Chapter 5
Gases and the Kinetic-Molecular Theory
5-1
기체와 분자운동론
• 기체상태의 물질
• 기체의 압력
• 기체 법칙
보일의 법칙, 샤를의 법칙, 아보가드로의 법칙
• 분압
• 기체분자운동론
온도와 에너지, 확산
• 실제기체 및 반데르발스 상태방정식
5-2
Gases and the Kinetic Molecular Theory
5.1 An Overview of the Physical States of Matter
5.2 Gas Pressure and Its Measurement
5.3 The Gas Laws and Their Experimental Foundations
5.4 Further Applications of the Ideal Gas Law
5.5 The Ideal Gas Law and Reaction Stoichiometry
5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior
5.7 Real Gases: Deviations from Ideal Behavior
5-3
기체는 왜 공부하는가?
역사적인 배경과 의미
1. Boyle’s Law – First Scientific Experiment, 1661
2. Charles’s Law – Definition of Temperature, 1780s
3. Avogadro’s Hypothesis –
4. Combined Ideal Gas Law and Kinetic Theories– First
Successful Scientific Law derived from purely
mathematical approach.
실용적인 배경과 의미
1. 기체는 열역학의 기초, 열역학 제1, 2 법칙을 설명
2. 화학 평형, 화학적 퍼텐셜의 기초
3. 화학반응의 근본적 이해를 도움- 반응동력학
5-4
Physical Characteristics of Gases
 Gases assume the volume and shape of their containers.
 Gases are the most compressible state of matter.
 Gases will mix evenly and completely when confined to the same
container.
 Gases have much lower densities than liquids and solids.
5-5
Force
Pressure = Area
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
bar = 105 Pa
5-6
psi = lb/in2 = 6894.757 Pa
Barometer
5-7
PRESSURE
Units and Measurement
Pressure = Force/Area
SI Units
Force = mass × acceleration
Force = kg-m/s2 = Newton
Pressure = Newton/m2 = Pascal
Customary Units
Pressure = atm, torr, mmHg, bar, psi
Relate SI to customary
1.01325 × 105 Pascal = 1 atm = 760 torr
bar = 105 Pascal
5-8
Common Units of Pressure
Unit
Atmospheric Pressure
Scientific Field
pascal(Pa);
1.01325×105Pa;
SI unit; physics, chemistry
atmosphere(atm)
1 atm*
chemistry
mm of mercury(Hg)
760 mm Hg*
chemistry, medicine, biology
torr
760 torr*
chemistry
14.7 psi
engineering
1.01325 bar
meteorology, chemistry,
physics
pounds per square
inch (psi or lb/in2)
bar
*This is an exact quantity; in calculations, we use as many significant figures as necessary.
5-9
Boyle’s Law(1662)
𝑝𝑉 = 𝑘
5-10
Charles’ Law (1780s)
Definition of Temperature
V = V0 +
at
V = aT
0
5-11
Temperature(K)
273.15
T (K) = t (0C) + 273.15
Avogadro’s Hypothesis
Equal volumes of gases contain the same
number of molecules at constant T,P
22.414 L of any gas
contains 6.022×1023
atoms (or molecules)
at 0 ℃ and 1 atm.
5-12
Avogadro’s Law(1811)
V ∝ number of moles (n)
V = constant × n
Constant temperature
Constant pressure
V1/n1 = V2/n2
3 H2(g) + N2(g) → 2 NH3(g)
5-13
3 volume + 1 volume → 2 volume
3 mol + 1 mol → 2 mol
3 molecules + 1 molecule → 2 molecules
Ideal Gas Equation
1
Boyle’s law: V ∝ (at constant n and T)
𝑃
Charles’ law: V ∝ T (at constant n and P)
Avogadro’s law: V ∝ n (at constant P and T)
V∝
𝑛𝑇
𝑃
V=
𝑛𝑇
constant×
𝑃
=R
𝑛𝑇
𝑃
R is the gas constant
PV = nRT
5-14
Dalton’s Law of Partial Pressures
V and T : constant
P1
5-15
P2
Ptotal = P1 + P2
Consider a case in which two gases, A and B, are in a
container of volume V.
nART
PA =
V
nA is the number of moles of A
nBRT
PB =
V
nB is the number of moles of B
PT = PA + PB
PA = XA PT
nA
XA =
nA + nB
PB = XB PT
Pi = xi PT
5-16
nB
XB =
nA + nB
mole fraction (xi) =
ni
nT
The Molar Mass of a Gas
𝑚𝑎𝑠𝑠 𝑃𝑉
𝑛=
=
𝑀
𝑅𝑇
𝑚𝑅𝑇
𝑀=
𝑉𝑃
𝑚
𝑑=
𝑉
𝑑𝑅𝑇
𝑀=
𝑃
5-17
Water Vapor Pressure Table
Temperature
(°C)
0.0
5.0
10.0
12.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.9
5-18
Pressure
(mmHg)
4.6
6.5
9.2
10.9
12.8
13.2
13.6
14.1
14.5
15.0
15.5
16.0
16.5
Temperature
(°C)
19.5
20.0
20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
26.0
Pressure
(mmHg)
17.0
17.5
18.1
18.6
19.2
19.8
20.4
21.1
21.7
22.4
23.1
23.8
25.2
Temperature
(°C)
27.0
28.0
29.0
30.0
35.0
40.0
50.0
60.0
70.0
80.0
90.0
95.0
100.0
Pressure
(mmHg)
26.7
28.3
30.0
31.8
42.2
55.3
92.5
149.4
233.7
355.1
525.8
633.9
760.0
T
(°C)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5-19
Pressure
(mmHg)
4.6
4.9
5.3
5.7
6.1
6.5
7.0
7.5
8.0
8.6
9.2
9.8
10.5
11.2
12.0
12.8
13.6
14.5
15.5
16.5
17.5
18.7
19.8
21.1
22.4
23.8
T
(°C)
Pressure
(mmHg)
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
25.2
26.7
28.3
30.0
31.8
33.7
35.7
37.7
39.9
42.2
44.6
47.1
49.7
52.4
55.3
58.3
61.5
64.8
68.3
71.9
75.7
79.6
83.7
88.0
92.5
T
(°C)
Pressure
(mmHg)
T
(°C)
Pressure
(mmHg)
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
97.2
102.1
107.2
112.5
118.0
123.8
129.8
136.1
142.6
149.4
156.4
163.8
171.4
179.3
187.5
196.1
205.0
214.2
223.7
233.7
243.9
254.6
265.7
277.2
289.1
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
301.4
314.1
327.3
341.0
355.1
369.7
384.9
400.6
416.8
433.6
450.9
468.7
487.1
506.1
525.8
546.0
567.0
588.6
610.9
633.9
657.6
682.1
707.3
733.2
760.0
stoichiometric relationships among the amount
(mol,n) of gaseous reactant or product and the
gas variables P, V, and T.
P,V,T
of gas A
ideal
gas
law
amount
(mol)
amount
(mol)
P,V,T
of gas A
of gas B
of gas B
molar ratio from
balanced equation
3 H2(g) + N2(g) → 2 NH3(g)
5-20
ideal
gas
law
Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be points;
that is, they possess mass but have negligible volume.
d(N2,g) = 0.00125 g/L (273°C)
d(N2,liq) = 0.808 g/mL (-195.8°C)
2. Gas molecules are in constant motion in random directions.
Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces
on one another.(no interaction)
5-21
V a
Avogadro’s Law
n
Ek = 1/2 mass x u 2
Ek = 1/2 mass x speed2
u 2 is the root-mean-square speed
urms =
√3RT
R = 8.314Joule/mol*K
M
Graham’s Law of Effusion
The rate of effusion of a gas is inversely related to the square root of its molar mass.
rate of effusion a
5-22
1
√M
Applying Graham’s Law of Effusion
PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4).
PLAN: The effusion rate is inversely proportional to the square root of the
molar mass for each gas. Find the molar mass of both gases and find
the inverse square root of their masses.
SOLUTION:
M of CH4 = 16.04g/mol
rate
He
rate
5-23
CH4
=
√
16.04
4.003
= 2.002
M of He = 4.003g/mol
Figure 5.18
Diffusion of a gas particle through a
space filled with other particles.
distribution of molecular speeds
mean free path
collision frequency
5-24
Molar Volume of Some Common Gases at 00C and 1 atm
Gas
He
H2
Ne
Ideal gas
Ar
N2
O2
CO
Cl2
NH3
5-25
Molar Volume
(L/mol)
22.435
22.432
22.422
22.414
22.397
22.396
22.390
22.388
22.184
22.079
Condensation Point
(0C)
-268.9
-252.8
-246.1
---185.9
-195.8
-183.0
-191.5
-34.0
-33.4
Figure 5.19
The behavior of several
real gases with increasing
external pressure.
5-26
Van der Waals Constants for Some Common Gases
Van der Waals
equation for n
moles of a real gas
n2a
(P  2 )(V  nb)  nRT
V
adjusts P up adjusts V down
a
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
NH3
H2O
5-27
atm·L2
mol2
0.034
0.211
1.35
2.32
4.19
0.244
1.39
1.36
6.49
3.59
2.25
4.17
5.46
b
L
mol
0.0237
0.0171
0.0322
0.0398
0.0511
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.0371
0.0305
Collisions of Gas Particles
5-28
Collisions of Gas Particles
5-29
The gas molecules in the container are in random motion
5-30
압력과 기체분자 운동
• 운동량 변화 ∆p = mvx – (–mvx) = 2mvx
• 분자의 왕복시간 ∆t = 2a/vx
• 분자 한 개가 면 A에 미치는 힘
f = ∆p/ ∆t= 2mvx/(2a/vx)= mvx2/a
• N개의 분자에 의해 면 A에 가해지는 압력
• P = Σf/a2 = m(vx12 + vx22 + ... + vxN2)/a3
=Nmvx2/V
PV=Nmvx2
5-31
압력과 기체분자 운동
•
공간 상에서 무작위로 움직이는 분자는 특별한 방향에 대한
선호도가 없을 것이므로
v x  v y  v z  v 2  v x  v x  v x  3v x
2
2
2
2
2
2
1 2
vx  v
3
2
1
2 1 2
2
PV  Nmv x  Nmv  N ( mv )
3
3 2
2
Gas Pressure in the Kinetic Theory
분자당 평균운동에너지
5-32
KE 
1 2
mv
2
2
운동에너지와 온도
PV 
•
1
2 1
Nmv 2  N ( mv 2 )
3
3 2
이상 기체방정식
PV  nRT
•
기체에 열이 가해지면 기체의 내부에너지와 온도가 증가!
1 2 3
mv  k BT
2
2
R
kB 
 1.38 10  23 J/K
NA
KE 
5-33
Mean Kinetic Energy per particle
Boltzmann constant
1
1
2
PV  Nmv  nM v 2
3
3
v2 
urms
PV = NkBT = nRT
3k BT 3RT

m
M
3k BT
3RT
 v2 

m
M
N = number of molecules,
kB = Boltzmann constant, J K-1,
n = number of moles = N/NA,
R = gas constant, 0.08204 l atm mol-1 K-1
T = temperature,
P = gas pressure,
V = volume,
NA = Avogadro’s number
5-34
Maxwell-Boltzmann Distribution
for Molecular Speeds
u~ u+du 사이의 속도를 가진 분자들의 수
 m 
f(u)  

 2πkT 
3/ 2
2
mu
4πu 2 exp( 
)
2kT
f(u) = speed distribution function,
m = molecular mass,
k = Boltzmann constant,
T = temperature, u = speed
5-35
분자속도와 에너지 분포
(Molecular velocity and energy distribution)
• 슈테른(Otto Stern)의 분자속도 측정
노벨물리학상 수상자(1943, w/ Gerlach)
Schematic diagram of a stern type experiment for determining the distribution
of molecular velocities  v ~ v+∆v 사이의 분자개수 ∆N측정가능!!
http://leifi.physik.uni-muenchen.de/web_ph12/originalarbeiten/stern/molekularstr.htm
5-36
Molecular speed Distribution of N2 gas
u mp 
2 RT
M
8 RT
u
M
u rms
3RT

M
Maxwell Boltzmann distribution of molecular speeds in nitrogen gas at two
temperatures. The ordinate is the fractional number of molecules per unit
speed interval in (km/s)-1.
5-37
Apparatus for studying molecular speed distribution
Chopper method
5-38
gravitation method
Chemistry in Action: Super Cold Atoms
Gaseous Rb Atoms
1.7 x 10-7 K(170 nK)
Bose-Einstein Condensate
1995, Eric Cornell and Carl Wieman
Three images showing formation of Bose-Einstein
condensate. picture from JILA
5-39
Gaseous Diffusion/Effusion
Diffusion of Ammonia and HCl
Effusion enrichment of UF6
Graham’s Law of Effusion
A bank of uranium gas centrifuges
Uranium Information Centre
5-40
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
5-41
HCl
36 g/mol
Kinetic-Molecular Theory for
Gaseous Behavior
• Principal Issues (잠재적 문제점)
– Negligible Volume and No interaction
• Hold only at low P, high T; for dilute gases
– Elastic Collisions
• Only in Newtonian mechanics is the reverse of an event as
likely as the event itself.
• In the real world you cannot “unscramble” eggs because of
entropy effects resulting from large ensembles of molecules
5-42
Deviations from Ideal Behavior
1 mole of ideal gas
PV = RT
PV = 1.0
Z=
RT
compression factor
5-43
Repulsive Forces
Attractive Forces
Real Gas
Effect of intermolecular forces on
the pressure exerted by a gas.
Has volume
Has interaction(usually
attractive)
5-44
Selected Values for a and b for the
van der Waals Equation
Gas
Van der Waals equation
nonideal gas
corrected
pressure
5-45
}
}
2
an
( P + V2 ) (V – nb) = nRT
corrected
volume
[(L2
a
· atm)/mol2]
b
[10-2 L/mol]
He
0.03412
2.370
Ne
0.211
1.71
Ar
1.34
3.22
Kr
2.32
3.98
Xe
4.19
2.66
H2
0.2444
2.661
N2
1.390
3.913
O2
1.360
3.183
CO2
3.592
4.267
CH4
2.25
4.28
CCl4
20.4
13.8
C2H2
4.390
5.136
Cl2
6.493
5.622
C4H10
14.47
12.26
C8H18
37.32
23.68
Hydrogen Economy & Fuel Cell
5-46
Ozone Depletion
Chlorine Destroys Ozone
but is not consumed in the process
5-47
5-48
Crutzen
Holland (The
Netherlands)
Max-Planck-Institute
for Chemistry
Mainz, Germany
1933 -
5-49
Molina
Rowland
USA (Mexico)
Department of Earth,
Atmospheric
and Planetary Sciences
and
Department of Chemistry,
MIT
Cambridge, MA, USA
1943 -
USA
Department of Chemistry,
University of California
Irvine, CA, USA
1927 -
Hindenberg
Crew: 40 to 61
Capacity: 50-72 passengers
Length: 245 m ,
Diameter: 41 m
Volume: 200,000 m³
Powerplant: 4 × Daimler-Benz diesel
engines,
890 kW (1,200 hp) each
기체: 수소(헬륨으로 계획되었으나, 수소로
대체)
부력:
5-50
Science of Hot Air Balloon
부력 = 밖 공기의 무게 – 안 공기의 무게
안 공기의 무게
= 공기의 밀도(ρ(Thot))× 풍선의 부피(V)×g
밖 공기의 무게
= 공기의 밀도(ρ(Tcold))× 풍선의 부피(V)×g
공기의 밀도
ρ(T) = w/V = Mp/RT
2008 National Balloon Classic in Indianola, Iowa.
부피(m3)
600 ~ 2800 ~ 17,000
5-51
Tcold ~ 20 ℃
Thot ~ 100 ℃
부력 ~ 700kg
Lake Nyos, the killer Lake of
Cameroon
5-52
Clarke, T., Taming Africa’s killer lake, Nature, V. 409,
p. 554-555, February 2001.
Lake Nyos, the killer Lake
Test of the de-gassing operation in 1995
5-53
Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p. 554-555, February 2001.
Air Bag Chemistry
5-54
Air Bag Chemistry
5-55
5-56
Air Bag Chemistry
5-57
Air Bag Chemistry
On ignition:
2 NaN3 → 2Na + 3N2
Secondary reactions:
10 Na + 2 KNO3 → K2O + 5 Na2O + N2
K2O + Na2O + SiO2 → K2Na2SiO4
5-58