path
taken.
003
10.0 points
Place the Following in order oF increasing en
tropy.
1.
gas, solid, and liqiud
2.
gas, liqiud, and solid
3.
solid, liquid, and gas
correct
4.
solid, gas, 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
(
ℓ
)
> S
(s)
.
004
10.0 points
Which substance has the higher molar en
tropy?
1.
They are the same
2.
Unable to determine
3.
Kr(g) at 298 K and 1.00 atm
correct
4.
Ne(g) at 298 K and 1.00 atm
Explanation:
Kr(g) is more massive and has more elemen
tary particles, hence a higher molar entropy.
005
10.0 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.

115 J
·
K
−
1
·
mol
−
1
3.

40.5 kJ
·
K
−
1
·
mol
−
1
4.
+40.5 kJ
·
K
−
1
·
mol
−
1
5.
+513 J
·
K
−
1
·
mol
−
1
Explanation:
Δ
H
vap
= 40500 J
·
mol
−
1
T
BP
= 352 K
Δ
S
cond
=
q
T
=
Δ
H
con
T
BP
=
Δ
H
vap
T
BP
=
40500 J
·
mol
−
1
352 K
= +115
.
057 J
·
mol
−
1
·
K
−
1
006
10.0 points
Assuming that the heat capacity oF an ideal
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View Full Documenttriesault (njt298) – H13: Thermo  2nd Law – McCord – (53740)
2
gas is independent of temperature, what is
the entropy change associated with lowering
the temperature of 2
.
25 mol of ideal gas atoms
from 102
.
13
◦
C to

35
.
27
◦
C at constant vol
ume?
Correct answer:

12
.
7991 J
/
K.
Explanation:
T
1
= 102
.
13
◦
C + 273 = 375
.
13 K
T
2
=

35
.
27
◦
C + 273 = 237
.
73 K
n
= 2
.
25 mol
R
= 8
.
314
J
K
·
mol
At constant volume,
dq
=
nC
V
dT
, so
dS
=
dq
T
=
V
dT
T
i
dS
=
V
i
dT
T
Δ
S
=
V
ln
p
T
2
T
1
P
For an ideal monatomic gas
C
V
=
3
2
R
, so
Δ
S
= (2
.
25 mol)
3
2
p
8
.
314
J
K
·
mol
P
×
ln
p
237
.
73 K
375
.
13 K
P
=

12
.
7991 J
/
K
.
007
10.0 points
What is the entropy change associated with
the isothermal compression of 6
.
16 mol of
ideal gas atoms from 9
.
13 atm to 12
.
65 atm?
Correct answer:

16
.
7005 J
/
K.
Explanation:
P
1
= 9
.
13 atm
P
2
= 12
.
65 atm
n
= 6
.
16 mol
R
= 8
.
314
J
K
·
mol
Because the process is isothermal, Δ
U
= 0,
so
q
=

w
, where
w
=

P dV
.
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 Fall '07
 Fakhreddine/Lyon
 Thermodynamics, Entropy, Correct Answer, mol

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