Cheung, Anthony – Homework 7 – Due: Oct 27 2006, midnight – Inst: McCord
1
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printout
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have
26
questions.
Multiplechoice questions may continue on
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Divers know that the pressure exerted by
the water increases about 100 kPa with every
10.2 m of depth.
This means that at 10.2 m
below the surface, the pressure is 201 kPa;
at 20.4 m below the surface, the pressure is
301 kPa; and so forth. If the volume of a bal
loon is 3
.
2 L at STP and the temperature of
the water remains the same, what is the vol
ume 54
.
82 m below the water’s surface?
Correct answer: 0
.
507854 L.
Explanation:
P
1
= 1 atm
Depth = 54
.
82 m
V
1
= 3
.
2 L
V
2
= ?
101.325 kPa = 1 atm
For
P
2
:
10.2 m
100 kPa
=
54
.
82 m
x
(10
.
2 m)(
x
) = (54
.
82 m)(100 kPa)
x
=
(54
.
82 m)(100 kPa)
10.2 m
= 537
.
451 kPa
P
2
= 101 kPa + 537
.
451 kPa
= 638
.
451 kPa
×
1 atm
101.325 kPa
= 6
.
30102 atm
Applying Boyle’s law,
P
1
V
1
=
P
2
V
2
V
2
=
P
1
V
1
P
2
=
(1 atm) (3
.
2 L)
6
.
30102 atm
= 0
.
507854 L
002
(part 1 of 1) 10 points
A gas is enclosed in a 10.0 L tank at 1200
mm Hg pressure.
Which of the following is
a reasonable value for the pressure when the
gas is pumped into a 5.00 L vessel?
1.
600 mm Hg
2.
0.042 mm Hg
3.
2400 mm Hg
correct
4.
24 mm Hg
Explanation:
V
1
= 10.0 L
V
2
= 5.0 L
P
1
= 1200 mm Hg
Boyle’s law relates the volume and pressure
of a sample of gas:
P
1
V
1
=
P
2
V
2
P
2
=
P
1
V
1
V
2
=
(1200 mm Hg)(10
.
0 L)
5 L
= 2400 mm Hg
003
(part 1 of 1) 10 points
A gas at 62
◦
C occupies 4
.
89 L. At what tem
perature will the volume be 3
.
6 L, assuming
the same pressure?
Correct answer:

26
.
3742
◦
C.
Explanation:
V
1
= 4
.
89 L
V
2
= 3
.
6 L
T
1
= 62
◦
C + 273 = 335 K
T
2
= ?
V
1
T
1
=
V
2
T
2
T
2
=
T
1
V
2
V
1
=
(335 K) (3
.
6 L)
4
.
89 L
= 246
.
626 K
◦
C = K

273 =

26
.
3742
◦
C
004
(part 1 of 1) 10 points
At standard temperature, a gas has a volume
of 316 mL. The temperature is then increased
to 135
◦
C, and the pressure is held constant.
What is the new volume?
Correct answer: 472
.
264 mL.
Explanation:
T
1
= 0
◦
C + 273 = 273 K
V
1
= 316 mL
T
2
= 135
◦
C + 273 = 408 K
V
2
= ?
V
1
T
1
=
V
2
T
2
V
2
=
V
1
T
2
T
1
=
(316 mL)(408 K)
273 K
= 472
.
264 mL
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Cheung, Anthony – Homework 7 – Due: Oct 27 2006, midnight – Inst: McCord
2
005
(part 1 of 1) 10 points
A sample of gas in a closed container at a
temperature of 76
◦
C and a pressure of 7 atm
is heated to 383
◦
C. What pressure does the
gas exert at the higher temperature?
Correct answer: 13
.
1576 atm.
Explanation:
T
1
= 76
◦
C + 273 = 349 K
P
1
= 7 atm
T
2
= 383
◦
C + 273 = 656 K
P
2
= ?
Applying the GayLussac law,
P
1
T
1
=
P
2
T
2
P
2
=
P
1
T
2
T
1
=
(7 atm) (656 K)
349 K
= 13
.
1576 atm
006
(part 1 of 1) 10 points
A gas at 658
×
10
4
Pa and 13
◦
C occupies a vol
ume of 754 cm
3
. At what temperature would
the gas occupy 957 cm
3
at 9
.
03
×
10
4
Pa?
Correct answer:

268
.
018
◦
C.
Explanation:
P
1
= 6
.
58
×
10
6
Pa
P
2
= 9
.
03
×
10
4
Pa
V
1
= 754 cm
3
T
1
= 13
◦
C + 273 = 286 K
V
2
= 957 cm
3
T
2
= ?
P
1
V
1
T
1
=
P
2
V
2
T
2
T
2
=
P
2
V
2
T
1
P
1
V
1
=
(9
.
03
×
10
4
Pa) (957 cm
3
) (286 K)
(6
.
58
×
10
6
Pa) (754 cm
3
)
= 4
.
9816 K =

268
.
018
◦
C
007
(part 1 of 1) 10 points
A sample of ideal gas occupies 250 mL at 25
◦
C
and 740 torr.
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 Fall '07
 Fakhreddine/Lyon
 Chemistry, mol

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