74
CHAPTER FIVE
GASES
Pressure
21.
4.75 cm
×
cm
mm
10
= 47.5 mm Hg or 47.5 torr;
47.5 torr
×
torr
760
atm
1
= 6.25
×
10
-2
atm
6.25
×
10
-2
atm
×
atm
Pa
10
_
1.013
5
= 6.33
×
10
3
Pa
22.
If the levels of Hg in each arm of the manometer are equal, then the pressure in the flask is equal
to atmospheric pressure.
When they are unequal, the difference in height in mm will be equal to
the difference in pressure in mm Hg between the flask and the atmosphere.
Which level is higher
will tell us whether the pressure in the flask is less than or greater than atmospheric.
a.
P
flask
< P
atm
; P
flask
= 760. - 118 = 642 torr
642 torr
×
torr
760
atm
1
= 0.845 atm
0.845 atm
×
atm
Pa
10
_
1.013
5
= 8.56
×
10
4
Pa
b.
P
flask
> P
atm
; P
flask
= 760. torr + 215 torr = 975 torr
975 torr
×
torr
760
atm
1
= 1.28 atm
1.28 atm
×
atm
Pa
10
_
1.013
5
= 1.30
×
10
5
Pa

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75
c.
P
flask
= 635 - 118 = 517 torr;
P
flask
= 635 + 215 = 850. torr
23.
Suppose we have a column of Hg 1.00 cm
×
1.00 cm
×
76.0 cm =
V = 76.0 cm
3
:
mass = 76.0 cm
3
×
13.59 g/cm
3
= 1.03
×
10
3
g
×
g
1000
kg
1
= 1.03 kg
F = mg = 1.03 kg
×
9.81 m/s
2
= 10.1 kg m/s
2
= 10.1 N
÷
ø
ö
ç
è
æ
m
cm
100
_
cm
N
10.1
=
area
Force
2
2
= 1.01
×
10
5
m
N
2
or 1.01
×
10
5
Pa
(Note:
76.0 cm Hg = 1 atm = 1.01
×
10
5
Pa.)
To exert the same pressure, a column of water will have to contain the same mass as the 76.0 cm
column of Hg.
Thus, the column of water will have to be 13.59 times taller or 76.0 cm
×
13.59 =
1.03
×
10
3
cm = 10.3 m.
24.
The pressure is proportional to the mass of the fluid.
The mass is proportional to the volume of the
column of fluid (or to the height of the column assuming the area of the column of fluid is
constant).
volume
mass
=
density
=
d
;
The volume of silicon oil is the same as the volume of Hg in Exercise
5.22.
V =
d
m
;
V
Hg
= V
oil
;
d
m
=
d
m
oil
oil
Hg
Hg
,
m
oil
=
d
d
m
Hg
oil
Hg
Since P is proportional to the mass of liquid:
P
oil
= P
Hg
÷
÷
ø
ö
ç
ç
è
æ
d
d
Hg
oil
= P
Hg
÷
ø
ö
ç
è
æ
13.6
1.30
= 0.0956 P
Hg
This conversion applies only to the column of silicon oil.
a.
P
flask
= 760. torr - (118
×
0.0956) torr = 760. - 11.3 =
749 torr

CHAPTER 5
GASES
76
749 torr
×
torr
760
atm
1
= 0.986 atm;
0.986 atm
×
atm
Pa
10
_
1.013
5
= 9.99
×
10
4
Pa
b.
P
flask
= 760. torr + (215
×
0.0956) torr = 760. + 20.6 =
781 torr
781 torr
×
torr
760
atm
1
= 1.03 atm;
1.03 atm
×
atm
Pa
10
_
1.013
5
= 1.04
×
10
5
Pa
25.
If we are measuring the same pressure, the height of the silicon oil column would be 13.6
÷
1.30 =
10.5 times the height of a mercury column.
The advantage of using a less dense fluid than
mercury is in measuring small pressures.
The height difference measured will be larger for the less
dense fluid.
Thus, the measurement will be more precise.

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CHAPTER 5
GASES
77
Gas Laws
26.
a.
PV = nRT
b.
PV = nRT
c.
PV = nRT
PV = Constant
P =
÷
ø
ö
ç
è
æ
V
nR
×
T = Const
×
T T =
÷
ø
ö
ç
è
æ
nR
P
×
V = Const
×
V
d.
PV = nRT
e.
P =
V
Constant
=
V
nRT
f.
PV = nRT
PV = Constant
P = Constant
×
V
1
T
PV
= nR = Constant

CHAPTER 5
GASES
78
Note:
The equation for a straight line is y = mx + b where y is the y-axis and x is the x-axis. Any
equation that has this form will produce a straight line with slope equal to m and y-