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Gases and the KineticMolecular Theory
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1
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
2
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
3
5.1 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.
4
The three states of matter: Bromine
5
5.2 Gas Pressure and Its Measurement
Effect of atmospheric pressure on objects
at the Earth’s surface.
Pinside = Poutside
Pinside = 0
6
Measuring Gas Pressure
Barometer - A device to measure atmospheric
pressure. Pressure is defined as force divided by
area. The force is the force of gravity acting on
the air molecules.
P=
F
A
Manometer - A device to measure gas
pressure in a closed container.
7
d
A mercury barometer
Torricelli
1608-1647
Is ∆h dependent on d?
8
closed-end
Two types of
manometer
open-end
9
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.
10
Sample Problem 5.1
PROBLEM:
Converting Units of Pressure
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,
∆h = 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:
291.4 mmHg
1torr
= 291.4 torr
1 mmHg
291.4 torr
1 atm
= 0.3834 atm
760 torr
0.3834 atm 101.325 kPa
1 atm
= 38.85 kPa
11
5.3 The Gas Laws and Their Experimental Foundations
Boyle’s Law - The relationship
between the volume and pressure of a
gas. (Temperature is kept constant.)
12
Charles’s Law - The
relationship between volume
and the temperature of a gas.
(Pressure is kept constant.)
13
Amontons’ Law
The relationship between the pressure and temperature
of a gas. (Volume is kept constant.)
14
V α
Boyle’s Law
VxP
1
P
= constant
n and T are fixed
V = constant / P
1627–1691
1837–1923
V α T
Charles’s Law
V
= constant
T
P and n are fixed
V = constant x T
1746–1823
P α T
Amontons’ Law
P
T
Combined gas law
= constant
V and n are fixed
P = constant x T
1663–1705
V α
T
P
V = constant x
T
PV
P
T
= constant
15
Gay-Lussac’s Law – Law of Combining Volumes
The ratio between the combining
volumes of gases and the product, if
gaseous, can be expressed in small
whole numbers.
Gay-Lussac
1778–1850
• Gay-Lussac’s law was discovered in 1809.
• This law played a major role in the development of
modern gas stoichiometry because in 1811, Avogadro used
Gay-Lussac’s Law to form Avogadro's hypothesis.
16
Avogadro’s Law - The volume of a gas is directly
proportionate to the amount of gas. (Pressure and
temperature are kept constant.)
An experiment to study the relationship between the
volume and amount of a gas.
1776–1856
V α n
V
n
= constant
V = constant x n
17
If the constant pressure is 1 atm and the constant temperature is 0o C,
1 mole of any gas has a volume of 22.4 L .
This is known as the Standard Molar Volume
18
The volume of 1 mol of an ideal gas compared
with some familiar objects.
He
22.4 L
13-in TV
5-gal
fish tank
18.9 L
21.6 L
7.5 L
19
THE IDEAL GAS LAW
PV = nRT
3 significant figures
R=
PV
nT
=
1atm x 22.414L
1mol x 273.15K
=
0.0821atm*L
mol*K
R is the universal gas constant
IDEAL GAS LAW
nRT
PV = nRT or V =
fixed n and T
Boyle’s Law
V=
constant
P
fixed n and P
fixed P and T
Charles’s Law
Avogadro’s Law
V=
constant X T
V=
constant X n
P
20
Sample Problem 5.2
PROBLEM:
Applying the Volume-Pressure Relationship
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, Hg 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
1cm3=1mL
unit
conversion
V1 in mL
103 mL=1L
V1 in L
gas law
calculation
xP1/P2
P1 = 1.12 atm
P2 = 2.64 atm
V1 = 24.8 cm3
V2 = unknown
L
24.8 cm3 1 mL
1 cm3 103 mL
P1V1
V2 in L
P and T are constant
P2V2
=
n1T1
P1V1 = P2V2
n2T2
1.12 atm
P1V1
V2 =
= 0.0248 L
= 0.0248 L
P2
= 0.0105 L
2.46 atm
21
Answer : 0.0105 L
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(torr)
P1 = 0.991atm
P2 = unknown
T1 = 230C
T2 = 1000C
P1V1
=
n1T1
P2V2
n2T2
P1
T1
=
P2
T2
0.991 atm 760 torr = 753 torr
1 atm
P2 = P1
T2
T1
= 753 torr
373K
= 949 torr
296K
22
Answer : 949 torr
Sample Problem 5.4
PROBLEM:
Applying the Volume-Amount Relationship
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.
SOLUTION:
n1(mol) of He
P and T are constant
x V2/V1
n1 = 1.10 mol
n2 = unknown
P1V1
n2(mol) of He
V1 = 26.2 dm3
V2 = 55.0 dm3
n1T1
subtract n1
V1
mol to be added
n1
=
V2
n2
n2 = n1
xM
g to be added
n2 = 1.10 mol
=
P2V2
n2T2
V2
V1
55.0 dm3
4.003 g He
= 9.24 g He
=
2.31
mol
26.2 dm3
mol He
23
Answer : 9.24 g He
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:
0.885kg
V = 438 L
T = 210C (convert to K)
n = 0.885 kg (convert to mol)
P = unknown
103 g
mol O2
kg
32.00 g O2
= 27.7 mol O2
24.7 mol x 0.0821
P=
nRT
V
atm*L
mol*K
=
210C + 273.15 = 294.15K
x 294.15K
= 1.53 atm
438 L
24
Answer : 1.53 atm
Sample Problem 5.6
PROBLEM:
Using Gas Laws to Determine a Balanced Equation
The piston-cylinders below depict a gaseous reaction carried out
at constant pressure. Before the reaction, the temperature is
150K; when it is complete, the temperature is 300K.
Which of the following balanced equations describes the reaction?
(1) A2 + B2
(3) A + B2
PLAN:
2AB
(2) 2AB + B2
AB2
(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:
Looking at the relationships, the equation that shows a
decrease in the number of moles of gas from 2 to 1 is
equation (3).
25
5.4 Further Applications of the Ideal Gas Law
The Density of a Gas
density = m/V
n = m/M
PV = nRT
PV = (m/M)RT
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.
26
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
0.0821
x 273.15K
mol*K
1.96 g mol CO2
6.022x1023 molecules
= 2.68x1022 molecules CO2/L
L
mol
44.01 g CO2
27
Answers : (a) 1.96 g/L, 2.68x1022 molecules CO2/L
Sample Problem 5.7
Calculating Gas Density
continued
(b)
44.01 g/mol x 1 atm
d=
0.0821
1.83g
mol CO2
L
44.01g CO2
= 1.83 g/L
atm*L x 293K
mol*K
6.022x1023 molecules
mol
= 2.50x1022 molecules CO2/L
28
Answers : (b) 1.83 g/L, 2.50x1022 molecules CO2/L
Calculating the Molar Mass, M, of a Gas
Dumas method Determining the molar
mass of an unknown
volatile liquid.
1800-1884
Since PV = nRT
Then n = PV / RT
And n = mass / M
So m / M = PV / RT
And M = mRT / PV
Or M = d RT / P
29
Sample Problem 5.8
PROBLEM:
Finding the Molar Mass of a Volatile Liquid
An organic chemist isolates from a petroleum sample a
colorless liquid with the properties of cyclohexane (C6H12). She
uses the Dumas method and obtains the following data to
determine its molar mass:
Volume of flask = 213mL
T = 100.00C
Mass of flask + gas = 78.416g
Mass of flask = 77.834g
P = 754 torr
Is the calculated molar mass consistent with the liquid being cyclohexane?
PLAN: Use unit conversions, mass of gas and density-M relationship.
SOLUTION:
m = (78.416 - 77.834)g = 0.582g
0.582g x 0.0821
M=
m RT
=
PV
atm*L
mol*K
x 373K
= 84.4g/mol
0.213L x 0.992atm
M of C6H12 is 84.16g/mol and the calculated value is within experimental error.
30
Answer : 84.4 g/mol
Partial Pressure of a Gas in a Mixture of Gases
Gases mix homogeneously.
Each gas in a mixture behaves as if it is the only gas present,
e.g. its pressure is calculated from PV = nRT with n equal to the
number of moles of that particular gas......called the Partial Pressure.
The gas pressure in a container is the sum of the partial pressures of
all of the gases present.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
1766–1844
P1= χ1 x Ptotal
χ1 =
Sample Problem 5.9
PROBLEM:
where χ1 is the mole fraction
n1
n1 + n2 + n3 +...
=
n1
ntotal
31
Applying Dalton’s Law of Partial Pressures
In a study of O2 uptake by muscle at high altitude, a physiologist
prepares an atmosphere consisting of 79 mol% N2, 17 mol%
16O and 4.0 mol% 18O . (The isotope 18O will be measured to
2,
2
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 χ 18 and P18 from Ptotal and mol% 18O2.
O2
O2
4.0 mol% 18O2
mol% 18O2
= 0.040
SOLUTION:
=
χ 18
O2
100
divide by 100
χ
18O
2
P18
multiply by Ptotal
O2
= χ 18 x Ptotal = 0.040 x 0.75 atm = 0.030 atm
O2
partial pressure P
18O
2
32
Answer : 0.030 atm
Collecting Gas over Water
Often in gas experiments the gas is collected “over water.” So
the gases in the container includes water vapor as a gas. The
water vapor’s partial pressure contributes to the total pressure
in the container.
33
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
) 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
34
Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water
PROBLEM:
Acetylene (C2H2), an important fuel in welding, is produced in
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 738torr and
the volume is 523mL. At the temperature of the gas (230C), the
vapor pressure of water is 21torr. 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 = 717 torr
2 2
Ptotal
P
atm
C2H2
717 torr
= 0.943 atm
P
PV
H2O
760
torr
n=
RT
0.943atm x 0.523L
=
n
= 0.0203 mol
n
C2H2
g
C2H2
atm*L
C2H2
x 296K
0.0821
xM
mol*K
26.04g C2H2
= 0.529 g C2H35
2
0.0203mol
Answer : 0.943 atm, 0.529 g C2H2
mol C2H2
5.5 The Ideal gas Law and Reaction Stoichiometry
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
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
ideal
gas
law
36
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
Cu(s) + H O(g)
SOLUTION: CuO(s) + H (g)
2
2
divide by M
mol of Cu
35.5 g Cu
mol Cu
1 mol H2
63.55 g Cu 1 mol Cu
molar ratio
0.559 mol H2 x 0.0821
mol of H2
atm*L
x
= 0.559 mol H2
498K
mol*K
= 22.6 L
use known P and T to find V
1.01 atm
L of H2
37
Answer : 22.6 L
Sample Problem 5.11
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 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g 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:
PV
n =
Cl2
RT
2K(s) + Cl2(g)
=
2KCl(s)
0.950atm x 5.25L
0.0821
atm*L
mol K
T = 293K
= 0.207mol
x 293K
mol*K
17.0g
P = 0.950atm
0.207mol Cl2
= 0.435mol K
2mol KCl
1mol Cl2
V = 5.25L
n = unknown
= 0.414mol
KCl formed
39.10g K
Cl2 is the limiting reactant.
0.414mol KCl
74.55g KCl
0.435mol K
2mol KCl
2mol K
= 0.435mol
KCl formed
= 30.9 g KCl
mol KCl
38
Answer : 0.414 mol KCl formed
5.6 The Kinetic-Molecular Theory
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.
Molecular View of the Gas Laws
39
Kinetic Molecular Theory of Gases
• The kinetic molecular theory of gases
proposed in 1734 by Bernoulli.
Bernoulli
1700–1782
• The pressure exerted by a gas on the
walls of its container is the sum of the
many collisions by individual molecules,
all moving independently of each other.
• This basic law leads to the equation of
state of real gases discovered by van
der Waals a century later.
40
A molecular description of Boyle’s Law
Boyle’s Law
V α
1
P
n and T are fixed
41
A molecular description of Charles’s Law
Charles’s Law
V α T
n and P are fixed
42
A molecular description of Dalton’s law of partial pressures
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
43
A molecular description of Avogadro’s Law
Avogadro’s Law
V α
n
P and T are fixed
44
Maxwell-Boltzmann Distribution
Law of Molecular Velocity
N (v) = 4π N (
− mv 2
2 2 k BT
m
)3 / 2 v e
2π k BT
1831–1879
1844–1906
N is the total number of
molecules and N(v)dv is the
number of molecules with speeds
between v and v+dv.
vp =
2 k BT
k T
= 1 . 41 B
most probable speed
m
m
v =
8 k BT
= 1 . 60
π m
k BT
m
mean
speed
∞
vrms
3k BT
k T
=
= 1.73 B root - mean - square speed
m
m
∫ N (v)dv = N
45
0
Average velocity is inversely proportional to the square root of molar
mass—applies to other transport phenomena such as diffusion,
thermal conduction, and nonturbulant flow.
Example
46
Kinetic-Molecular Theory
Gas particles are in motion and have a molecular speed, u.
But they are moving at various speeds, some very slow, some very fast,
but most near the average (mean) speed of all of the particles, u(avg) .
Since kinetic energy is defined as
½ mass x (speed) 2, we can define
the average kinetic energy,
Ek(avg) = ½mu 2(avg)
Ek = ½mu 2
< Ek > = ½m< u 2 >
Since energy is a function of
temperature, the average energy
and, hence, average molecular
speed will increase with
temperature.
47
An increase in temperature results in an increase in
average molecular kinetic energy.
Average kinetic energy, Ek(avg) = ½mu 2(avg)
The relationship between average kinetic energy and temperature
is given as, Ek(avg) = 3/2 (R/NA) x T
(where R is the gas constant in energy units, 8.314 J mol-1 K-1
NA is Avogadro’s number, and T temperature in K.)
To have the same average kinetic energy, heavier atoms must have smaller speeds.
The root-mean-square speed, u(rms) , is the speed where a molecule has the average
kinetic energy.
Ek(avg) = ½mu 2(avg) = 3/2 (R/NA) x T
(u2(avg))½ = u(rms)
The relationship between u(rms) and molar mass is:
u(rms) = (3RT/M ) ½
So the speed (or rate of movement) is: rate α 1 / (M ) ½
48
Relationship between molar mass and molecular speed
49
Effusion and Diffusion
Effusion
The process by which a gas
escapes from its container
through a tiny hole into an
evacuated space.
Diffusion
The movement of one gas
through another.
50
Graham’s Law of Effusion
Effusion is the process by which a gas in a closed container moves
through a pin-hole into an evacuated space. This rate is proportional to
the speed of a molecule so.....
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 α 1 / (M ) ½
1805–1869
So doing two identical effusion experiments measuring the rates of two
gases, one known, one unknown, allows the molecular mass of the
unknown to be determined.
rate A
( MB / MA ) ½
=
rate B
51
Sample Problem 5.13
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
CH4
=
√
16.04
M of He = 4.003g/mol
= 2.002
4.003
52
Answer : 2.002
Mean Free Path and Collision Frequency
• Two spherical molecules of diameter d will collide if their
paths are within a distance d of each other.
• The mean distance traveled by molecules between
collisions is called the mean free path (λ).
• The average number of collisions per second that each
molecule undergoes is called the collision frequency (z).
Mean free path =
1
2π d 2 nV
nV is the number of molecules
per unit volume (N/V).
Collision frequency =
Relative mean speed,
mean speed
= 2πd 2ν nV
mean free path
53
vrel speed one molecule approaches another
Range of approaches, but typical
approach is from side
vrel = 2 v
vrel
 8kT 
=

 πµ 
0.5
k = Boltzmann constant =
µ = reduced mass =
If mA = mB , µ =
vrel
 8RT 
= 2

 πM 
R
NA
mA mB
mA + mB
m
2
0.5
= 2v
54
5.7 Real Gases: Deviations from Ideal Gases
Table 5.4 Molar Volume of Some Common Gases at STP
(00C and 1 atm)
Molar Volume
(L/mol)
Gas
He
H2
Ne
Ideal gas
Ar
N2
O2
CO
Cl2
NH3
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
55
Effects of Extreme Conditions on Gas Behavior
PV = nRT
PV
=1
nRT
56
The effect of intermolecular attractions on measured
gas pressure
57
The effect of molecular volume on measured gas volume
58
Van der Waals Equation
n2a
(P + 2 )(V − nb) = nRT
V
adjusts P up adjusts V down
1837–1923
Table 5.5 Van der Waals Constants for Some Common Gases
a
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
NH3
H2O
PIdeal
atm*L2
L
b
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
n
= PReal + a 
V 
VIdeal = VReal − nb
59
2
Effect of stickiness (i.e.
probability of collision ∝
square of number density).
Effect of finite
molecular volume
PV = nRT

n2 
 P + a 2 (V − nb ) = nRT
V 

nRT
n2
P=
−a 2
V − nb
V
Molecular volume
Molecular interactions
60

n2 
 P + a 2 (V − nb ) = nRT
V 

nRT
n2
RT
a
P=
−a 2 =
− 2
V − nb
V
Vm − b Vm
8.314
8.314
61
62
T/Tc = 1.5
Critical isotherm
T/Tc = 1.0
63
Chemistry in Planetary Science
Variations in pressure, temperature, and
composition of the Earth’s atmosphere.
64
65
66