segal (mas6582) – H09: Gas Laws – mccord – (50970)
1
This printout should have 22 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 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
balloon is 2
.
3 L at STP and the temperature
oF the water remains the same, what is the
volume 54
.
93 m below the water’s surFace?
Correct answer: 0
.
364405 L.
Explanation:
P
1
= 1 atm
Depth = 54
.
93 m
V
1
= 2
.
3 L
V
2
= ?
101.325 kPa = 1 atm
±or
P
2
:
10.2 m
100 kPa
=
54
.
93 m
x
(10
.
2 m)(
x
) = (54
.
93 m)(100 kPa)
x
=
(54
.
93 m)(100 kPa)
10.2 m
= 538
.
529 kPa
P
2
= 101 kPa + 538
.
529 kPa
= 639
.
529 kPa
×
1 atm
101.325 kPa
= 6
.
31166 atm
Applying Boyle’s law,
P
1
V
1
=
P
2
V
2
V
2
=
P
1
V
1
P
2
=
(1 atm) (2
.
3 L)
6
.
31166 atm
= 0
.
364405 L
002
10.0 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.
0.042 mm Hg
2.
600 mm Hg
3.
24 mm Hg
4.
2400 mm Hg
correct
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
10.0 points
At standard temperature, a gas has a volume
oF 230 mL. The temperature is then increased
to 138
◦
C, and the pressure is held constant.
What is the new volume?
Correct answer: 346
.
264 mL.
Explanation:
T
1
= 0
◦
C + 273 = 273 K
V
1
= 230 mL
T
2
= 138
◦
C + 273 = 411 K
V
2
= ?
V
1
T
1
=
V
2
T
2
V
2
=
V
1
T
2
T
1
=
(230 mL)(411 K)
273 K
= 346
.
264 mL
004
10.0 points
A sample oF gas in a closed container at a
temperature oF 96
◦
C and a pressure oF 8 atm
is heated to 272
◦
C. What pressure does the
gas exert at the higher temperature?
Correct answer: 11
.
8157 atm.
Explanation:
T
1
= 96
◦
C + 273 = 369 K
P
1
= 8 atm
T
2
= 272
◦
C + 273 = 545 K
P
2
= ?
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View Full Documentsegal (mas6582) – H09: Gas Laws – mccord – (50970)
2
Applying the GayLussac law,
P
1
T
1
=
P
2
T
2
P
2
=
P
1
T
2
T
1
=
(8 atm) (545 K)
369 K
= 11
.
8157 atm
005
10.0 points
A gas at 1
.
65
×
10
6
Pa and 13
◦
C occu
pies a volume of 406 cm
3
.
At what tem
perature would the gas occupy 555 cm
3
at
2
.
88
×
10
6
Pa?
Correct answer: 409
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 Fall '08
 wandelt
 Chemistry, mol

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