Physical Chemistry: Thermodynamics, Structure, and Change Tenth Edition Solution Manual

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Foundations
A Matter
Answers to discussion questions
A.2
Metals conduct electricity, have luster, and they are malleable and ductile.
Nonmetals do not conduct electricity and are neither malleable nor ductile.
Metalloids typically have the appearance of metals but behave chemically like a nonmetal.
1
IA
2
IIA
3
IIIB
4
IVB
5
VB
6
VIB
7
VIIB
8
VIIIB
9
VIIIB
10
VIIIB
11
IB
12
IIB
13
IIIA
14
IVA
15
VA
16
VIA
17
VIIA
18
VIIIA
1
H
1.008
Periodic Table of the Elements 2
He
4.003
3
Li
6.941
4
Be
9.012
5
B
10.81
6
C
12.01
7
N
14.01
8
O
16.00
9
F
19.00
10
Ne
20.18
11
Na
22.99
12
Mg
24.31
13
Al
26.98
14
Si
28.09
15
P
30.97
16
S
32.07
17
Cl
35.45
18
Ar
39.95
19
K
39.10
20
Ca
40.08
21
Sc
44.96
22
Ti
47.88
23
V
50.94
24
Cr
52.00
25
Mn
54.94
26
Fe
55.85
27
Co
58.93
28
Ni
58.69
29
Cu
63.55
30
Zn
65.38
31
Ga
69.72
32
Ge
72.59
33
As
74.92
34
Se
78.96
35
Br
79.90
36
Kr
83.80
37
Rb
85.47
38
Sr
87.62
39
Y
88.91
40
Zr
91.22
41
Nb
92.91
42
Mo
95.94
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3
55
Cs
132.9
56
Ba
137.3
57
La
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222)
87
Fr
(223)
88
Ra
226
89
Ac
(227)
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
145
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
237
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
Transition metals
Lanthanoids
Actinoids
1
IA
2
IIA
3
IIIB
4
IVB
5
VB
6
VIB
7
VIIB
8
VIIIB
9
VIIIB
10
VIIIB
11
IB
12
IIB
13
IIIA
14
IVA
15
VA
16
VIA
17
VIIA
18
VIIIA
1
H
1.008
Periodic Table of the Elements 2
He
4.003
3
Li
6.941
4
Be
9.012
5
B
10.81
6
C
12.01
7
N
14.01
8
O
16.00
9
F
19.00
10
Ne
20.18
11
Na
22.99
12
Mg
24.31
13
Al
26.98
14
Si
28.09
15
P
30.97
16
S
32.07
17
Cl
35.45
18
Ar
39.95
19
K
39.10
20
Ca
40.08
21
Sc
44.96
22
Ti
47.88
23
V
50.94
24
Cr
52.00
25
Mn
54.94
26
Fe
55.85
27
Co
58.93
28
Ni
58.69
29
Cu
63.55
30
Zn
65.38
31
Ga
69.72
32
Ge
72.59
33
As
74.92
34
Se
78.96
35
Br
79.90
36
Kr
83.80
37
Rb
85.47
38
Sr
87.62
39
Y
88.91
40
Zr
91.22
41
Nb
92.91
42
Mo
95.94
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

NH H
H
Cs
132.9
Ba
137.3
La
138.9
Hf
178.5
Ta
180.9
W
183.9
Re
186.2
Os
190.2
Ir
192.2
Pt
195.1
Au
197.0
Hg
200.6
Tl
204.4
Pb
207.2
Bi
209.0
Po
(209)
At
(210)
Rn
(222)
87
Fr
(223)
88
Ra
226
89
Ac
(227)
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
145
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
237
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
A.4 Valence-shell electron pair repulsion theory (VSEPR theory) predicts molecular
shape with the concept that regions of high electron density (as represented by single bonds,
multiple bonds, and lone pair) take up orientations around the central atom that maximize their
separation. The resulting positions of attached atoms (not lone pairs) are used to classify the
shape of the molecule. When the central atom has two or more lone pair, the molecular geometry
must minimize repulsion between the relatively diffuse orbitals of the lone pair. Furthermore, it
is assumed that the repulsion between a lone pair and a bonding pair is stronger than the
repulsion between two bonding pair, thereby, making bond angles smaller than the idealized
bond angles that appear in the absence of a lone pair.
Solutions to exercises
A.1(b)
A.2(b)
(i) chemical formula and name: CaH2 , calcium hydride
ions: Ca2+ and H–
oxidation numbers of the elements: calcium, +2; hydrogen, –1
(ii) chemical formula and name: CaC 2 , calcium carbide
ions: Ca2+ and C22– (a polyatomic ion)
oxidation numbers of the elements: calcium, +2; carbon, –1
(iii) chemical formula and name: LiN3 , lithium azide
ions: Li + and N3– (a polyatomic ion)
oxidation numbers of the elements: lithium, +1; nitrogen, –⅓
A.3(b)
(i) Ammonia, NH3 , illustrates a molecule with one lone pair on the central atom.
Example Element Ground-state Electronic Configuration
(i) Group 3 Sc, scandium [Ar]3d1 4s 2
(ii) Group 5 V, vanadium [Ar]3d3 4s 2
(iii) Group 13 Ga, gallium [Ar]3d10 4s2 4p1

H O
H
H F
Xe FF
(ii) Water, H2 O, illustrates a molecule with two lone pairs on the central atom.
(iii) The hydrogen fluoride molecule, HF, illustrates a molecule with three lone pairs on the
central atom. Xenon difluoride has three lone pairs on both the central atom and the two
peripheral atoms.
A.4(b)
(i) Ozone, O3 . Formal charges (shown in circles) may be indicated.
(ii) ClF3+
(iii) azide anion, N3–
A.5(b) The central atoms in XeF 4 , PCl 5 , SF4 , and SF6 are hypervalent.
A.6(b) Molecular and polyatomic ion shapes are predicted by drawing the Lewis structure and
applying the concepts of VSEPR theory.
(i) H2 O2 Lewis structure:
O O
O
O O
O
F Cl F
F
N N N
H O O H

Orientations caused by repulsions between two lone pair and two bonding pair around
each oxygen atom:
Molecular shape around each oxygen atom: bent (or angular) with bond angles somewhat
smaller than 109.5º
(ii) FSO3– Lewis structure:
(Formal charge is circled.)
Orientations around the sulfur are caused by repulsions between one lone pair, one
double bond, and two single bonds while orientations around the oxygen to which
fluorine is attached are caused by repulsions between two lone pair and two single bonds:
Molecular shape around the sulfur atom is trigonal pyramidal with bond angles somewhat
smaller than 109.5º while the shape around the oxygen to which fluorine is attached is
bent (or angular) with a bond angle somewhat smaller than 109.5º.
(iii) KrF2 Lewis structure:
Orientations caused by repulsions between three lone pair and two bonding pair:
H
O O
H
SSOO OO
OO
FF
S O
F
O
O
F Kr F
Kr
F
F

Molecular shape: linear with a 180º bond angle.
(iv) PCl4+ Lewis structure:
(Formal charge is shown in a circle.)
Orientations caused by repulsions between four bonding pair (no lone pair):
Molecular shape: tetrahedral and bond angles of 109.5º
A.7(b)
(i) Nonpolar or weakly polar toward the slightly more electronegative carbon.
(ii) (c)
A.8(b)
(i) O3 is a bent molecule that has a small dipole as indicated by consideration of electron
densities and formal charge distributions.
(ii) XeF2 is a linear, nonpolar molecule.
(iii) NO2 is a bent, polar molecule.
(iv) C 6 H14 is a nonpolar molecule.
A.9(b) In the order of increasing dipole moment: XeF2 ~ C 6 H14 , NO2 , O3
A.10(b)
(i) Pressure is an intensive property.
(ii) Specific heat capacity is an intensive property.
(iii) Weight is an extensive property.
(iv) Molality is an intensive property.
A.11(b)
(i) 1 mol
5.0 g 0.028 mol [A.3]
180.16 g
m
n M
 
= = = 
 
(ii) 23
22
A
6.0221 10 molecules
0.028 mol 1.7 10 molecules
mol
N nN  ×
= = = × 
 
A.12(b)
Cl P Cl
Cl
Cl
P
Cl Cl
Cl Cl
C H
N Cl
δ
+ δ
−
P S
δ+ δ−

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(i) 78.11 g
10.0 mol 781. g [A.3]
mol
m n M  
= = = 
 
(ii) gravity on Mars Marsweight F m g= =
( ) ( )2 21 kg
781. g 3.72 m s 2.91 kg m s 2.91 N
1000 g
− − 
= × × = = 
 
A.13(b) F mg
p A A
= =
( ) ( )2 2
6 2 6
2 4 2 5
60 kg 9.81 m s 1 cm 1 bar
3 10 N m 3 10 Pa
2 cm 10 m 10 Pa
30 bar 10 bar
−
−
−
×    
= = × = ×   
  
= ±
A.14(b) ( ) 1 atm
30 bar 10 bar 30 atm 10 atm
1.01325 bar
 
± = ± 
 
A.15(b)
(i) 5
51.01325 10 Pa
222 atm 225 10 Pa
1 atm
 × = × 
 
(ii) 1.01325 bar
222 atm 225 bar
1 atm
  = 
 
A.16(b) / C / K 273.15 90.18 273.15 182.97 [A.4]T
θ ° = − = − = −
182.97 C
θ = − °
A.17(b) The absolute zero of temperature is 0 K and 0 ºR. Using the scaling relationship 1 ºF / 1
ºR (given in the exercise) and knowing the scaling ratios 5 ºC / 9 ºF and 1 K / 1 ºC, we find the
scaling factor between the Kelvin scale and the Rankine scale to be:
1 F 5 C 1 K 5 K
1 R 9 F 1 C 9 R
° °     
× × =     
° ° ° °     
The zero values of the absolute zero of temperature on both the Kelvin and Rankine scales and
the value of the scaling relationship implies that:
( ) ( )R R
5 9/ K / R or / R / K
9 5
T T
θ
θ= × ° ° = ×
Normal freezing point of water:
( ) ( )R
R
9 9/ R / K 273.15 491.67
5 5
491.67 R
T
θ
θ
° = × = × =
= °

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A.18(b) 1 mol
0.325 g 0.0161 mol
20.18 g
n  
= × = 
 
( ) ( ) ( )1 1 3
3 3 3
4
0.0161 mol 8.314 J K mol 293.15 K dm
[A.5] 2.00 dm 10 m
1.96 10 Pa 19.6 kPa
nRT
p V
− −
−
 
= =  
 
= × =
A.19(b) [A.5] mRT
pV nRT M
= =
( )( ) ( )3 1 1
1 1
3
where is the mass density [A.2]
0.6388 kg m 8.314 J K mol 373.15 K 0.124 kg mol 124 g mol
16.0 10 Pa
mRT RT
M pV p
ρ
ρ
− − −
− −
= =
= = =
×
The molecular mass is four times as large as the atomic mass of phosphorus (30.97 g mol –1 ) so
the molecular formula is P 4 .
A.20(b) 1 mol
7.05 g 0.220 mol [A.3]
32.00 g
n  
= × = 
 
( ) ( ) ( )1 1 3
3 6 3
6
0.220 mol 8.314 J K mol 373.15.15 K cm
[A.5] 100. cm 10 m
6.83 10 Pa 6.83 MPa
nRT
p V
− −
−
 
= =  
 
= × =
A.21(b) 2 2O CO0.25 mole and 0.034 molen n= =
( ) ( ) ( )2
2
1 1 3
O
O 3 6 3
0.25 mol 8.314 J K mol 283.15 K cm
[A.5] 5.9 MPa
100. cm 10 m
n RT
p V
− −
−
 
= = = 
 
Since the ratio of CO2 moles to O2 moles is 0.034/0.25, we may scale the oxygen partial pressure
by this ratio to find the partial pressure of CO2 .
( )2 2 2CO O CO
0.034 5.9 MPa 0.80 MPa [1.6] 6.7 MPa
0.25
p p p p
 
= × = = + = 
 
B Energy
Answers to discussion questions
B.2 All objects in motion have the ability to do work during the process of slowing. That is,
they have energy, or, more precisely, the energy possessed by a body because of its motion is its
kinetic energy, Ek . The law of conservation of energy tells us that the kinetic energy of an object
equals the work done on the object in order to change its motion from an initial (i) state of vi = 0
to a final (f) state of vf = v. For an object of mass m travelling at a speed v, 212k [B.8]E m= v .
The potential energy, Ep or more commonly V, of an object is the energy it possesses as a result

Loading page 8...

( )
( )
0 0
Mars
d d
t t t
t g t
t g t
=
= =
=
=
∫ ∫
v
v v
v
of its position. For an object of mass m at an altitude h close to the surface of the Earth, the
gravitational potential energy is
Eqn B.11 assigns the gravitational potential energy at the surface of the Earth, V(0), a value
equal zero and g is called the acceleration of free fall.
The Coulomb potential energy describes the particularly important electrostatic interaction
between two point charges Q1 and Q2 separated by the distance r:
( ) 1 2
0
0
in a vacuum [B.14, is the vacuum permittivity]
4π
Q Q
V r r
ε
ε
=
and
Eqn B.14 assigns the Coulomb potential energy at infinite separation, V(∞), a value equal to
zero. Convention assigns a negative value to the Coulomb potential energy when the interaction
is attractive and a positive value when it is repulsive. The Coulomb potential energy and the
force acting on the charges are related by the expression F = −dV/dr.
B.4 Quantized energies are certain discrete values that are permitted for particles confined to
a region of space.
The quantization of energy is most important—in the sense that the allowed energies are widest
apart—for particles of small mass confined to small regions of space. Consequently, quantization
is very important for electrons in atoms and molecules. Quantization is important for the
electronic states of atoms and molecules and for both the rotational and vibrational states of
molecules.
B.6 The Maxwell distribution of speeds indicates that few molecules have either very low or
very high speeds. Furthermore, the distribution peaks at lower speeds when either the
temperature is low or the molecular mass is high. The distribution peaks at high speeds when
either the temperature is high or the molecular mass is low.
Solutions to exercises
B.1(b) a = d𝑣/dt = g so d𝑣 = g dt . The acceleration of free fall is constant near the surface
of the Mars.
(i)
( ) 2
[B.11] where 9.81 m sV h mgh g −
= =
( ) 1 2
r
r 0
in a medium that has the relative permittivity (formerly, dielectric constant)
4π
Q Q
V r r
ε
ε ε
=
( ) ( ) ( )
( ) ( )
2 1
22 11 1
2 2k
1.0 s 3.72 m s 1.0 s 3.72 m s
0.0010 kg 3.72 m s 6.9 mJE m
− −
−
= × =
= = × =
v
v

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(ii)
B.2(b) The terminal velocity occurs when there is a balance between the force exerted by the
pull of gravity, mg = Vparticleρg = 4 / 3πR3ρg, and the force of frictional drag, 6πηRs. It will be in the
direction of the gravitational pull and have the magnitude sterminal.
3
terminal
2
terminal
4 π 6π
3
2
9
R g Rs
R g
s
ρ η
ρ
η
=
=
B.3(b) The harmonic oscillator solution x(t) = A sin(ωt) has the characteristics that
1/ 2 2
f f
d
( ) cos( ) where ( / ) or
d
x
t A t k m m k
t
ω ω
ω
ω= = = =v
xmin = x(t=nπ/ω, n=0,1,2...) = 0 and xmax = x(t=(n+½)π/ω, n=0,1,2...) = A
At xmin the harmonic oscillator restoration force (Hooke’s law, Fx = –kf x, Brief illustration B.2)
is zero and, consequently, the harmonic potential energy, V, is a minimum that is taken to equal
zero while kinetic energy, Ek , is a maximum. As kinetic energy causes movement away from
xmin , kinetic energy continually converts to potential energy until no kinetic energy remains at
xmax where the restoration force changes the direction of motion and the conversion process
reverses. We may easily find an expression for the total energy E(A) by examination of either
xmin or xmax.
Analysis using xmin :
2 2 2 21 1 1
k k,max max f2 2 2
0E E V E m mA k A
ω= + = + = = =v
We begin the analysis that uses xmax, by deriving the expression for the harmonic potential
energy.
f
( )
f0 0
21
f2
d d [B.10] d
d d
( )
x
V x x
V F x k x x
V k x x
V x k x
= − =
=
=
∫ ∫
Thus, 2 21 1
max max f k max f2 2
( ) and 0V V x k A E E V V k A= = = + = + =
B.4(b) 21
e2 where is the displacement from equilibrium [Brief illustration B.3]w kx x R R= = −
( ) ( )21 12 19 191
2 510 N m 20 10 m 1.02 10 N m 1.02 10 Jw − − − −
= × × = × = ×
B.5(b) 2 1
k 6 4where 2 for C H and 76.03 g mol for the major isotopesE ze z M
φ + −
= ∆ = =
1/ 2
21
A2
2
or where /
ze
m ze m M N
m
φ
φ ∆ 
= ∆ = = 
 
v v
( ) ( ) ( )
( ) ( )
2 1
22 11 1
2 2k
3.0 s 3.72 m s 3.0 s 11.2 m s
0.0010 kg 11.2 m s 63 mJE m
− −
−
= × =
= = × =
v
v

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( ) ( )
( ) ( )
1/ 2 1/2 1/219 3
5 5
1 23 1
1/22 2
5 5 1
2 2 1.6022 10 C 20 10 V C V J
3.2 10 3.2 10
kg kg0.07603 kg mol / 6.022 10 mol
kg m s
3.2 10 3.2 10 m s
kg
−
− −
−
−
 × × × × ×    
 = = × = ×    ×     
 
= × = × 
 
v
( )k 2 20 kV 40 keVE E ze e
φ= = ∆ = × =
B.6(b) The work needed to separate two ions to infinity is identical to the Coulomb potential
drop that occurs when the two ions are brought from an infinite separation, where the interaction
potential equals zero, to a separation of r.
In a vacuum:
( ) ( )
( )
( ) ( )
2 2
1 2
0 0 0 0
219
18
12 1 2 1 12
2 2 4
[B.14]
4π 4π 4π π
1.6022 10 C 3.69 10 J
π 8.85419 10 J C m 250 10 m
e eQ Q e e
w V r r r r
ε
ε
ε ε
−
−
− − − −
× −  
= − = − = − = =  
   
×
= = ×
× × ×
In water:
( ) ( )
( )
( ) ( ) ( )
2 2
1 2
0
219
20
12 1 2 1 12
2 2 [B.15] where 78 for water at 25 C
4π 4π π π
1.6022 10 C 4.73 10 J
π 78 8.85419 10 J C m 250 10 m
r
r
e eQ Q e e
w V r r r r
ε
ε
ε
ε ε ε
−
−
− − − −
× −  
= − = − = − = = = °  
   
×
= = ×
× × × ×
B.7(b) We will model a solution by assuming that the NaCl pair consists of the two point charge
ions Na+ and Cl – . The electric potential will be calculated along the line of the ions.
( )
+
+ +
Na Cl
0 0 0Na Cl Na Cl
1 1
[2.17] [B.16]
4π 4π 4π
ee e
r r r r
φ φ φ
ε
ε
ε
−
− −
 −
= + = + = − 
 
 
When +
Na Cl
r r −= , the electric potential equals zero in this model. Likewise, +
Na Cl
r r −= at every point
both on the line perpendicular to the internuclear line and crossing the internuclear line at the
mid-point so electric potential equals zero at every point on that perpendicular line.

Loading page 11...

( )1 1 1 1
s m
1 mol
/ 28.24 J K mol 1.228 J K g
22.99 g
C C M − − − − 
= = × = 
 
( )1 1 1 1
m s
63.55 g
0.384 J K g 24.4 J K mol
mol
C C M − − − − 
= = × = 
 
B.8(b)
( ) ( ) ( )
( )
( ) ( )
ethanol
3 1
ethanol ethanol
ethanol ethanol
ethanol ethano
energy dissipated by the electric circuit
[B.20]
1.12 A 12.5 V 172 s 2.41 10 C s V s 2.41 kJ
[B.21]
/
U
I t
U nC T
U U
T nC mC M
φ
−
∆ =
= ∆ ∆
= × × = × =
∆ = ∆
∆ ∆
∆ = = ( ) ( ) ( )
3
1 1 1
l
2.41 10 J
150 g 111.5 J K mol / 46.07 g mol
6.64 K = 6.64 C
− − −
×
= ×
= °
B.9(b) 1
50.0 kJ
[B.21] 8.67 K or 8.67 C
5.77 kJ K
U
T C −
∆
∆ = = = °
B.10(b) 1 mol
10.0 g 0.555 mol
18.01 g
n  
= × = 
 
( ) ( ) ( )
m
1 1
[B.21]
0.555 mol 75.2 J K mol 10.0 K 417 J
U C T nC T
− −
∆ = ∆ = ∆
= =
B.11(b)
B.12(b)
B.13(b) ( ) ( )5 1 6 3
1
m m m 3 3
1.00 10 Pa 18.02 g mol 10 m
[B.23] 1.81 J mol
0.997 g cm cm
pM
H U pV
ρ
− −
−
−
× ×  
− = = = = 
 
B.14(b) 2 2H O(l) H O(s)S S>
B.15(b) 2 2H O(l, 100 C) H O(l, 0 C)S S° °>
B.16(b) In a state of static equilibrium there is no net force or torque acting on the system and,
therefore, there is no resultant acceleration. Examples:
When holding an object in a steady position above the ground, there is a balance between the
downward gravitational force pulling on the object downward and the upward force of the hold.
Release the object and it falls.
A movable, but non-moving, piston within a cylinder may be at equilibrium because of equal
pressures on each side of the piston. Increase the pressure on one side of the piston and it moves
away from that side.
In the Bohr atomic model of 1913 there is a balance between the electrostatic attraction of an
electron to the nucleus and the centrifugal force acting on the orbiting electron. Should the

Loading page 12...

electron steadily lose kinetic energy, it spirals into the nucleus.
B.17(b) ( )/ /
e e [B.25a]i j i jkT kTi
j
N
N
ε ε
ε− − −∆
= =
(i)
( ) ( ) ( ) ( ){ }19 1 23 1
2.0 eV 1.602 10 J eV / 1.381 10 J K 200 K 512
1
e 4.2 10
N
N
− − − −
− × × × × −
= = ×
(ii)
( ) ( ) ( ) ( ){ }19 1 23 1
2.0 eV 1.602 10 J eV / 1.381 10 J K 2000 K 62
1
e 9.2 10
N
N
− − − −
− × × × × −
= = ×
B.18(b) ( )upper / 0
lower
lim lim e [B.25a] e 1kT
T T
N
N
ε−∆ −
→∞ →∞
  = = = 
 
In the limit of the infinite temperature both the upper and the lower state are equally occupied.
B.19(b) ( )( )34 9 1 24
upper lower 6.626 10 J s 10.0 10 s 6.63 10 Jh
ε ε ε ν − − −
∆ = − = = × × = ×
( ) ( ) ( ){ }24 23 1
6.63 10 J / 1.381 10 J K 293 Kupper /
lower
e [B.25a] e 0.998klT
N
N
ε − − −
− × × ×−∆
= = =
The ratio Nupper/Nlower indicates that the two states are equally populated. A large fraction of gas-
phase molecules will be in excited rotational states.
B.20(b) Rates of chemical reaction typically increase with increasing temperature because more
molecules have the requisite speed and corresponding kinetic energy to promote excitation and
bond breakage during collisions at the high temperatures.
B.21(b) ( )1/ 2
mean / [B.26]T M∝v
( )
( )
( )
( )
( )
( )
1/ 21/ 2
mean 2 2 2
1/ 2
mean 1 11
1/ 2
mean
mean
/
/
303 K 303 K 1.02
293 K 293 K
T T M T
T TT M
 
= =  
 
 
= = 
 
v
v
v
v
B.22(b) ( )1/ 2
mean / [2.26]T M∝v
( )
( )
( )
( )
( )
( )
1/ 21/ 2
mean 2 2 1
1/ 2
mean 1 21
1/ 21
mean 2
1
mean 2
/
/
H 401.2 g mol 14.11
Hg 2.016 g mol
M T M M
M MT M
−
−
 
= =  
 
 
= = 
 
v
v
v
v
B.23(b) A gaseous helium atom has three translational degrees of freedom (the components of
motion in the x, y, and z directions). Consequently, the equipartition theorem assigns a mean

Loading page 13...

m 3 for water vapourU RT=
m 3 for Pb(s)U RT=
energy of 3 2 kT to each atom. The molar internal energy is
( ) ( )
( )
1 1 13 3 3
m A2 2 2
1
m m
= 8.3145 J mol K 303 K 3.78 kJ mol
1 mol 3.78 kJ
10.0 g 9.45 kJ
4.00 g mol
U N kT RT
U nU mM U
− − −
−
= = =
  
= = = =  
  
B.24(b) A solid state lead atom has three vibrational quadratic degrees of freedom (the
components of vibrational motion in the x, y, and z directions). Its potential energy also has a
quadratic form in each direction because V ∝ (x – xeq )2 . There is a total of six quadratic degrees
of freedom for the atom because the atoms have no translational or rotational motion.
Consequently, the equipartition theorem assigns a mean energy of 6 2 3kT kT= to each atom.
This is the law of Dulong and Petit. The molar internal energy is
( ) ( )
( )
1 1 1
m A
1
m m
3 3 = 3 8.3145 J mol K 293 K 7.31 kJ mol
1 mol 7.31 kJ
10.0 g 0.353 kJ
207.2 g mol
U N kT RT
U nU mM U
− − −
−
= = =
  
= = = =  
  
B.25(b) See exercise B.23(b) for the description of the molar internal energy of helium.
( ) ( )
3
2 1 1 1 1m 3 3
,m 2 2 8.3145 J mol K 12.47 J mol KV
RTU
C R
T T
− − − −
∂∂
= = = = =
∂ ∂
B.26(b)
(i) Water, being a bent molecule, has three quadratic translational and three quadratic
rotational degrees of freedom. So,
( ) ( )1 1 1 1m
,m
3 3 3 8.3145 J mol K 24.94 J mol KV
RTU
C R
T T
− − − −∂∂
= = = = =
∂ ∂
(ii) See exercise B.24(b) for the description of the molar internal energy of lead.
( ) ( )1 1 1 1m
,m
3 3 3 8.3145 J mol K 24.94 J mol KV
RTU
C R
T T
− − − −∂∂
= = = = =
∂ ∂
C Waves
Answers to discussion questions
C.2 The sound of a sudden ‘bang’ is generated by sharply slapping two macroscopic objects
together. This creates a sound wave of displaced air molecules that propagates away from the
collision with intensity, defined to be the power transferred by the wave through a unit area
normal to the direction of propagation. Thus, the SI unit of intensity is the watt per meter squared
( 2
W m− ) and ‘loudness’ increases with increasing intensity. The ‘bang’ creates a shell of
compressed air molecules that propagates away from the source as a shell of higher pressure and

Loading page 14...

density. This longitudinal impulse propagates when gas molecules escape from the high pressure
shell into the adjacent, lower pressure shell. Molecular collisions quickly cause momentum
transfer from the high density to the low density shell and the effective propagation of the high
density shell. The regions over which pressure and density vary during sound propagation are
much wider than the molecular mean free path because sound is immediately dissipated by
molecular collisions in the case for which pressure and density variations are of the order of the
mean free path.
Solutions to exercises
C.1(b) 8 1
8 1
benzene
r
3.00 10 m s
[C.4] 1.97 10 m s
1.52
c
c n
−
−×
= = = ×
C.2(b) 6
1 2
1 1 10 μm
[C.5] 2.78 μm
3600 cm 10 cm
λ ν −
 
= = = 
 
8 1
14 1 14
6
3.00 10 m s
[C.1] 1.08 10 s 1.08 10 Hz
2.78 10 m
c
ν λ
−
−
−
×
= = = × = ×
×
Integrated activities
F.2 The plots of Problem F.1 indicate that as temperature increases the peak of the Maxwell–
Boltzmann distribution shifts to higher speeds with a decrease in the fraction of molecules that
have low speeds and an increase in the fraction that have high speeds. Thus, justifying summary
statements like 'temperature is a measure of the average molecular speed and kinetic energy of
gas molecules', 'temperature is a positive property because molecular speed is a positive
quantity', 'the absolute temperature of 0 K is unobtainable because the area under the plots of
Problem F.1 must equal 1'.

Loading page 15...

1 The properties of gases
1A The perfect gas
Answers to discussion questions
1A.2 The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it
occupied alone the same container as the mixture at the same temperature. Dalton’s law is a
limiting law because it holds exactly only under conditions where the gases have no effect
upon each other. This can only be true in the limit of zero pressure where the molecules of
the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect
gases; for real gases, the law is only an approximation.
Solutions to exercises
1A.1(b) The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be
p = nRT
V
All quantities on the right are given to us except n, which can be computed from the given
mass of Ar.
n = 25 g
39.95 −1
g mol
= 0.626 mol
so p = (0.626 mol) × (8.31 × 10−2 dm 3 bar K−1 mol−1 ) × (30 + 273) K
1.5 dm 3 = 10.5bar
So no, the sample would not exert a pressure of 2.0 bar.
1A.2(b) Boyle’s law [1A.4a] applies.
pV = constant so pfVf = piVi
Solve for the initial pressure:
(i) pi = pfVf
Vi
= (1.97 bar) × (2.14 dm 3 )
(2.14 + 1.80) dm 3 = 1.07 bar
(ii) The original pressure in Torr is
pi = (1.07 bar) × 1 atm
1.013 bar



 × 760 Torr
1 atm



 = 803 Torr
1A.3(b) The relation between pressure and temperature at constant volume can be derived from the
perfect gas law, pV = nRT [1A.5]
so p ∝ T and pi
Ti
= pf
Tf
The final pressure, then, ought to be
pf = piTf
Ti
= (125 kPa) × (11 + 273)K
(23 + 273)K = 120 kPa
1A.4(b) According to the perfect gas law [1.8], one can compute the amount of gas from pressure,
temperature, and volume.
pV = nRT
so n = pV
RT = (1.00 atm) × (1.013 × 105 Pa atm−1 ) × (4.00 × 10 3 m 3 )
(8.3145 J K−1 mol−1 ) × (20 + 273)K = 1.66 × 105 mol
Once this is done, the mass of the gas can be computed from the amount and the molar
mass:
m = (1.66 × 105 mol) × (16.04 −1
g mol ) = 2.67 × 10 6 g = 2.67 × 10 3 kg
1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the
total pressure
1

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Physical Chemistry: Thermodynamics, Structure, and Change Tenth Edition Solution Manual

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