padilla (tp5647) – Homework 9 – Sutcliffe – (53770)
1
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25
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001
10.0 points
A gas has a volume of 2.0 L at a pressure of
3.00 atm. What pressure is needed to change
the volume to 6.00 L at constant temperature?
1.
2.0 atm
2.
1.0 atm
correct
3.
9.0 atm
4.
4.0 atm
Explanation:
V
1
= 2.00 L
P
1
= 3 atm
V
2
= 6.00 L
Boyle’s law,
P
1
V
1
=
P
2
V
2
, relates the vol
ume and pressure of a sample of gas.
P
2
=
P
1
V
1
V
2
=
(3.00 atm) (2.00 L)
6.00 L
= 1.0 atm
002
10.0 points
At constant temperature if the volume of a
gas sample is tripled the pressure will
1.
be cut in half.
2.
remain the same.
3.
be tripled.
4.
be one third the original.
correct
5.
be increased by a factor of 6.
Explanation:
3
V
1
=
V
2
Boyle’s Law relates the volume and pressure
of a sample of gas:
P
1
V
1
=
P
2
V
2
P
1
V
1
=
P
2
(3
V
1
)
P
1
= 3
P
2
P
2
=
1
3
P
1
003
10.0 points
A certain quantity of a gas occupies 61.3 mL
at 68
◦
C. If the pressure remains constant,
what would be the volume of the gas at
128
◦
C?
1.
72 mL
correct
2.
92 mL
3.
32 mL
4.
52 mL
Explanation:
V
1
= 61
.
3 mL
T
1
= 68
◦
C = 341 K
T
2
= 128
◦
C = 401 K
Charles’s law
V
1
T
1
=
V
2
T
2
relates the volume
and absolute (Kelvin) temperature of a sam
ple of gas:
V
2
=
V
1
T
2
T
1
=
(61
.
3mL) (401 K)
341 K
= 72 L
004
10.0 points
If a sample of gas is heated from 100
◦
C to
300
◦
C at constant pressure, the volume will
?
by a factor of
?
.
1.
increase; about three
2.
increase; about 1.5
correct
3.
decrease; about 1.5
4.
No other answer is correct, because the
volume cannot change when the pressure is
constant.
5.
decrease; about three
Explanation:
T
1
= 100
◦
C + 273 = 373 K
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padilla (tp5647) – Homework 9 – Sutcliffe – (53770)
2
T
2
= 300
◦
C + 273 = 573 K
The temperature increases by a factor of
573
373
= 1
.
5.
According to Charles’ Law, the
volume of the gas would increase by the same
factor.
005
10.0 points
A sample of a gas occupies 460 mL at 70.0
◦
C
and 1.00 atmosphere.
At what temperature
would the gas occupy 650 mL at the same
pressure?
1.

30.0
◦
C
2.
435
◦
C
3.
99
◦
C
4.
212
◦
C
correct
5.
154
◦
C
6.
188
◦
C
7.
111
◦
C
Explanation:
V
1
= 460 mL
V
2
= 650 mL
T
1
= 70.0
◦
C + 273 = 343 K
Charles’s law relates the volume and ab
solute (Kelvin) temperature of a sample of
gas:
V
1
T
1
=
V
2
T
2
T
2
=
V
2
T
1
V
1
=
(650 mL) (343 K)
460 mL
= 485 K
◦
C = 485 K

273 = 212
◦
C
006
10.0 points
The normal respiratory rate for a human be
ing is 15.0 breaths per minute.
The average
volume of air for each breath is 505 cm
3
at
20
◦
C and 9
.
95
×
10
4
Pa. What is the volume
of air at STP that an individual breathes in
one day? Give your answers in cubic meters.
1.
9
.
98 m
3
/
day
correct
2.
505 cm
3
/
day
3.
505 m
3
/
day
4.
None of these
5.
100 m
3
/
day
6.
9
.
95 cm
3
/
day
Explanation:
P
1
= 9
.
95
×
10
4
Pa
P
2
= 1
.
01325
×
10
5
Pa
T
1
= 20
◦
C + 273 = 293 K
V
1
= 505 cm
3
T
2
= 0
◦
C + 273 = 273 K
V
2
= ?
P
1
V
1
T
1
=
P
2
V
2
T
2
V
2
=
P
1
V
1
T
2
P
2
T
1
=
(9
.
95
×
10
4
Pa)(505 cm
3
)(273 K)
(1
.
01325
×
10
5
Pa)(293 K)
= 0
.
000462054 m
3
parenleftbigg
15 breaths
min
parenrightbiggparenleftbigg
60 min
hour
parenrightbiggparenleftbigg
24 hours
day
parenrightbigg
= 21600 breaths
/
day
parenleftbigg
21600 breaths
day
parenrightbiggparenleftbigg
0
.
000462054 m
3
breath
parenrightbigg
= 9
.
98037 m
3
/
day
007
10.0 points
At STP a gas occupies 121 mL. How many
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 Spring '08
 SUTCLIFFE
 Mole, Padilla

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