CHAPTER 1
1.1
=0.981 Pa=0.0981 Pa1.2 Pinitial=Pfinal(Vfinal/Vinitial) =(6 atm)(2 L/4 L) =3 atm1.3 (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 K1.5 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= bar202 65
1
101 325
. 0 002
,
Pa.atm
Pa= atm=×981 10. −^2 − 2
kg
msP
mg
AVg
A()(.)(.)
== =
ρ 1001981 kg m−−33 2m m s
((. m)001^2=×981 10. −^1 − 2
kg
msP
mg
AVg
A()(.)(.)
(
== =
ρ 101981 kg m−−^33 m m s^2
001 .)m^2P
F
A
==
Force
AreaAnswers to problems
Answers to problems
9781405124362_5_end.qxd 4/29/08 9:17 Page 439