Chapter 10
10
Thermal Physics
PROBLEM SOLUTIONS
10.1
10.2
(a)
TF
9
T
5 C
(b)
TC
5
T
9 F
(c)
TF
9
T
5 C
32
9
5
273.15
5
98.6
9
32
9
T
5
32
32
32
273.15
460°F
37.0°C
9
5
32
173.15
32
280°F
When the volume of a low density gas is held constant, pressure and temperature are related by a linear equation
P
AT
B , where A and B are constants to be determined. For the given constant-volume gas thermometer,
P
0.700 atm when T
100°C
P
0.512 atm when T
0°C
From Equation [2], B
A
0.700 atm
0.512 atm
If P
A 0
0.700 atm 0.512 atm
100°C
[1]
B
1.88
10
3
[2]
atm °C
1.88
10
3
atm °C T
0.0400 atm , then
T
B
0.512 atm . Substituting this result into Equation [1] yields
so, the linear equation for this thermometer is: P
(a)
A 100°C
P
B
A
0.0400 atm 0.512 atm
1.88 10-3 atm °C
Page 10.1
251°C
0.512 atm .
Chapter 10
(b)
If T
450°C , then
P
10.3
(a)
(b)
10.4
(a)
TF
9
T
5 C
TK
TC
TF
9
T
5 C
TK
TC
TC
5
T
9 F
1.88
10
9
5
32
273.15
3
196
32
196
273.15 K
9
37.0
5
32
atm °C 450°C
273.15
37.0
32
5
134
9
32
0.512 atm
1.36 atm
321°F
77 K
98.6°F
273.15 K
32
310 K
56.7°C
and
TC
(b)
TK
5
9
79.8
TC
273.15
TK
TC
32
56.7
62.1°C
273.15 K
330 K
and
10.5
Start with TF
TC
5
T
9 F
273.15
62.1°C
273.15 K
211 K
40°F and convert to Celsius.
32
5
9
40
32
40°C
Since Celsius and Fahrenheit degrees of temperature change are different sizes, this is the only temperature with the
same numeric value on both scales.
10.6
Since we have a linear graph, we know that the pressure is related to the temperature as P
and B are constants. To find A and B, we use the given data:
Page 10.2
A
BTC , where A
Chapter 10
0.900 atm
A
B
80.0°C
1.635 atm
A
B 78.0°C
[1]
and
[2]
Solving Equations [1] and [2] simultaneously, we find:
A
1.27 atm
B
4.65
and
10
3
atm °C
Therefore,
P
(a)
1.27 atm
10
3
atm °C TC
At absolute zero the gas exerts zero pressure P
TC
(b)
4.65
4.65
1.27 atm
10 3 atm °C
1.27 atm
0
P
1.27 atm
9
T
5 C
Apply TF
TF
1
TF
2
4.65
0 and
1.27 atm
At the boiling point of water, TC
10.7
273°C
At the freezing point of water, TC
P
0 , so
100 C , so
10
3
atm °C 100°C
32 to two different Celsius temperatures, TC
9
T
5 C
1
1.74 atm
1
and TC
2
, to obtain
32
[1]
32
[2]
and
9
T
5 C
2
Page 10.3
Chapter 10
Subtracting Equation [1] from [2] yields
TF
TF
2
9
5
1
TC
TC
2
1
or
TF
10.8
10.9
(a)
95
TC
Using the result of Problem 10.7 above,
(b)
TK
TF
9
T
5 C
TC,out
273.15
9
5
32
43
TC,in
32
77
5
9
TC
273.15
32 °F
5
9 57.0 °C
TF
TC,out
TC,in
TC
31.7°C .
31.7 K
109°F
Yes . The normal body temperature is 98.6°F , so this patient has a high fever and needs immediate attention.
10.10
(a)
Since temperature differences on the Rankine and Fahrenheit scales are identical, the temperature readings on
the two thermometers must differ by no more than an additive constant (i.e., TR
TF
constant ). To
evaluate this constant, consider the temperature readings on the two scales at absolute zero. We have
TR
0°R at absolute zero, and
9
T
5 C
TF
32
9
5
273.15
32
459.67°F
Substituting these temperatures in our Fahrenheit to Rankine conversion gives
0
459.67
giving TR
(b)
TF
constant
or
constant
459.67
459.67 .
We start with the Kelvin temperature and convert to the Rankine temperature in several stages, using the
Fahrenheit to Rankine conversion from part (a) above.
TK
TC
273.15
5
T
9 R
491.67
5
T
9 F
32
273.15
273.15
5
T
9 R
5
9
TR
5
491.67
9
Page 10.4
459.67
32
273.15
273.15
5
T
9 R
273.15
273.15
5
T
9 R
Chapter 10
10.11
T
The increase in temperature is
35°C
20°C
55°C
Thus,
L
10.12
(a)
L0
T
10
6
1
°C
518 m 55°C
0.31 m
31 cm
L
As the temperature drops by 20°C, the length of the pendulum changes by
19
(b)
11
10
6
°C
1
1.3 m
20°C
Thus, the final length of the rod is L
1.3 m
From the expression for the period, T
2
4.9
4
10
m
L0
T
0.49 mm
0.49 mm .
L g , we see that as the length decreases the period decreases.
Thus, the pendulum will swing too rapidly and the clock will run fast .
10.13
L
We choose the radius as our linear dimension. Then, from
T
TC
20.0°C
L
L0
L0
2 .21 cm
130
10
6
L0
T ,
2 .20 cm
°C
1
2 .20 cm
35.0°C
or
TC
10.14
(a)
The diameter is a linear dimension, so we consider the linear expansion of steel:
d
(b)
55.0°C
d0 1
T
2.540 cm 1
11
V
If the volume increases by 1%, then
(1.0
10
6
°C
1
100°C
10 2) V0 . Then, using
is the volume expansion coefficient, we find
T
V V0
1.0
3 11
10
10
6
2
°C
1
3.0
Page 10.5
102 °C
25°C
V
2.542 cm
V0
T , where
3
Chapter 10
10.15
From
L
L
L0
L0
T , the final value of the linear dimension is L
L0
L0
T . To remove the
ring from the rod, the diameter of the ring must be at least as large as the diameter of the rod. Thus, we require that
LBrass
LAl ,
L0
or
L0
Brass
Brass
T
Brass
L0
Al
Al
L0
Al
T
This gives
L0
T
L0
Brass
(a)
If L0
L0
Al
Brass
L0
Al
Brass
Al
10.01 cm ,
Al
10.01
T
10-6 °C
19
1
10.00
10.00
24
10
6
°C
1
199°C
10.01
so
T
(b)
If L0
19
and T
m
V
10.17
(a)
T
20.0° C
199° C
179°C which is attainable
10.02 cm
Al
T
10.16
T0
6
10
T0
°C
T
1
10.00
10.00
24
V
V0
m
V0
6
°C
1
396°C
10.02
m V0
T
1
11.3
0
T
1+3 29
10
0
T
1
T
3 , gives
Using the result of problem 10.16, with
1
10
376°C, which is below absolute zero and unattainable .
m
V0
10.02
6
103 kg m3
°C
1
Page 10.6
90°C
0°C
11.2
103 kg m3
Chapter 10
(b)
No. Although the density of gold would be less on a warm day, the mass of the bar would be the same, regardless of its temperature, and that is what you are paying for. (Note that the volume of the bar increases
with increasing temperature, whereas its density decreases. Its mass, however, remains constant.)
10.18
When the temperature drops by 10.0°C, the wire will attempt to contract by
L
Cu L0
T
17
6
10
1
°C
10.0 m 10.0 °C
1.70
10
3
m
If the ends of the wire are held stationary, and the wire is not allowed to contract, it will develop an additional tension force
F sufficient to keep it stretched by an additional 1.70
F
quired additional tension force is
10
3
m beyond its natural length. The re-
YCu A ( L L0) , where YCu is Young’s modulus for copper and A is the
cross-sectional area of the wire. Thus,
F
11
1010 Pa 2.40
10
5
1.7 10 3 m
10.0 m
m2
449 N
The total tension in the wire, after the decrease in temperature, is then
F
10.19
F0
F
75.0 N
449 N
524 N
The difference in Celsius temperature in the underground tank and the tanker truck is
TC
5
9
5
95.0
9
TF
52.0
23.9°C
If V52°F is the volume of gasoline that fills the tank at 52.0°F, the volume this quantity of gas would occupy on the
tanker truck at 95.0°F is
V95°F
V52°F
1.00
10.20
V
V52°F
103 m3
1
V52°F
9.6
T
10
4
V52°F 1
°C
1
23.9°C
Consider a regular solid with initial volume given by V0
T
1.02
103 m 3
A0 L0 at temperature T0. Here, A is the cross-sectional
area and L is the length of the regular solid.
As the temperature undergoes a change
A
A
A0
A0
T
2 A0
T
T
T0 , the change in the cross-sectional area is
T , giving A
A0
Page 10.7
2 A0
T . Similarly, the new length will be
Chapter 10
L
L0
L0
T , so the new volume is
V
A0
2 A0
2
The term involving
V
A0 L0
3
10.21
T
L0
L0
T
A0 L0
3 A0 L0
T
2
2A L
0 0
T
2
is negligibly small in comparison to the other terms, so
3 A0 L0
T
V0
3 V0
T . This is of the form
V
V
V0
V0 ( T ) , where
.
[Note that some rules concerning significant figures are deliberately violated in this solution to better illustrate the
method of solution.]
When the temperature of a material is raised, the linear dimensions of any cavity in that material expands as if it
were filled with the surrounding material. Thus, the final value of the inner diameter of the ring will be given by
L
L0
L0
Dinner
10.22
T as
2.168 cm
1.42
10
5
°C
1
2.168 cm 100°C
15.0°C
2.171 cm
[Note that some rules concerning significant figures are deliberately violated in this solution to better illustrate the
method of solution.]
Let L be the final length of the aluminum column. This will also be the final length of the quantity of tape now
stretching from one end of the column to the other. In order to determine what the scale reading now is, we need to
find the initial length this quantity of tape had at 21.2°C (when the scale markings were presumably painted on the
tape).
Thus, we let this initial length of tape be (L0)tape and require that
L
L0
tape
1
steel
T
L0
column
1
which gives
L0
L0
tape
column
1
1
Al
steel
T
T
or
Page 10.8
Al
T
Chapter 10
18.700 m 1
L0
10.23
tape
1
24
11
6
10
The initial length of the band is L0
10
6
1
°C
2 r0
1
°C
2
29.4°C
21.2°C
29.4°C
21.2°C
5.0
3
10
m
18.702 m
3.1
10
2
m . The amount this length would
contract, if allowed to do so, as the band cools to 37°C is
L
L0
T
17.3
10
6
°C
1
3.1
10
2
m 80°C
37°C
2 .3
10
5
m
Since the band is not allowed to contract, it will develop a tensile stress given by
Stress
L
Y
L0
5
2 .3
3.1
10
10
m 0.50
10
L
T , or
1010 Pa
18
2
m
m
108 Pa
1.3
and the tension in the band will be
F
10.24
A Stress
4.0
10
3
The expansion of the pipeline will be
L
11
10
6
1
°C
1300
L0
103 m
3
m
108 Pa
1.3
35°C
73°C
1.5
102 N
2.7
10 3 m
1.5 km
This is accommodated by accordion-like expansion joints placed in the pipeline at periodic intervals.
10.25
Both the drum and the carbon tetrachloride expand as the temperature rises by
T
20.0°C . Since the drum was
completely filled when the temperature was 10.0°C, the initial volume for the drum and the carbon tetrachloride are
the same. From
V
V0
Vspillage
Vcarbon
Vspillage
5.81
3
T , where
tetrachloride
Vsteel
is the coefficient of volume expansion, we obtain
3
carbon
tetrachloride
drum
steel
V0
T
or
10
4
°C
1
3 11
10
6
°C
Page 10.9
1
50.0 gal 20.0°C
0.548 gal
Chapter 10
10.26
(a)
m0
(b)
V
0V
V0
V
3.80
102 kg m 3 10.0 gal
7.30
V0
V0
T
V0 1
1
9.60
10 3 m 3
1 gal
27.7 kg
T
or
1.000 m3
V
(c)
Gasoline having a mass of m
of V
10.27
m20
(e)
m
°C
1
20.0°C
1.02 m 3
102 kg occupies a volume of V0
7.30
1.000 m3 at 0°C and a volume
7.30 102 kg
1.02 m3
m
V
20V
10 2 kg m 3 10.0 gal
7.16
m0
m20
716 kg m 3
27.7 kg
27.2 kg
3.80
10 3 m 3
1 gal
27.2 kg
0.5 kg
(a)
The gap width is a linear dimension, so it increases in “thermal enlargement” as the temperature goes up.
(b)
At 190°C, the length of the piece of steel that is missing, or has been removed to create the gap, is
L
L0
L
10.28
4
1.02 m3 at 20.0°C. The density of gasoline at 20.0°C is then
20
(d)
10
L
L0 1
1.600 cm 1
T
11
. This gives
10
6
°C
If allowed to do so, the concrete would expand by
(a)
1
190°C
L
L0
30.0°C
1.603 cm
T .
Since it is not permitted to expand, the concrete experiences a compressive stress of
Stress
Y
L
L0
Y
T
7.00
109 Pa 12
Page 10.10
10
6
°C
1
30.0°C
Chapter 10
or Stress
(b)
106 Pa .
2.5
Since this stress is less than the compressive strength of concrete, the sidewalk
will not fracture .
10.29
(a)
nRT , we find P T
From the ideal gas law, PV
nR V . Thus, if both n and V are constant as the gas is
heated, the ratio P T is constant giving
Pf
Pi
Ti
Tf
(b)
10.30
Tf
or
Ti
Pf
Pi
Pf V f
2 Pi
Ti
TC
(b)
ni
PV
i i
RTi
(c)
MCO2
(d)
mi
ni MCO2
7.83 mol 44.0 g mol
(e)
Tf
Ti
292 K
nf
Ti
2Vi
Ti
(f)
3Pi
Pi
900 K
627°C
If both pressure and volume double as n is held constant, the ideal gas law gives
Tf
(a)
300 K
PV
i i
273.15
9.50
PV
i i
19.0
4Ti
273.15 K
105 Pa
4 300 K
1 200 K
292 K
20.0 L 103 cm3 1 L 1 m3 106 cm3
8.31 J mol K 292 K
12.0
T
mf
MCO2
2 16.0
mi
m
MCO2
g
mol
224 K
927°C
44.0 g mol
345 g
516 K
345 g 82.0 g
44.0 g mol
5.98 mol
Neglecting any change in volume of the tank, V f
Page 10.11
Vi , and we have
7.83 mol
Chapter 10
10.31
(g)
Pf
(a)
n
Pf V f
n f R Tf
Pi Vi
ni R Ti
nf
Tf
ni
Ti
PV
RT
5.98 mol
7.83 mol
Pi
n f Tf
Pf
516 K
292 K
10 5 Pa atm 1.0
1.013
Pi
niTi
10
105 Pa
9.50
6
m3
4.2
8.31 J mol K 293 K
10
1.28
5
106 Pa
mol
Thus,
N
(b)
n NA
4.2
10
5
mol
6.02
molecules
mol
1023
1019 molecules
2.5
Since both V and T are constant,
n2
n1
P2 V2
RT2
P2
P1
P1 V 1 RT 1
or
n2
10.32
P2
n
P1 1
1.0 10 11 Pa
1.013 105 Pa
The volume of helium in each balloon is Vb
The total volume of the helium at P2
V2
P1
V
P2 1
150 atm
1.20 atm
4.2
10
4 r3 3 .
1.20 atm will be
0.100 m 3
12 .5 m 3
Thus, the number of balloons that can be filled is
Page 10.12
5
mol
4.1
10
21
mol
Chapter 10
N
10.33
12.5 m 3
V2
Vb
4 3 0.150 m
884 balloons
3
The initial and final absolute temperatures are
Ti
TC,i
273
25.0
273 K
298 K
and
Tf
TC, f
The volume of the tank is assumed to be unchanged, or V f
nf
10.34
75.0
273 K
Pf Vf
n f R Tf
Pi Vi
ni R Ti
Vi . Also, since two-thirds of the gas is withdrawn,
Pf
nf
Tf
ni
Ti
1
3
Pi
348 K
298 K
11.0 atm
4.28 atm
If the volume and the temperature are both constant, the ideal gas law gives
n f RT f
Pi Vi
ni RTi
Pf
nf
or
so the amount of gas to be withdrawn is
Pi
n
ni
5.00 atm
25.0 atm
ni
nf
1.50 mol
1.50 mol
0.300 mol
0.300 mol
1.20 mol .
With n held constant, the ideal gas law gives
V1
V2
P2
P1
T1
T2
0.030 atm
1.0 atm
300 K
200 K
4.5
4 3 r 3 , V1 V2
Since the volume of a sphere is V
10
r1 r2
2
3
Thus,
r1
10.36
348 K
ni 3 . Thus, from the ideal gas law,
Pf Vf
10.35
273
V1
V2
13
r2
4.5
10
2
13
20 m
7.1 m
The mass of the gas in the balloon does not change as the temperature increases. Thus,
Page 10.13
Chapter 10
f
m Vf
i
m Vi
Vi
Vf
or
f
i
Vi
Vf
From the ideal gas law with both n and P constant, we find Vi Vf
f
10.37
i
Ti
Tf
273 K
373 K
0.179 kg m3
0.131 kg m3
The pressure 100 m below the surface is found, using P1
P1
1.013
105 Pa
Ti Tf and now have
Patm
gh , to be
103 kg m3 9.80 m s2 100 m
106 Pa
1.08
The ideal gas law, with T constant, gives the volume at the surface as
V2
10.38
(a)
P1
V
P2 1
1.08 106 Pa
1.013 105 Pa
P1
V
Patm
We assume the density
can be written as
1.50 cm 3
16.0 cm 3
naV b Mc where n is the number of moles, V is the volume,
and M is the molecular weight in kilograms per mole, while a, b, and c are constants to be determined by dimensional analysis. In terms of mass (M), length (L), time (T), and number of moles (N), the fundamental
mass / volume
units of density are
molecular weight M
kg mol
ML 3 , those of n are n
N , for volume V
L 3 , and
MN 1 . In terms of basic units, our assumed equation for density be-
comes
n
a
V
b
M
c
or
M1L
3
Na L3
b
MN
1
c
Na c L3b Mc
and equating the powers of each of the basic units on the two sides of the equation gives:
1
c
c
1;
3
3b
b
1;
0
so our expression for density, derived by dimensional analysis, is
a
c
a
c
1
n1V 1M1 , or
nM
where M is in kilograms per mole.
V
(b)
From the ideal gas law, PV
nRT , or P
n V RT . But, from the result of part (a), we may write
Page 10.14
Chapter 10
nV
M , so the ideal gas law may be written in terms of the density of the gas as
P
RT
M
where M is in kilograms per mole.
(c)
For carbon dioxide, M
44 g mol
P
105 Pa 1 atm
90.0 atm 1.013
44
3
10
9.12
kg mol . Then, if the pressure is
106 Pa , and T
102 K , the density of the at-
7.00
mosphere on Venus is
(d)
Since
8.31 J mol K
10
3
7.00
kg mol
10 2 K
69.0 kg m 3
Mshell
Vshell
2.00
4
atmosphere ,
shell
102 kg
1.00 m
3
47.7 kg m 3
3
this shell would rise in the atmosphere on Venus .
The average kinetic energy of the molecules of any ideal gas at 300 K is
KE
10.40
44
The density of the evacuated steel shell would be
shell
10.39
106 Pa
9.12
PM
RT
1
mv2
2
3
k T
2 B
3
1.38
2
10
23
J
K
300 K
6.21
10
21
J
Since the sample contains three times Avogadro’s number of molecules, there must be 3 moles of gas present. The
ideal gas law then gives
P
nRT
V
3 mol 8.31 J mol K
0.200 m
293 K
9.13
3
10 5 Pa
The force this gas will exert on one face of the cubical container is
F
10.41
PA
9.13
105 Pa 0.200 m
2
3.65
104 N
36.5 kN
One mole of any substance contains Avogadro’s number of molecules and has a mass equal to the molar mass, M.
Thus, the mass of a single molecule is m
M NA .
Page 10.15
Chapter 10
For helium, M
m
4.00 g mol
4.00
4.00 10 3 kg mol
6.02 1023 molecule mol
10
3
kg mol , and the mass of a helium molecule is
6.64
10
27
kg molecule
Since a helium molecule contains a single helium atom, the mass of a helium atom is
matom
10.42
6.64
10
27
kg
3RT M .
The rms speed of molecules in a gas of molecular weight M and absolute temperature T is vrms
625 m s for molecules in oxygen (O2), for which M
Thus, if vrms
32.0 g mol
32.0
10
3
kg mol , the
temperature of the gas is
T
10.43
10
3
2
kg mol 625 m s
501 K
3 8.31 J mol K
The average translational kinetic energy per molecule in an ideal gas at absolute temperature T is
3kBT 2 , where kB
KEmolecule
is T
TC
273.15
KE molecule
10.44
32.0
2
M vrms
3R
(a)
77.0
3
1.38
2
1.38
273.15 K
10
23
10
23
J K is Boltzmann’s constant. Thus, if the absolute temperature
350 K , we have
J K 350 K
7.25
10
21
J
The volume occupied by this gas is
V
7.00 L 103 cm3 1 L 1 m3 106 cm3
7.00
10
3
m3
Then, the ideal gas law gives
T
(b)
PV
nR
1.60
106 Pa 7.00
10
3
m3
3.50 mol 8.31 J mol K
385 K
The average kinetic energy per molecule in this gas is
KE molecule
3
k T
2 B
3
1.38
2
10
23
J K 385 K
Page 10.16
7.97
10
21
J
Chapter 10
(c)
You would need to know the mass of the gas molecule to find its average speed, which in
turn requires knowledge of the molecular weight of the gas .
10.45
Consider a time interval of 1.0 min = 60 s, during which 150 bullets bounce off Superman’s chest. From the impulse–momentum theorem, the magnitude of the average force exerted on Superman is
Fav
150
I
p bullet
t
150 m v
t
150 8.0
t
3
10
v0
kg
400 m s
400 m s
16 N
60 s
10.46
From the impulse–momentum theorem, the average force exerted on the wall is
Fav
Fav
N
I
N m v
p molecule
t
t
1023
5.0
v0
or
t
4.68
10
26
kg
300 m s
300 m s
14 N
1.0 s
The pressure on the wall is then
10.47
14 N
10 4 cm 2
2
8.0 cm
1 m2
Fav
A
P
10 4 N m 2
1.8
As the pipe undergoes a temperature change
T
46.5°C
18 kPa
18.0°C
28.5°C , the expansion of the horizontal
segment is
Lx
L0 x
17
T
10
6
°C
1
28.0 cm 28.5°C
1.36
10
2
cm
0.136 mm
The expansion of the vertical section is
Ly
L0 y
T
17
10
6
°C
1
134 cm 28.5°C
Page 10.17
0.649 mm
Chapter 10
The total displacement of the pipe elbow is
L2x
L
L2y
0.136 mm
2
0.649 mm
2
0.663 mm
at
tan
1
Ly
Lx
tan
1
0.649 mm
0.136 mm
78.2
or
L
10.48
0.663 mm at 78.2° below the horizontal
(a)
L
L0
(b)
D
D0
(c)
T
T
V0
3 V0
10
9.0
The initial volume is V0
V
10.49
9.0
6
10
°C
6
1
°C
20 cm 75 °C
1
D02
L0
4
1.0 cm 75 °C
4
1.0 cm
2
20 cm
1.4
10
6.8
2
10
cm
4
cm
16 cm 3 .
T
T
3 9.0
10
6
°C
1
16 cm 3
75 °C
3.2
10
2
The number of moles of CO2 present is
n
6.50 g
44.0 g mol
0.148 mol
Thus, at the given temperature and pressure, the volume will be
V
nRT
P
0.148 mol 8.31 J mol K 293 K
1.013
105 Pa
Page 10.18
3.55
10
3
m3
3.55 L
cm 3
Chapter 10
10.50
(a)
The sketch at the right shows a
thermometer marked with the
Fahrenheit scale on one side and
Réaumer’s scale on the other.
Note that 80 RE scale divisions
span the range from the freezing
to the boiling of water, but 180 F
divisions are required to do this.
Then, note that the current temperature is TRE RE divisions
above the freezing point of water,
but is TF
32 F divisions
above this same level.
Since the range between the current temperature and the freezing of water is the same fraction of the span
between freezing and boiling points of water on both sides of the thermometer, we must have
TRE
80
(b)
or
4
T
9 F
14.2
80
T
180 F
TRE
Normal body temperature TF
TRE
10.51
TF 32
180
4
T
9 F
32
32
98.6°F on the Réaumer scale is
4
98.6
9
14.2 °RE
29.6 °RE
The ideal gas law will be used to find the pressure in the tire at the higher temperature. However, one must always
be careful to use absolute temperatures and absolute pressures in all ideal gas law calculations.
The initial absolute pressure is Pi
Pi,gauge
The initial absolute temperature is Ti
and the final absolute temperature is Tf
Ti,C
Patm
2.5 atm
273.15
Tf ,C
273.15
15
1.0 atm
3.5 atm .
273.15 K
45
273.15 K
288 K .
318 K .
The ideal gas law, with volume and quantity of gas constant, gives the final absolute pressure as
Page 10.19
Chapter 10
Pf Vf
n f R Tf
Pi V i
ni R Ti
Tf
Pf
Pi
Ti
The final gauge pressure in the tire is Pf ,gauge
10.52
318 K
288 K
3.5 atm
Patm
3.9 atm
Pf
3.9 atm
1.0 atm
2.9 atm .
When air trapped in the tube is compressed, at constant temperature, into a cylindrical volume 0.40-m long, the
ideal gas law gives its pressure as
V1
P
V2 1
P2
L1
P
L2 1
1.5 m
0.40 m
1.013
10 5 Pa
3.8
This is also the water pressure at the bottom of the lake. Thus, P
h
10.53
P2
3.8
Patm
g
1.013
103 kg m 3
10 5 Pa
10 5 Pa
Patm
gh gives the depth of the lake as
28 m
9.80 m s2
The mass of CO2 produced by three astronauts in 7.00 days is m
3 1.09 kg d 7.00 d
22.9 kg , and the
number of moles of CO2 available is
n
m
M
44.0
22 .9 kg
10 3 kg mol
520 mol
The recycling process will generate 520 moles of methane to be stored. In a volume of V
and at temperature T
P
10.54
(a)
nRT
V
45.0°C
150 L
0.150 m3
228 K , the pressure of the stored methane is
520 mol 8.31 J mol K 228 K
0.150 m 3
6.57
106 Pa
6.57 MPa
The piston in this vertical cylinder has three forces acting on it. These are: (1) a downward gravitational force,
mg, the piston’s own weight; (2) a downward pressure force, Fd
above the piston; and (3) an upward pressure force, Fu
P0 A , due to the atmospheric pressure
PA , due to the absolute pressure of the gas trapped
inside the cylinder. Since the piston is in equilibrium, Newton’s second law requires
Fy
0
Fu
mg
Fd
0
or
Page 10.20
PA
mg
P0 A
[1]
Chapter 10
From the ideal gas law, the absolute pressure of the trapped gas is
nRT
V
P
nRT
Ah
[2]
Substituting Equation [2] into [1] yields
nRT
Ah
(b)
A
mg
P0 A
nRT
mg P0 A
h
or
From Equation [1] above, the absolute pressure inside the cylinder is P
mg A
P0 where
P0 is atmospheric pressure. This is greater than atmospheric pressure because mg A
(c)
0.
Observe from the result of part (a) above, if the absolute temperature T increases,
the equilibrium value of h also increases .
10.55
(a)
As the acetone undergoes a change in temperature
T
20.0
35.0 °C
15.0 °C , the final volume
will be
Vf
V0
V
V0
100 mL 1
(b)
V0
1.50
T
V0 1
10
1
4
°C
T
15.0 °C
99.8 mL
When acetone at 35°C is poured into the Pyrex flask that was calibrated at 20°C, the volume of the flask temporarily expands to be larger than its calibration markings indicate. However, the coefficient of volume expansion for Pyrex [
[
1.5
10
4
3
9.6
10
6
(°C) 1] is much smaller than that of acetone
(°C) 1] . Hence, the temporary increase in the volume of the flask will be much smaller than
the change in volume of the acetone as the materials cool back to 20°C, and this change in volume of the
flask has negligible effect on the answer .
10.56
If Pi is the initial gauge pressure of the gas in the cylinder, the initial absolute pressure is Pi,abs
Pi
Patm , where
Pf is atmospheric pressure. Likewise, the final absolute pressure in the cylinder is , where is the final gauge pressure. The initial and final masses of gas in the cylinder are mi
ni M and m f
moles of gas present and M is the molecular weight of this gas. Thus, m f mi
Page 10.21
n f M , where n is the number of
n f ni .
Chapter 10
We assume the cylinder is a rigid container whose volume does not vary with internal pressure. Also, since the
temperature of the cylinder is constant, its volume does not expand nor contract. Then, the ideal gas law (using absolute pressures) with both temperature and volume constant gives
Pf ,abs V
n f RT
mf
Pi,abs V
ni RT
mi
Pf ,abs
mf
or
Pi,abs
mi
and in terms of gauge pressures,
mf
10.57
(a)
Pf
Patm
Pi
Patm
The volume of the liquid expands by
Vflask
3
Voverflow
V0
Vflask
Voverflow
A
3 (3.2
glass
10
6
V0
V0
3
V0
A
For a mercury thermometer,
3
Vliquid
T and the volume of the glass flask expands by
T . The amount of liquid that must overflow into the capillary is
Vliquid
h
(b)
mi
3
1.82
Hg
(°C) 1)
T . The distance the liquid will rise into the capillary is then
9.6
T
10
10
6
4
°C
°C
1
1
and (assuming Pyrex glass),
. Thus, the expansion of the mercury is
almost 20 times the expansion of the flask , making it a rather good approximation to neglect the expansion of the flask.
10.58
(a)
The initial absolute pressure in the tire is
P1
Patm
P1
gauge
1.00 atm
and the final absolute pressure is P2
1.80 atm
2 .80 atm
3.20 atm .
The ideal gas law, with volume constant, gives
T2
P2
T
P1 1
3.20 atm
2.80 atm
300 K
343 K
Page 10.22
Chapter 10
(b)
When the quantity of gas varies, while volume and temperature are constant, the ideal gas law gives
n3 n2
P3 P2 . Thus, when air is released to lower the absolute pressure back to 2.80 atm, we have
n3
n2
2 .80 atm
3.20 atm
0.875
At the end, we have 87.5% of the original mass of air remaining, or
12.5% of the original mass was released.
10.59
After expansion, the increase in the length of one span is
L
L.0
12
T
10
6
1
°C
giving a final length of L
125 m 20.0°C
L0
L
125 m
0.0300 m
0.0300 m
From the Pythagorean theorem,
y
10.60
(a)
L2
L20
0.0300 m2
125
From the ideal gas law,
P2V2
T2
125 m
PV
1 1
, or
T1
P2
P1
2
V2
V1
2.74 m
T2
T1
The initial conditions are
P1
1 atm, V1
5.00 L
5.00
10
3
m3, and T1
20.0°C
293 K
The final conditions are
P2
1 atm
F
=1 atm
A
k h
, V2
A
V1
Thus,
1
k h
A 1 atm
1
A h
V1
523 K
293 K
Page 10.23
A h, and T2
250 °C
523 K
Chapter 10
or
1
103 N m
2 .00
0.0100
m2
h
105
1.013
N
m2
1
0.0100 m2
h
3
m3
5.00
10
523
293
Simplifying and using the quadratic formula yields
(b)
10.61
(a)
P2
h
0.169 m
1 atm
k h
A
1.013
105 Pa
16.9 cm
2.00
103 N m 0.169 m
0.0100 m 2
10 5 Pa
1.35
The two metallic strips have the same length L0 at the initial temperature T0. After the temperature has
changed by
L1
T
T
T0 , the lengths of the two strips are
L0 1
T
2
T
The lengths of the circular arcs are related to their radii by L1
r1
1
and
L2
L0 1
and L2
radians.
Thus,
r
r2
r1
L2
L1
L0
T
2
1
L0
T
or
2
1
r
(b)
As seen in the above result,
(c)
If
T
0 , then
0 if either T
0 or
1
2
.
is negative so the bar bends in the opposite direction .
Page 10.24
r2 , where
is measured in
Chapter 10
10.62
(a)
If the bridge were free to expand as the temperature increased by
T
20°C , the increase in length would
be
L
(b)
L0
T
6
10
°C
1
250 m 20°C
6.0
10
2
m
6.0 cm
Combining the defining equation for Young’s modulus,
Y
Stress Strain
with the expression,
Stress
(c)
12
When
T
L
L
L
Y
Stress
L
LL
T , for the linear expansion when the temperature changes by
L
Y
T
Y
L
T yields
T
20°C , the stress in the specified bridge would be
Stress
Y
T
12
10
6
°C
1
2.0
1010 Pa 20°C
Since this considerably less than the maximum stress, 2.0
106 Pa
4.8
107 Pa , that concrete can withstand,
the bridge will not crumble .
10.63
(a)
As the temperature of the pipe increases, the original 5.0-m length between the water heater and the floor
above will expand by
L
L0
T
17
10
6
°C 5.0 m 46°C
20°C
2.21
10
3
m
If this expansion occurs in a series of 18 “ticks,” the expansion per tick is
movement per tick
(b)
L 18
2.21
10
3
m 18
1.23
10
4
m
0.12 mm
When the pipe is stuck in the hole, the floor exerts a friction force on the pipe preventing it from expanding.
Just before a “tick” occurs, the pipe is compressed a distance of 0.123 mm. The force required to produce this
compression is given by the equation defining Young’s modulus, Y
Page 10.25
F A
L L , as
Chapter 10
F
10.64
L
L
YA
11
1010 Pa 3.55
10
5
1.23
m2
10 4 m
5.0 m
96 N
0°C 273 K , while the temperature of container 2 is raised to
Let container 1 be maintained at T1 T0
T2 100°C 373 K . Both containers have the same constant volume, V, and the same initial pressures,
P0
P0
2
P0 . As the temperature of container 2 is raised, gas flows from one container to the other until the
1
final pressures are again equal, P2
n2
n1
n0
2
From the ideal gas law, n
PV
RT1
PV
RT2
P0V
RT0
n0
P1
P . The total mass of gas is constant, so
[1]
1
PV RT , so Equation [1] becomes
P0V
,
RT0
or
P
1
T1
1
T2
2 P0
T0
Thus,
P
2 P0
T0
T1T2
T1 T2
2 1.00 atm
273
273 373
273 373
Page 10.26
1.15 atm