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Chapter 19
559
P19.39
PV
nRT
=
:
m
m
n
n
PV
RT
RT
P
P
f
i
f
i
ff
f
i
ii
f
i
==
=
so
mm
P
P
fi
f
i
=
F
H
G
I
K
J
∆
mmm m
PP
P
ifi
if
i
=−
=
−
F
H
G
I
K
J
=
−
F
H
G
I
K
J
=
12 0
26 0
439
.
.
.
kg
41.0 atm
atm
41.0 atm
kg
P19.40
My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20
293
°=
C
K. Think
of the air as 80.0% N
2
and 20.0% O
2
.
Avogadro’s number of molecules has mass
0 800 28 0
0 200 32 0
0 028 8
..
.
af
bg
g mol
g mol
kg mol
+=
Then
PV
nRT
m
M
RT
F
H
G
I
K
J
gives
m
PVM
RT
×
⋅
=
100 10
384
00288
8314
293
45 4
5
.
.
.
Nm
m
kgmo
l
Jmo
lK
K
kg ~10 kg
23
2
ej
e
j
*P19.41
The CO
2
is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molar
mass is
M
=+
=
12 0
2 16 0
44 0
.
g mol
g mol
g mol
. The quantity of gas in the cylinder is
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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