122
CHAPTER 6
GASES
PRACTICE EXAMPLES
1A
The pressure measured by each liquid must be the same. They are related
through
P = g h d
Thus, we have the following
g h
DEG
d
DEG
=
g h
Hg
d
Hg
.
The
g
’s
cancel; we substitute known values: 9.25 m
DEG
×1.118 g/cm
3
DEG
=
h
Hg
× 13.6
g/cm
3
Hg
3
Hg
3
1.118g/cm
9.25m
0.760mHg,
= 0.760m Hg =760.mmHg
13.6g/cm
hP
1B
The solution is found through the expression relating density and height:
TEG
TEG
Hg
Hg
hd
We substitute known values and solve for triethylene glycol’s density:
9.14 m
TEG
×
d
TEG
= 757 mmHg ×13.6 g/cm
3
Hg
.
Using unit conversions, we get
d
TEG
=
33
0.757 m
13.6 g/cm
1.13 g/cm
9.14 m
2A
We know that
P
gas
=
P
bar
+
Δ
P
with
P
bar
= 748.2 mmHg.
We are told that
Δ
P
=
7.8 mmHg. Thus,
P
gas
= 748.2 mmHg + 7.8 mmHg = 756.0 mmHg.
2B
The difference in pressure between the two levels must be the same, just
expressed in different units. Hence, this problem is almost a repetition of Practice
Example 61.
h
Hg
=748.2 mmHg – 739.6 mmHg=8.6 mmHg.
Again we have
g h
g
d
g
=
g h
Hg
d
Hg
.
This becomes
h
g
× 1.26 g/cm
3
glycerol
=
8.6 mmHg ×13.6 g/cm
3
Hg
h
g
3
mmHg
gcmH
g
gcmg
l
y
c
e
r
o
l
mm glycerol
86
136
126
93
3
.
./
3A
A =
r
2
(here
r
= ½(2.60 cm
×
1 m
100 cm
) = 0.0130 m)
0.0130 m)
2
= 5.31 × 10
4
m
2
F = m × g = (1.000 kg)(9.81 m s
2
) = 9.81 kg m s
2
= 9.81 N
P =
A
F
=
4
2
9.81 N
5.31
10 m
= 18475 N m
2
or 1.85 × 10
4
Pa
P (torr) = 1.85 × 10
4
Pa × = 139 torr
3B
Final pressure = 100 mb.
101, 325 Pa
100 mb
1013.25 mb
= 1.000 × 10
4
Pa
The area of the cylinder is unchanged from that in Example 63, (1.32 × 10
3
m
2
).
P =
F
A
= 1.000 × 10
4
Pa =
3
2
F
1.32×10 m
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View Full DocumentChapter 6: Gases
123
Solving for F, we find F = 13.2 (Pa)m
2
= 13.2 (N m
2
)m
2
= 13.2 N
F = m × g = 13.2 kg m s
2
=
2
m
9.81 m s
total mass = mass of cylinder + mass added weight = m =
g
F
=
2
2
13.2 kg m s
9.81 m s
= 1.35 kg
An additional 350 grams must be added to the top of the 1.000 kg (1000 g) red
cylinder to increase the pressure to 100 mb.
It is not necessary to add a mass
with the same cross sectional area.
The pressure will only be exerted over the
area that is the base of the cylinder on the surface beneath it.
4A
The ideal gas equation is solved for volume. Conversions are made within the equation.
3
3
3
3
1 mol NH
0.08206L atm
20.2g NH
25
273 K
17.03g NH
mol K
24.4 L NH
1atm
752 mmHg
760 mmHg
nRT
V
P
4B
The amount of Cl g
2
b
g
is 0.193 mol
Cl
2
and the pressure is 0.980 atm (0.993
barr × (1 atm/1.01325 barr) = 0.980 atm).
This information is substituted into
the ideal gas equation after it has been solved for temperature.
11
0.980atm 7.50 L
464K
0.193mol 0.08206Latmmol K
PV
T
nR
5A
The ideal gas equation is solved for amount and the quantities are substituted.
10.5atm 5.00L
2.11molHe
0.08206L atm
30.0
273.15 K
mol K
PV
n
RT
5B
n
=
PV
RT
=
7
3
3
1
1
1 atm
1000L
6.67
10
Pa
3.45 m
101325 Pa
1m
(0.08206 L atm K mol ) (25
273.15)K
= 9.28 × 10
10
moles of N
2
molecules of N
2
= 9.28 × 10
10
mol N
2
×
23
2
2
6.022
10 molecules of N
1 mole N
molecules of N
2
= 5.59 × 10
14
molecules N
2
6A
The general gas equation is solved for volume, after the constant amount in moles is cancelled.
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 Fall '11
 hammedmirza
 mol, mol Ne

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