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184
Chapter 18
18.16.
IDENTIFY:
Fp
A
=
and
p
Vn
R
T
=
SET UP:
For a cube,
/
.
VA L
=
EXECUTE:
(a)
The force of any side of the cube is
()
(
)
,
An
R
T
V
R
TL
==
=
since the ratio of area to
volume is
/
1/ .
AV
L
=
For
20.0 C
293.15 K,
°
T
4
(3 mol) (8.3145 J mol K) (293.15 K)
3.66 10 N.
0.200 m
nRT
F
L
⋅
=×
(b)
For
100.00 C
373.15 K,
T
=°
=
4
(3 mol)(8.3145 J mol K)(373.15 K)
4.65 10 N.
0.200 m
nRT
F
L
⋅
=
×
EVALUATE:
When the temperature increases while the volume is kept constant, the pressure increases and
therefore the force increases. The force increases by the factor
21
/.
TT
18.17.
IDENTIFY:
Example 18.4 assumes a temperature of 0 C
°
at all altitudes and neglects the variation of
g
with
elevation. With these approximations,
/
0
.
Mgy RT
pp
e
−
=
SET UP:
ln(
)
.
x
ex
−
=−
For air,
3
28.8 10 kg/mol.
M
−
EXECUTE:
We want
y
for
0
0.90
p
p
=
so
/
0.90
Mgy RT
e
−
=
and
ln(0.90)
850 m.
RT
y
Mg
=
EVALUATE:
This is a commonly occurring elevation, so our calculation shows that 10% variations in
atmospheric pressure occur at many locations.
18.18.
IDENTIFY:
From Example 18.4, the pressure at elevation
y
above sea level is
/
0
.
Mgy RT
e
−
=
SET UP:
The average molar mass of air is
3
M
−
EXECUTE:
At an altitude of 100 m,
32
1
(28.8 10 kg mol)(9.80 m s )(100 m)
0.01243,
(8.3145 J mol K)(273.15 K)
Mgy
RT
−
×
⋅
and the percent
decrease in pressure is
0.01243
0
1
1
0.0124 1.24%.
e
−
−=
−
=
=
At an altitude of 1000 m,
2
0.1243
Mgy RT
=
and
the percent decrease in pressure is
0.1243
1
0.117 11.7%.
e
−
−==
EVALUATE:
These answers differ by a factor of
(11.7%) (1.24%)
9.44,
=
which is less than 10 because the
variation of pressure with altitude is exponential rather than linear.
18.19.
IDENTIFY:
0
Myg RT
e
−
=
from Example 18.4. Eq.(18.5) says
p
ρ
MRT
.
=
Example 18.4 assumes a constant
273 K, so
and
Tp
ρ
=
are directly proportional and we can write
0
.
Mgy RT
ρρ
e
−
=
SET UP:
From Example 18.4,
1.10
Mgy
RT
=
when
8863 m.
y
=
EXECUTE:
For
100 m,
y
=
0.0124,
Mgy
RT
=
so
0.0124
00
0.988
.
e
ρ
−
The density at sea level is 1.2% larger
than the density at 100 m.
EVALUATE:
The pressure decreases with altitude.
tot
,
m
p
VR
T
M
=
so when the pressure decreases and
T
is
constant the volume of a given mass of gas increases and the density decreases.
18.20.
IDENTIFY:
/
0
Mgy RT
e
−
=
from Example 18.4 gives the variation of air pressure with altitude. The density
of the air is
,
pM
RT
=
so
is proportional to the pressure
p
. Let
0
be the density at the surface, where the
pressure is
0
.
p
SET UP:
From Example 18.4,
41
(28.8 10 kg/mol)(9.80 m/s )
1.244 10 m .
(8.314 J/mol K)(273 K)
Mg
RT
−
−
−
×
×
⋅
EXECUTE:
3
(1.244 10
m
)(1.00 10 m)
0.883
.
p
pe
p
−−
−×
×
constant,
M
pR
T
so
0
0
p
p
=
and
0
0.883
.
p
p
⎛⎞
⎜⎟
⎝⎠
The density at an altitude of 1.00 km is 88.3% of its value at the surface.
EVALUATE:
If the temperature is assumed to be constant, then the decrease in pressure with increase in altitude
corresponds to a decrease in density.
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View Full DocumentThermal Properties of Matter
185
18.21.
IDENTIFY:
Use Eq.(18.5) and solve for
p
.
SET UP:
/
pMR
T
ρ
=
and
/
p
RT
M
=
( 56.5
273.15) K
216.6 K
T
=−
+
=
For air
3
28.8 10 kg/mol
M
−
=×
(Example 18.3)
EXECUTE:
3
4
3
(8.3145 J/mol K)(216.6 K)(0.364 kg/m )
2.28 10 Pa
p
−
⋅
==
×
×
EVALUATE:
The pressure is about onefifth the pressure at sealevel.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Force

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