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baum (awb464) – 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 3
.
6 L at STP and the temperature
oF the water remains the same, what is the
volume 38
.
5 m below the water’s surFace?
Correct answer: 0
.
762398 L.
Explanation:
P
1
= 1 atm
Depth = 38
.
5 m
V
1
= 3
.
6 L
V
2
= ?
101.325 kPa = 1 atm
±or
P
2
:
10.2 m
100 kPa
=
38
.
5 m
x
(10
.
2 m)(
x
) = (38
.
5 m)(100 kPa)
x
=
(38
.
5 m)(100 kPa)
10.2 m
= 377
.
451 kPa
P
2
= 101 kPa + 377
.
451 kPa
= 478
.
451 kPa
×
1 atm
101.325 kPa
= 4
.
72194 atm
Applying Boyle’s law,
P
1
V
1
=
P
2
V
2
V
2
=
P
1
V
1
P
2
=
(1 atm) (3
.
6 L)
4
.
72194 atm
= 0
.
762398 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.
24 mm Hg
2.
600 mm Hg
3.
2400 mm Hg
correct
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
003
10.0 points
At standard temperature, a gas has a volume
oF 215 mL. The temperature is then increased
to 119
◦
C, and the pressure is held constant.
What is the new volume?
Correct answer: 308
.
718 mL.
Explanation:
T
1
= 0
◦
C + 273 = 273 K
V
1
= 215 mL
T
2
= 119
◦
C + 273 = 392 K
V
2
= ?
V
1
T
1
=
V
2
T
2
V
2
=
V
1
T
2
T
1
=
(215 mL)(392 K)
273 K
= 308
.
718 mL
004
10.0 points
A sample oF gas in a closed container at a
temperature oF 90
◦
C and a pressure oF 5 atm
is heated to 277
◦
C. What pressure does the
gas exert at the higher temperature?
Correct answer: 7
.
57576 atm.
Explanation:
T
1
= 90
◦
C + 273 = 363 K
P
1
= 5 atm
T
2
= 277
◦
C + 273 = 550 K
P
2
= ?
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View Full Document baum (awb464) – 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
=
(5 atm) (550 K)
363 K
= 7
.
57576 atm
005
10.0 points
A gas at 1
.
59
×
10
6
Pa and 26
◦
C occu
pies a volume of 442 cm
3
.
At what tem
perature would the gas occupy 507 cm
3
at
3
.
48
×
10
6
Pa?
Correct answer: 477
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This note was uploaded on 11/21/2010 for the course CHEM 50970 taught by Professor Dr.mccord during the Fall '10 term at University of Texas at Austin.
 Fall '10
 Dr.McCord

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