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Chapter 5
Gases and the Kinetic-Molecular Theory
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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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.
5-3
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Figure 5.1
5-4
The three states of matter.
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Figure 5.2
5-5
Effect of atmospheric pressure on objects
at the Earth’s surface.
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Figure 5.3
5-6
A mercury barometer.
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Table 5.1 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.
5-7
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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,
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:
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
5-8
= 38.85 kPa
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Figure 5.4
The relationship between the volume
and pressure of a gas.
Boyle’s Law
5-9
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Figure 5.5
The relationship between the
volume and temperature of a
gas.
Charles’s Law
5-10
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V a
Boyle’s Law
VxP
V a T
V
= constant
T
Amontons’s Law
T
Combined gas law
5-11
P
= constant
Charles’s Law
P
1
P a T
= constant
V a
n and T are fixed
V = constant / P
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 5.6
5-12
An experiment to study the relationship between the
volume and amount of a gas.
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Figure 5.7
5-13
Standard molar volume.
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Figure 5.8
5-14
The volume of 1 mol of an ideal gas compared with some
familiar objects.
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THE IDEAL GAS LAW
Figure 5.9
PV = nRT
3 significant figures
PV
R=
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
5-15
P
fixed n and P
fixed P and T
Charles’s Law
Avogadro’s Law
V=
constant X T
V=
constant X n
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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, 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:
V1 in cm3
1cm3=1mL
V1 in mL
unit
conversion
103 mL=1L
V1 in L
xP1/P2
gas law
calculation
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
n1T1
V2 =
P1V1
P2
5-16
P and T are constant
SOLUTION:
=
P2V2
= 0.0248 L
P1V1 = P2V2
n2T2
1.12 atm
= 0.0248 L
2.46 atm
= 0.0105 L
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Sample Problem 5.3
PROBLEM:
Applying the Pressure-Temperature 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)
P2 = unknown
T1 = 230C
T2 = 1000C
P1V1
=
n1T1
P2V2
n2T2
0.991 atm 760 torr = 753 torr
1 atm
P2 = P1
5-17
P1 = 0.991atm
T2
T1
= 753 torr
373K
296K
= 949 torr
P1
T1
=
P2
T2
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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.
n1(mol) of He
P and T are constant
SOLUTION:
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
n2 = n1
xM
g to be added
5-18
55.0 dm3
P1V1
n1T1
=
P2V2
n2T2
V2
V1
4.003 g He
= 9.24 g He
n2 = 1.10 mol
= 2.31 mol
3
26.2 dm
mol He
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Sample Problem 5.5
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.
PROBLEM:
PLAN:
Solving for an Unknown Gas Variable at Fixed
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=
5-19
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 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.
New figures go here.
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:
5-20
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
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.
5-21
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Sample Problem 5.7
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).
PROBLEM:
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
(a)
d=
0.0821
5-22
x 1atm
1.96 g
mol CO2
L
44.01 g CO2
atm*L
mol*K
6.022x1023 molecules
mol
= 1.96 g/L
x 273.15K
= 2.68x1022 molecules CO2/L
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Sample Problem 5.6
Calculating Gas Density
continued
(b)
44.01 g/mol x 1 atm
d=
0.0821
5-23
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
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Sample Problem 5.8
PROBLEM:
Finding the Molar Mass of a Volatile Liquid
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
P = 754 torr
Mass of flask + gas = 78.416 g Mass of flask = 77.834 g
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
=
0.582 g x 0.0821
atm*L
mol*K
0.213 L x 0.992 atm
5-24
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
where c1 is the mole fraction
c1 =
5-25
n1
n1 + n2 + n3 +...
=
n1
ntotal
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Sample Problem 5.9
PROBLEM:
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.
18
PLAN: Find the c 18Oand P18O from Ptotal and mol% O2.
2
mol%
18O
2
SOLUTION:
2
divide by 100
c
18O
P18
2
multiply by Ptotal
partial pressure P
18O
5-26
2
O2
c
18O
=
2
4.0 mol% 18O2
= 0.040
100
= c 18 x Ptotal = 0.040 x 0.75 atm = 0.030 atm
O2
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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
5-27
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Table 5.2 Vapor Pressure of Water (P
H2O
T(0C)
0
5
10
11
12
13
14
15
16
18
20
22
24
26
28
5-28
) 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
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Figure 5.10
5-29
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 (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.
P
SOLUTION:
C2H2 = (738-21)torr = 717torr
Ptotal
P
C2H2
atm
P
= 0.943atm
PV
717torr
H2O
n=
760torr
RT
n
g
C2H2
C2H2
5-30
xM
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Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water
continued
n
C2H2
=
0.943atm x
0.0821
atm*L
0.523L
= 0.0203mol
x 296K
mol*K
0.0203mol
26.04g C2H2
mol C2H2
5-31
= 0.529 g C2H2
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Figure 15.11
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
5-32
amount
(mol)
amount
(mol)
P,V,T
of gas A
of gas B
of gas B
molar ratio from
balanced equation
ideal
gas
law
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Sample Problem 5.11
Using Gas Variables to Find Amounts 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
SOLUTION:
divide by M
mol of Cu
35.5 g Cu
CuO(s) + H2(g)
mol Cu
0.559 mol H2 x 0.0821
use known P and T to find V
L of H2
5-33
1 mol H2
63.55 g Cu 1 mol Cu
molar ratio
mol of H2
Cu(s) + H2O(g)
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:
2K(s) + Cl2(g)
2KCl(s)
P = 0.950 atm
V = 5.25 L
PV
0.950 atm x 5.25L
T = 293K
n = unknown
n =
=
= 0.207 mol
Cl2
RT
atm*L
0.0821
x 293K
mol*K
2 mol KCl
mol
K
0.207
mol
Cl
= 0.414 mol
2
17.0g
= 0.435 mol K
1 mol Cl2
KCl formed
39.10 g K
2 mol KCl
Cl2 is the limiting reactant.
0.435 mol K
= 0.435 mol
2 mol K
KCl formed
74.55 g KCl
0.414 mol KCl
= 30.9 g KCl
mol KCl
5-34
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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.
5-35
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Figure 5.12
5-36
Distribution of molecular speeds at three temperatures.
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Figure 5.13
5-37
A molecular description of Boyle’s Law.
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Figure 5.14
5-38
A molecular description of Dalton’s law of partial pressures.
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Figure 5.15
5-39
A molecular description of Charles’s Law.
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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
5-40
1
√M
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Figure 5.16
5-41
A molecular description of Avogadro’s Law.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 5.17
Relationship between molar mass and molecular speed.
Ek = 3/2 (R/NA) T
5-42
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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 (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
5-43
CH4
=
√
16.04
4.003
= 2.002
M of He = 4.003g/mol
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Figure 5.18
Diffusion of a gas particle through a
space filled with other particles.
distribution of molecular speeds
mean free path
collision frequency
5-44
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Table 5.3 Molar Volume of Some Common Gases at STP
(00C and 1 atm)
Gas
He
H2
Ne
Ideal gas
Ar
N2
O2
CO
Cl2
NH3
5-45
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 5.19
The behavior of several
real gases with increasing
external pressure.
5-46
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Figure 5.20
5-47
The effect of intermolecular attractions on
measured gas pressure.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 5.21
5-48
The effect of molecular volume on measured gas volume.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 5.54 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
5-49
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