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Chapter 5
Gases and the Kinetic-Molecular Theory
แก๊สก ับทฤษฎีจลน์โมเลกุล
5-1
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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
5-2
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Table 5.1
Some Important Industrial Gases
Name (Formula)
Origin and Use
Methane (CH4)
natural deposits; domestic fuel
Ammonia (NH3)
from N2+H2; fertilizers, explosives
Chlorine (Cl2)
electrolysis of seawater; bleaching and
disinfecting
Oxygen (O2)
liquefied air; steelmaking
Ethylene (C2H4)
high-temperature decomposition of
natural gas; plastics
Atmosphere-Biosphere Redox Interconnections
5-3
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An Overview of the Physical States of Matter
The Distinction of Gases from Liquids and Solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gases have relatively low viscosity.
4. Most gases have relatively low densities under normal conditions.
5. Gases are miscible.
5-4
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Figure 5.1
5-5
The three states of matter.
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Figure 5.2
5-6
Effect of atmospheric pressure on objects
at the Earth’s surface.
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Figure 5.3
5-7
A mercury barometer.
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Figure 5.4
Two types of
manometer
closed-end
open-end
5-8
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Table 5.2 Common Units of Pressure
Unit
Atmospheric Pressure
Scientific Field
pascal(Pa);
kilopascal(kPa)
1.01325x105Pa;
101.325 kPa
SI unit; physics, chemistry
atmosphere(atm)
1 atm*
chemistry
millimeters of
mercury(Hg)
760 mm Hg*
chemistry, medicine, biology
torr
760 torr*
chemistry
14.7lb/in2
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
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Converting Units of Pressure
Sample Problem 5.1
PROBLEM:
A geochemist heats a limestone (CaCO3) sample and collects
the CO2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature,
Dh = 291.4 mm Hg. Calculate the CO2 pressure in torrs,
atmospheres, and kilopascals.
PLAN: Construct conversion factors to find the other units of pressure.
SOLUTION:
1torr
291.4 mmHg x
1 mmHg
1 atm
291.4 torr x
= 291.4 torr
= 0.3834 atm
760 torr
0.3834 atm x
101.325 kPa
1 atm
5-10
= 38.85 kPa
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Figure 5.5
The relationship between the
volume and pressure of a gas.
Boyle’s Law
5-11
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Figure 5.6
The relationship between
the volume and temperature
of a gas.
Charles’s Law
5-12
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V a
Boyle’s Law
1
P
V x P = constant
V
= constant
T
Amontons’s Law
P
T
combined gas law
5-13
V = constant / P
V a T
Charles’ s Law
P a T
= constant
V a
n and T are fixed
P and n are fixed
V = constant x T
V and n are fixed
P = constant x T
T
P
V = constant x
T
PV
P
T
= constant
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Figure 5.7
5-14
An experiment to study the relationship between the
volume and amount of a gas.
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Figure 5.8
5-15
Standard molar volume.
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Figure 5.9
5-16
The volume of 1 mol of an ideal gas compared with some
familiar objects.
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THE IDEAL GAS LAW
Figure 5.10
R = universal gas constant
PV = nRT
PV
R=
nT
=
3 significant figures
1atm x 22.414L
0.0821atm*L
=
mol*K
1mol x 273.15K
IDEAL GAS LAW
PV = nRT
V=
nRT
P
fixed n and T
fixed n and P
fixed P and T
Boyle’s Law
Charles’s Law
Avogadro’s Law
V=
constant
P
5-17
or
V=
constant x T
V=
constant x n
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Applying the Volume-Pressure Relationship
Sample Problem 5.2
PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies
24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases
the pressure on the trapped air to 2.64 atm. Assuming constant
temperature, what is the new volume of air (in L)?
PLAN:
SOLUTION:
V1 in cm3
P and T are constant
P1 = 1.12 atm
P2 = 2.64 atm
V1 = 24.8 cm3
V2 = unknown
1cm3=1mL
V1 in mL
unit
conversion
103 mL=1L
24.8 cm3 x
1 mL
1
V1 in L
xP1/P2
V2 in L
V2 =
P1V1
P2
5-18
P1V1
gas law
calculation
n1T1
=
=
cm3
x
L
103 mL
P2V2
= 0.0248 L
P1V1 = P2V2
n2T2
0.0248 L x
1.12 atm
2.46 atm
= 0.0105 L
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Sample Problem 5.3
PROBLEM:
Applying the Temperature-Pressure Relationship
A steel tank used for fuel delivery is fitted with a safety valve
that opens when the internal pressure exceeds 1.00x103 torr.
It is filled with methane at 230C and 0.991 atm and placed in
boiling water at exactly 1000C. Will the safety valve open?
PLAN:
SOLUTION:
T1 and T2(0C)
P1(atm)
1atm=760torr
P1(torr)
K=0C+273.15
T1 and T2(K)
x T2/T1
P2 = unknown
T1 = 230C
T2 = 1000C
P1V1
=
n1T1
P2(torr)
0.991 atm x
P2 =
5-19
P1 = 0.991atm
P1T2
T1
P2V2
n2T2
760 torr = 753 torr
1 atm
= 753 torr x
373K
296K
= 949 torr
P1
T1
=
P2
T2
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Applying the Volume-Amount Relationship
Sample Problem 5.4
PROBLEM:
A scale model of a blimp rises when it is filled with helium to a
volume of 55 dm3. When 1.10 mol of He is added to the blimp,
the volume is 26.2 dm3. How many more grams of He must be
added to make it rise? Assume constant T and P.
PLAN: We are given initial n1 and V1 as well as the final V2. We have to find
n2 and convert it from moles to grams.
n1(mol) of He
SOLUTION:
P and T are constant
x V2/V1
n1 = 1.10 mol
n2 = unknown
n2(mol) of He
V1 = 26.2 dm3
V2 = 55.0 dm3
subtract n1
mol to be added
V1
n1
=
V2
n2
xM
g to be added
n2 = 1.10 mol x
n2 =
=
n 1T1
P2V2
n2T2
V2 n 1
V1
55.0 dm3
26.2 dm3
= 2.31 mol x
4.003 g He
mol He
= 9.24 g He
5-20
P1V1
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Sample Problem 5.5
PROBLEM:
PLAN:
Solving for an Unknown Gas Variable at Fixed
Conditions
A steel tank has a volume of 438 L and is filled with 0.885 kg
of O2. Calculate the pressure of O2 at 210C.
V, T and mass, which can be converted to moles (n), are given. We
use the ideal gas law to find P.
SOLUTION:
0.885 kg x
V = 438 L
T = 210C (convert to K)
n = 0.885 kg (convert to mol)
P = unknown
103 g
kg
x
mol O2
32.00 g O2
= 27.7 mol O2
27.7 mol x 0.0821
P=
5-21
nRT
V
atm*L
mol*K
=
438 L
210C + 273.15 = 294.15K
x 294.15K
= 1.53 atm
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Sample Problem 5.6
PROBLEM:
Using Gas Laws to Determine a Balanced Equation
The piston-cylinders depict a gaseous reaction carried out at
constant pressure. Before the reaction, the temperature is
150K; when it is complete, the temperature is 300K and the
volume does not change.
Which of the following balanced equations describes the reaction?
(1) A2 + B2
(3) A + B2
PLAN:
AB2
(2) 2AB + B2
(4) 2AB2
2AB2
A2 + 2B2
We know P, T, and V, initial and final, from the pictures. Note
that the volume doesn’t change even though the temperature is
doubled. With a doubling of T then, the number of moles of gas
must have been halved in order to maintain the volume.
SOLUTION:
5-22
2AB
Looking at the relationships, the equation that
shows a decrease in the number of moles of gas
from 2 to 1 is equation (3).
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The Density of a Gas
density = m/V
n = m/M
PV = nRT
PV = (m/M)RT
Density = m/V = M x P/ RT
•The density of a gas is directly proportional to its molar mass.
•The density of a gas is inversely proportional to the temperature.
5-23
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Sample Problem 5.7
PROBLEM:
Calculating Gas Density
To apply a green chemistry approach, a chemical engineer
uses waste CO2 from a manufacturing process, instead of
chlorofluorocarbons, as a “blowing agent” in the
production of polystyrene containers. Find the density (in
g/L) of CO2 and the number of molecules (a) at STP (00C
and 1 atm) and (b) at room conditions (20.0C and 1.00 atm).
PLAN: Density is mass/unit volume; substitute for volume in the ideal gas
equation. Since the identity of the gas is known, we can find the molar
mass. Convert mass/L to molecules/L with Avogadro’s number.
MxP
d = mass/volume
PV = nRT
V = nRT/P
d=
RT
SOLUTION:
44.01 g/mol x 1atm
= 1.96 g/L
d=
(a)
atm*L
x
0.0821
x 273.15K
mol*K
1.96 g
mol CO2
6.022x1023 molecules
x
x
= 2.68x1022 molecules CO2/L
L
44.01 g CO2
mol
5-24
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Calculating Gas Density
Sample Problem 5.6
continued
(b)
44.01 g/mol
d=
= 1.83 g/L
0.0821
1.83g
L
5-25
x
x 1 atm
mol CO2
44.01g CO2
x
atm*L x 293K
mol*K
6.022x1023 molecules
mol
= 2.50x1022 molecules CO2/L
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Figure 5.11
Determining the molar mass
of an unknown volatile liquid.
based on the method of
J.B.A. Dumas (1800-1884)
5-26
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Finding the Molar Mass of a Volatile Liquid
Sample Problem 5.8
PROBLEM:
An organic chemist isolates a colorless liquid from a
petroleum sample. She uses the Dumas method and
obtains the following data:
Volume of flask = 213 mL
T = 100.00C
Mass of flask + gas = 78.416 g
Mass of flask = 77.834 g
P = 754 torr
Calculate the molar mass of the liquid.
PLAN: Use unit conversions, mass of gas, and density-M relationship.
SOLUTION:
M=
m = (78.416 - 77.834) g = 0.582 g
m RT
VP
5-27
0.582 gx 0.0821
=
atm*L
mol*K
0.213 L x 0.992 atm
x 373K
= 84.4 g/mol
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Mixtures of Gases
•Gases mix homogeneously in any proportions.
•Each gas in a mixture behaves as if it were the only gas present.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
P1 = c1 x Ptotal
c1 =
5-28
where c1 is the mole fraction
n1
n1 + n2 + n3 +...
=
n1
ntotal
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Sample Problem 5.9
PROBLEM:
In a study of O2 uptake by muscle at high altitude, a physiologist
prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2,
and 4.0 mol% 18O2. (The isotope 18O will be measured to
determine the O2 uptake.) The pressure of the mixture is 0.75atm
to simulate high altitude. Calculate the mole fraction and partial
pressure of 18O2 in the mixture.
PLAN: Find the c
18
mol%
18O
Applying Dalton’s Law of Partial Pressures
O2
and P18
O2
from Ptotal and mol% 18O2.
SOLUTION:
2
c 18
O2
=
4.0 mol% 18O2
= 0.040
100
divide by 100
c 18
P18
O2
multiply by Ptotal
partial pressure P
18O
5-29
2
O2
= c 18
O2
x Ptotal = 0.040 x 0.75 atm = 0.030 atm
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The Molar Mass of a Gas
n=
mass
M
=
PV
RT
M=
m RT
VP
d=
m
V
d RT
M=
P
5-30
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Table 5.3 Vapor Pressure of Water (P
H2O
T(0C)
0
5
10
11
12
13
14
15
16
18
20
22
24
26
28
5-31
) at Different T
P (torr)
T(0C)
P (torr)
4.6
6.5
9.2
9.8
10.5
11.2
12.0
12.8
13.6
15.5
17.5
19.8
22.4
25.2
28.3
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
31.8
42.2
55.3
71.9
92.5
118.0
149.4
187.5
233.7
289.1
355.1
433.6
525.8
633.9
760.0
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Figure 5.12
5-32
Collecting a water-insoluble gaseous reaction
product and determining its pressure.
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Sample Problem 5.10
Calculating the Amount of Gas Collected Over Water
PROBLEM: Acetylene (C H ), an important fuel in welding, is produced in
2 2
the laboratory when calcium carbide (CaC2) reaction with water:
CaC2(s) + 2H2O(l)
C2H2(g) + Ca(OH)2(aq)
For a sample of acetylene that is collected over water, the total
gas pressure (adjusted to barometric pressure) is 738 torr and
the volume is 523 mL. At the temperature of the gas (230C), the
vapor pressure of water is 21 torr. How many grams of
acetylene are collected?
PLAN: The difference in pressures will give us the P for the C2H2. The ideal
gas law will allow us to find n. Converting n to grams requires the
molar mass, M.
SOLUTION: PC H = (738-21)torr = 717torr
2 2
Ptotal
P
C2H2
atm
P
= 0.943 atm
PV
717
torr
x
H2O
n=
760torr
RT
n
g
C2H2
C2H2
xM
5-33
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Calculating the Amount of Gas Collected Over Water
Sample Problem 5.9
continued
n =
PV
RT
0.943 atm x
n
=
C2H2
0.0821
atm*L
mol*K
0.0203 mol x
26.04 g C2H2
mol C2H2
5-34
0.523L
= 0.0203 mol
x 296K
= 0.529 g C2H2
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Figure 15.13
Summary of the stoichiometric relationships among the
amount (mol,n) of gaseous reactant or product and the gas
variables pressure (P), volume (V), and temperature (T).
P,V,T
amount (mol)
amount (mol)
P,V,T
of gas A
of gas A
of gas B
of gas B
ideal
gas
law
5-35
molar ratio from
balanced equation
ideal
gas
law
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Sample Problem 5.11
Using Gas Variables to Find Amount of
Reactants and Products
PROBLEM:
Dispersed copper in absorbent beds is used to react with
oxygen impurities in the ethylene used for producing
polyethylene. The beds are regenerated when hot H2
reduces the metal oxide, forming the pure metal and H2O.
On a laboratory scale, what volume of H2 at 765 torr and
2250C is needed to reduce 35.5 g of copper(II) oxide?
PLAN: Since this problem requires stoichiometry and the gas laws, we have
to write a balanced equation, use the moles of Cu to calculate mols
and then volume of H2 gas.
mass (g) of Cu
divide by M
SOLUTION:
35.5 g Cu x
mol Cu
molar ratio
0.0821
0.559 mol H2 x
use known P and T to find V
5-36
L of H2
x
63.55 g Cu
mol of Cu
mol of H2
CuO(s) + H2(g)
Cu(s) + H2O(g)
1 mol H2
1 mol Cu
atm*L
x
mol*K
1.01 atm
= 0.559 mol H2
498K
= 22.6 L
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Sample Problem 5.12 Using the Ideal Gas Law in a Limiting-Reactant Problem
PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group
7A(17)] to form ionic metal halides. What mass of potassium
chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293K
reacts with 17.0 g of potassium?
PLAN: After writing the balanced equation, we use the ideal gas law to find the
number of moles of reactants, the limiting reactant and moles of product.
SOLUTION:
nCl =
2
PV
RT
2K(s) + Cl2(g)
2KCl(s)
T = 293K
0.950 atm x 5.25 L
=
0.0821
atm*L
17.0 g x
=
=
0.207 mol
Cl2 is the limiting reactant.
0.435 mol K
39.10 g K
0.207 mol Cl2 x
5-37
0.414 mol KCl x
n = unknown
x 293K
mol*K
mol K
P = 0.950 atm V = 5.25 L
2 mol KCl
= 0.414 mol KCl formed
1 mol Cl2
74.55 g KCl
mol KCl
= 30.9 g KCl
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Postulates of the Kinetic-Molecular Theory
Postulate 1: Particle Volume
Because the volume of an individual gas particle is so
small compared to the volume of its container, the gas
particles are considered to have mass, but no volume.
Postulate 2: Particle Motion
Gas particles are in constant, random, straight-line
motion except when they collide with each other or
with the container walls.
Postulate 3: Particle Collisions
Collisions are elastic therefore the total kinetic
energy(Ek) of the particles is constant.
5-38
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Figure 5.14
5-39
Distribution of molecular speeds at three temperatures.
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Figure 5.15
5-40
A molecular description of Boyle’s Law.
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Figure 5.16
5-41
A molecular description of Dalton’s law of partial pressures.
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Figure 5.17
5-42
A molecular description of Charles’s Law.
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Avogadro’s Law
Ek = 1/2 mass x speed2
V a
n
Ek = 1/2 mass x u 2
u 2 is the root-mean-square speed
urms =
√3RT
R = 8.314 Joule/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
5-43
1
√M
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Figure 5.18
5-44
A molecular description of Avogadro’s Law.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 5.19
Relationship between molar mass and molecular speed.
Ek = 3/2 (R/NA) T
5-45
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Sample Problem 5.13
Applying Graham’s Law of Effusion
PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH ).
4
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
5-46
CH4
=
√
16.04
4.003
= 2.002
M of He = 4.003g/mol
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Figure 5.20
Diffusion of a gas particle through a
space filled with other particles.
distribution of molecular speeds
mean free path
collision frequency
5-47
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Figure B5.1
5-48
Variations in pressure, temperature, and
composition of the Earth’s atmosphere.
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5-49
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5-50
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 5.4 Molar Volume of Some Common Gases at STP (00C and 1 atm)
Gas
He
H2
Ne
Ideal gas
Ar
N2
O2
CO
Cl2
NH3
5-51
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
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Figure 5.21
The behavior of several
real gases with increasing
external pressure.
5-52
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Figure 5.22
5-53
The effect of intermolecular attractions on
measured gas pressure.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 5.23
5-54
The effect of molecular volume on measured gas volume.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 5.5 Van der Waals Constants for Some Common Gases
n2a
(P  2 )(V  nb)  nRT
V
Van der Waals
equation for n
moles of a real gas
adjusts P up adjusts V down
a
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
NH3
H2O
5-55
atm*L2
b
L
mol2
mol
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
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