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In doing the numerical calculations for the exercises and problems for this chapter, the
values of the idealgas constant have been used with the precision given on page 501 of
the text,
K.
mol
atm
L
0.08206
K
mol
J
3145
.
8
⋅
⋅
=
⋅
=
R
Use of values of these constants with either greater or less precision may introduce
differences in the third figures of some answers.
18.1:
a)
mol.
3
.
56
)
mol
kg
10
400
(
kg)
225
.
0
(
3
tot
=
×
=
=

M
m
n
b) Of the many ways
to find the pressure, Eq. (18.3) gives
Pa.
10
6.81
atm
67.2
L)
(20.0
K)
K)(291.15
mol
atm
L
06
mol)(0.082
3
.
56
(
6
×
=
=
⋅
⋅
=
=
V
nRT
p
18.2:
a)
The final temperature is four times the initial Kelvin temperature, or 4(314.15
K) –273.15=
C
983
°
to the nearest degree.
kg.
10
24
.
5
)
K
15
.
314
)(
K
atm/mol
L
08206
.
0
(
)
L
60
.
2
)(
atm
30
.
1
)(
kg/mol
10
00
.
4
(
b)
4
3
tot


×
=
⋅
⋅
×
=
=
=
RT
MpV
nM
m
18.3:
For constant temperature, Eq. (18.6) becomes
atm.
96
.
0
)
390
.
0
110
.
0
)(
atm
40
.
3
(
)
(
2
1
1
2
=
=
=
V
V
p
p
18.4:
a) Decreasing the pressure by a factor of onethird decreases the Kelvin
temperature by a factor of onethird, so the new Celsius temperatures is 1/3(293.15 K) –
273.15=
C
175
°

rounded to the nearest degree.
b) The net effect of the two changes is
to keep the pressure the same while decreasing the Kelvin temperature by a factor of one
third, resulting in a decrease in volume by a factor of onethird, to 1.00 L.
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View Full Document 18.5:
Assume a room size of 20 ft X 20 ft X 20 ft
C.
20
of
re
temperatu
a
Assume
.
m
113
ft
4000
3
3
°
=
=
V
molecules
10
8
.
2
mol
4685
)
K
293
)(
K
J/mol
315
.
8
(
)
m
113
)(
Pa
10
01
.
1
(
so
27
3
5
×
=
=
=
⋅
×
=
=
=
A
nN
N
RT
pV
n
nRT
pV
b)
3
19
3
6
27
cm
molecules/
10
5
.
2
cm
10
113
molecules
10
8
.
2
×
=
×
×
=
V
N
18.6:
The temperature is
K.
15
.
295
C
0
.
22
=
°
=
T
(a) The average molar mass of air is
so
mol,
kg
10
8
.
28
3

×
=
M
kg.
10
07
.
1
K)
K)(295.15
mol
atm
L
08206
.
0
(
mol)
kg
10
L)(28.8
atm)(0.900
00
.
1
(
3
3
tot


×
=
⋅
⋅
×
=
=
=
M
RT
pV
nM
m
(b) For Helium
mol,
kg
10
00
.
4
3

×
=
M
so
kg.
10
49
.
1
K)
K)(295.15
mol
atm
L
08206
.
0
(
mol)
kg
10
L)(4.00
atm)(0.900
00
.
1
(
4
3
tot


×
=
⋅
⋅
×
=
=
=
M
RT
pV
nM
m
18.7:
From Eq. (18.6),
C.
503
K
776
)
cm
Pa)(499
10
01
.
1
(
)
cm
Pa)(46.2
10
(2.821
K)
15
.
300
(
3
5
3
6
1
1
2
2
1
2
°
=
=
×
×
=
=
V
p
V
p
T
T
18.8:
a)
(
29 (
29 (
29
(
29 (
29
kg.
373
.
0
K
15
.
310
K
mol
J
3145
.
8
m
0750
.
0
Pa
10
013
.
4
mol
kg
10
0
.
32
3
5
3
tot
=
⋅
×
×
=
=

RT
MpV
m
b) Using the final pressure of
Pa
10
813
.
2
5
×
and temperature of
kg,
0.275
K,
15
.
295
=
′
m
so the mass lost is
kg
098
.
0
where extra figures were kept in the intermediate calculation
of
tot
m
.
18.9:
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .
 Spring '08
 Hickman

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