Chapter 5
Gases and the Kinetic-Molecular Theory
5-1
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
5-2
An Overview of the Physical States of Matter
Distinction Between Gases and Liquids/Solids (condensed phases)
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gas have relatively low viscosity.
4. Most gases have relatively low densities under
normal conditions.
5. Gases are miscible.
5-3
States of Matter
Figure 5.1
5-4
A Mercury Barometer
Pressure = force/area
A device used to
measure
atmospheric
pressure
Figure 5.3
5-5
Two Types of Manometer
Figure 5.4
5-6
Table 5.2 Common Units of Pressure
Unit
5-7
Atmospheric Pressure
Scientific Field
pascal (Pa)
kilopascal (kPa)
1.01325 x 105 Pa;
101.325 kPa
SI unit; physics, chemistry;
(1 Pa = 1 N/m2)
atmosphere (atm)
1 atm
chemistry
millimeters of
mercury (Hg)
760 mm Hg
chemistry, medicine, biology
torr
760 torr
chemistry
pounds per square
inch (psi or lb/in2)
14.7 lb/in2
engineering
bar
0.01325 bar
meteorology, chemistry,
physics
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:
1 torr
291.4 mm Hg x
291.4 torr x
1 mm Hg
1 atm
= 291.4 torr
= 0.3834 atm
760 torr
0.3834 atm x
101.325 kPa
1 atm
5-8
= 38.85 kPa
Three laws (Boyle’s, Charles’s and Avogadro’s) are combined to
describe a universal relationship among the key gas variables
(volume, pressure, temperature, amount). This universal
relationship is known as the Ideal Gas Law.
Let’s examine the three individual laws first, and then see how
they are combined to generate the Ideal Gas Law.
5-9
Relationship between volume and pressure of a gas
Boyle’s Law
Figure 5.5
5-10
Boyle’s Law
V
a
1
P
n and T are fixed
(volume is inversely proportional to pressure)
PV = constant
or
V = constant / P
P = pressure
V = volume
n = number of moles of gas
T = temperature
5-11
Relationship between volume and temperature of a gas
Charles’s Law
Figure 5.6
5-12
V a
Boyle’s Law:
Charles’s Law:
V
Amonton’s Law:
T
5-13
V a T
P a T
= constant
Combined Gas Law:
(Boyle’s + Charles’s)
P
= constant
T
P
1
V a
T
P
n and T are fixed
P and n are fixed
V = constant x T
V and n are fixed
P = constant x T
V = constant x
T
PV
P
T
= constant
An experiment to study the relationship
between volume and amount of a gas
Figure 5.7
Avogadro’s Law
V a n
(P and T fixed)
At fixed T and P, equal volumes of any ideal gas contain
equal numbers of particles (or moles).
5-14
Standard Molar Volume
STP
0 oC (273.15 K)
1 atm (760 torr)
Standard
Molar
Volume:
22.4141 L or
22.4 L
5-15
Figure 5.8
THE IDEAL GAS LAW
PV = nRT
PV
R=
nT
=
1 atm x 22.414 L
1 mol x 273.15 K
0.0821atm L
mol K
=
nRT
PV = nRT or V =
At fixed n and T:
Boyle’s Law
V=
constant
P
5-16
At fixed n and P:
Charles’s Law
V = constant x T
P
At fixed P and T:
Avogadro’s Law
V = constant x n
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
SOLUTION:
n and T are constant
P1 = 1.12 atm
P2 = 2.64 atm
V1 = 24.8 cm3
V2 = unknown
1 cm3 = 1 mL
V1 in mL
103 mL
1 mL
=1L
24.8
V1 in L
P1V1
x P1/P2
V2 in L
n1T1
V2 =
P1V1
P2
5-17
cm3
x
=
1
cm3
P2V2
L
x
103 mL
or
n2T2
= 0.0248 L x
1.12 atm
2.64 atm
= 0.0248 L
P1V1 = P2V2
= 0.0105 L
Sample Problem 5.3
PROBLEM:
Applying the Temperature-Pressure Relationship
A 1 L steel tank is fitted with a safety valve that opens if the
internal pressure exceeds 1.00 x 103 torr. It is filled with helium
at 23 oC and 0.991 atm and placed in boiling water at exactly
100 oC. Will the safety valve open?
PLAN:
P1 (atm)
SOLUTION:
T1 and T2 (oC)
1 atm = 760 torr
P1 (torr)
P1 = 0.991 atm
K = oC + 273.15
T1 and T2 (K)
n1T1
P2 (torr)
P2 = unknown
T1 = 23 oC
P1V1
x T2/T1
=
P2V2
T2 = 100
or
n2T2
T2
T1
= 753 torr x
P1
T1
0.991 atm x 760 torr
1 atm
P2 = P1
5-18
n and V are constant
373 K
=
oC
P2
T2
= 753 torr
= 949 torr
296 K
(valve will not open)
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 (V2). When 1.10 mol of He (n1) are added to
the blimp, the volume is 26.2 dm3 (V1). How many more grams of
He must be added to make it rise? Assume constant T and P.
PLAN: We are given the initial n1 and V1 and the final V2. We need to find
n2 and convert it from moles to grams.
n1(mol) of He
x V2 / V1
n2(mol) of He
subtract n1
mol to be added
n1 = 1.10 mol
V1 = 26.2
V1
n1
xM
g He to add
P and T are constant
SOLUTION:
n2 = 1.10 mol x
dm3
=
V2
n2
55.0 dm3
26.2
n2 = unknown
P1V1
dm3
n1T1
V2 = 55.0
or
n2 = n1
n2T2
V2
V1
= 2.31 mol
dm3
2.31 mol - 1.10 mol = 1.21 mol x
5-19
=
P2V2
4.003 g He
mol He
= 4.84 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 21 oC.
V, T and mass, which can be converted to moles (n), are given. Use
the ideal gas law to find P.
SOLUTION:
0.885 kg x
V = 438 L
T = 21 oC (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=
5-20
nRT
V
atm L
mol K
=
21 oC + 273.15 = 294 K
x 294 K
= 1.53 atm
438 L
The Density of a Gas
PV = nRT or PV = m/M x RT where m = mass and M = molar mass
m/V = d = (M x P)/RT where d = density
5-21
Sample Problem 5.6
Calculating the Density of a Gas
Calculate the density (in g/L) of carbon dioxide and the number
of molecules per liter (a) at STP (0 oC and 1 atm) and (b) at
ordinary room conditions (20. oC 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, the molar mass can be
determined. Convert mass/L to molecules/L using Avogadro’s number.
d = mass/volume
SOLUTION:
PV = nRT
44.01 g/mol
(a)
L
5-22
x
mol CO2
44.01 g CO2
x
atm L
MxP
RT
= 1.96 g/L
x 273 K
mol K
6.022 x 1023 molecules
mol
d =
x 1 atm
d=
0.0821
1.96 g
V = nRT/P
= 2.68 x 1022 molecules CO2/L
Sample Problem 5.6
(continued)
44.01 g/mol
(b)
x 1 atm
d=
= 1.83 g/L
0.0821
atm L
x 293 K
mol K
1.83 g
L
5-23
x
mol CO2
44.01 g CO2
x
6.022 x 1023 molecules
mol
= 2.50 x 1022 molecules CO2/L
The Molar Mass of a Gas
n=
mass
M
=
PV
RT
M=
mRT
PV
d=
m
V
d RT
M=
P
5-24
Determining the molar
mass of an unknown
volatile liquid
[based on the method of J.B.A.
Dumas (1800 -1884)]
M = mRT/PV
or
M = dRT/P
Figure 5.11
5-25
Sample Problem 5.7
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 = 213 mL
T = 100.0 oC
Mass of flask + gas = 78.416 g
Mass of flask = 77.834 g
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=
d=
m = (78.416 - 77.834)g = 0.582 g of gas
m RT
=
0.582 g x 0.0821
atm L
mol K
VP
0.213 L
MxP
RT
mRT
M=
x 373 K
= 84.4 g/mol
x 0.992 atm
M of C6H12 is 84.16 g/mol - the calculated value is within experimental error
5-26
VP
Dalton’s Law of Partial Pressures
In a mixture of unreacting gases, the total pressure is equal to the
sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + P3 + ...
where
P1 = c1 x Ptotal
c1 =
5-27
and c1 is the mole fraction
n1
n1 + n2 + n3 +...
=
n1
ntotal
Sample Problem 5.8
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 0.75
atm 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% 18O2
2
c
SOLUTION:
divide by 100
c
18O
P18
2
multiply by Ptotal
partial pressure P
18O
2
5-28
O2
= c
18O
18O
=
2
4.0 mol% 18O2
= 0.040
100
x Ptotal = 0.040 x 0.75 atm = 0.030 atm
2
Figure 5.12
5-29
Collecting a water-insoluble gaseous reaction
product and determining its pressure
Sample Problem 5.9
PROBLEM:
Calculating the Amount of Gas Collected Over Water
Acetylene (C2H2), an important fuel in welding, is produced in
the laboratory when calcium carbide (CaC2) reacts 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 (23 oC),
the vapor pressure of water is 21 torr. How many grams of
acetylene are collected?
PLAN: The difference in pressures will give P for C2H2. The ideal gas law
allows a determination of n. Converting n to grams requires the
molar mass, M.
SOLUTION: PC H = (738 - 21) torr = 717 torr
2 2
Ptotal
P
C2H2
atm
-P
PV
717
torr
x
H2O
= 0.943 atm
n=
760 torr
RT
n
g
C2H2
C2H2
xM
5-30
Sample Problem 5.9
n
C2H2
(continued)
=
0.943 atm x 0.523 L
0.0821
atm L
= 0.0203 mol C2H2
x
296 K
mol K
0.0203 mol x
26.04 g C2H2
mol C2H2
5-31
= 0.529 g C2H2
Summary of the stoichiometric relationships
between 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
Figure 15.13
5-32
ideal
gas
law
Sample Problem 5.10
Using Gas Variables to Find Amount of
Reactants and Products
PROBLEM:
A laboratory-scale method for reducing a metal oxide is to heat it
with H2. The pure metal and H2O are products. What volume of
H2 at 765 torr and 225 oC is needed to form 35.5 g of Cu from
copper(II) oxide?
PLAN: This problem requires stoichiometry and the gas laws; write a
balanced equation and use the moles of Cu to calculate moles and
then volume of H2 gas.
mass (g) of Cu
SOLUTION:
divide by M
mol of Cu
35.5 g Cu x
CuO(s) + H2(g)
mol Cu
63.55 g Cu
molar ratio
mol of H2
0.559 mol H2 x 0.0821
use known P and T to find V
L of H2
5-33
x
Cu(s) + H2O(g)
1 mol H2
1 mol Cu
atm L
x
mol K
1.01 atm
= 0.559 mol H2
498 K
= 22.6 L
Using the Ideal Gas Law in a Limiting Reactant
Sample Problem 5.11
Problem
PROBLEM: The alkali metals (Group 1A) react with the halogens (Group 7A) to
form ionic metal halides. What mass of potassium chloride forms when
5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of
potassium?
PLAN: Write a balanced equation, and use the ideal gas law to find the number
of moles of reactants, the limiting reactant and the moles of product.
2K(s) + Cl2(g)
SOLUTION:
PV
n =
Cl2
RT
=
0.950 atm x 5.25 L
0.0821
atm L
P = 0.950 atm
2KCl(s)
T = 293 K
= 0.207 mol Cl2
V = 5.25 L
n = unknown
x 293 K
mol K
17.0 g x
mol K
= 0.435 mol K
0.207 mol Cl2 x
39.10 g K
0.435 mol K x
2 mol KCl
1 mol Cl2
2 mol KCl
2 mol K
0.414 mol KCl x
5-34
74.55 g KCl
mol KCl
= 30.9 g KCl
= 0.414 mol
KCl formed
= 0.435 mol
KCl formed
Cl2 is the limiting reactant
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
Distribution of molecular speeds at three temperatures
The most probable
speed increases as
the temperature
increases
The average kinetic
energy, Ek, is
proportional to the
absolute temperature
Figure 5.14
5-36
For N2 gas
A molecular description of Boyle’s Law
Figure 5.15
5-37
A molecular description of Dalton’s law of partial pressures
Figure 5.16
5-38
A molecular description of Charles’s Law
Figure 5.17
5-39
A molecular description of Avogadro’s Law
Figure 5.18
5-40
Why do equal numbers of molecules of two different gases, such
as O2 and H2, occupy the same volume (c.f. standard molar volume)?
At constant T, two gases possess the same kinetic energy; thus,
the heavier gas must be moving more slowly.
Ek = 1/2 massx u2
Ek = 1/2 mass x speed2
urms =
√ 3RT
M
=
u2 is the average of the squares of
the molecular speeds; its square root
equals urms
root-mean-square speed; a molecule moving
at this speed has the average kinetic energy
R = 8.314 joule/mol K
urms
5-41
a
1
√M
Relationship between molar mass and molecular speed
Figure 5.19
At a given temperature, gases with lower molar masses
have higher most probable speeds
5-42
EFFUSION: the process by which a gas escapes from its container
through a tiny hole into an evacuated space
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
1
√M
(related to the rms speed)
rateA/rateB = MB1/2/MA1/2
The same relationships pertain to gaseous diffusion rates!
5-43
Sample Problem 5.12
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 of each gas. Find the molar mass of both gases and find
the inverse square root of their masses.
SOLUTION:
rate
He
rate
5-44
M of He = 4.003 g/mol
M of CH4 = 16.04 g/mol
CH4
=
√
16.04
4.003
= 2.002
Diffusion of a gas
particle through a
space filled with
other particles
Distribution of Molecular Speeds
Mean Free Path: the average distance
a molecule travels between collisions at
a given T and P
Collision Frequency: the average number
of collisions per second (has implications
for chemical reaction rates)
5-45
Figure 5.20
Real Gases
Molecules are not
points of mass.
There are attractive and
repulsive forces between
molecules.
Real gases approach ideal
behavior at high T and low P.
5-46
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Figure 5.21
The behavior of
several real gases
with increasing
external pressure
At moderately high P: intermolecular attractions dominate
At very high P: molecular volume effects dominate
5-47
The effect of intermolecular attractions on measured gas pressure
Figure 5.22
5-48
The effect of molecular volume on measured gas volume
Figure 5.23
5-49
The van der Waals equation for n moles of a real gas
(adjusts
V down)
(P + n2a/V2) (V - nb) = nRT
(adjusts
P up)
a and b are the
van der Waals constants
5-50
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