nanni (arn437) – HW 02 Gas Laws – vanden bout – (50985)
1
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printout
should
have
23
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
Questions 2, 3 and 7 are not multiple choice.
For questions 2 and 3, your answer should
have five decimal places. For question 7, your
answer should have four decimal places.
001
10.0points
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.
2400 mm Hg
correct
2.
600 mm Hg
3.
24 mm Hg
4.
0.042 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
002
10.0points
A sample of gas in a closed container at a
temperature of 85
◦
C and a pressure of 8 atm
is heated to 267
◦
C. What pressure does the
gas exert at the higher temperature?
Correct answer: 12
.
067 atm.
Explanation:
T
1
= 85
◦
C + 273 = 358 K
P
1
= 8 atm
T
2
= 267
◦
C + 273 = 540 K
P
2
= ?
Applying the GayLussac law,
P
1
T
1
=
P
2
T
2
P
2
=
P
1
T
2
T
1
=
(8 atm) (540 K)
358 K
= 12
.
067 atm
003
10.0points
A flask containing 170 cm
3
of hydrogen was
collected under a pressure of 20
.
7 kPa. What
pressure would have been required for the vol
ume of the gas to have been 72 cm
3
, assuming
the same temperature?
Correct answer: 48
.
875 kPa.
Explanation:
V
1
= 170 cm
3
V
2
= 72 cm
3
P
1
= 20
.
7 kPa
P
2
= ?
Applying Boyle’s law,
P
1
V
1
=
P
2
V
2
P
2
=
P
1
V
1
V
2
=
(20
.
7 kPa)(170 cm
3
)
72 cm
3
= 48
.
875 kPa
004
10.0points
A 5.00L sample of a gas exerts a pressure of
1040 torr at 50.0
◦
C. In what volume would
the same sample exert a pressure of 1.00 atm
at 50.0
◦
C?
1.
0.581 L
2.
10.5 L
3.
3.33 L
4.
6.84 L
correct
5.
1.95 L
Explanation:
P
1
= (1040 torr)
parenleftbigg
1 atm
760 torr
parenrightbigg
= 1
.
36842 atm
P
2
= 1 atm
V
1
= 5 L
Since the number of moles and the temper
ature of the gas sample are held constant, the
change in the volume is determined by Boyle’s
Law:
P
1
V
1
=
P
2
V
2
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nanni (arn437) – HW 02 Gas Laws – vanden bout – (50985)
2
V
2
=
P
1
V
1
P
2
=
(1
.
36842 atm)(5 L)
1 atm
= 6
.
84211 L
005
10.0points
Consider
2 Al + 6 HCl
→
2 AlCl
3
+ 3 H
2
,
the reaction of Al with HCl to produce hy
drogen gas. This reaction has a yield of 82.5
percent.
How much HCl are needed to pro
duce 14.0 L of H
2
at 351 K and 1.11 atm?
1.
1.08 mol
2.
1.31 mol
correct
3.
0.655 mol
4.
0.890 mol
5.
0.540 mol
6.
0.223 mol
7.
0.446 mol
Explanation:
V
= 14 L
T
= 351 K
P
= 1
.
11 atm
p
= 82
.
5%
P V
=
n R T
n
=
P V
R T
=
(1
.
11 atm)(14 L)
(0
.
0821
L
·
atm
mol
·
K
)(351 K)
= 0
.
539263 mol H
2
.
For 82.5% yield,
n
H
2
=
0
.
539263 mol H
2
0
.
825
= 0
.
539263 mol H
2
.
n
H
2
= (0
.
539263 mol H
2
)
6 mol HCl
3 mol H
2
= 1
.
61779 mol HCl
.
006
10.0points
The tires of a car get hot after the car has
been driven, resulting in an increase in the
tires’ pressure.
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 Fall '07
 Fakhreddine/Lyon
 mol, gas particles

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