Chap 17 Kinetic Theory of Gases
Table of Contents 目錄內容
17.1 Probability Density for Molecular Speeds of Gas Molecules
17.2 Velocity Distribution in One Direction
17.3 Maxwell Distribution of Speeds
17.4 Types of Average Speeds
17.5 Pressure of an Ideal Gas
17.6 Collisions with a Surface and Effusion
17.7 Collisions of Hard-Sphere Molecules
17 8 Mean
17.8
M
F
Free P
Path
th
17.9 Effects of Molecular Interactions on Collisions
17.10 Transport Phenomena in Gases
17.11 Calculation of Transport Coefficients
1
氣體動力論
• 描述理想氣體分子最簡單的運動模式可以分子的隨機運動來表示.
• 氣體動力論:僅限單原子氣體.利用分子運動模型導出氣體的速度分佈。
說明氣體分子平均動能與溫度之間的關係‧
• 本章所討論的方法,可用於各種粒子在所有的流體中運動型態的一般
技巧.我們將專注於物質的傳輸性質(transport properties).
• 傳輸性質是指傳輸物質,能量,或其他性質的能力,有四個例子
– 擴散 (diffusion),
– 熱傳導(Thermal conduction)
– 電導(Electric conduction)
– 黏度(Viscosity)
• 另外包含逸散現象(Effusion)的討論.
2
kinetic model of a perfect gas
• 理想氣體動力學微觀模型的四個假設
1) 分子本身佔有的空間體積可忽略不計，因為在一般情形分子的線度
比分子之間的平均距離小得多。
2) 分子在不停地隨機運動著，分子之間及分子與容器壁之間不斷地進
行著彈性碰撞。
3) 除了碰撞的瞬間外，分子之間、分子與容器壁之間均無相互作用。
系統的能量只包括分子運動的動能。
4) 分子運動遵從古典力學規律
此模型可推出常體積下的壓力表示式: pV = (1/3)nMc2 = (1/3)Nmc2,
其中M = mNA 表分子量(molar mass), c 表分子均平方根速度
(root mean square speed )。
3
Chap 17 Kinetic Theory of Gases
• 氣體分子的位置向量 : r = ix + jy, + kz
• 位置向量對時間的偏微分(υX = dx/ dt ), 即為粒子的速率向量
(velocity)
υ = i υX + j υy + k υZ
• 速率向量的大小即為速度υ(speed), υ是純量 : 依 Pythagorean 定理
υ = |υ| = (υX 2 + υy 2 + υz 2 )1/2
4
Probability Density for Molecular Speeds of Gas Molecules
• Figure 17.1 Velocity Vector of a
particle in velocity space. The
length s of the vector υ that
represents the speed and
direction of a particle can be
calculated from the components
υX , υy and υZ by use of the
Pythagorean theorem, given in
equation 17.3:
υ = |υ| = (υX 2 +υX 2 +υX 2 ) 1/2
5
Velocity Distribution Function in Three-Dimension
1
υ
Velocity Space
To evaluate the probability
that a molecule has a speed
i the
in
h range υ to υ + dυ
d , we
evaluate the total probability
that the molecule will have a
speed that is anywhere on
the surface of a sphere of
radius υ=(υX2 +υy2 +υz2)1/2
by summing the probabilities
that it is in a volume element
dυX dυy dυz at a distance υ
from the origin.
6
Probability Density for Molecular Speeds of Gas Molecules
• Figure 17.2 Point
representing the heads of
velocity vectors for the
plane. Note
molecules in a p
that very few molecules
have very high speeds, that
is large absolute values of
the velocity vector, that the
distribution is isotropic,
that is, the same in each
direction.
7
Velocity Distribution in One Direction
• Figure 17.3 The volume
element in velocity space is
infinitesimal in size, and it
has the density of points at
the end of a specific
velocity vector υ.
8
Probability Density for Velocity in 3D
• Probability density for velocity in 3D
– f (υX,υy,υZ) dυXdυydυZ : probability of finding a molecule
with a velocity in the range υX to υX+ dυX, υy to υy+dυy and
υZ to υZ+dυZ.
– f (υX,υy,υZ) : probability density function, the probability
per unit volume at a point of velocity space.
f (υx ,υy ,υz )dυx dυy dυz = f (υx )dυx ⋅ f (υy )dυy ⋅ f (υz )dυz
∞ ∞ ∞
∫ ∫ ∫ f (υx ,υy ,υz )dυx dυy dυz = 1
-∞-∞-∞
∞
∫- f (υ) dυ =1
with f (υ) = f (υx ,υ ,υ ) = f (υx ) ⋅f (υy )⋅f (υ )
or
∞
y
z
z
9
Boltzmann Distribution
• Ludwig Boltzmann developed a very general idea about how
energy was distributed among systems consisting of many
particles. He said that the number of particles with energy E
( E/kT) , where
would
ld b
be proportionall to the
h value
l e(-E/kT)
h
¾ T is the absolute temperature and
¾ k is a universal constant known as Boltzmann's constant
(1.38x10-23 J K-1)
• For any state s of the system, the probability of occurrence
of that state is P (s ) :
exp (− ε (s )/ kT )
P (s ) =
∑ exp (− ε i (s )/ kT
)
i
10
Boltzmann Distribution
• The fraction of molecules with velocity components υX,υy,υZ
is proportional to an exponential function of their kinetic
energy E = ½ mυX + ½ mυy + ½ mυZ = εX + εy + εz
F = K ⋅e
= K ⋅e
= Ke
⎛ E ⎞
⎜−
⎟
⎝ kT ⎠
(
)
− mυ X2 + mυy2 + mυ Z2 2 kT
−mυ X2
2 kT
e
−mυy2 2 kT
e
−mυ Z2 2 kT
11
Boltzmann Distribution
• The kinetic energy in one dimension as :
EX
1
2
= m υX
2
• The velocity
y distribution function in one direction ((1D)) :
⎛ m υ 2X ⎞
⎟⎟
f (υ X ) = const ⋅ exp ⎜⎜ −
⎝ 2kT ⎠
• According to Boltzmann law :
exp (− m υ 2X / 2 kT )
f (υ X ) =
2
(
exp
−
m
υ
∑
X (i ) / 2 kT
)
i
• The fraction of molecules between the velocity limits dυX :
f (υ X ) d υ X =
exp (− m υ / 2 kT ) d υ
∫ exp (− m υ / 2 kT ) d υ
∞
−∞
2
X
X
2
X
X
12
Velocity Distribution in One Direction
4 Probability of Speed
To calculate the probability
that a molecule will have a
speed in the range υ1 to υ2,
we integrate the distribution
between those two limits; the
integral is equal to the area of
the curve between the limits,
as shown shaded here.
13
The Integration Constant of Distribution Function
• In one-dimensional
⎛ mυ X2 ⎞
⎟⎟
as ∫ f (υx ) d υx = 1 ; f (υ X ) = const ⋅ exp⎜⎜ −
-∞
⎝ 2kT ⎠
1/ 2
1
⎛ m
⎞
=⎜
const = ∞
⎟
⎝ 2 π kT ⎠
exp (− m υ X2 / 2kT ) d υ X
∞
∫
−∞
1/ 2
⎛ m ⎞
f (υ x ) dυ X = ⎜
⎟
⎝ 2 π kT ⎠
⎛ mυ X
exp ⎜⎜ −
⎝ 2kT
2
⎞
⎟⎟dυ X
⎠
14
Standard Integral Formula
• Standard Integral Formula:
∫
∞
x
0
∫
∞
0
e
2 n +1
2n − ax2
x e
∫
∞
∫
∞
x
e
x e
2n
2
n!
dx =
n +1
2a
(a > 0, n : a
1⋅ 3 ⋅ 5K(2n − 1) π
dx =
2n+1
a2n +1
2n +1 −ax 2
−∞
−∞
−ax
−ax 2
(a > 0;
positive integer )
n : positive integer )
dx = 0
∞
dx = 2 ∫ x e
0
2n
−ax 2
dx
15
Table 17.1
Table 17.1 Definite Integrals Occurring in the Kinetic Theory of Gases
n
Integral
0
1
1/2
1⎛ π ⎞
n
2
∫0 x exp(-ax ) dx 2 ⎜⎝ a ⎟⎠
∞
∞
∫x
-∞
n
exp(-ax
2
) dx
1
2a
3
1/2
1⎛ π ⎞
⎜ 3⎟
4⎝a ⎠
1
2
2a
1/2
1/2
⎛π ⎞
⎜ ⎟
⎝a⎠
2
0
1⎛ π ⎞
⎜ 3⎟
2⎝ a ⎠
4
5
1/2
3⎛ π ⎞
⎜ 5⎟
8⎝a ⎠
1
a2
1/2
0
3⎛ π ⎞
⎜ 5⎟
4⎝a ⎠
0
16
Probability Density for Molecular Speeds of Gas Molecules
Example 17.1 Probability density in a specific direction
Calculate the probability density for υX of O2 molecules at 300 K at 0,
300, and 600 m s-1.
1/ 2
2
⎛
⎞
M
M
υ
⎛
⎞
X
⎟
⎟ exp ⎜⎜ −
Ans: At 300 m s-1, f (υ X ) = ⎜
⎟
2
RT
⎝ 2π RT ⎠
⎝
⎠
= [0.032 kg mol-1/2π(8.3145 J K-1 mol-1)(300 K)]1/2 x
exp[-(0.032 kg mol-1)(300 m s-1) / 2(8.3145 J K-1 mol-1)(300 K)]
= 8.022x10-4 s m-1
The p
probability
y densities at 0 and 600 m s-1 are 1.429x10-3 s m-1
and 1.419x10-4 s m-1, in agreement with Fig. 17.4.
17
Velocity Distribution in One Direction
• Figure 17.4 Probability density for the velocity υX of oxygen
molecules in x direction at 100, 300, 500 and 1000 K.
18
Average Kinetic Energy in One Direction
• the one dimensional mean velocity <υX >
∞
υ X = ∫ υ X f (υ X )dυ X = 0
−∞
• the average kinetic energy in the x direction <εX >
εX
1
1 ⎛ kT ⎞ 1
2
= m υX = m ⎜
⎟ = kT
2
2 ⎝m ⎠ 2
19
Probability Density for Molecular Speeds of Gas Molecules
Example 17.2 Average square of the velocity in a specific direction
Using the distribution function for velocities in the x direction show
that : υ 2 = kT
X
m
Ans: The average is the integral of υX2 multiplied by the distribution
function:
1/ 2
∞
⎛ mυ X ⎞
⎛ m ⎞ ∞ 2
2
2
υ X = ∫ υ X f (υ X )dυ X = ⎜
υ X exp ⎜ dυ X
⎟
⎟
∫
-∞
⎝ 2kT ⎠
⎝ 2πkT ⎠ -∞
Using the value of the definite integral given in Table 17.1,
υX
2
1/ 2
⎛ m ⎞
=⎜
⎟
⎝ 2 π kT ⎠
π1/ 2
2(m/ 2kT
)
3/ 2
kT
=
m
20
Velocity Distribution Function in 3D
• The joint probability in 3D distribution function is:
f (υx ,υy ,υz ) dυx dυy dυz = f (υ X )dυ X ⋅ f (υ y )dυ y ⋅ f (υZ )dυZ
3/2
⎡ m
⎤
2
2
2
(υ X + υ y + υZ )⎥ dυ Xdυ ydυZ
exp ⎢ −
⎦
⎣ 2kT
3/2
⎛ mυ 2 ⎞
exp ⎜ −
⎟ dυ xdυ y dυz
⎝ 2kT ⎠
⎛ m ⎞
=⎜
⎟
⎝ 2πkT ⎠
⎛ m ⎞
=⎜
⎟
⎝ 2πkT ⎠
• The velocity distribution function F (υ) in three-dimension (3D) :
3/2
2
υ
m
⎛
⎞
⎛ m ⎞
2
F (υ ) dυ = 4π ⎜
⎟ υ exp ⎜ −
⎟ dυ
⎝ 2 π kT ⎠
⎝ 2kT ⎠
21
Maxwell Distribution of Speeds
• Figure 17.5 (a) Calculation
of the probability of velocity
component in the range υ
t υ + dυ. (b) V
to
Volume
l
element dV in spherical
coordinates.
22
Maxwell Distribution of Speeds
• The volume element in spherical coordinates:
υX = υ·sinθ cosΦ
υy = υ·sinθ sinΦ
υZ = υ·cosθ
• The differential volume of the shell:
d V = d υX d υy d υZ
Now
dV = dυ·(υ·sinθ·dΦ )·(υ·dθ )
= υ 2 dυ·sinθ dθ·dΦ
23
Maxwell Distribution of Speeds
• The volume element in spherical coordinates:
dV = dυ·(υ·sinθ·dΦ )·(υ·dθ ) = υ 2 dυ·sinθ dθ·dΦ
• Integrate for velocity space (υ → υ + dυ ) we have a spherical
shell with a thickness of dυ.
• The probability that a speed υ on the shell is F (υ ) dυ:
F (υ ) dυ = ∫ sinθ dθ ⋅ ∫ dφ ⋅ ∫ f (υx ,υy ,υz ) υ 2 dυ
π
2π
∞
0
0
0
3/2
m ⎞
2⎛
= 4πυ ⎜
⎟
⎝ 2 π kT ⎠
⎛ mυ 2 ⎞
exp ⎜⎟dυ
⎝ 2kT ⎠
24
Maxwell Distribution of Speeds
• James Clerk Maxwell was a famous English physicist who used
Boltzmann's ideas and applied them to the particles of an ideal gas
to produce the distribution bearing both men's names. He also used
the usual formula for kinetic energy, that E=(1/2)mυ2. The
di ib i
distribution
iis b
best shown
h
as a graph
h which
hi h shows
h
how
h
many
particles have a particular speed in the gas. A graph is shown below.
25
Maxwell Distribution of Speeds
• The Maxwell-Boltzmann distribution shows how many
particles in a gas have the different speeds.
• For a reaction to take place, molecules must collide with
sufficient energy to create a transition state
state. This usually
only happens for the fastest molecules, so if the MaxwellBoltzmann distribution tells us how many molecules have
energies or speeds above a certain level, that will be useful
to us.
26
Maxwell Distribution of Speeds
5 Maxwell Distribution
The distribution of
molecular speeds with
temperature and molar
mass. Note that the most
probable speed
(corresponding to the peak
of the distribution)
increases with temperature
and with decreasing molar
d simultaneously,
i
lt
l
mass and,
the distribution becomes
broader.
27
Maxwell Distribution of Speeds
• Figure 17.6 Probability density of various speeds υ for oxygen
molecules at 100, 300, 500, and 1000 K calculated using the
3/ 2
equation:
⎛ mυ 2 ⎞
m
⎞
2⎛
⎟⎟
F (υ ) = 4 πυ ⎜
⎟ exp ⎜⎜ −
⎝ 2 π kT ⎠
⎝ 2kT ⎠
28
Translational Energy Distribution
• As ε=mυ2/2, and υ= (2ε/m )1/2, we have dυ = dε/(2mε)1/2
• The probability density as a function of kinetic energy ε :
3/2
⎛ 2ε ⎞⎛ m ⎞
F (ε ) dε = 4π ⎜
⎟⎜
⎟
⎝ m ⎠⎝ 2πkT ⎠
2π ε 1/2 -ε kT
=
e
dε
3/2
(πkT )
• The average kinetic energy <ε>:
2π
ε = ∫ εF (ε ) dε =
3/2
0
(πkT )
∞
∫
⎛ ε ⎞ dε
exp ⎜ - ⎟
1/2
⎝ kT ⎠ (2m ε )
∞
0
ε
3/2
e
- ε /kT
dε
2
2π
3
⎛ kT ⎞
1/2
=
3
⎟ (πkT ) = kT
3/2 ⎜
2
(πkT ) ⎝ 2 ⎠
29
Translational Energy Distribution
2π ε 1/ 2
⎛ ε ⎞
F (ε ) =
exp ⎜ −
⎟
3/ 2
(π kT )
⎝ kT ⎠
• Figure 17.7 Plot of probability density F(ε) versus ε for the
translational energy of an ideal gas molecule at 300 K.
30
Types of Average Speeds
• The most probable speed υmp is the speed at the maximum of
F(υ ):
3/ 2
dF(υ)
⎛ m ⎞
=0=⎜
⎟
dυ
⎝ 2 π kT ⎠
1/ 2
⎛ 2kT ⎞
⎛ 2RT
υmp = ⎜
⎟ =⎜
⎝ M
⎝ m ⎠
e
-mυ 2/ 2kT
1/ 2
⎡
2 ⎛ mυ ⎞ ⎤
⎢8 πυ + 4 πυ ⎜- kT ⎟⎥
⎝
⎠⎦
⎣
⎞
⎟
⎠
• The mean speed <υ> is calculated as the average of υ using the
probability distribution F (υ):
3/ 2
2
∞
υ = ∫ υF (υ ) dυ
0
1/ 2
⎛ 8kT ⎞
υ =⎜
⎟
⎝ πm ⎠
⎛ m ⎞
υ = 4 π⎜
⎟
⎝ 2 π kT ⎠
1/ 2
⎛ 8RT ⎞
=⎜
⎟
⎝ πM ⎠
∫
∞
0
⎛ mυ
υ exp⎜⎜⎝ kT
3
⎞
⎟⎟dυ
⎠
1/ 2
⎛ 4⎞
= ⎜ ⎟ υmp ≈ 1.128 υmp
⎝π⎠
31
Types of Average Speeds
• The root-mean-square speed <υ2>1/2 is the square root of <υ2>:
3/ 2
⎡
∞
⎛ mυ
⎛ m ⎞
4
⎤
⎡
υ
υ exp⎜⎜= ∫ υ F (υ )dυ
= ⎢4 π⎜
⎟
∫
⎢⎣ 0
⎥⎦
⎢⎣ ⎝ 2 π kT ⎠ 0
⎝ kT
1/ 2
1/ 2
1/ 2
⎛ 3kT ⎞
⎛ 3⎞
⎛ 3RT ⎞
2 1/2
υ
=⎜
⎟ =⎜
⎟ = ⎜ ⎟ υmp ≈ 1.225 υmp
⎝ M ⎠
⎝2⎠
⎝ m ⎠
2 1/2
∞
1/ 2
2
2
1/2
⎞ ⎤
⎟⎟dυ ⎥
⎠ ⎥⎦
• The speed of sound is due to a reversible and adiabatic expansion
and contraction of gas with PV γ = constant :
V
V
γPV
υS = ==
ρ (∂V ∂P )S
ρ (-V γ P ) ρV
2
1/ 2
⎛ γkT ⎞
υS = ⎜
⎟
⎝ m ⎠
1/ 2
⎛ γRT ⎞
=⎜
⎟
⎝ M ⎠
1/ 2
⎛ 5RT ⎞
≈⎜
⎟
⎝ 3M ⎠
for ideal gas ≈ 0.913 υmp
32
PV γ = 1
V γ= P
-1
1/γ
V = P -1/γ
⎛ 1 ⎞ − γ1 − 1 ⎛ −1
⎛ ∂V ⎞
= ⎜⎜
⎜
⎟ = ⎜⎜ − ⎟⎟ P
⎝ ∂P ⎠S ⎝ γ ⎠
⎝γ P
⎛V
⎛ ∂V ⎞
⎜
⎟ = − ⎜⎜
⎝ ∂P ⎠S
⎝γ P
⎞ − γ1 ⎛ −1
⎟⎟ P = ⎜⎜
⎠
⎝γ P
⎞
⎟⎟V
⎠
⎞
⎟⎟
⎠
1 ⎛ ∂V ⎞
1
1
− ⎜
=
⎟ =
V ⎝ ∂P ⎠S γ P ρ υS2
γ P γRT
υS =
=
ρ
M
2
33
Types of Average Speeds
• The reduced form of distribution function in 3-D :
3/ 2
M ⎞
2⎛
F (υ )dυ = 4πυ ⎜
⎟
⎝ 2π RT ⎠
let u = υ υmp
⎛ M υ2 ⎞
⎟⎟dυ
exp ⎜⎜⎝ 2RT ⎠
1// 2
1
⎛ 2RT ⎞
=υ ⎜
⎟
⎝ M ⎠
1/ 2
2
2
2
υ
RT
RT
M
⎞
⎛
2
then dυ = ⎜
u2, = - u2
⎟ du, υ =
2RT
M
⎝ M ⎠
⎛ 2RT ⎞
F (u ) du = 4 π u ⎜
⎟
⎝ M ⎠
2
F (u ) du =
4
π
3/ 2
⎛ M ⎞
⎟
⎜
⎝ 2 π RT ⎠
3/ 2
exp
p (-u 2 ) du
u 2e -u 2du
34
Types of Average Speeds
6 Speeds
A summary of the conclusions
that can be deduced from the
Maxwell distribution for
molecules of molar mass M at a
temperature T:
υmp/(2RT/M)1/2
<υ>/(2RT/M)1/2
υrms/(2RT/M)1/2
υmp is the most probable speed,
<υ> is the mean speed, and
υrms is the root mean square
speed.
35
Types of Average Speeds
Example 17.3 Various speeds for hydrogen molecules
Calculate the most probable speed υmp, the mean speed <υ>,
and the root-mean-square speed <υ2>1/2 for hydrogen molecules
at 0 ℃
Ans: υ
mp
1/2
⎛ 8 RT ⎞
υ =⎜
⎟
⎝ πM ⎠
2
υ
⎡ 2(8.3145)(273.15)⎤
=⎢
-3
⎥
2.016x10
⎣
⎦
1/2
1/2
⎛ 2RT ⎞
=⎜
⎟
⎝ M ⎠
1/2
= 1.50x103 m s-1
⎡ 8(8.3145)(273.15) ⎤
=⎢
-3 ⎥
3.1416
2.016x10
⎣
⎦
1/2
(
1/2
⎛ 3RT ⎞
=⎜
⎟
⎝ M ⎠
)
⎡3(8.3145)(273.15)⎤
=⎢
⎥
2 016x10-3
2.016x10
⎣
⎦
1/2
= 1.69x103 m s-1
= 1.84x103 m s-1
The root-mean-square speed of a hydrogen molecules at 0 ℃ is
6620 km h-1, but at ordinary pressure a molecule travels only an
exceedingly short distance before colliding with another molecule
and changing direction. υmp < <υ> < <υ2>1/2
36
Table 17.2
Table 17.2 Various Types of Average Speeds of Gas Molecules at 298 K
G
Gas
<υ2>1/2/ m s-11
<υ>/
/ m s-11
υmp / m s-11
H2
1920
1769
1568
O2
482
444
394
CO2
411
379
336
CH4
681
627
556
37
Doppler Broadening of Spectral Lines
• 都普勒效應：一個運動中的原子發出的光，如果運動方向以υX 的速度
接近觀察者（接受器），則在觀察者看來，其頻率ν 較靜止原子所發的
頻率νo 高，反之變低。
υX ⎞
⎛
ν ≈ ν o ⎜1 +
⎟
c ⎠
⎝
or υ X = c (ν - ν o )/ν o
• 此觀察頻率ν 的分佈與分子的速率分佈函數成正比:
e
-m υ X2 / 2 kT
= e
-mc
2
(ν-ν o )2 / 2ν o2kT
• 若頻率ν 依Gaussian函數分佈,則
p (ν ) d ν = (2 π σ ν
)
2 - 1/ 2
e
- (ν-ν o
)2 / 2 σ ν2
dν
• 依此函數其中頻率分佈的標準偏差σν即代表都普勒變寬的值
σν
⎛ ν o kT
= ⎜⎜
2
⎝ mc
2
⎞
⎟⎟
⎠
1/ 2
38
Pressure of an Ideal Gas
• Figure 17.8 Specular
reflection of a molecule
with a wall in the yz plane.
The x component
p
of the
velocity has its sign
reversed by the collision,
but υy and υZ are
unchanged if the impinging
molecule is in the xy plane.
F
ma x
m ⎛ dυ x ⎞
1 ⎛ dpx ⎞
P =
=
=
⎜
⎟ =
⎜
⎟
A
A ⎝ dt ⎠
A ⎝ dt ⎠
A
39
Pressure of an Ideal Gas
7 Pressure from Collision
The pressure of a gas arises from
the
h impact
i
off its
i molecules
l
l on the
h
walls. In an elastic collision of a
molecule with a wall perpendicular
to the x-axis, the x-component of
velocity is reversed but the y- and
z-components are unchanged.
Force per collision
<Fx>= (dpx/dt ) = (-2 mυx) (υx/2x) = - mυx2 /x
40
Pressure of an Ideal Gas
• A single particle in container with volume V =xyz
– Moving in velocity υ, with component +υx normal to wall yz.
– The change
g of momentum by
y reflection on each collision is
Δpx = m Δυx = m (-υx -υx ) = - mυx - mυx = -2 mυx
– In a time Δt a particle with +υx will travel a distance υxΔt
– In a time Δt any molecule with +υx will hit the surface if it is
within the effective volume:
– Effective volume above the surface is Veff = |υx Δt | A
41
Pressure of an Ideal Gas
8 Effective Volume
A molecule will reach the
wall on the right within an
interval Δt if it is within a
distance υxΔt of the wall
and traveling to the right.
42
Pressure of an Ideal Gas
Figure 17.9 For a container with n
mole particle in volume of V, the
number density ρ of particles is
ρ n NA /V
ρ=
• The number of molecules in the
volume Veff = AυX Δt is
ρVeff = (n NA/V ) ( AυX Δt )
• Molecule with velocity +υX will
strike on wall of area A. Half of
molecules
l
l iin th
the volume
l
V eff with
ith
υX ≧ 0 will strike this surface in
time Δt.
#/ Δt =ρVeff / 2Δt = n NAA υX / 2V
43
Pressure of an Ideal Gas
• Rate of momentum change is the force Fx
Fx =Δpx/ Δt = n NA A υX /2V •(2 m υX )
= n NA m AυX2/V = n M AυX2/V
• The pressure is the force divided by the area,
Pressure = <Fx> / A = n M <υx2>/V
• For N particles with distribution of velocities in 3D
< υ2 > is the mean square velocity
< υ 2 > = (<
( υX2 >+ <υy2 >+ <υZ2 >)) = 3<
3 υX2 >
44
Pressure of an Ideal Gas
• Use average <υ2 > for molecules travel with various
velocities:
P = n M <υ2> /3V
or PV = n M <υ2> /3
• or PV = n M <υ2>/3 = (1/3) n M c2
• where c is the root mean square speed of molecules
υrms = (<υX2 >+<υy2 >+<υZ2 >)1/2 =<υ2 >1/2 =(3RT/M )1/2
• The root mean square speed υrms of molecules of a gas is
– p
proportional
p
to the square
q
root of the temperature
p
(T )1/2
– inversely proportional to the square root of the molar
mass (M )-1/2 .
45
Pressure of an Ideal Gas
• Probability of particles having a velocity in range of υX to υX + dυX and
within the effective volume Veff above the surface A in time dt is
N f (υX) dυX (Veff / V ) = N f (υX) d υX (υX dt A /V )
• The average force in x-direction
x direction <Fx>
A
FX = N ∫ dυ X (2 mυ X )f (υ X )υ X
0
V
1/2
∞
NA
⎛ m ⎞
2 -mυ / 2kT
υ
=
2m ⎜
dυ X
⎟ ∫0 X e X
V
⎝ 2πkT ⎠
3/2
1/2
∞
NA
⎛ m ⎞ ⎛ 2kT ⎞
2 -u 2
=
2m ⎜
u
e du
⎟ ∫0
⎟ ⎜
V
2 π kT ⎠ ⎝ m ⎠
⎝ 2π
NA
=
kT
V
• Pressure is the force per unit area
P = <Fx>/A = NkT/V = nRT /V
∞
2
46
Collisions with a Surface and Effusion
• Effusion is the escape of gas from a gas container through a
small hole(s).
• The surface collision frequency zw is the number of collision
per unit area of a surface per unit time when the pressure is P.
This is also know as the flux JN.
• The flux ( J ) of a property is the quantity that passes per unit
area per unit time. The number of collision with wall per unit
area per unit time in terms of number flux JN are
JN = ∫
∞
0
⎛N
⎜
⎝V
1/ 2
⎞
⎛ N ⎞⎛ kT ⎞
⎟
⎟f (υx ) υxdυx = ⎜ ⎟⎜
⎠
⎝ V ⎠⎝ 2 π m ⎠
ρυ
P
=
=
4
2 π mkT
47
Collisions with a Surface and Effusion
• Knudsen method for determination of the vapor pressure of
liquid and solids is based upon the loss of mass by effusion
of vapor through a small hole in a container.
• the Knudsen method for effusion of vapor
M
AP t
Δw =
2 π RT
48
Collisions with a Surface and Effusion
Example 17.4 Vapor pressure from an effusion measurement
The vapor pressure of solid beryllium was measured in 1948 using
a Knudsen cell. The effusion hole was 0.318 cm in diameter, and
tthey
ey found
ou d a mass
ass loss
oss o
of 9
9.54
5 mg
g in 60
60.1 min at a te
temperature
pe atu e o
of
1457 K. What is the vapor pressure? (4Be=9.012 g mol-1)
-6
23
-1
Δw Δw N A ⎡
⎤
(
)(
)
9.54
x
10
kg
6
.
022
x
10
mol
Ans: J N =
=
mtA
MtA = ⎢⎢ (9.012x 10- 3 kg mol -1 ) (60x 60.1 s )π (0.159x10-2 m)2 ⎥⎥
⎣
⎦
= 2.23x1022 m-2 s-1
P = JN (2πm k T )1/2
= (2.23x1022
- 3 kg mol -1 ) (1.381 x10 - 23 J K -1 ) (1457 K ) ⎤
(
⎡
2
π
9
.
012
x
10
m-2 s-2⎢ )x
⎥
23 mol -1 )
(
6
.
022
x
10
⎦
⎣
=0.968 Pa = 0.968x10-5 bar
49
Graham’s law of effusion
• According to the Maxwell-Boltzmann distribution, the mean
molecular speed is proportional to 1/m½ .
• The Graham’s law of effusion:
“The rate of effusion is inversely proportional to the square
root of the molar mass of the effusing gas.”
50
Collisions of Hard-Sphere Molecules
• Figure 17.10 Potential
energy diagram for two
hard-sphere molecules with
diameters d1 and d2 .
51
Collisions of Hard-Sphere Molecules
• Figure 17.11 Collisions of
hard-sphere molecules. If
molecules of type 2 are
stationary,
y, a molecule of
type 1 will collide in unit
time Δt with all molecules
of type 2 that have their
centers in a cylinder of
volume πd122 υ1 Δt.
υ 1 Δt
52
Effusion
• The collision cross-section is σ=πd 2, the area swept out by
a molecule and within which the presence of the center of
another molecule counts as a collision.
• The excluded volume into which the centre of a molecule
cannot penetrate is 8ν, where ν is the volume of the
molecule itself. Here d1 = d2 .
53
Collisions of Hard-Sphere Molecules
23 The excluded volume into
which the centre of a
molecule cannot penetrate
is 8ν, where ν is the volume
of the molecule itself. Here
d1 = d2 .
54
Table Collision cross-sections
πd
2
/nm2
Argon, Ar
0.36
Methane, CH4
0.46
Benzene C6 H6
Benzene,
0 88
0.88
Carbon dioxide, CO2
0.52
Helium, He
0.21
Nitrogen, N2
0.43
Ethene, C2 H4
0.64
H d
Hydrogen
molecule,
l
l H2
0 27
0.27
Oxygen molecule, O2
0.40
Sulfur dioxide, SO2
0.58
55
Collisions of Hard-Sphere Molecules
10 Swept Volume
In an interval Δt a molecule of diameter d sweeps out a tube of
diameter 2d and length ̿crel Δt . As it does so it encounters other
molecules with centres that lie within the tube,
tube and each such
encounter counts as one collision. In practice, the tube is not
straight, but changes direction at each collision. Nevertheless,
the volume swept out is the same, and this straightened version
of the tube can be used as a basis of the calculation.
56
Collisions of Hard-Sphere Molecules
• Figure 17.12 The mean
relative speed <υ12> of
molecules 1 and 2 can be
calculated from a right
g
triangle involving the mean
speeds of molecules of
types 1 and 2. According to
the Pythagorean theorem,
<υ12>2 = <υ1>2 + <υ2>2
57
Collisions of Hard-Sphere Molecules
11 mean relative speed
A simplified version of the argument
to show that the mean relative speed
of molecules in a gas is related to the
mean speed.
d Wh
When the
h molecules
l
l are
moving in the same direction, the
mean relative speed is zero; it is 2υ
when the molecules are approaching
each other. A typical mean direction
of approach is from the side, and the
mean speed of approach is then 21/2υ.
The last direction of approach is the
most characteristic, so the mean
speed
d off approach
h can be
b expected
d
to be about 21/2υ. This value is
confirmed by more detailed
calculation.
58
Collisions with a surface and effusion
• The mean relative speed <⌡12> is
υ12
2
⎛ 8kT
= ⎜⎜
⎝ πμ
⎞ ⎛ 8kT ⎞ ⎛ 1
1⎞
2
⎟⎟ = υ1 + υ 2
+
⎟⎟ = ⎜
⎟ ⎜⎜
⎠ ⎝ π ⎠ ⎝ m1 m2 ⎠
2
• The collision frequency z12 of molecule of type 1 with
molecules of type 2 is
z12
1/2
⎛ 8kT ⎞
N 2 (Aυ12Δt /V )
2
=
= ρ2 π d12 ⎜⎜
⎟⎟
Δt
⎝ πμ ⎠
2
= ρ2 π d12
υ12
υ12 = ∫ f (υ1) f (υ2 ) υ12dυ1dυ2
59
Collisions with a surface and effusion
Example 17.5 Mean relative speed of two different molecules
What is the mean relative speed of hydrogen molecules with
respect to oxygen molecules at 298.15 K.
A
Ans:
Th molecular
The
l
l masses are
2.16x10-3 kg mol-1
-27
m1 =
=
3.348x10
kg
23
-1
6.022x10 mol
32.00x10-3 kg mol-1
-27
m2 =
=
53.14x10
kg
23
-1
6.022x10 mol
μ = [(3.348x10-27 )-1 + (53.14x10-27 )-1]-1 = 3.150x10-27 kg
1/2
υ12
⎛ 8kT ⎞
= ⎜⎜
⎟⎟
⎝ πμ ⎠
⎡ 8(1.381x10
1 381x10 J K )(298 K )⎤
=⎢
⎥
-27
(
)
π
3.150x10
kg
⎣
⎦
-23
-1
1/2
= 1824 m s-1
Note that the mean relative speed is closer to the mean speed of
molecular hydrogen (1920 m s-1) than to that of molecular oxygen
(482 m s-1)
60
Collisions with a surface and effusion
• The hard sphere collision frequency
z12 = ρ2 π d122 ∫ f (υ1 ) f (υ2 ) υ12 dυ1dυ2
• In the collision frequency z12 , the product f (υ1) f (υ2) is
converted from the sum of kinetic energies to the kinetic
energy of the center of mass plus the relative kinetic energy.
z12 = ρ2 π d122 ∫ f (υ12 ) υ12 dυ12
where
3/2
⎛ μ ⎞
2 − μυ 2/ 2kT
f (υ12 ) = 4 π ⎜
⎟ υ12 e 12
⎝ 2 π kT ⎠
and the reduced mass is μ = m1 m2 /(m1 + m2)
61
Collision Density
• For same type of molecules, the collision frequency z11 is
z11 = 21/2ρπd 2<υ>
since μ
μ=m/2
/ and <υ11>=(8kT/πμ)
(
/ μ)1/2 =21/2 <υ>
• The collision density is the number of collisions per unit time
per unit volume, (mol m-3 s-1)
– Z12 is for molecules of type 1 with molecules of type 2
(multiply z12 by the number density ρ1 )
Z12 = ρ1 ρ2 πd122 <υ12>
i ffor molecules
l
l off type
t
1 with
ith other
th molecules
l
l off type
t
– Z11 is
1 (multiply z11 by the number density ρ1/2)
Z11 = ρ1 ρ1 πd112<υ11>/2 = 2-1/2ρ2πd 2 <υ>
62
Example 17.6 Collision frequency and collision density
Collisions with a surface and effusion
For molecular oxygen at 25 ℃, calculate the collision frequency z11
and the collision density Z11 at a pressure of 1 bar. The collision
diameter of oxygen is 0.361 nm or 3.61x10-10 m.
Ans: The molecular mean speed is
1/ 2
1/ 2
-1
-1
⎡ 8(8.3145 J K mol )(298 K )⎤
⎛ 8RT ⎞
-1
υ =⎜
=
=
444
m
s
⎟
⎢
⎥
-3
-1
(
)
π
32
x
10
k
kg
mol
l
⎝ πM ⎠
⎣
⎦
The number density is given by
N PN A (1 bar )(6.022x 1023 mol -1 )(103 L m - 3 )
ρ= =
=
= 2.43x 1025 m - 3
(0.083145 L bar K -1 mol -1 )(298 K )
V
RT
The collision frequency is given by
z11 = 2ρ π d υ = (1.414)(2.43x10 m
2
25
-3
) π (3.61x10
-10
m) (444 m s-1) = 6.24x109 s-1
2
The collision density is given by
1 2
Z11 =
ρ πd
2
2
υ = (0.707 ) (2.43x1025 m-3 )2π (3.61x10-10 m)2 (444 m s-1)
34
-3 -1
(
7.58x10
m
s )
34
-3 -1
= 7.58x10 m s =
= 1.26x108 mol L-1 s-1
6.022x1023 mol-1
63
Collisions with a surface and effusion
Example 17.7 Relation between Collision frequency and collision
density
Above we have explicit expressions for the collision frequency z12
between molecules of type 1 and type 2, z11 between molecules of
the same type, collision density Z12 between molecules of type 1
and type 2, and Z11 between molecules of the same type. What are
the relations between z12 and Z12 and between z11 and Z11.
Ans: Comparing the equation in the text, we see that
Z12 = z12 (N1/V) = ρ1 z12
so that the number of collisions per unit volume per unit time
between molecules of type 1 and type 2 is equal to the density ρ1
of molecules of type 1 times the frequency of collisions between
molecules of type 1 and type 2.
Z11 = ρ z11 /2
The collision frequency is divided by 2 to avoid double counting.
64
Collision Frequencies and Collision Densities
Table 17.3 Collision Frequencies and Collision Densities at 298 K
z11 / s-1
Z11 / mol L-1 s-1
Gas
1 bar
10-6 bar
1 bar
10-6 bar
H2
14.13x109
14.13x103
2.85x108
2.85x10-4
CH4
11.60x109
11.60x103
2.08x108
2.08x10-4
O2
6.24x109
6.24x103
1.26x108
1.26x10-4
CO2
8.81x109
8.81x103
1.58x108
1.58x10-4
65
Mean Free Path
• The mean free path λ is the average distance traveled
between collisions. It is computed by dividing the average
distance traveled per unit time by the collision frequency.
υ
1
λ=
= 1/ 2
z 11 2 ρ π d 2
• As ρ=N/V=nNA/V=PNA/RT=P/kT, and
λ=
kT
21/ 2 π d 2P
t t temperature,
t
t
the
th mean free
f
path
th is
i inversely
i
l
• At constant
proportional to the pressure P.
66
Collisions with a surface and effusion
Example 17.8 Calculate the mean free path
For oxygen at 25 ℃, the collision diameter is 0.361 nm. What is the
mean free path at (2) 1 bar pressure and (b) 0.1 Pa pressure?
A
Ans:
( )F
(a)
From E
Example
l 17.7,
17 7 ρ=2.43x10
2 43 1025 m-33 att 1 bar,
b
and
d using
i
υ
1
λ=
=
z 11
2ρπd 2
λ = [(1.414) (2.43x10-25m-3 )π (3.61x10-10 m)2 ]-1 = 7.11x10-8 m
(b)
PN A (0.1 bar ) (6.022x 1023 mol - 1 )
19
-3
ρ=
2
.
43
x
10
m
=
=
(8.3145 J K -1 mol -1 )(298 K )
RT
λ = [(1.414) (2.43x10-19m-3 )π (3.61x10-10 m)2 ]-1 = 7.11x10-2 m = 7.11 cm
67
Effects of Molecular Interactions on Collisions
16 Potential Energy
The general form of an
intermolecular potential energy
curve At long range the
curve.
interaction is attractive, but at
close range the repulsions
dominate.
68
Effects of Molecular Interactions on Collisions
15 Potential Energy
The variation of the potential
energy of two molecules on their
separation High positive potential
separation.
energy (at very small separations)
indicates that the interactions
between them are strongly
repulsive at these distances. At
intermediate separations, where
the potential energy is negative,
the attractive interactions
dominate. At large separations (on
the right) the potential energy is
zero and there is no interaction
between the molecules.
69
Effects of Molecular Interactions on Collisions
17 Potential Energy
The Lennard-Jones potential,
the relation of the parameters
to the features of the curve,
curve
and the two contributions. Note
that 21/6 = 1.122 . . . .
Ex. Lennard-Jones potential:
12
6
⎡⎛ σ ⎞
⎛σ ⎞ ⎤
Φ (r ) = 4ε ⎢⎜ ⎟ − ⎜ ⎟ ⎥
⎝ r ⎠ ⎥⎦
⎢⎝ r ⎠
⎣
ε: well depth
σ: closest distance when Φ(σ)=0
70
Effects of Molecular Interactions on Collisions
21 Potential Energy
The basic arrangement of a
molecular beam apparatus.
The atoms or molecules
emerge from a heated source,
and pass through the velocity
selector, a train of rotating
disks. The scattering occurs
from the target gas (which
might take the form of
another beam), and the flux of
particles entering the detector
set at some angle is recorded.
(a) The definition of the solid
angle,
l dΩ, for
f scattering.
i
(b) The definition of the
impact parameter, b, as the
perpendicular separation of
the initial paths of the
particles.
71
Effects of Molecular Interactions on Collisions
23 Potential Energy
Three typical cases for the
collisions of two hard
spheres:
(a) b = 0, giving backward
scattering;
(b) b > RA + RB, giving
forward scattering;
(c) 0 < b < RA + RB, leading
to scattering into one
direction on a ring
g of
possibilities. (The target
molecule is taken to be so
heavy that it remains
virtually stationary.)
72
Effects of Molecular Interactions on Collisions
• Figure 17.13 Collision of
two spherically
p
y symmetrical
y
molecules with an impact
parameter b. The diagram
is drawn so that the center
of gravity of the system
does not shift during the
collision. The angle of
deflection is χ.
73
Effects of Molecular Interactions on Collisions
26 The dark central zone
represents the repulsive
core; the fuzzy outer zone
represents the long-range
attractive potential. The
interference of paths
leading to rainbow
scattering. The rainbow
angle, θr, is the maximum
scattering angle reached as
b is decreased.
74
Effects of Molecular Interactions on Collisions
• The extent of scattering may depend on the relative speed of
approach as well as the impact parameter b. Two paths leading
to the same destination will interfere quantum mechanically; in
this case they give rise to quantum oscillations in the forward
direction. Interference between the numerous paths at that
angle modifies the scattering intensity markedly.
75
Effects of Molecular Interactions on Collisions
• Figure 17.14 Collision
trajectories for a pair of
molecules interacting by a
Lennard-Jones 6-12
potential.
76
Transport Processes
• Four types of transport process:
(a) diffusion, the spreading of
one species into another; (b)
thermal conduction,, when
molecules with different
energies of thermal motion
(represented by the arrows)
spread into each others' regions;
(c) electrical conduction, when
ions migrate under the influence
of an electric field; (d) viscosity,
h molecules
l
l with
ith different
diff
t
when
linear momenta (represented by
the arrows) migrate.
77
Transport properties
•
Flux is the transport property with respect to the composition,
temperature, or potential, or velocity when there are not in uniform.
•
Flux of a property is often proportional to the gradient of another
property.
Flow density = Constant x Gradient
• Flow density in quantity per area per second.
• Constant is a phenomenological coefficient.
• Gradient is the driving force of transportation. A change of
properties with distance over limiting range of independent
variables.
1 Matter : diffusion without bulk flow.
1.
flow
2. Heat : energy conduction without convection.
3. Electricity : charge motion without change the potential
4. Momentum : a viscous flow without non-newtonian
motion.
78
Transport properties
• Fick’s first law of diffusion:
Matter flows in decreasing concentration direction.
Jiz = -D (dci /dz )
Jiz : flux in mol m-2 s-1
dci /dz : concentration gradient along z-axis in mol m-4
D : diffusion coefficient in m2 s-1
• Fourier’s law :
Heat flows in decreasing temperature direction.
qZ = -κ(d
(dT /d
dz )
qZ : thermal flux in J m-2 s-1
dT/dz : temperature gradient along z-axis in K m-1
κ: thermal conductivity coefficient in J m-1 s-1 K-1
79
Transport properties
• Einstein-Stoker law :
Momentum flows in decreasing velocity direction.
F = -η(d
η( vy/
/dz )
F : force per unit area to move the plate in kg m-1 s-2
dvy/dz : y-velocity gradient along z-axis in m s-1/m
η: viscosity in kg m-1 s-1 (Pascal•s)
• Viscosity in 1 Pa•s if a force of 1 N is required to move a
plane of 1m2 at a velocity of 1 m/s.
• 1 Pa•s = 1 kg m-1 s-1 = 10 poise
80
Transport Phenomena in Gases
• Figure 17.15 Schematic diagrams of
apparatus for measurements of
irreversible properties. (a) In the
measurement of the diffusion
coefficient D, the sliding separator is
withdrawn so that a substance in
chamber A can diffuse into B. (b) In
the measurement of the thermal
conductivity κ, the rate of heat
transfer from an axial hot wire is
measured. (c) In the measurement of
th viscosity
the
i
it η off a gas, the
th outer
t
cylinder is rotated and the torsion on
the inner cylinder is determined from
the twist in the suspension wire.
81
Newtonian Flow
• Newtonian flow is a fluid in a state that it flows as in a series
of streaming layers, or laminas. The lamina adjacent to the
wall is stationary and successive layers having increasingly
greater speeds.
g
p
82
• The viscosity of a fluid arises
from the transport of linear
momentum. In this
illustration the fluid is
undergoing laminar flow,
and particles bring their
initial momentum when they
enter a new layer. If they
arrive with a high xcomponent of momentum
they accelerate the layer; if
th arrive
they
i with
ith a low
l
xcomponent of momentum
they retard the layer.
83
Viscosity Gradient in a Fluid
• Figure 17.16 Viscosity
gradient in a fluid due to a
shearing action.
84
Calculation of Transport Coefficients
• Figure 17.17 Planes
constructed at distance ±λ
(the mean free p
path) from
the origin. The concentration
gradient is in the z direction.
85
Calculation of Transport Coefficients
• Consider the diffusion of molecules in a concentration gradient in
the z direction. Imaging the we are at z=0 and we construct
planes parallel to the xy plane at z = ±λ, where λ is the mean free
path. Now let us calculate the flux of p
p
particles across z=0 due to
the molecules above (z>0) and below (z<0). The flux across z=0
from above is
⎡
⎛ dρ ⎞⎤ υ
J + = ⎢ρo + λ ⎜
⎟⎥
⎝ dz ⎠⎦ 4
⎣
whereρ̊ is number density of particles in the plane at z=0. Similarly,
the flux across z=0 due to the molecules below z=0 is
⎡
⎛ dρ ⎞⎤ υ
J - = ⎢ρo - λ ⎜
⎟⎥
⎝ dz ⎠⎦ 4
⎣
86
Calculation of Transport Coefficients
• The net flux of particles across the plane z=0 is then
1
⎛ dρ ⎞
J = - υ λ⎜
⎟
2
⎝ dz ⎠
⎛ dci ⎞
• Compare with equation: J Z (i ) = -D ⎜
⎟ to obtain
⎝ dz ⎠
1/2
1
1 8kT ⎛ kT ⎞
kT ⎛ 1 ⎞ ⎛ kT ⎞
Da = υ λ =
⎜
⎟=⎜
⎜
⎟=
⎟
2
2 π m ⎝ 2σ P ⎠
π m ⎝σ ρ ⎠ ⎝ π m ⎠
• Da indicates approximated diffusion coefficient
1
ρ πd 2
• The exact theoretical expression
p
for the diffusion coefficient
1/2
of hard spheres is
3π
D=
8
⎛ kT ⎞
⎟
⎜
⎝ πm ⎠
1
3π
=
λ
υ
ρ π d 2 16
• Highly simplified model still yields a good qualitative results.
87
Diffusion properties
• Larger molecules with larger collision cross section σ will
diffuse more slowly than the smaller ones.
• Everything diffuses more slowly as pressure is increased at
constant T (Since λ ∝1/P, D ∝ 1/P), and
• Everything diffuses faster as T is increased (D ∝ T3/2 at
constant P, D ∝ T1/2 at constant V ). (Winn, Physical
Chemistry, 1995)
1
1 8kT ⎛ kT ⎞
Da = υ λ =
⎜
⎟
2
2 π m ⎝ 2σ P ⎠
88
Diffusion Coefficients at 25 ℃
Table A7
89
Transport properties
• A similar model for thermal conductivity of hard spheres
yields the approximate value κa:
1
1 CV
κ a = λ υ CV [A ] =
λυ ρ
3
3 NA
1/2
⎛ 8kT ⎞
υ =⎜
⎟
⎝π m ⎠
kT
V
1
1
=
=
=
and λ =
2σP
2σ n N A
2σ N A [A ]
2σ ρ
1/2
⎛ 8kT ⎞
υ λ [A ] = ⎜
⎟
⎝π m ⎠
1/2
⎛ 1 ⎞ ⎛ 4RT ⎞
⎟
⎜
⎜ 2σN ⎟ = ⎜⎝ π M ⎟⎠
A ⎠
⎝
⎛ 1 ⎞
⎜⎜
⎟⎟
⎝ σN A ⎠
1/ 2
1/2
CV ⎛ 4RT ⎞
2C RT ⎞
1
κa =
⎜
⎟ = V ⎛⎜
⎟
3σ NA ⎝ πM ⎠
3N A ⎝ π M ⎠ π d 2
90
Transport properties
• The exact theoretical expression for the thermal conductivity
of hard spheres is
1/2
25 πCV ⎛ kT ⎞
κ=
⎜
⎟
32 N A ⎝ π m ⎠
1
25π
=
λ υ CV [A ]
2
πd
64
• Because λ ∝ 1/P and [A] ∝ P, the thermal conductivity κ of gas
is independent of the pressure. Here the presence of many
energy carrier also limited their mean free paths and cannot
carry the energy over a long distance.
• As a result, κ ∝ <υ>CV .
91
Table 17.4 Viscosity and Thermal Conductivity of Gases
at 273.2K and 1 bar and Calculated Molecular Diameters
Gas
η/
10-5 kg m-1 s-1
He
1.85
Ne
2.97
Ar
2.11
H2
κ/
10-2 J K-1 m-1 s-1
14.3
Gas Diameter, d/nm
From η
From κ
0.218
0.218
0.258
0.258
1.63
0.364
0.365
0.845
16.7
0.272
0.269
O2
1 92
1.92
2 42
2.42
0 360
0.360
0 358
0.358
CO2
1.36
1.48
0.464
0.458
CH4
1.03
3.04
0.414
0.405
4.60
92
Exercise Use the experimental value of the thermal conductivity of
oxygen gas to estimate the collision cross-section of O2 molecules
at 273.2 K.
1/2
25 πCV ⎛ kT ⎞
1
25π
-1 m-1 s-1
[
]
κ=
=
λ
υ
C
A
⎜
⎟
Solution:
J
K
V
32 N A ⎝ π m ⎠ π d 2 64
Assumed ideal gas, C̿V (O2, 298 K)=5R/2 J K-1 mol-1
The thermal conductivity is proportional to λ, hence κ ∝ σ-1
1/2
⎛
⎞
8RT ⎞
25π
1
⎛
⎜
⎟⎜
κ=
⎟ CV
⎜
⎟
64 ⎝ 2σ N A ⎠ ⎝ πM ⎠
1/2
⎛
⎞ ⎛ 5R ⎞ ⎛ 8RT ⎞
25π
⎟⎜
σ = ⎜⎜
⎟⎜
⎟
⎟
π
2
M
⎠⎝
⎠
⎝ 64 2 κ N A ⎠ ⎝
1/2
25π x5x8.3145
⎛
⎞⎛ 8x8.3145x2 73.2 ⎞
=⎜
⎟⎜
⎟
-3
23
⎠
⎝ 128 2x0.0242x6. 022x10 ⎠⎝ π 32.0x10
= 5.26x10
-19
m2
(0.526
nm2 )
93
Transport properties
• The approximate model for the viscosity of hard spheres
yields :
1/ 2
1
2 ⎛ kT ⎞
m
ηa = ρ υ mλ = ⎜
⎟
3
3 ⎝ πm ⎠ πd 2
• The exact expression for the viscosity of hard spheres is
1/2
5π ⎛ kT ⎞
η=
⎜
⎟
16 ⎝ πm ⎠
m
5π m λ υ
=
2
πd
σ
32 2
94
Viscosity of a gas
Example 17.9 Viscosity of a gas
Calculate the viscosity of molecular oxygen at 273.2 K and 1 bar.
The molecular diameter is 0.360 nm.
A
Ans:
U i
Using
the
th exactt equation
ti
for
f hard
h d spheres,
h
we find
fi d
mO2
32.00x10-3 kg mol-1
-26
=
=
5.314x10
kg
23
-1
6.022x10 mol
1/ 2
5 π ⎛ kT ⎞
m
η=
⎜
⎟
16 ⎝ π m ⎠ π d 2
1/2
23
1
5π ⎡ (1.381x10 J K ) (273.2 K )⎤
5.314x10-26 kg
=
2
⎥ π (0.360x10
-9
16 ⎢⎣
π (5.314x10
5 314x10-26 kg )
)
0
360x10
m
⎦
= 1.926x 10- 5 kg m - 1s -1
95
Transport properties
• From the hard spheres model of the viscosity:
1/ 2
5π ⎛ kT ⎞
η=
⎟
⎜
16 ⎝ πm ⎠
m
σ
• the viscosity of gas is independent of the pressure as λ
∝ 1/P and [A] ∝ P, implying that η <υ>, but
independent of pressure. Because <υ> ∝ T1/2. The
viscosity coefficient η is proportional to T1/2.
• But the viscosity of a liquid decreases with the
temperature because intermolecular interactions must
be overcome.
96
The pressure dependence of the viscosity
• The experimental results
for the pressure
dependence of the viscosity
of argon.
g
Here η
ηis constant
from 0.01 atm to 20 atm.
97
The temperature dependence of the viscosity
• The experimental results for
the temperature dependence of
the viscosity of argon. The
dotted line in the g
graph
p is the
calculated value using
σ=22x10-20 m2. implying a
collision diameter of 260 pm,
in contrast to the van der Waals
diameter of 335 pm obtained
from the density of the solid.
Fitting the observed and
l l t d curves iis one way off
calculated
determining the collision
cross-section.
98
Table A1 Transport properties of perfect gases
Property
Transported
quantity
i
Diffusion
Matter
Thermal
Energy
conductivity
Viscosity
Simple kinetic
theory
h
3π
D=
8
1/ 2
⎛ kT ⎞
⎜
⎟
⎝π m⎠
Units
1
ρ πd 2
1/ 2
25 πCV ⎛ kT ⎞
κ=
⎜
⎟
32 N A ⎝ π m ⎠
Momentum
1/2
5 π ⎛ kT ⎞
η=
⎜
⎟
16 ⎝ π m ⎠
1
πd 2
m
πd 2
m2 s-1
J K-1 m-1 s-1
kg m-11 s-11
99