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582
Heat and the First Law of Thermodynamics
*P20.14
Vessel one contains oxygen according to
PV
nRT
=
:
n
PV
RT
c
==
××
⋅
=
−
1 75 1 013 10
16 8 10
8314
1194
53
..
.
.
.
Pa
m
Nm mol K 300 K
mol
3
ej
.
Vessel two contains this much oxygen:
n
h
=
=
−
2 25 1 013 10 22 4 10
8314450
1365
.
.
.
a
f
mol
mol.
(a)
The gas comes to an equilibrium temperature according to
mc T
mc T
nMcT
cf
hf
∆∆
af
di
cold
hot
K
K
=−
−+
−=
300
450
0
The molar mass
M
and specific heat divide out:
1 194
358 2
1 365
614 1
0
972 3
380
.
TT
T
ff
f
−+−=
K
K
2.559
K
(b)
The pressure of the whole sample in its final state is
P
nRT
V
+×
=×
=
−
2559
16 8
10
206 10
204
3
5
.
.
mol 8.314 J 380 K
mol K 22.4
m
Pa
atm
3
.
Section 20.3
Latent Heat
P20.15
The heat needed is the sum of the following terms:
Q
needed
heat to reach melting point
heat to melt
heat to reach melting point
heat to vaporize
heat to reach 110 C

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