19
Temperature
CHAPTER OUTLINE
19.1
19.2
19.3
19.4
19.5
Temperature and the Zeroth
Law of Thermodynamics
Thermometers and the
Celsius Temperature Scale
The Constant-Volume Gas
Thermometer and the
Absolute Temperature Scale
Thermal Expansion of
Solids and Liquids
Macroscopic Description of
an Ideal Gas
ANSWERS TO QUESTIONS
Q19.1
Two objects in thermal equilibrium need not be in contact.
Consider the two objects that are in thermal equilibrium in
Figure 19.1(c). The act of separating them by a small distance
does not affect how the molecules are moving inside either
object, so they will still be in thermal equilibrium.
Q19.2
The copper’s temperature drops and the water temperature
rises until both temperatures are the same. Then the metal and
the water are in thermal equilibrium.
Q19.3
The astronaut is referring to the temperature of the lunar
surface, specifically a 400°F difference. A thermometer would
register the temperature of the thermometer liquid. Since there
is no atmosphere in the moon, the thermometer will not read a
realistic temperature unless it is placed into the lunar soil.
Q19.4
Rubber contracts when it is warmed.
Q19.5
Thermal expansion of the glass bulb occurs first, since the wall of the bulb is in direct contact with
the hot water. Then the mercury heats up, and it expands.
Q19.6
If the amalgam had a larger coefficient of expansion than your tooth, it would expand more than the
cavity in your tooth when you take a sip of your ever-beloved coffee, resulting in a broken or
cracked tooth! As you ice down your now excruciatingly painful broken tooth, the amalgam would
contract more than the cavity in your tooth and fall out, leaving the nerve roots exposed. Isn’t it nice
that your dentist knows thermodynamics?
Q19.7
The measurements made with the heated steel tape will be too short—but only by a factor of
5 × 10 −5 of the measured length.
Q19.8
(a)
One mole of H 2 has a mass of 2.016 0 g.
(b)
One mole of He has a mass of 4.002 6 g.
(c)
One mole of CO has a mass of 28.010 g.
Q19.9
The ideal gas law, PV = nRT predicts zero volume at absolute zero. This is incorrect because the
ideal gas law cannot work all the way down to or below the temperature at which gas turns to
liquid, or in the case of CO 2 , a solid.
549
550
Temperature
Q19.10
Call the process isobaric cooling or isobaric contraction. The rubber wall is easy to stretch. The air
inside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wall
moves in, just maintaining equality of pressure outside and inside. The air is nearly an ideal gas to
start with, but PV = nRT soon fails. Volume will drop by a larger factor than temperature as the
water vapor liquefies and then freezes, as the carbon dioxide turns to snow, as the argon turns to
slush, and as the oxygen liquefies. From the outside, you see contraction to a small fraction of the
original volume.
Q19.11
Cylinder A must be at lower pressure. If the gas is thin, it will be at one-third the absolute pressure
of B.
Q19.12
At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong
latches hold them together, but they would explode apart if you tried to open the hot cooker.
Q19.13
(a)
The water level in the cave rises by a smaller distance than the water outside, as the trapped
air is compressed. Air can escape from the cave if the rock is not completely airtight, and also
by dissolving in the water.
(b)
The ideal cave stays completely full of water at low tide. The water in the cave is supported
by atmospheric pressure on the free water surface outside.
(a)
(b)
FIG. Q19.13
Q19.14
Absolute zero is a natural choice for the zero of a temperature scale. If an alien race had bodies that
were mostly liquid water—or if they just liked its taste or its cleaning properties—it is conceivable
that they might place one hundred degrees between its freezing and boiling points. It is very
unlikely, on the other hand, that these would be our familiar “normal” ice and steam points, because
atmospheric pressure would surely be different where the aliens come from.
Q19.15
As the temperature increases, the brass expands. This would effectively increase the distance, d,
from the pivot point to the center of mass of the pendulum, and also increase the moment of inertia
of the pendulum. Since the moment of inertia is proportional to d 2 , and the period of a physical
I
pendulum is T = 2π
, the period would increase, and the clock would run slow.
mgd
Q19.16
As the water rises in temperature, it expands. The excess volume would spill out of the cooling
system. Modern cooling systems have an overflow reservoir to take up excess volume when the
coolant heats up and expands.
Q19.17
The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar,
both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be a
certain temperature at which the inner diameter of the lid has expanded more than the top of the
jar, and the lid will be easier to remove.
Chapter 19
Q19.18
The sphere expands when heated, so that it no longer fits through the
ring. With the sphere still hot, you can separate the sphere and ring by
heating the ring. This more surprising result occurs because the thermal
expansion of the ring is not like the inflation of a blood-pressure cuff.
Rather, it is like a photographic enlargement; every linear dimension,
including the hole diameter, increases by the same factor. The reason
for this is that the atoms everywhere, including those around the inner
circumference, push away from each other. The only way that the
atoms can accommodate the greater distances is for the
circumference—and corresponding diameter—to grow. This property
was once used to fit metal rims to wooden wagon and horse-buggy
wheels. If the ring is heated and the sphere left at room temperature,
the sphere would pass through the ring with more space to spare.
551
FIG. Q19.18
SOLUTIONS TO PROBLEMS
Section 19.1
Temperature and the Zeroth Law of Thermodynamics
No problems in this section
Section 19.2
Thermometers and the Celsius Temperature Scale
Section 19.3
The Constant-Volume Gas Thermometer and the Absolute Temperature Scale
P19.1
Since we have a linear graph, the pressure is related to the temperature as P = A + BT , where A and
B are constants. To find A and B, we use the data
a
f
1.635 atm = A + a78.0° CfB
0.900 atm = A + −80.0° C B
(1)
(2)
Solving (1) and (2) simultaneously,
we find
A = 1.272 atm
and
B = 4.652 × 10 −3 atm ° C
Therefore,
P = 1.272 atm + 4.652 × 10 −3 atm ° C T
(a)
P = 0 = 1.272 atm + 4.652 × 10 −3 atm ° C T
At absolute zero
which gives
e
j
e
j
T = −274° C .
(b)
At the freezing point of water P = 1.272 atm + 0 = 1.27 atm .
(c)
And at the boiling point P = 1.272 atm + 4.652 × 10 −3 atm ° C 100° C = 1.74 atm .
e
ja
f
552
P19.2
Temperature
P1V = nRT1
and P2 V = nRT2
imply that
P19.3
P19.4
P2 T2
=
P1 T1
a
fa
f
(a)
P2 =
0.980 atm 273 K + 45.0 K
P1T2
=
= 1.06 atm
273 + 20.0 K
T1
(b)
T3 =
T1 P3
P1
a
f
a293 K fa0.500 atmf = 149 K =
=
0.980 atm
−124° C
FIG. P19.2
a
f
9
9
TC + 32.0° F = −195.81 + 32.0 = −320° F
5
5
(a)
TF =
(b)
T = TC + 273.15 = −195.81 + 273.15 = 77.3 K
(a)
To convert from Fahrenheit to Celsius, we use
TC =
and the Kelvin temperature is found as
T = TC + 273 = 310 K
(b)
In a fashion identical to that used in (a), we find TC = −20.6° C
T = 253 K
and
P19.5
P19.6
b
FG 212° F − 32.0° F IJ =
H 100° C − 0.00° C K
(a)
∆T = 450° C = 450° C
(b)
∆T = 450° C = 450 K
810° F
a
f
100° C = aa60.0° Sf + b
Subtracting, 100° C = aa75.0° Sf
Require
0.00° C = a −15.0° S + b
a
a = 1.33 C° S° .
f
Then 0.00° C = 1.33 −15.0° S C°+ b
b = 20.0° C .
b
g
So the conversion is TC = 1.33 C° S° TS + 20.0° C .
P19.7
(a)
T = 1 064 + 273 = 1 337 K melting point
T = 2 660 + 273 = 2 933 K boiling point
(b)
g a
f
5
5
TF − 32.0 = 98.6 − 32.0 = 37.0° C
9
9
∆T = 1 596° C = 1 596 K . The differences are the same.
Chapter 19
Section 19.4
P19.8
553
Thermal Expansion of Solids and Liquids
α = 1.10 × 10 −5 ° C −1 for steel
e
a
j
f
∆L = 518 m 1.10 × 10 −5 ° C −1 35.0° C − −20.0° C = 0.313 m
P19.9
The wire is 35.0 m long when TC = −20.0° C .
b
∆L = Liα T − Ti
g
a f
a f for Cu.
∆L = a35.0 mfe1.70 × 10 aC°f jc35.0° C − a −20.0° C fh =
−1
α = α 20.0° C = 1.70 × 10 −5 C°
−1
−5
a
fe
ja
+3.27 cm
f
P19.10
∆L = Liα ∆T = 25.0 m 12.0 × 10 −6 C° 40.0° C = 1.20 cm
P19.11
For the dimensions to increase, ∆L = αLi ∆T
a
fa
1.00 × 10 −2 cm = 1.30 × 10 −4 ° C −1 2.20 cm T − 20.0° C
f
T = 55.0° C
*P19.12
P19.13
*P19.14
ja
e
fa
f
∆L = αLi ∆T = 22 × 10 −6 C° 2.40 cm 30° C = 1.58 × 10 −3 cm
a
fa
f
a
fa
f
(a)
∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 30.0 cm 65.0° C = 0.176 mm
(b)
∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 1.50 cm 65.0° C = 8.78 × 10 −4 cm
(c)
∆V = 3αVi ∆T = 3 9.00 × 10 −6 ° C −1
e
F
jGH 30.0aπ fa4 1.50f
Ia
JK
2
f
cm3 65.0° C = 0.093 0 cm3
The horizontal section expands according to ∆L = αLi ∆T .
ja
e
fa
f
∆x = 17 × 10 −6 ° C −1 28.0 cm 46.5° C − 18.0° C = 1.36 × 10 −2 cm
The vertical section expands similarly by
ja
e
FIG. P19.14
fa
f
∆y = 17 × 10 −6 ° C −1 134 cm 28.5° C = 6.49 × 10 −2 cm .
The vector displacement of the pipe elbow has magnitude
∆r = ∆x 2 + ∆ y 2 =
a0.136 mmf + a0.649 mmf
2
2
= 0.663 mm
and is directed to the right below the horizontal at angle
θ = tan −1
FG ∆y IJ = tan FG 0.649 mm IJ = 78.2°
H 0.136 mm K
H ∆x K
−1
∆r = 0.663 mm to the right at 78.2° below the horizontal
554
P19.15
Temperature
b
g
b
L Al 1 + α Al ∆T = LBrass 1 + α Brass ∆T
L Al − LBrass
∆T =
LBrassα Brass − L Alα Al
(a)
∆T =
g
a10.01 − 10.00f
a10.00fe19.0 × 10 j − a10.01fe24.0 × 10 j
−6
−6
∆T = −199° C so T = −179° C. This is attainable.
∆T =
(b)
a10.02 − 10.00f
a10.00fe19.0 × 10 j − a10.02fe24.0 × 10 j
−6
−6
∆T = −396° C so T = −376° C which is below 0 K so it cannot be reached.
P19.16
g a50.0° Cf
jb
e
∆A = 2 17.0 × 10 −6 ° C −1 0.080 0 m
∆A = 2αAi ∆T :
(a)
2
∆A = 1.09 × 10 −5 m 2 = 0.109 cm 2
(b)
The length of each side of the hole has increased. Thus, this represents an increase in the
area of the hole.
b
g
e
e
P19.17
∆V = β − 3α Vi ∆T = 5.81 × 10 −4 − 3 11.0 × 10 −6
P19.18
(a)
a
f
jjb50.0 galga20.0f =
0.548 gal
a
5.050 cm = 5.000 cm 1 + 24.0 × 10 −6 ° C −1 T − 20.0° C
L = Li 1 + α∆T :
f
T = 437° C
(b)
L Al = LBrass for some ∆T , or
We must get
b
g
b
Li , Al 1 + α Al ∆T = Li , Brass 1 + α Brass ∆T
e
g
j
e
j
5.000 cm 1 + 24.0 × 10 −6 ° C −1 ∆T = 5.050 cm 1 + 19.0 × 10 −6 ° C −1 ∆T
Solving for ∆T , ∆T = 2 080° C ,
T = 3 000° C
so
This will not work because aluminum melts at 660° C .
P19.19
b
a
g
f
(a)
V f = Vi 1 + β∆T = 100 1 + 1.50 × 10 −4 −15.0 = 99.8 mL
(b)
∆Vacetone = βVi ∆T
b
b
∆Vflask = βVi ∆T
g
g
acetone
Pyrex
b
= 3αVi ∆T
g
Pyrex
for same Vi , ∆T ,
∆Vacetone β acetone
1.50 × 10 −4
1
=
=
=
−6
∆Vflask
β flask
×
6
.
40
10 −2
3 3.20 × 10
e
j
The volume change of flask is
about 6% of the change in the acetone’s volume .
Chapter 19
P19.20
(a),(b)
555
The material would expand by ∆L = αLi ∆T ,
∆L
= α∆T , but instead feels stress
Li
F Y∆L
=
= Yα∆T = 7.00 × 10 9 N m 2 12.0 × 10 −6 C°
A
Li
e
a f a30.0° Cf
j
−1
= 2.52 × 10 6 N m 2 . This will not break concrete.
P19.21
(a)
b
g
∆V = Vt β t ∆T − VAl β Al ∆T = β t − 3α Al Vi ∆T
e
j
e
ja
= 9.00 × 10 −4 − 0.720 × 10 −4 ° C −1 2 000 cm3 60.0° C
∆V = 99.4 cm3
(b)
f
overflows.
The whole new volume of turpentine is
ja
e
f
2 000 cm3 + 9.00 × 10 −4 ° C −1 2 000 cm3 60.0° C = 2 108 cm3
so the fraction lost is
99.4 cm3
= 4.71 × 10 −2
3
2 108 cm
and this fraction of the cylinder’s depth will be empty upon cooling:
a
f
4.71 × 10 −2 20.0 cm = 0.943 cm .
*P19.22
The volume of the sphere is
VPb =
a
f
4 3 4
π r = π 2 cm
3
3
3
= 33.5 cm3 .
The amount of mercury overflowing is
e
j
overflow = ∆VHg + ∆VPb − ∆Vglass = β Hg VHg + β PbVPb − β glassVglass ∆T
where Vglass = VHg + VPb is the initial volume. Then
e
j e
j
e
j e
1
1
L
O
= Ma182 − 27f10
118 cm + a87 − 27f10
33.5 cm P 40° C = 0.812 cm
C°
C°
N
Q
j
overflow = β Hg − β glass VHg + β Pb − β glass VPb ∆T = β Hg − 3α glass VHg + 3α Pb − 3α glass VPb ∆T
−6
P19.23
In
−6
3
3
F Y∆L
=
require ∆L = αLi ∆T
A
Li
F
= Yα∆T
A
F
500 N
=
∆T =
−4
2
AYα
2.00 × 10 m 20.0 × 10 10 N m 2 11.0 × 10 −6 C°
e
∆T = 1.14° C
je
je
j
3
556
*P19.24
Temperature
Model the wire as contracting according to ∆L = αLi ∆T and then stretching according to
∆L Y
F
stress = = Y
= αLi ∆T = Yα∆T .
A
Li Li
e
1
45° C = 396 N
C°
j
(a)
F = YAα∆T = 20 × 10 10 N m 2 4 × 10 −6 m 2 11 × 10 −6
(b)
∆T =
3 × 10 8 N m 2
stress
=
= 136° C
Yα
20 × 10 10 N m 2 11 × 10 −6 C°
e
j
To increase the stress the temperature must decrease to 35° C − 136° C = −101° C .
(c)
*P19.25
The original length divides out, so the answers would not change.
The area of the chip decreases according to
∆A = γA1 ∆T = A f − Ai
b
a
g
A f = Ai 1 + γ∆T = Ai 1 + 2α∆T
f
The star images are scattered uniformly, so the number N of stars that fit is proportional to the area.
a
f
ja
e
f
Then N f = N i 1 + 2α∆T = 5 342 1 + 2 4.68 × 10 −6 ° C −1 −100° C − 20° C = 5 336 star images .
Section 19.5
P19.26
P19.27
Macroscopic Description of an Ideal Gas
a
fe
n=
(b)
N = nN A
(a)
Initially, PV
i i = n i RTi
a
= a 2.99 molfe6.02 × 10
j
fa
f
j
a1.00 atmfV = n Ra10.0 + 273.15f K
P b0. 280V g = n Ra 40.0 + 273.15f K
23
molecules mol = 1.80 × 10 24 molecules
i
Finally, Pf V f = n f RT f
f
i
i
i
0.280 Pf
giving
313.15 K
=
1.00 atm 283.15 K
Pf = 3.95 atm
or
Pf = 4.00 × 10 5 Pa abs. .
Dividing these equations,
(b)
je
9.00 atm 1.013 × 10 5 Pa atm 8.00 × 10 −3 m 3
PV
=
= 2.99 mol
8.314 N ⋅ mol K 293 K
RT
(a)
a f
P a1.02fb0.280V g = n Ra85.0 + 273.15f K
After being driven
d
i
i
Pd = 1.121Pf = 4.49 × 10 5 Pa
P19.28
a fa f
a fa f
3 150 0.100
3 PV
=
= 884 balloons
3
4π r P ′ 4π 0.150 3 1.20
If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will be
full of helium at 1.20 atm when the last balloon is inflated. The number of balloons is then reduced
0.100 m3 3
to to 884 −
= 877 .
3
4π 0.15 m
PV = NP ′V ′ =
e
a
4 3
π r NP ′ :
3
j
f
N=
Chapter 19
P19.29
The equation of state of an ideal gas is PV = nRT so we need to solve for the number of moles to find N.
e
ja
fa
ga
fa
f
1.01 × 10 5 N m 2 10.0 m 20.0 m 30.0 m
PV
=
= 2. 49 × 10 5 mol
n=
RT
8.314 J mol ⋅ K 293 K
b
e
5
N = nN A = 2.49 × 10 mol 6.022 × 10
*P19.30
(a)
PV
i i = n i RTi =
23
f
j
molecules mol = 1.50 × 10 29 molecules
mi
RTi
M
e
j
3
4.00 × 10 −3 kg 1.013 × 10 5 N 4π 6.37 × 10 6 m mole ⋅ K
MPV
i i
=
mi =
RTi
8.314 Nm 50 K
mole
m2 3
= 1.06 × 10 21 kg
(b)
Pf V f
PV
i i
=
n f RT f
ni RTi
F 1.06 × 10 kg + 8.00 × 10 kg I T
GH
JK 50 K
1.06 × 10 kg
F 1 IJ = 56.9 K
T = 100 K G
H 1.76 K
nRT F 9.00 g I F 8.314 J I F
773 K
I
=G
P=
G
J
G
J
V
H 18.0 g mol K H mol K K H 2.00 × 10 m JK =
21
2 ⋅1 =
20
f
21
f
P19.31
P19.32
P19.33
557
−3
a
1.61 MPa = 15.9 atm
fa f
(a)
T2 = T1
P2
= 300 K 3 = 900 K
P1
(b)
T2 = T1
P2 V2
= 300 2 2 = 1 200 K
P1V1
∑ Fy = 0 :
3
a fa f
b
g
ρ out gV − ρ in gV − 200 kg g = 0
bρ
out
ge
− ρ in 400 m
3
j = 200 kg
The density of the air outside is 1.25 kg m3 .
n
P
=
From PV = nRT ,
V RT
The density is inversely proportional to the temperature, and the density
of the hot air is
jFGH 283T K IJK
e1.25 kg m jFGH 1 − 283T K IJK e400 m j = 200 kg
e
ρ in = 1.25 kg m3
Then
in
3
3
in
283 K
1−
= 0.400
Tin
283 K
0.600 =
Tin = 472 K
Tin
FIG. P19.33
558
*P19.34
P19.35
Temperature
Consider the air in the tank during one discharge process. We suppose that the process is slow
enough that the temperature remains constant. Then as the pressure drops from 2.40 atm to
1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 L
to 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changes
from 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of water
comes out. Were it not for male pattern dumbness, each person could more efficiently use his device
by starting with the tank half full of water.
(a)
PV = nRT
e
je
j
b
ga f
m = nM = a 41.6 molfb 28.9 g molg =
1.013 × 10 5 Pa 1.00 m3
PV
n=
=
= 41.6 mol
RT
8.314 J mol ⋅ K 293 K
(b)
1.20 kg , in agreement with the tabulated density of
3
1.20 kg m at 20.0°C.
*P19.36
e
j
2
The void volume is 0.765Vtotal = 0.765π r 2 = 0.765π 1.27 × 10 −2 m 0.2 m = 7.75 × 10 −5 m3 . Now for
the gas remaining PV = nRT
n=
P19.37
(a)
b
n=
PV = nRT
e
j
ga
−5
5
2
3
PV 12.5 1.013 × 10 N m 7.75 × 10 m
=
= 3.96 × 10 −2 mol
RT
8.314 Nm mole K 273 + 25 K
f
PV
RT
a
fe
3
−3
5
PVM 1.013 × 10 Pa 0.100 m 28.9 × 10 kg mol
=
m = nM =
RT
8.314 J mol ⋅ K 300 K
b
f
ga
j
m = 1.17 × 10 −3 kg
P19.38
e
j
(b)
Fg = mg = 1.17 × 10 −3 kg 9.80 m s 2 = 11.5 mN
(c)
F = PA = 1.013 × 10 5 N m 2 0.100 m
(d)
The molecules must be moving very fast to hit the walls hard.
ja
e
At depth,
P = P0 + ρgh
At the surface,
P0 V f = nRT f :
Therefore
and
2
= 1.01 kN
PVi = nRTi
P0 V f
bP + ρghgV
F T I FG P + ρgh IJ
GH T JK H P K
F 293 K IJ FG 1.013 × 10
= 1.00 cm G
H 278 K K GH
V f = Vi
Vf
f
f
0
i
=
Tf
Ti
0
i
0
3
V f = 3.67 cm3
5
e
je
ja
f IJ
JK
Pa + 1 025 kg m 3 9.80 m s 2 25.0 m
1.013 × 10 5 Pa
Chapter 19
P19.39
mf
PV = nRT :
mi
=
nf
m f = mi
so
Pf V f RTi Pf
=
RT f PV
Pi
i i
=
ni
FP I
GH P JK
f
i
∆m = mi − m f = mi
P19.40
559
F P − P I = 12.0 kgFG 41.0 atm − 26.0 atm IJ =
GH P JK
H 41.0 atm K
i
f
4.39 kg
i
My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20° C = 293 K . Think
of the air as 80.0% N 2 and 20.0% O 2 .
Avogadro’s number of molecules has mass
a0.800fb28.0 g molg + a0.200fb32.0 g molg = 0.028 8 kg mol
F mI
PV = nRT = G J RT
H MK
PVM e1.00 × 10 N m je38.4 m jb0.028 8 kg molg
=
= 45.4 kg
m=
RT
b8.314 J mol ⋅ K ga293 K f
Then
5
gives
*P19.41
2
3
~ 10 2 kg
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 g mol + 2 16.0 g mol = 44.0 g mol . The quantity of gas in the cylinder is
m sample
6.50 g
=
= 0.148 mol
n=
M
44.0 g mol
b
Then
PV = nRT
gives
V=
g
b
ga
nRT 0.148 mol 8.314 J mol ⋅ K 273 K + 20 K
=
P
1.013 × 10 5 N m 2
e
je
je
j
f FG 1 N ⋅ m IJ F 10 L I =
H 1 J K GH 1 m JK
3
3
3.55 L
P19.42
10 −9 Pa 1.00 m 3 6.02 × 10 23 molecules mol
PVN A
=
= 2.41 × 10 11 molecules
N=
RT
8.314 J K ⋅ mol 300 K
P19.43
P0 V = n1 RT1 =
b
FG m IJ RT
H MK
F m IJ RT
P V = n RT = G
HMK
P VM F 1
1I
− J
m −m =
G
R HT T K
0
2
1
2
1
1
2
2
2
0
1
2
ga
f
560
P19.44
Temperature
(a)
Initially the air in the bell satisfies P0 Vbell = nRTi
or
a
f
P0 2.50 m A = nRTi
(1)
When the bell is lowered, the air in the bell satisfies
a
f
Pbell 2.50 m − x A = nRT f
(2)
where x is the height the water rises in the bell. Also, the pressure in the bell, once it is
lowered, is equal to the sea water pressure at the depth of the water level in the bell.
a
a
f
f
Pbell = P0 + ρg 82.3 m − x ≈ P0 + ρg 82.3 m
(3)
The approximation is good, as x < 2.50 m. Substituting (3) into (2) and substituting nR from
(1) into (2),
a
fa
f
P0 + ρg 82.3 m 2.50 m − x A = P0 Vbell
Tf
Ti
.
Using P0 = 1 atm = 1.013 × 10 5 Pa and ρ = 1.025 × 10 3 kg m3
L
O
fMM TT FGH1 + ρga82P.3 mf IJK PP
N
Q
LM 277.15 K F e1.025 × 10 kg m je9.80 m s ja82.3 mf I
G1 +
JJ
= a 2.50 mfM1 −
293.15 K G
1.013 × 10 N m
K
H
MN
a
x = 2.50 m 1 −
−1
f
0
0
3
3
−1
2
2
5
OP
PP
Q
x = 2.24 m
(b)
If the water in the bell is to be expelled, the air pressure in the bell must be raised to the
water pressure at the bottom of the bell. That is,
a
f
Pbell = P0 + ρg 82.3 m
5
e
ja
je
f
= 1.013 × 10 Pa + 1.025 × 10 3 kg m 3 9.80 m s 2 82.3 m
5
Pbell = 9.28 × 10 Pa = 9.16 atm
Additional Problems
P19.45
The excess expansion of the brass is
b
a f a
f
∆a ∆Lf = 2.66 × 10 m
∆ ∆L = 19.0 − 11.0 × 10 −6
−4
(a)
The rod contracts more than tape to
a length reading 0.950 0 m − 0.000 266 m = 0.949 7 m
(b)
g
a° Cf a0.950 mfa35.0° Cf
∆Lrod − ∆Ltape = α brass − α steel Li ∆T
0.950 0 m + 0.000 266 m = 0.950 3 m
−1
Chapter 19
P19.46
At 0°C, 10.0 gallons of gasoline has mass,
ρ=
from
m
V
jb
e
m = ρV = 730 kg m3 10.0 gal
80 m I
= 27.7 kg
gFGH 0.003
1.00 gal JK
3
The gasoline will expand in volume by
b
ga
f
∆V = βVi ∆T = 9.60 × 10 −4 ° C −1 10.0 gal 20.0° C − 0.0° C = 0.192 gal
At 20.0°C,
10.192 gal = 27.7 kg
10.0 gal = 27.7 kg
F 10.0 gal I = 27.2 kg
GH 10.192 gal JK
The extra mass contained in 10.0 gallons at 0.0°C is
27.7 kg − 27.2 kg = 0.523 kg .
P19.47
Neglecting the expansion of the glass,
∆h =
∆h =
V
β∆T
A
b0.250 cm 2g e1.82 × 10
π e 2.00 × 10 cmj
3
4
3π
−3
2
−4
ja
f
° C −1 30.0° C = 3.55 cm
FIG. P19.47
P19.48
(a)
The volume of the liquid increases as ∆V = Vi β∆T . The volume of the flask increases as
∆Vg = 3αVi ∆T . Therefore, the overflow in the capillary is Vc = Vi ∆T β − 3α ; and in the
b
capillary Vc = A∆h .
Therefore, ∆h =
(b)
Vi
β − 3α ∆T .
A
b
g
b g
For a mercury thermometer
β Hg = 1.82 × 10 −4 ° C −1
and for glass,
3α = 3 × 3.20 × 10 −6 ° C −1
Thus
β − 3α ≈ β
or
α << β .
g
561
562
P19.49
Temperature
The frequency played by the cold-walled flute is fi =
v
v
=
.
λ i 2 Li
When the instrument warms up
ff =
fi
v
v
v
=
=
=
.
λ f 2L f 2Li 1 + α∆T 1 + α∆T
a
f
The final frequency is lower. The change in frequency is
FG
H
IJ
K
1
1 + α∆T
α∆T
v
≈
α∆T
1 + α∆T
2 Li
∆f = f i − f f = f i 1 −
FG
H
IJ
K
a f
b343 m sge24.0 × 10 C°ja15.0° Cf =
∆f ≈
2a0.655 mf
∆f =
v
2 Li
−6
0.094 3 Hz
This change in frequency is imperceptibly small.
P19.50
(a)
P0 V P ′V ′
=
T
T′
V ′ = V + Ah
P ′ = P0 +
k
kh
A
FG P + kh IJ aV + Ahf = P V FG T ′ IJ
H AK
HTK
e1.013 × 10 N m + 2.00 × 10 N m hj
e5.00 × 10 m + e0.010 0 m jhj
= e1.013 × 10 N m je5.00 × 10
0
5
2
−3
3
5
3
FIG. P19.50
2
−3
2
m3
KI
J
jFGH 523
293 K K
2 000 h 2 + 2 013 h − 397 = 0
(b)
20°C
0
5
h=
250°C
h
−2 013 ± 2 689
= 0.169 m
4 000
e
ja
2.00 × 10 3 N m 0.169
kh
5
= 1.013 × 10 Pa +
P′ = P +
A
0.010 0 m 2
P ′ = 1.35 × 10 5 Pa
f
Chapter 19
P19.51
(a)
ρ=
m
m
and dρ = − 2 dV
V
V
For very small changes in V and ρ, this can be expressed as
∆ρ = −
m ∆V
= − ρβ∆T .
V V
The negative sign means that any increase in temperature causes the density to decrease
and vice versa.
(b)
*P19.52
For water we have β =
1.000 0 g cm 3 − 0.999 7 g cm 3
∆ρ
=
= 5 × 10 −5 ° C −1 .
3
ρ∆T
1.000 0 g cm 10.0° C − 4.0° C
ja
e
f
The astronauts exhale this much CO 2 :
n=
m sample
M
=
F
GH
Ia
JK
gFGH
I
JK
1.09 kg
1 000 g
1 mol
= 520 mol .
3 astronauts 7 days
astronaut ⋅ day 1 kg
44.0 g
fb
Then 520 mol of methane is generated. It is far from liquefaction and behaves as an ideal gas.
P=
P19.53
(a)
b
f
ga
nRT 520 mol 8.314 J mol ⋅ K 273 K − 45 K
= 6.57 × 10 6 Pa
=
V
150 × 10 −3 m3
We assume that air at atmospheric pressure is above the
piston.
mg
+ P0
A
In equilibrium
Pgas =
Therefore,
nRT mg
=
+ P0
hA
A
h=
or
nRT
mg + P0 A
where we have used V = hA as the volume of the gas.
(b)
From the data given,
h=
e
20.0 kg 9.80 m s
= 0.661 m
b
ga
0.200 mol 8.314 J K ⋅ mol 400 K
2
j e
5
+ 1.013 × 10 N m
2
je
f
0.008 00 m 2
j
FIG. P19.53
563
564
P19.54
Temperature
The angle of bending θ, between tangents to the two ends of the strip, is
equal to the angle the strip subtends at its center of curvature. (The angles
are equal because their sides are perpendicular, right side to the right side
and left side to left side.)
(a)
Li + ∆L1 = θ r1
The definition of radian measure gives
FIG. P19.54
Li + ∆L 2 = θ r2
and
b
∆L 2 − ∆L1 = θ r2 − r1
By subtraction,
g
α 2 Li ∆T − α 1 Li ∆T = θ ∆r
θ=
2
g
− α 1 Li ∆T
∆r
b
g
(b)
In the expression from part (a), θ is directly proportional to ∆T and also to α 2 − α 1 .
Therefore θ is zero when either of these quantities becomes zero.
(c)
The material that expands more when heated contracts more when cooled, so the bimetallic
strip bends the other way. It is fun to demonstrate this with liquid nitrogen.
(d)
θ=
b
g
2 α 2 − α 1 Li ∆T
2 ∆r
=
ee
j
ja
fa f
2 19 × 10 −6 − 0.9 × 10 −6 ° C −1 200 mm 1° C
0.500 mm
= 1.45 × 10 −2 = 1. 45 × 10 −2 rad
P19.55
bα
FG 180° IJ =
H π rad K
0.830°
From the diagram we see that the change in area is
∆A = ∆w + w∆ + ∆w∆ .
Since ∆ and ∆w are each small quantities, the product ∆w∆ will
be very small. Therefore, we assume ∆w∆ ≈ 0.
Since
∆w = wα∆T
we then have
∆A = wα∆T + wα∆T
and
∆ = α∆T ,
FIG. P19.55
and since A = w , ∆A = 2αA∆T .
The approximation assumes ∆w∆ ≈ 0, or α∆T ≈ 0 . Another way of stating this is α∆T << 1 .
Chapter 19
P19.56
L
Ti = 2π i
g
(a)
Li =
so
Ti2 g
4π 2
565
a1.000 sf e9.80 m s j = 0.248 2 m
=
2
2
4π 2
b
f
ga
∆L = αLi ∆T = 19.0 × 10 −6 ° C −1 0.284 2 m 10.0° C = 4.72 × 10 −5 m
0.248 3 m
Li + ∆L
= 2π
= 1.000 095 0 s
g
9.80 m s 2
T f = 2π
∆T = 9.50 × 10 −5 s
e
In one week, the time lost is time lost = 1 week 9.50 × 10 −5 s lost per second
(b)
b
time lost = 7.00 d week
gFGH 861.00400ds IJK FGH 9.50 × 10
−5
time lost = 57.5 s lost
P19.57
z
I = r 2 dm
a f b ga
f
I aT f
= a1 + α∆T f
I bT g
I aT f − I bT g
≈ 2α∆T
I bT g
r T = r Ti 1 + α∆T
and since
2
for α∆T << 1 we find
i
i
thus
i
(a)
(b)
P19.58
(a)
With α = 17.0 × 10 −6 ° C −1 and
∆T = 100° C
we find for Cu:
∆I
= 2 17.0 × 10 −6 ° C −1 100° C = 0.340%
I
With
α = 24.0 × 10 −6 ° C −1
and
∆T = 100° C
we find for Al:
∆I
= 2 24.0 × 10 −6 ° C −1 100° C = 0.480%
I
e
B = ρgV ′
B=
(b)
e
P ′ = P0 + ρgd
ρgP0 Vi
=
P′
ja
f
ja
f
P ′V ′ = P0 Vi
ρgP0 Vi
bP + ρgdg
0
Since d is in the denominator, B must decrease as the depth increases.
(The volume of the balloon becomes smaller with increasing pressure.)
(c)
b
af
af
g
ρgP0 Vi P0 + ρgd
P0
1 Bd
=
=
=
ρgP0Vi P0
2 B0
P0 + ρgd
P0 + ρgd = 2 P0
d=
1.013 × 10 5 N m 2
P0
=
= 10.3 m
ρg 1.00 × 10 3 kg m3 9.80 m s 2
e
je
j
j
s lost
s
IJ
K
566
*P19.59
Temperature
The effective coefficient is defined by ∆Ltotal = α effective Ltotal ∆T where ∆Ltotal = ∆LCu + ∆LPb and
Ltotal = LCu + LPb = xLtotal + 1 − x Ltotal . Then by substitution
a f
b
g
α Cu LCu ∆T + α Pb LPb ∆T = α eff LCu + LPb ∆T
a f
gx = α
α Cu x + α Pb 1 − x = α eff
bα
Cu
x=
*P19.60
eff
20 × 10
−6
17 × 10
−6
− α Pb
1 C° − 29 × 10 −6 1 C°
1 C° − 29 × 10
−6
1 C°
9
= 0.750
12
=
(a)
No torque acts on the disk so its angular momentum is constant. Its moment of inertia
decreases as it contracts so its angular speed must increase .
(b)
I iω i = I f ω f =
1
1
1
1
2
MRi2ω i = MR 2f ω f = M Ri + Riα∆T ω f = MRi2 1 − α ∆T
2
2
2
2
−2
ω f = ω i 1 − α ∆T
P19.61
− α Pb
=
25.0 rad s
e1 − e17 × 10
−6
j
1 C° 830° C
j
2
=
2
ωf
25.0 rad s
= 25.7 rad s
0.972
After expansion, the length of one of the spans is
a
f
a
f
L f = Li 1 + α∆T = 125 m 1 + 12 × 10 −6 ° C −1 20.0° C = 125.03 m .
L f , y, and the original 125 m length of this span form a right triangle with y as the altitude. Using the
Pythagorean theorem gives:
a125.03 mf
2
a
f
= y 2 + 125 m
2
yielding y = 2.74 m .
P19.62
a
f
After expansion, the length of one of the spans is L f = L 1 + α∆T . L f , y, and the original length L of
this span form a right triangle with y as the altitude. Using the Pythagorean theorem gives
L2f = L2 + y 2 ,
Since
P19.63
(a)
or
y = L2f − L2 = L
2
Let m represent the sample mass. The number of moles is n =
Then, ρ =
m
m
RT or PM = RT .
M
V
m
PM
.
=
V
RT
e
a f
− 1 = L 2α∆T + α∆T
2
y ≈ L 2α∆T .
α∆T << 1 ,
So PV = nRT becomes PV =
(b)
a1 + α∆T f
jb
g
1.013 × 10 5 N m 2 0.032 0 kg mol
PM
ρ=
=
= 1.33 kg m 3
RT
8.314 J mol ⋅ K 293 K
b
ga
f
m
m
and the density is ρ = .
V
M
Chapter 19
P19.64
(a)
(b)
FG nR IJ T
H PK
From PV = nRT , the volume is:
V=
Therefore, when pressure is held constant,
dV nR V
=
=
dT
P T
Thus,
β≡
At T = 0° C = 273 K , this predicts
β=
FG 1 IJ dV = FG 1 IJ V , or β =
H V K dT H V K T ′
1
T
1
= 3.66 × 10 −3 K −1
273 K
β He = 3.665 × 10 −3 K −1 and β air = 3.67 × 10 −3 K −1
Experimental values are:
They agree within 0.06% and 0.2%, respectively.
P19.65
For each gas alone, P1 =
N 1 kT
N kT
N kT
and P2 = 2
and P3 = 3 , etc.
V
V
V
For all gases
b
g
P1V1 + P2 V2 + P3 V3 … = N 1 + N 2 + N 3 … kT and
bN
1
g
+ N 2 + N 3 … kT = PV
Also, V1 = V2 = V3 = … = V , therefore P = P1 + P2 + P3 … .
P19.66
(a)
Using the Periodic Table, we find the molecular masses of the air components to be
b g
a f
b g
M N 2 = 28.01 u , M O 2 = 32.00 u , M Ar = 39.95 u
b
g
and M CO 2 = 44.01 u .
Thus, the number of moles of each gas in the sample is
75.52 g
= 2.696 mol
b g 28.01
g mol
23.15 g
nbO g =
= 0.723 4 mol
32.00 g mol
1.28 g
na Ar f =
= 0.032 0 mol
39.95 g mol
n N2 =
2
b g
n CO 2 =
0.05 g
= 0.001 1 mol
44.01 g mol
The total number of moles is n 0 = ∑ ni = 3.453 mol . Then, the partial pressure of N 2 is
b g
P N2 =
2.696 mol
1.013 × 10 5 Pa = 79.1 kPa .
3.453 mol
e
j
Similarly,
b g
P O 2 = 21.2 kPa
continued on next page
567
a f
P Ar = 940 Pa
b
g
P CO 2 = 33.3 Pa
568
Temperature
(b)
Solving the ideal gas law equation for V and using T = 273.15 + 15.00 = 288.15 K , we find
V=
Then, ρ =
(c)
a
fb
100 × 10 −3 kg
m
=
= 1.22 kg m3 .
V 8.166 × 10 −2 m3
The 100 g sample must have an appropriate molar mass to yield n 0 moles of gas: that is
a f
M air =
*P19.67
f
ga
3.453 mol 8.314 J mol ⋅ K 288.15 K
n 0 RT
= 8.166 × 10 −2 m3 .
=
P
1.013 × 10 5 Pa
100 g
= 29.0 g mol .
3.453 mol
Consider a spherical steel shell of inner radius r and much smaller thickness t, containing helium at
pressure P. When it contains so much helium that it is on the point of bursting into two
hemispheres, we have Pπ r 2 = 5 × 10 8 N m 2 2π rt . The mass of the steel is
e
ρ sV = ρ s 4π r 2 t = ρ s 4π r 2
Pr
10 9 Pa
j
. For the helium in the tank, PV = nRT becomes
m
4
P π r 3 = nRT = He RT = 1 atmVballoon .
3
M He
The buoyant force on the balloon is the weight of the air it displaces, which is described by
m
4
1 atmVballoon = air RT = P π r 3 . The net upward force on the balloon with the steel tank hanging
M air
3
from it is
+ m air g − m He g − m s g =
M air P 4π r 3 g M He P 4π r 3 g ρ s P 4π r 3 g
−
−
3 RT
3 RT
10 9 Pa
The balloon will or will not lift the tank depending on whether this quantity is positive or negative,
M air − M He
ρ
− 9 s . At 20°C this quantity is
which depends on the sign of
3 RT
10 Pa
b
=
g
a28.9 − 4.00f × 10 kg mol − 7 860 kg m
3b8.314 J mol ⋅ K g293 K
10 N m
−3
9
3
2
= 3.41 × 10 −6 s 2 m 2 − 7.86 × 10 −6 s 2 m 2
where we have used the density of iron. The net force on the balloon is downward so the helium
balloon is not able to lift its tank.
Chapter 19
P19.68
With piston alone:
T = constant, so PV = P0 V0
or
P Ahi = P0 Ah0
569
b g b g
Fh I
P=P G J
Hh K
With A = constant,
0
0
i
P = P0 +
But,
mp g
A
FIG. P19.68
where m p is the mass of the piston.
P0 +
Thus,
hi =
which reduces to
mp g
= P0
A
h0
1+
mp g
P0 A
FG h IJ
Hh K
0
i
50.0 cm
=
1+
e
20.0 kg 9.80 m s 2
a
= 49.81 cm
j
1.013 × 10 5 Pa π 0.400 m
f
2
With the man of mass M on the piston, a very similar calculation (replacing m p by m p + M ) gives:
h′ =
h0
em
1+
p +M
P0 A
jg
50.0 cm
=
1+
e
95.0 kg 9.80 m s 2
a
j
1.013 × 10 5 Pa π 0.400 m
= 49.10 cm
f
2
Thus, when the man steps on the piston, it moves downward by
∆h = hi − h ′ = 49.81 cm − 49.10 cm = 0.706 cm = 7.06 mm .
(b)
P = const, so
(a)
(b)
z
Ti
dL
= αdT :
L
a
Ahi Ah ′
=
T
Ti
hi
49.81
= 293 K
= 297 K
T = Ti
49.10
h′
or
FG IJ
H K
giving
P19.69
V V′
=
T Ti
Ti
f
L f = 1.00 m e
αdT =
z
Li
Li
FG
H
IJ
K
FG IJ = α∆T ⇒
H K
Lf
dL
⇒ ln
L
Li
a
2.00 × 10 −5 °C −1 100 °C
f
L f = Li eα∆T
= 1.002 002 m
a
f
L ′f = 1.00 m 1 + 2.00 × 10 −5 ° C −1 100° C = 1.002 000 m :
a
f
L f = 1.00 m e
a
2.00 ×10 −2 °C −1 100 °C
a
f
(or 24°C)
L f − L ′f
Lf
= 2.00 × 10 −6 = 2.00 × 10 −4%
= 7.389 m
f
L ′f = 1.00 m 1 + 0.020 0° C −1 100° C = 3.000 m :
L f − L ′f
Lf
= 59. 4%
570
P19.70
Temperature
At 20.0°C, the unstretched lengths of the steel and copper wires are
a f a
f
a f a−20.0° Cf = 1.999 56 m
L a 20.0° C f = a 2.000 mf 1 + 17.0 × 10 aC°f a −20.0° C f = 1.999 32 m
Ls 20.0° C = 2.000 m 1 + 11.0 × 10 −6 C°
−1
−6
c
−1
Under a tension F, the length of the steel and copper wires are
LM
N
Ls′ = Ls 1 +
F
YA
OP
Q
LM
N
L c′ = L c 1 +
s
F
YA
OP
Q
where Ls′ + L c′ = 4.000 m .
c
Since the tension, F, must be the same in each wire, solve for F:
F=
b L ′ + L ′ g − bL
s
c
Ls
Ys As
+
s + Lc
Lc
Yc A c
g.
When the wires are stretched, their areas become
e
= π e1.000 × 10
j
mj
As = π 1.000 × 10 −3 m
Ac
−3
2
2
e
ja f
1 + e17.0 × 10 ja −20.0 f
1 + 11.0 × 10 −6 −20.0
2
−6
2
= 3.140 × 10 −6 m 2
= 3.139 × 10 −6 m 2
Recall Ys = 20.0 × 10 10 Pa and Yc = 11.0 × 10 10 Pa . Substituting into the equation for F, we obtain
F=
b
4.000 m − 1.999 56 m + 1.999 32 m
1.999 56 m
e20.0 × 10
10
je
j
Pa 3.140 × 10 −6 m 2 + 1.999 32 m
g
e11.0 × 10
10
je
F = 125 N
To find the x-coordinate of the junction,
b
L
gMM 20.0 × 10
N e
Ls′ = 1.999 56 m 1 +
125 N
10
je
N m 2 3.140 × 10 −6
Thus the x-coordinate is −2.000 + 1.999 958 = −4.20 × 10 −5 m .
OP
= 1.999 958 m
m jP
Q
2
j
Pa 3.139 × 10 −6 m 2
Chapter 19
P19.71
e
j e7.86 × 10
(a)
µ = π r 2 ρ = π 5.00 × 10 −4 m
(b)
f1 =
2
1
v
T
and v =
so f1 =
µ
2L
2L
3
j
kg m3 = 6.17 × 10 −3 kg m
T
µ
b g = e6.17 × 10 ja2 × 0.800 × 200f
2
Therefore, T = µ 2Lf1
(c)
−3
2
= 632 N
First find the unstressed length of the string at 0°C:
FG
H
L = L natural 1 +
e
A = π 5.00 × 10 −4
Therefore,
IJ
K
mj
T
L
so L natural =
+
AY
1 T AY
2
= 7.854 × 10 −7 m 2 and Y = 20.0 × 10 10 Pa
T
632
=
= 4.02 × 10 −3 , and
−7
10
AY
7.854 × 10
20.0 × 10
e
je
j
L natural =
a0.800 mf
e1 + 4.02 × 10 j
−3
= 0.796 8 m .
a
f
The unstressed length at 30.0°C is L30 °C = L natural 1 + α 30.0° C − 0.0° C ,
b
g e
ja f
LM1 + T ′ OP , where T ′ is the tension in the string at 30.0°C,
N AY Q
L L − 1OP = e7.854 × 10 je20.0 × 10 jLM 0.800 − 1OP = 580 N .
T ′ = AY M
NL
Q
N 0.797 06 Q
or L30 °C = 0.796 8 m 1 + 11.0 × 10 −6 30.0 = 0.797 06 m .
Since L = L30 °C
−7
10
30 °C
To find the frequency at 30.0°C, realize that
a
f1′
T′
=
so f1′ = 200 Hz
f1
T
*P19.72
f
580 N
= 192 Hz .
632 N
Some gas will pass through the porous plug from the reaction chamber 1 to the reservoir 2 as the
reaction chamber is heated, but the net quantity of gas stays constant according to
n i1 + n i 2 = n f 1 + n f 2 .
Assuming the gas is ideal, we apply n =
PV
to each term:
RT
b g
b g
a f a f a f a f
F 5 IJ = P FG 1 + 4 IJ
1 atmG
P = 1.12 atm
H 300 K K H 673 K 300 K K
Pf V0
Pf 4V0
Pi 4V0
PV
i 0
+
=
+
300 K R 300 K R
673 K R 300 K R
f
f
571
572
P19.73
Temperature
Let 2θ represent the angle the curved rail subtends. We have
a
Li + ∆L = 2θR = Li 1 + α∆T
and
sin θ =
Thus,
θ=
Li
2
R
=
a
f
Li
2R
f a
f
Li
1 + α∆T = 1 + α∆T sin θ
2R
FIG. P19.73
a
f
b
g
and we must solve the transcendental equation
θ = 1 + α∆T sin θ = 1.000 005 5 sin θ
Homing in on the non-zero solution gives, to four digits,
θ = 0.018 16 rad = 1.040 5°
Now,
h = R − R cos θ =
a
Li 1 − cos θ
2 sin θ
f
This yields h = 4.54 m , a remarkably large value compared to ∆L = 5.50 cm .
*P19.74
(a)
Let xL represent the distance of the stationary line below the
top edge of the plate. The normal force on the lower part of the
plate is mg 1 − x cos θ and the force of kinetic friction on it is
µ k mg 1 − x cos θ up the roof. Again, µ k mgx cos θ acts down the
roof on the upper part of the plate. The near-equilibrium of the
plate requires ∑ Fx = 0
a f
a f
a f
motion
f kt
f kb
xL
temperature rising
− µ k mgx cos θ + µ k mg 1 − x cos θ − mg sin θ = 0
FIG. P19.74(a)
−2 µ k mgx cos θ = mg sin θ − µ k mg cos θ
2 µ k x = µ k − tan θ
x=
1 tan θ
−
2 2µ k
and the stationary line is indeed below the top edge by xL =
(b)
With the temperature falling, the plate contracts faster than the
roof. The upper part slides down and feels an upward frictional
force µ k mg 1 − x cos θ . The lower part slides up and feels
downward frictional force µ k mgx cos θ . The equation ∑ Fx = 0
is then the same as in part (a) and the stationary line is above
L
tan θ
the bottom edge by xL =
.
1−
µk
2
a f
FG
H
(c)
IJ
K
Start thinking about the plate at dawn, as the temperature
starts to rise. As in part (a), a line at distance xL below the top
edge of the plate stays stationary relative to the roof as long as
the temperature rises. The point P on the plate at distance xL
above the bottom edge is destined to become the fixed point
when the temperature starts falling. As the temperature rises,
this point moves down the roof because of the expansion of the
central part of the plate. Its displacement for the day is
continued on next page
IJ
K
FG
H
tan θ
L
1−
.
2
µk
motion
f kt
f kb
xL
temperature falling
FIG. P19.74(b)
xL
xL
P
FIG. P19.74(c)
Chapter 19
f
ga
L L F tanθ I OPbT − T g
− α gML − 2 G 1 −
NM 2 H µ JK PQ
F L tanθ IJ bT − T g .
= bα − α gG
H µ K
b
= bα
573
∆L = α 2 − α 1 L − xL − xL ∆T
2
1
2
1
h
k
k
h
c
c
At dawn the next day the point P is farther down the roof by the distance ∆L . It represents
the displacement of every other point on the plate.
bα
(d)
−α1
2
θI
gFGH L tan
bT − T g = FGH 24 × 10
µ JK
k
h
c
−6
IJ
K
1
1 1.20 m tan 18.5°
− 15 × 10 −6
32° C = 0.275 mm
C°
C°
0. 42
If α 2 < α 1 , the diagram in part (a) applies to temperature falling and the diagram in part (b)
applies to temperature rising. The weight of the plate still pulls it step by step down the
roof. The same expression describes how far it moves each day.
(e)
ANSWERS TO EVEN PROBLEMS
P19.2
(a) 1.06 atm ; (b) −124° C
P19.32
(a) 900 K; (b) 1 200 K
P19.4
(a) 37.0° C = 310 K ; (b) −20.6° C = 253 K
P19.34
see the solution
P19.6
TC = 1.33 C° S° TS + 20.0° C
P19.36
3.96 × 10 −2 mol
P19.8
0.313 m
P19.38
3.67 cm3
P19.10
1.20 cm
P19.40
between 10 1 kg and 10 2 kg
P19.12
15.8 µ m
P19.42
2.41 × 10 11 molecules
P19.14
0.663 mm to the right at 78.2° below the
horizontal
P19.44
(a) 2.24 m; (b) 9.28 × 10 5 Pa
P19.46
0.523 kg
P19.48
(a) see the solution; (b) α << β
P19.50
(a) 0.169 m; (b) 1.35 × 10 5 Pa
P19.52
6.57 MPa
P19.54
(a) θ =
; (b) see the solution;
∆r
(c) it bends the other way; (d) 0.830°
b
g
2
P19.16
(a) 0.109 cm ; (b) increase
P19.18
(a) 437°C ; (b) 3 000°C ; no
P19.20
(a) 2.52 × 10 6 N m 2 ; (b) no
P19.22
0.812 cm3
P19.24
(a) 396 N; (b) −101° C ; (c) no change
P19.26
(a) 2.99 mol ; (b) 1.80 × 10 24 molecules
P19.56
(a) increase by 95.0 µs ; (b) loses 57.5 s
P19.28
884 balloons
P19.58
P19.30
(a) 1.06 × 10 21 kg ; (b) 56.9 K
(a) B = ρgP0 Vi P0 + ρgd
(c) 10.3 m
bα
2
g
− α 1 Li ∆T
b
g
−1
up; (b) decrease;
574
Temperature
P19.60
(a) yes; see the solution; (b) 25.7 rad s
P19.62
y ≈ L 2α∆T
P19.64
(a) see the solution;
(b) 3.66 × 10 −3 K −1 , within 0.06% and 0.2%
of the experimental values
P19.66
a
f
12
(a) 79.1 kPa for N 2 ; 21.2 kPa for O 2 ;
940 Pa for Ar; 33.3 Pa for CO 2 ;
(b) 81.7 L; 1.22 kg m3 ; (c) 29.0 g mol
P19.68
(a) 7.06 mm; (b) 297 K
P19.70
125 N ; −42.0 µm
P19.72
1.12 atm
P19.74
(a), (b), (c) see the solution; (d) 0.275 mm;
(e) see the solution