CHAPTER 1
1.1
= 0.981 Pa
= 0.0981 Pa
1.2 P initial = P final (V final /V initial ) = (6 atm)(2 L/4 L) = 3 atm
1.3 (a)
(b)
1.4 Temperature (K) = temperature (°C) + 273.15°
(a) 0°C = 273.15 K
(b) −270°C = 3.15 K
(c) +100°C = 373.15 K
1.5 The absolute temperature is a measure of the average kinetic energy
of the molecules.
202 6
1 01325
101 325
0 0020265
.
.
,
.
Pa
bar
Pa
bar
=
202 65
1
101 325
0 002
.
,
.
Pa
atm
Pa
atm
=
=
×
−
−
.
9 81 10
2
2
kg
m s
P
mg
A
Vg
A
(
) ( .
)( .
)
=
=
=
−
−
ρ
1
00 1
98 1
3
3
2
kg m
m
m s
( ( .
)
0 01
2
m
=
×
−
−
.
9 81 10
1
2
kg
m s
P
mg
A
Vg
A
(
) ( . )( .
)
(
=
=
=
−
−
ρ
1
01
98 1
3
3
2
kg m
m
m s
0 0 1
2
. )
m
P
F
A
=
=
Force
Area
Answers to problems
9781405124362_5_end.qxd 4/29/08 9:17 Page 439
1.1
= 0.981 Pa
= 0.0981 Pa
1.2 P initial = P final (V final /V initial ) = (6 atm)(2 L/4 L) = 3 atm
1.3 (a)
(b)
1.4 Temperature (K) = temperature (°C) + 273.15°
(a) 0°C = 273.15 K
(b) −270°C = 3.15 K
(c) +100°C = 373.15 K
1.5 The absolute temperature is a measure of the average kinetic energy
of the molecules.
202 6
1 01325
101 325
0 0020265
.
.
,
.
Pa
bar
Pa
bar
=
202 65
1
101 325
0 002
.
,
.
Pa
atm
Pa
atm
=
=
×
−
−
.
9 81 10
2
2
kg
m s
P
mg
A
Vg
A
(
) ( .
)( .
)
=
=
=
−
−
ρ
1
00 1
98 1
3
3
2
kg m
m
m s
( ( .
)
0 01
2
m
=
×
−
−
.
9 81 10
1
2
kg
m s
P
mg
A
Vg
A
(
) ( . )( .
)
(
=
=
=
−
−
ρ
1
01
98 1
3
3
2
kg m
m
m s
0 0 1
2
. )
m
P
F
A
=
=
Force
Area
Answers to problems
9781405124362_5_end.qxd 4/29/08 9:17 Page 439