Atkins CHAPTER_24 ( SOLUTIONS MANUAL )

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Part 3: Change
24 Molecules in motion
Solutions to exercises
Discussion questions
E24.1(b) Diffusion is the migration of particles (molecules) down a concentration gradient. Diffusion can be
interpreted at the molecular level as being the result of the random jostling of the molecules in a fluid.
The motion of the molecules is the result of a series of short jumps in random directions, a so-called
random walk.
In the random walk model of diffusion, although a molecule may take many steps in a given time,
it has only a small probability of being found far from its starting point because some of the steps lead
it away from the starting point but others lead it back. As a result, the net distance traveled increases
only as the square root of the time. There is no net flow of molecules unless there is a concentration
gradient in the fluid, alse there are just as many molecules moving in one direction as another. The
rate at which the molecules spread out is proportional to the concentration gradient. The constant of
proportionality is called the diffusion coefficient.
On the molecular level in a gas, thermal conduction occurs because of random molecular motions
in the presence of a temperature gradient. Across any plane in the gas, there is a net flux of energy
from the high temperature side, because molecules coming from that side carry a higher average
energy per molecule across the plane than those coming from the low temperature side. In solids, the
situation is more complex as energy transport occurs through quantized elastic waves (phonons) and,
in metals, also by electrons. Conduction in liquids can occur by all the mechanisms mentioned.
At the molecular (ionic) level, electrical conduction in an electrolytic solution is the net migration
of ions in any given direction. When a gradient in electrical potential exists in a conductivity cell there
will be a greater flow of positive ions in the direction of the negative electrode than in the direction of
the positive electrode, hence there is a net flow of positive charge toward the region of low electrical
potential. Likewise a net flow of negative ions in the direction of the positive electrode will occur. In
metals, only negatively charged electrons contribute to the current.
To see the connection between the flux of momentum and the viscosity, consider a fluid in a
state of Newtonian flow, which can be imagined as occurring by a series of layers moving past one
another (Fig. 24.11 of the text). The layer next to the wall of the vessel is stationary, and the velocity of
successive layers varies linearly with distance, z, from the wall. Molecules ceaselessly move between
the layers and bring with them the x-component of linear momentum they possessed in their original
layer. A layer is retarded by molecules arriving from a more slowly moving layer because they have
a low momentum in the x-direction. A layer is accelerated by molecules arriving from a more rapidly
moving layer. We interpret the net retarding effect as the fluid’s viscosity.
E24.2(b) According to the Grotthuss mechanism, there is an effective motion of a proton that involves the
rearrangement of bonds in a group of water molecules. However, the actual mechanism is still highly
contentious. Attention now focuses on the H9O4+ unit in which the nearly trigonal planar H3O+
ion is linked to three strongly solvating H2O molecules. This cluster of atoms is itself hydrated, but
the hydrogen bonds in the secondary sphere are weaker than in the primary sphere. It is envisaged
that the rate-determining step is the cleavage of one of the weaker hydrogen bonds of this secondary
sphere (Fig. 24.19a of the text). After this bond cleavage has taken place, and the released molecule
has rotated through a few degrees (a process that takes about 1 ps), there is a rapid adjustment of bond
lengths and angles in the remaining cluster, to form an H5O2+ cation of structure H2O · · ·H+ · · ·OH2
(Fig. 24.19b). Shortly after this reorganization has occurred, a new H9O4+ cluster forms as other
molecules rotate into a position where they can become members of a secondary hydration sphere,
386 INSTRUCTOR’S MANUAL
but now the positive charge is located one molecule to the right of its initial location (Fig. 24.19c).
According to this model, there is no coordinated motion of a proton along a chain of molecules,
simply a very rapid hopping between neighbouring sites, with a low activation energy. The model
is consistent with the observation that the molar conductivity of protons increases as the pressure is
raised, for increasing pressure ruptures the hydrogen bonds in water.
E24.3(b) Because the drift speed governs the rate at which charge is transported, we might expect the conductiv-
ity to decrease with increasing solution viscosity and ion size. Experiments confirm these predictions
for bulky ions, but not for small ions. For example, the molar conductivities of the alkali metal ions
increase from Li+ to Cs+ (Table 24.6) even though the ionic radii increase. The paradox is resolved
when we realize that the radius a in the Stokes formula is the hydrodynamic radius (or “Stokes
radius”) of the ion, its effective radius in the solution taking into account all the H2O molecules it
carries in its hydration sphere. Small ions give rise to stronger electric fields than large ones, so small
ions are more extensively solvated than big ions. Thus, an ion of small ionic radius may have a large
hydrodynamic radius because it drags many solvent molecules through the solution as it migrates.
The hydrating H2O molecules are often very labile, however, and NMR and isotope studies have
shown that the exchange between the coordination sphere of the ion and the bulk solvent is very
rapid.
The proton, although it is very small, has a very high molar conductivity (Table 24.6)! Proton and
17O-NMR show that the times characteristic of protons hopping from one molecule to the next are
about 1.5 ps, which is comparable to the time that inelastic neutron scattering shows it takes a water
molecule to reorientate through about 1 rad (1–2 ps).
Numerical exercises
E24.4(b) (a) The mean speed of a gas molecule is
c =
(
8RT
πM
)1/2
so
c(He)
c(Hg)
=
(
M(Hg)
M(He)
)1/2
=
(
200.59
4.003
)1/2
= 7.079
(b) The mean kinetic energy of a gas molecule is 12mc2, where c is the root mean square speed
c =
(
3RT
M
)1/2
So 12 mc
2 is independent of mass, and the ratio of mean kinetic energies of He and Hg is 1
E24.5(b) (a) The mean speed can be calculated from the formula derived in Example 24.1.
c =
(
8RT
πM
)1/2
=
(
8 × (8.314 J K−1mol−1)× (298 K)
π × (28.02 × 10−3 kg mol−1)
)1/2
= 4.75 × 102 m s−1
(b) The mean free path is calculated from λ = kT
21/2σp
[24.14]
with σ = πd2 = π × (3.95 × 10−10 m)2 = 4.90 × 10−19 m2
Then, λ = (1.381 × 10
−23 J K−1)× (298 K)
21/2 × (4.90 × 10−19 m2)× (1 × 10−9 Torr)×
(
1 atm
760 Torr
)
×
(
1.013×105 Pa
1 atm
)
= 4 × 104 m
MOLECULES IN MOTION 387
(c) The collision frequency could be calculated from eqn 31, but is most easily obtained from eqn 32,
since λ and c have already been calculated z = c
λ
= 4.75 × 10
2 m s−1
4.46 × 104 m = 1 × 10
−2 s−1
Thus there are 100 s between collisions, which is a very long time compared to the usual timescale
of molecular events. The mean free path is much larger than the dimensions of the pumping
apparatus used to generate the very low pressure.
E24.6(b) p = kT
21/2σλ
[24.14]
σ = πd2, d =
(σ
π
)1/2 =
(
0.36 nm2
π
)1/2
= 0.34 nm
p = (1.381 × 10
−23 J K−1)× (298 K)
(21/2)× (0.36 × 10−18 m2)× (0.34 × 10−9 m) = 2.4 × 10
7 Pa
This pressure corresponds to about 240 atm, which is comparable to the pressure in a compressed
gas cylinder in which argon gas is normally stored.
E24.7(b) The mean free path is
λ= kT
21/2σp
= (1.381 × 10
−23 J K−1)× (217 K)
21/2[0.43 × (10−9 m)2] × (12.1 × 103 Pa atm−1) = 4.1 × 10
−7 m
E24.8(b) Obtain data from Exercise 24.7(b)
The expression for z obtained in Exercise 24.8(a) is z =
(
16
πmkT
)1/2
σp
Substituting σ = 0.43 nm2, p = 12.1 × 103 Pa, m = (28.02 u), and T = 217 K we obtain
z = 4 × (0.43 × 10
−18 m2)× (12.1 × 103 Pa)
[π × (28.02)× (1.6605 × 10−27 kg)× (1.381 × 10−23 J K−1)× (217 K)]1/2
= 9.9 × 108 s−1
E24.9(b) The mean free path is
λ = kT
21/2σp
= (1.381 × 10
−23 J K−1)× (25 + 273)K
21/2[0.52 × (10−9 m)2]p =
5.50 × 10−3 m Pa
p
(a) λ = 5.50 × 10
−3 m Pa
(15 atm)× (1.013 × 105 Pa atm−1) = 3.7 × 10
−9 m
(b) λ = 5.50 × 10
−3 m Pa
(1.0 bar)× (105 Pa bar−1) = 5.5 × 10
−8 m
(c) λ = 5.50 × 10
−3 m Pa
(1.0 Torr)×
(
1.013×105 Pa atm−1
760 Torr atm−1
) = 4.1 × 10−5 m
388 INSTRUCTOR’S MANUAL
E24.10(b) The fraction F of molecules in the speed range from 200 to 250 m s−1 is
F =
∫ 250 m s−1
200 m s−1
f (v)dv
where f (v) is the Maxwell distribution. This can be approximated by
F ≈ f (v)�v = 4π
(
M
2πRT
)3/2
v2 exp
(
−Mv2
2RT
)
�v,
with f (v) evaluated in the middle of the range
F ≈ 4π
(
44.0 × 10−3 kg mol−1
2π(8.3145 J K−1mol−1)× (300 K)
)3/2
× (225 m s−1)2
× exp
(
−(44.0 × 10−3 kg mol−1)× (225 m s−1)2
2(8.3145 J K−1mol−1)× (300 K)
)
× (50 m s−1),
F ≈ 9.6 × 10−2
Comment. The approximation we have employed, taking f (v) to be nearly constant over a narrow
range of speeds, may not be accurate enough, for that range of speeds includes about 10 per cent of
the molecules.
Numerical exercises
E24.11(b) The number of collisions is
N = ZWAt = pAt
(2πmkT )1/2
= (111 Pa)× (3.5 × 10
−3 m)× (4.0 × 10−3 m)× (10 s)
{2π × (4.00 u)× (1.66 × 10−27 kg u−1)× (1.381 × 10−23 J K−1)× (1500 K)}1/2
= 1.1 × 1021
E24.12(b) The mass of the sample in the effusion cell decreases by the mass of the gas which effuses out of it.
That mass is the molecular mass times the number of molecules that effuse out
�m = mN = mZWAt = mpAt
(2πmkT )1/2
= pAt
( m
2πkT
)1/2 = pAt ( M
2πRT
)1/2
= (0.224 Pa)× π × ( 12 × 3.00 × 10−3 m)2 × (24.00 h)× (3600 s h−1)
×
{
300 × 10−3 kg mol−1
2π × (8.3145 J K−1 mol−1)× (450 K)
}1/2
= 4.89 × 10−4 kg
MOLECULES IN MOTION 389
E24.13(b) The flux is
J = −κ dT
dz
= −1
3
λCV,m〈v〉[X]dTdz
where the minus sign indicates flow toward lower temperature and
λ = 1√
2Nσ
, 〈v〉 =
(
8kT
πm
)1/2
=
(
8RT
πM
)1/2
, and [M] = n/V = N/NA
So J = −2CV,m
3σNA
(
RT
πM
)1/2 dT
dz
= −
(
2 × (28.832 − 8.3145) J K−1 mol−1
3 × [0.27 × (10−9 m)2] × (6.022 × 1023 mol−1)
)
×
(
(8.3145 J K−1 mol−1)× (260 K)
π × (2.016 × 10−3 kg mol−1)
)1/2
× (3.5 K m−1)
= 0.17 J m−2 s−1
E24.14(b) The thermal conductivity is
κ = 1
3
λCV,m〈v〉[X] = 2CV,m3σNA
(
RT
πM
)1/2
so σ = 2CV,m
3κNA
(
RT
πM
)1/2
= (0.240 mJ cm−2 s−1)× (K cm−1)−1 = 0.240 × 10−1 J m−1 s−1 K−1
so σ =
(
2 × (29.125 − 8.3145) J K−1 mol−1
3 × (0.240 × 10−1 J m−1 s−1 K−1)× (6.022 × 1023 mol−1)
)
×
(
(8.3145 J K−1 mol−1)× (298 K)
π × (28.013 × 10−3 kg mol−1)
)1/2
= 1.61 × 10−19 m2
E24.15(b) Assuming the space between sheets is filled with air, the flux is
J = −κ dT
dz
= [(0.241 × 10−3 J cm−2 s−1)× (K cm−1)−1] ×
( [50 − (−10)] K
10.0 cm
)
= 1.45 × 10−3 J cm−2 s−1.
So the rate of energy transfer and energy loss is
JA = (1.45 × 10−3 J cm−2 s−1)× (1.50 m2)× (100 cm m−1)2 = 22 J s−1
390 INSTRUCTOR’S MANUAL
E24.16(b) The time dependence of the pressure of a gas effusing without replenishment is
p = p0e−t/τ where τ ∝
√
m
The time t it takes for the pressure to go from any initial pressure p0 to a prescribed fraction of that
pressurefp0 is
t = τ ln fp0
p0
= τ ln f
so the time is proportional to τ and therefore also to
√
m. Therefore, the ratio of times it takes two
different gases to go from the same initial pressure to the same final pressure is related to their molar
masses as follows
t1
t2
=
(
M1
M2
)1/2
and M2 = M1
(
t2
t1
)2
So Mfluorocarbon = (28.01 g mol−1)×
(
82.3 s
18.5 s
)2
= 554 g mol−1
E24.17(b) The time dependence of the pressure of a gas effusion without replenishment is
p = p0e−t/τ so t = τ lnp0/p
where τ = V
A0
(
2πm
kT
)1/2
= V
A0
(
2πM
RT
)1/2
=
(
22.0 m3
π × (0.50 × 10−3 m)2
)
×
(
2π × (28.0 × 10−3 kg mol−1)
(8.3145 J K−1 mol−1)× (293 K)
)1/2
= 2.4 × 105 s
so t = (8.6 × 105 s) ln 122 kPa
105 kPa
= 1.5 × 104 s
E24.18(b) The coefficient of viscosity is
η = 13λmN〈v〉 =
2
3σ
(
mkT
π
)1/2
so σ = 2
3η
(
mkT
π
)1/2
= 1.66µP = 166 × 10−7 kg m−1 s−1
so σ =
(
2
3 × (166 × 10−7 kg m−1 s−1)
)
×
(
(28.01 × 10−3 kg mol−1)× (1.381 × 10−23 J K−1)× (273 K)
π × (6.022 × 1023 mol−1)
)1/2
= 3.00 × 10−19 m2
E24.19(b) The rate of fluid flow through a tube is described by
dV
dt
= (p
2
in − p2out)πr4
16lηp0
so pin =
(
16lηp0
πr4
dV
dt
+ p2out
)1/2
Several of the parameters need to be converted to MKS units
r = 12 (15 × 10−3 m) = 7.5 × 10−3 m
and
dV
dt
= 8.70 cm3 × (10−2 m cm−1)3 s−1 = 8.70 × 10−6 m3 s−1.
MOLECULES IN MOTION 391
Also, we have the viscosity at 293 K from the table. According to the T 1/2 temperature dependence,
the viscosity at 300 K ought to be
η(300 K) = η(293 K)×
(
300 K
293 K
)1/2
= (176 × 10−7 kg m−1 s−1)×
(
300
293
)1/2
= 1.78 × 10−7 kg m−1 s−1
pin =
{(
16(10.5 m)× (178 × 10−7 kg m−1 s−1)× (1.00 × 105 Pa)
π × (7.5 × 10−3 m)4
)
×(8.70 × 10−6 m3 s−1)+ (1.00 × 105 Pa)2
}1/2
= 1.00 × 105 Pa
Comment. For the exercise as stated the answer is not sensitive to the viscosity. The flow rate is so
low that the inlet pressure would equal the outlet pressure (to the precision of the data) whether the
viscosity were that of N2 at 300 K or 293 K—or even liquid water at 293 K!
E24.20(b) The coefficient of viscosity is
η = 13λmN〈v〉 =
2
3σ
(
mkT
π
)1/2
=
(
2
3[0.88 × (10−9 m)2]
)
×
(
(78.12 × 10−3 kg mol−1)× (1.381 × 10−23 J K−1)T
π × (6.022 × 1023 mol−1)
)1/2
= 5.72 × 10−7 × (T /K)1/2 kg m−1 s−1
(a) At 273 K η = (5.72 × 10−7)× (273)1/2 kg m−1 s−1 = 0.95 × 10−5 kg m−1 s−1
(b) At 298 K η = (5.72 × 10−7)× (298)1/2 kg m−1 s−1 = 0.99 × 10−5 kg m−1 s−1
(c) At 1000 K η = (5.72 × 10−7)× (1000)1/2 kg m−1 s−1 = 1.81 × 10−5 kg m−1 s−1
E24.21(b) The thermal conductivity is
κ = 13λCV,m〈v〉[X] =
2CV,m
3σNA
(
RT
πM
)1/2
(a) κ =
(
2 × [(20.786 − 8.3145) J K−1 mol−1]
3[0.24 × (10−9 m)2] × (6.022 × 1023 mol−1)
)
×
(
(8.3145 J K−1 mol−1)× (300 K)
π(20.18 × 10−3 kg mol−1)
)1/2
= 0.0114 J m−1 s−1 K−1
392 INSTRUCTOR’S MANUAL
The flux is
J = −κ dT
dz
= (0.0114 J m−1 s−1 K−1)×
(
(305 − 295)K
0.15 m
)
= 0.76 J m−2 s−1
so the rate of energy loss is
JA = (0.76 J m−2 s−1)× (0.15 m)2 = 0.017 J s−1
(b) κ =
(
2 × [(29.125 − 8.3145) J K−1 mol−1]
3[0.43 × (10−9 m)2] × (6.022 × 1023 mol−1)
)
×
(
8.3145 J K−1 mol−1)× (300 K)
π(28.013 × 10−3 kg mol−1)
)1/2
= 9.0 × 10−3 J m−1 s−1 K−1
The flux is
J = −κ dT
dz
= (9.0 × 10−3 J m−1 s−1 K−1)×
(
(305 − 295)K
0.15 m
)
= 0.60 J m−2 s−1
so the rate of energy loss is
JA = (0.60 J m−2 s−1)× (0.15 m)2 = 0.014 J s−1
E24.22(b) The rate of fluid flow through a tube is described by
dV
dt
= (p
2
in − p2out)πr4
16lηp0
so the rate is inversely proportional to the viscosity, and the time required for a given volume of
gas to flow through the same tube under identical pressure conditions is directly proportional to the
viscosity
t1
t2
= η1
η2
so η2 =
η1t2
t1
ηCFC =
(208µP)× (18.0 s)
72.0 s
= 52.0µP = 52.0 × 10−7 kg m−1 s−1The coefficient of viscosity is
η = 13λmN〈v〉 =
(
2
3σ
)
×
(
mkT
π
)1/2
=
(
2
3πd2
)
×
(
mkT
π
)1/2
so the molecular diameter is
d =
(
2
3πη
)1/2
×
(
mkT
π
)1/4
=
(
2
3π(52.0 × 10−7 kg m−1 s−1)
)1/2
×
(
(200 × 10−3 kg mol−1)× (1.381 × 10−23 J K−1)× (298 K)
π × (6.022 × 1023 mol−1)
)1/4
= 9.23 × 10−10 m = 923 pm
MOLECULES IN MOTION 393
E24.23(b) κ = 13λCV,m〈v〉[X] =
2CV,m
3σNA
(
RT
πM
)1/2
=
(
2 × (29.125 − 8.3145) J K−1 mol−1
3[0.43 × (10−9 m)2] × (6.022 × 1023 mol−1)
)
×
(
(8.3145 J K−1 mol−1)× (300 K)
π × (28.013 × 10−3 kg mol−1)
)1/2
= 9.0 × 10−3 J m−1 s−1 K−1
E24.24(b) The diffusion constant is
D = 13λ〈v〉 =
2(RT )3/2
3σpNA(πM)1/2
= 2[(8.3145 J K
−1 mol−1)× (298 K)]3/2
3[0.43 × (10−9 m)2]p(6.022 × 1023 mol−1)×
{
π(28.013 × 10−3 kg mol−1)
}1/2
= 1.07 m
2 s−1
p/Pa
The flux due to diffusion is
J = −D d[X]
dx
= −D d
dx
( n
V
)
= −
(
D
RT
)
dp
dx
where the minus sign indicates flow from high pressure to low. So for a pressure gradient of
0.10 atm cm−1
J =
(
D/(m2 s−1)
(8.3145 J K−1 mol−1)× (298 K)
)
× (0.20 × 105 Pa m−1)
= (8.1 mol m−2 s−1)× (D/(m2 s−1))
(a) D = 1.07 m
2 s−1
10.0
= 0.107 m2 s−1
and J = (8.1 mol m−2 s−1)× (0.107) = 0.87 mol m−2 s−1
(b) D = 1.07 m
2 s−1
100 × 103 = 1.07 × 10
−5 m2 s−1
and J = (8.1 mol m−2 s−1)× (1.07 × 10−5) = 8.7 × 10−5 mol m−2 s−1
(c) D = 1.07 m
2 s−1
15.0 × 106 = 7.13 × 10
−8 m2 s−1
and J = (8.1 mol m−2 s−1)× (7.13 × 10−8) = 5.8 × 10−7 mol m−2 s−1
E24.25(b) Molar ionic conductivity is related to mobility by
λ = zuF = (1)× (4.24 × 10−8 m2 s−1V−1)× (96 485 C mol−1)
= 4.09 × 10−3 S m2 mol−1
E24.26(b) The drift speed is given by
s = uE = u�φ
l
= (4.01 × 10
−8 m2 s−1 V−1)× (12.0 V)
1.00 × 10−2 m = 4.81 × 10
−5 m s−1
394 INSTRUCTOR’S MANUAL
E24.27(b) The limiting transport number for Cl− in aqueous NaCl at 25◦ C is
t◦− =
u−
u+ + u− =
7.91
5.19 + 7.91 = 0.604
(The mobilities are in 10−8 m2 s−1 V−1.)
E24.28(b) The limiting molar conductivity of a dissolved salt is the sum of that of its ions, so
)◦m(MgI2) = λ(Mg2+)+ 2λ(I−) = )◦m(Mg(C2H3O2)2)+ 2)◦m(NaI)− 2)◦m(NaC2H3O2)
= (18.78 + 2(12.69)− 2(9.10))mS m2 mol−1 = 25.96 mS m2 mol−1
E24.29(b) Molar ionic conductivity is related to mobility by
λ = zuF so u = λ
zF
F−: u = 5.54 × 10
−3 S m2 mol−1
(1)× (96 485 C mol−1) = 5.74 × 10
−8 m2 V−1 s−1
Cl−: u = 7.635 × 10
−3 S m2 mol−1
(1)× (96 485 C mol−1) = 7.913 × 10
−8 m2 V−1 s−1
Br−: u = 7.81 × 10
−3 S m2 mol−1
(1)× (96 485 C mol−1) = 8.09 × 10
−8 m2 V−1 s−1
E24.30(b) The diffusion constant is related to the mobility by
D = uRT
zF
= (4.24 × 10
−8 m2 s−1 V−1)× (8.3145 J K−1 mol−1)× (298 K)
(1)× (96 485 C mol−1)
= 1.09 × 10−9 m2 s−1
E24.31(b) The mean square displacement for diffusion in one dimension is
〈x2〉 = 2Dt
In fact, this is also the mean square displacement in any direction in two- or three-dimensional
diffusion from a concentrated source. In three dimensions
r2 = x2 + y2 + z2 so 〈r2〉 = 〈x2〉 + 〈y2〉 + 〈z2〉 = 3〈x2〉 = 6Dt
So the time it takes to travel a distance
√
〈r2〉 is
t = 〈r
2〉
6D
= (1.0 × 10
−2 m)2
6(4.05 × 10−9 m2 s−1) = 4.1 × 10
3 s
E24.32(b) The diffusion constant is related to the viscosity of the medium and the size of the diffusing molecule
as follows
D = kT
6πηa
so a = kT
6πηD
= (1.381 × 10
−23 J K−1)× (298 K)
6π(1.00 × 10−3 kg m−1 s−1)× (1.055 × 10−9 m2 s−1)
a = 2.07 × 10−10 m = 207 pm
MOLECULES IN MOTION 395
E24.33(b) The Einstein–Smoluchowski equation related the diffusion constant to the unit jump distance and
time
D = λ
2
2τ
so τ = λ
2
2D
If the jump distance is about one molecular diameter, or two effective molecular radii, then the jump
distance can be obtained by use of the Stokes–Einstein equation
D = kT
6πηa
= kT
3πηλ
so λ = kT
3πηD
and τ = (kT )
2
18(πη)2D3
= [(1.381 × 10
−23 J K−1)× (298 K)]2
18[π(0.387 × 10−3 kg m−1 s−1)]2 × (3.17 × 10−9 m2 s−1)3
= 2.00 × 10−11 s = 20 ps
E24.34(b) The mean square displacement is (from Exercise 24.31(b))
〈r2〉 = 6Dt so t = 〈r
2〉
6D
= (1.0 × 10
−6 m)2
6(1.0 × 10−11 m2 s−1) = 1.7 × 10
−2 s
Solutions to problems
Solutions to numerical problems
P24.3 〈X〉 = 1
N
∑
i
NiXi [See Problem 24.2]
(a) 〈h〉 = 1
53
{1.80 m + 2 × (1.82 m)+ · · · + 1.98 m} 1.89 m
(b) 〈h2〉 = 1
53
{
(1.80 m)2 + 2 × (1.82 m)2 + · · · + (1.98 m)2
}
= 3.57 m2√
〈h2〉 = 1.89 m
P24.4 κ = 13λcCV,m[A] [24.28]
c =
(
8kT
πm
)1/2
[24.7] ∝ T 1/2
Hence, κ ∝ T 1/2CV,m, so κ
′
κ
=
(
T ′
T
)1/2
×
(
C′V,m
CV,m
)
At 300 K, CV,m ≈ 32R + R = 52R At 10 K, CV,m ≈ 32R [rotation not excited]
Therefore,
κ ′
κ
=
(
300
10
)1/2
×
(
5
3
)
= 9.1
P24.7 The atomic current is the number of atoms emerging from the slit per second, which is ZWA with
A = 1 × 10−7 m2. We use
ZW = p
(2πmkT )1/2
[24.15]
= p/Pa
[(2π)× (M/g mol−1)× (1.6605 × 10−27 kg)× (1.381 × 10−23 J K−1)× (380 K)]1/2
= (1.35 × 1023 m−2 s−1)×
(
p/Pa
(M/g mol−1)1/2
)
396 INSTRUCTOR’S MANUAL
(a) Cadmium:
ZWA = (1.35 × 1023 m−2 s−1)× (1 × 10−7 m2)×
(
0.13
(112.4)1/2
)
= 2 × 1014 s−1
(b) Mercury:
ZWA = (1.35 × 1023 m−2 s−1)× (1 × 10−7 m2)×
(
152
(200.6)1/2
)
= 1 × 1017 s−1
P24.10 c = κ
)m
[24.98] ≈ κ
)◦m
[c small, conductivity of water allowed for in the data]
c ≈ 1.887 × 10
−6 S cm−1
138.3 S cm2 mol−1
[Exercise 24.28(a)]
≈ 1.36 × 10−8 mol cm−3 = solubilit y = 1.36 × 10−5 M
P24.12 t (H+) = u(H
+)
u(H+)+ u(Cl−) [24.61] =
3.623
3.623 + 0.791 = 0.82
When a third ion is present we use
t (H+) = I (H
+)
I (H+)+ I (Na+)+ I (Cl−) [24.58]
For each I , I = zuνcFAE = constant × cu. Hence, when NaCl is added
t (H+) = c(H
+)u(H+)
c(H+)u(H+)+ c(Na+)u(Na+)+ c(Cl−)u(Cl−)
= (1.0 × 10
−3)× (3.623)
(1.0 × 10−3)× (3.623)+ (1.0)× (0.519)+ (1.001)× (0.791) = 0.0028
P24.14 t+ =
(
zcAF
I
)
×
( x
�t
)
[Problem 24.13]
The density of the solution is 0.682 g cm−3; the concentration c is related to the molality m by
c/(mol L−1) = ρ/(kg L−1)×m/(mol kg−1)
which holds for dilute solutions such as these.
A = πr2 = π × (2.073 × 10−3 m)2 = 1.350 × 10−5 m2
czAF
I�t
= (1.350 × 10
−5 m2)× (9.6485 × 104 C mol−1)
(5.000 × 10−3 A)× (2500 s) × c = (0.1042 m
2 mol−1)× c
= (0.1042 m2 mol−1)× ρ ×m = (0.1042 m2 mol−1)× (682 kg m−3)×m
= (71.06 kg m−1 mol−1)×m = (0.07106 kg mm−1 mol−1)×m
and so t+ = (0.07106 kg mm−1 mol−1)× x ×m
In the first solution t+ = (0.07106 kg mm−1 mol−1) × (286.9 mm) × (0.01365 mol kg−1) =
0.278
In the second solution t+ = (0.07106 kg mm−1 mol−1) × (92.03 mm) × (0.04255 mol kg−1) =
0.278
MOLECULES IN MOTION 397
Therefore, t (H+) = 0.28, a value much less than in pure water where t (H+) = 0.63. Hence, the
mobility is much less relative to its counterion, NH−2 .
P24.17 If diffusion is analogous to viscosity [Section 24.5, eqn 24.36] in that it is also an activation energy
controlled process, then we expect
D ∝ e−Ea/RT
Therefore, if the diffusion constant is D at T and D′ at T ′,
Ea = −
R ln
(
D′
D
)
(
1
T ′ − 1T
) = − (8.314 J K−1 mol−1)× ln
(
2.89
2.05
)
1
298 K − 1273 K
= 9.3 kJ mol−1
That is, the activation energy for diffusion is 9.3 kJ mol−1
P24.19 〈x2〉 = 2Dt [24.91], D = kT
6πaη
[24.83]
Hence, η = kT
6πDa
= kT t
3πa〈x2〉 =
1.381 × 10−23 J K−1)× (298.15 K)× t
(3π)× (2.12 × 10−7 m)× 〈x2〉
= (2.06 × 10−15 J m−1)×
(
t
〈x2〉
)
and therefore η/(kg m−1s−1) = 2.06 × 10
−11(t/s)
(〈x2〉/cm2)
We draw up the following table
t /s 30 60 90 120
108〈x2〉/cm2 88.2 113.4 128 144
103η/(kg m−1 s−1) 0.701 1.09 1.45 1.72
Hence, the mean value is 1.2 × 10−3kgm−1 s−1 .
P24.21 The viscosity of a perfect gas is
η = 13Nmλc =
mc
3σ
√
2
= 2
3σ
(
mkT
π
)1/2
so σ = 2
3η
(
mkT
π
)1/2
The mass is
m = 17.03 × 10
−3 kg mol−1
6.022 × 1023 mol−1 = 2.828 × 10
−26 kg
(a) σ = 2
3(9.08 × 10−6 kg m−1 s−1)
×
(
(2.828 × 10−26 kg)× (1.381 × 10−23 J K−1)× (270 K)
π
)1/2
= 4.25 × 10−19 m2 = πd2 so d =
(
4.25 × 10−19 m2
π
)1/2
= 3.68 × 10−10 m
398 INSTRUCTOR’S MANUAL
(b) σ = 2
3(17.49 × 10−6 kg m−1 s−1)
×
(
(2.828 × 10−26 kg)× (1.381 × 10−23 J K−1)× (490 K)
π
)1/2
= 2.97 × 10−19 m2 = πd2 so d =
(
2.97 × 10−19 m2
π
)1/2
= 3.07 × 10−10 m
Comment. The change in diameter with temperature can be interpreted in two ways. First, it shows
the approximate nature of the concept of molecular diameter, with different values resulting from
measurements of different quantities. Second, it is consistent with the idea that, at higher temperatures,
more forceful collisions contract a molecule’s perimeter.
P24.22 The diffusion constant of an ion in solution is related to the mobility of the ion and to its radius in
separate relations
D = uRT
zF
= kT
6πηa
so a = zFk
6πηuR
= ze
6πηu
a = (1)× (1.602 × 10
−19 C)
6π(0.93 × 10−3 kg m−1 s−1)× (1.1 × 10−8 m2 V−1 s−1) = 8.3 × 10
−10 m = 830 pm
Solutions to theoretical problems
P24.25 Write the mean velocity initially as a; then in the emerging beam 〈vx〉 = K
∫ a
0
vxf (vx) dvx where
K is a constant which ensures that the distribution in the emergent beam is also normalized. That is,
1 = K
∫ a
0
f (vx) dvx = K
( m
2πkT
)1/2 ∫ a
0
e−mv
2
x/2kT dvx
This integral cannot be evaluated analytically but it can be related to the error function by defining
x2 = mv
2
x
2kT
which gives dvx =
(
2kT
m
)1/2
dx. Then
1 = K
( m
2πkT
)1/2 (2kT
m
)1/2 ∫ b
0
e−x
2
dx [b = (m/2kT )1/2 × a]
= K
π1/2
∫ b
0
e−x
2
dx = 12 Kerf(b)
where erf (z) is the error function [Table 12.2]: erf(z) = 2
π1/2
∫ z
0
e−x
2
dx
Therefore, K = 2
erf(b)
The mean velocity of the emerging beam is
〈vx〉 = K
( m
2πkT
)1/2 ∫ a
0
vxe
−mv2x/2kT dvx = K
( m
2πkT
)1/2 (−kT
m
)∫ a
0
d
dvx
(e−mv
2
x/2kT dvx
MOLECULES IN MOTION 399
= −K
(
kT
2mπ
)1/2
(e−ma
2/2kT − 1)
Now use a = 〈vx〉initial =
(
2kT
mπ
)1/2
This expression for the average magnitude of the one-dimensional velocity in the x direction may be
obtained from
〈vx〉 = 2
∫ ∞
0
vxf (vx)dvx = 2
∫ ∞
0
vx
( m
2πkT
)1/2
e−mv
2
x/2kT dvx
=
( m
2πkT
)1/2 (2kT
m
)
=
(
2kT
mπ
)1/2
It may also be obtained very quickly by setting a = ∞ in the expression for 〈vx〉 in the emergent
beam with erf(b) = erf(∞) = 1.
Substituting a =
(
2kT
mπ
)1/2
into 〈vx〉 in the emergent beam e−ma2/2kT = e−1/π and erf(b) =
erf
(
1
π1/2
)
Therefore, 〈vx〉 =
(
2kT
mπ
)1/2
× 1 − e
−1/π
erf
(
1
π1/2
)
From tables of the error function (expanded version of Table 12.2), or from readily available software,
or by interpolating Table 12.2.
erf
(
1
π1/2
)
= erf(0.56) = 0.57 and e−1/π = 0.73
Therefore, 〈vx〉 = 0.47〈vx〉initial
P24.27 The most probable speed, c∗, was evaluated in Problem 24.23 and is
c∗ = v(most probable) =
(
2kT
m
)1/2
Consider a range of speeds �v around c∗ and nc∗, then with v = c∗
f (nc∗)
f (c∗)
= (nc
∗)2e−mn2c∗2/2kT
c∗2e−mc∗2/2kT
[24.4] = n2e−(n2−1)mc∗2/2kT = n2e(1−n2)
Therefore,
f (3c∗)
f (c∗)
= 9 × e−8 = 3.02 × 10−3 f (4c
∗)
f (c∗)
= 16 × e−15 = 4.9 × 10−6
P24.28 The current Ij carried by an ion j is proportional to its concentration cj , mobility uj , and charge
number |zj |. [Justification 24.9] Therefore
Ij = Acjuj zj
where A is a constant. The total current passing through a solution is
I =
∑
j
Ij = A
∑
j
cjuj zj
400 INSTRUCTOR’S MANUAL
The transport number of the ion j is therefore
tj =
Ij
I
= Acjuj zj
A
∑
j cjuj zj
= cjuj zj∑
j cjuj zj
If there are two cations in the mixture
t ′
t ′′
= c
′u′z′
c′′u′′z′′
= c
′u′
c′′u′′
if z′ = z′′
P24.29
∂c
∂t
= D ∂
2c
∂x2
[24.84] with c = n0e
−x2/4Dt
A(πDt)1/2
[24.88]
or c = a
t1/2
e−bx
2/t
then
∂c
∂t
= −
(
1
2
)
×
(
a
t3/2
)
e−bx
2/t +
(
a
t1/2
)
×
(
bx2
t2
)
e−bx
2/t = − c
2t
+ bx
2
t2
c
∂c
∂x
=
(
a
t1/2
)
×
(−2bx
t
)
e−bx
2/t
∂2c
∂x2
= −
(
2b
t
)
×
(
a
t1/2
)
e−bx
2/t +
(
a
t1/2
)
×
(
2bx
t
)2
e−bx
2/t = −
(
2b
t
)
c +
(
2bx
t
)2
c
= −
(
1
2Dt
)
c +
(
bx2
Dt2
)
c
= 1
D
∂c
∂t
as required
Initially the material is concentrated at x = 0. Note that c = 0 for x > 0 when t = 0 on
account of the very strong exponential factor (e−bx2/t → 0 more strongly than 1
t1/2
→ ∞). When
x = 0, e−x2/4Dt = 1. We confirm the correct behaviour by noting that 〈x〉 = 0 and 〈x2〉 = 0 at
t = 0 [24.90], and so all the material must be at x = 0 at t = 0.
P24.31 Draw up the following table based on the third and last equations of Justification 24.12
N 4 6 8 10 20
P(6λ)Exact 0 0.016 0.0313 0.0439 0.0739
P(6λ)Approx. 0.004 0.162 0.0297 0.0417 0.0725
N 30 40 60 100
P(6λ)Exact 0.0806 0.0807 0.0763 0.0666
P(6λ)Approx. 0.0799 0.0804 0.0763 0.0666
The points are plotted in Fig. 24.1.
The discrepancy is less than 0.1 per cent when N > 60
MOLECULES IN MOTION 401
0.10
0.05
0
100806040200
Figure 24.1
Solutions to applications
P24.33 The work required for a mass, m, to go from a distance r from the centre of a planet of mass m′ to
infinity is
w =
∫ ∞
r
F dr
where F is the force of gravity and is given by Newton’s law of universal gravitation, which is
F = Gmm
′
r2
G is the gravitational constant (not to be confused with g). Then
w′ =
∫ ∞
r
Gmm′
r2
dr = Gmm
′
r
Since according to Newton’s second law of motion, F = mg, we may make the identification
g = Gm
′
r2
Thus, w = grm. This is the kinetic energy that the particle must have in order to escape the planet’s
gravitational attraction at a distance r from the planet’s centre; hence w = 12mv2 = mgr
ve = (2g Rp)1/2 [Rp = radius of planet]
which is the escape velocity.
(a) ve = [(2)× (9.81 m s−2)× (6.37 × 106 m)]1/2 = 11.2 km s−1
(b) g(Mars) = m(Mars)
m(Earth)
× R(Earth)
2
R(Mars)2
× g(Earth) = (0.108)×
(
6.37
3.38
)2
× (9.81 m s−2)
= 3.76 m s−2
Hence, ve = [(2)× (3.76 m s−2)× (3.38 × 106 m)]1/2 = 5.0 km s−1
402 INSTRUCTOR’S MANUAL
Since c =
(
8RT
πM
)1/2
, T = πMc
2
8R
and we can draw up the following table
10−3 T/K H2 He O2
Earth 11.9 23.7 190 [c = 11.2 km s−1]
Mars 2.4 4.8 38 [c = 5.0 km s−1]
In order to calculate the proportion of molecules that have speeds exceeding the escape velocity,
ve, we must integrate the Maxwell distribution [24.4] from ve to infinity.
P =
∫ ∞
ve
f (v)dv =
∫ ∞
ve
4π
( m
2πkT
)3/2
v2e−mv
2/2kT dv
[
M
R
= m
k
]
This integral cannot be evaluated analytically and must be expressed in terms of the error function.
We proceed as follows.
Defining β = m
2kT
and y2 = βv2 gives v = β−1/2y, v2 = β−1y2, ve = β−1/2ye,
ye = β1/2ve, and dv = β−1/2dy
P = 4π
(
β
π
)3/2
β−1β−1/2
∫ ∞
β1/2ve
y2e−y
2
dy = 4
π1/2
∫ ∞
β1/2ve
y2e−y
2
dy
= 4
π1/2
[∫ ∞
0
y2e−y
2
dy −
∫ β1/2ve
0
y2e−y
2
dy
]
The first integral can be evaluated analytically; the second cannot.∫ ∞
0
y2e−y
2
dy = π
1/2
4
, hence
P = 1 − 2
π1/2
∫ β1/2ve
0
ye−y
2
(2y dy) = 1 − 2
π1/2
∫ β1/2ve
0
y d(−e−y2)
This integral may be evaluated by parts
P = 1 − 2
π1/2
[
y(−e−y2)∣∣∣∣β
1/2ve
0
−
∫ β1/2ve
0
(−e−y2) dy
]
P = 1 + 2
(
β
π
)1/2
vee
−βv2e − 2
π1/2
∫ β1/2ve
0
e−y
2
dy = 1 + 2
(
β
π
)1/2
vee
−βv2e − erf(β1/2ve)
= erfc(β1/2ve)+ 2
(
β
π
)1/2
vee
−βv2e [erfc(z) = 1 − erf(z)]
From β = m
2kT
= M
2RT
and ve = (2gRp)1/2
β1/2ve =
(
MgRp
RT
)1/2
MOLECULES IN MOTION 403
For H2 on Earth at 240 K
β1/2ve =
(
(0.002016 kg mol−1)× (9.807 m s−2)× (6.37 × 106 m)
(8.314 J K−1 mol−1)× (240 K)
)1/2
= 7.94
P = erfc(7.94)+ 2
(
7.94
π1/2
)
e−(7.94)
2 = (2.9 × 10−29)+ (3.7 × 10−27) = 3.7 × 10−27
at 1500 K
β1/2ve =
(
(0.002016 kg mol−1)× (9.807 m s−2)× (6.37 × 106 m)
(8.314 J K−1 mol−1)× (1500 K)
)1/2
= 3.18
P = erfc(3.18)+ 2
(
3.18
π1/2
)
e−(3.18)
2 = (6.9 × 10−6)+ (1.46 × 10−4) = 1.5 × 10−4
For H2 on Mars at 240 K
β1/2ve =
(
(0.002016 kg mol−1)× (3.76 m s−2)× (3.38 × 106 m)
(8.314 J K−1 mol−1)× (240 K)
)1/2
= 3.58
P = erfc(3.58)+ 2
(
3.58
π1/2
)
e−(3.58)
2 = (4.13 × 10−7)+ (1.10 × 10−5) = 1.1 × 10−5
at 1500 K, β1/2ve = 1.43
P = erfc(1.43)+ (1.128)× (1.43)× e−(1.43)2 = 0.0431 + 0.209 = 0.25
For He on Earth at 240 K
β1/2ve =
(
(0.004003 kg mol−1)× (9.807 m s−2)× (6.37 × 106 m)
(8.314 J K−1 mol−1)× (240 K)
)1/2
= 11.19
P = erfc(11.2)+ (1.128)× (11.2)× e−(11.2)2 = 0 + (4 × 10−54) = 4 × 10−54
at 1500 K, β1/2ve = 4.48
P = erfc(4.48)+ (1.128)× (4.48)× e−(4.48)2 = (2.36 × 10−10)+ (9.71 × 10−9)
= 1.0 × 10−8
For He on Mars at 240 K
β1/2ve =
(
(0.004003 kg mol−1)× (3.76 m s−2)× (3.38 × 106 m)
(8.314 J K−1 mol−1)× (240 K)
)1/2
= 5.05
P = erfc(5.05)+ (1.128)× (5.05)× e−(5.05)2 = (9.21 × 10−13)+ (4.79 × 10−11)
= 4.9 × 10−11
at 1500 K, β1/2ve = 2.02
P = erfc(2.02)+ (1.128)× (2.02)× e−(2.02)2 = (4.28 × 10−3)+ (0.0401) = 0.044
For O2 on Earth it is clear that P ≈ 0 at both temperatures.
404 INSTRUCTOR’S MANUAL
For O2 on Mars at 240 K, β1/2ve = 14.3
P = erfc(14.3)+ (1.128)× (14.3)× e−(14.3)2 = 0 + (2.5 × 10−88) = 2.5 × 10−88 ≈ 0
at 1500 K, β1/2ve = 5.71
P = erfc(5.71)+ (1.128)× (5.71)× e−(5.71)2 = (6.7 × 10−6)+ (4.46 × 10−14)
= 4.5 × 10−14
Based on these numbers alone, it would appear that H2 and He would be depleted from the
atmosphere of both Earth and Mars only after many (millions?) years; that the rate on Mars, though
still slow, would be many orders of magnitude larger than on Earth; that O2 would be retained on
Earth indefinitely; and that the rate of O2 depletion on Mars would be very slow (billions of years?),
though not totally negligible. The temperatures of both planets may have been higher in past times
than they are now.
In the analysis of the data, we must remember that the proportions, P , are not rates of depletion,
though the rates should be roughly proportional to P .
The results of the calculations are summarized in the following table
240 K 1500 K
H2 He O2 H2 He O2
P (Earth) 3.7 × 10−27 4 × 10−54 0 1.5 × 10−4 1.0 × 10−8 0
P (Mars) 1.1 × 10−5 4.9 × 10−11 0 0.25 0.044 4.5 × 10−14