Chapter 9
The Gaseous State
• The Behavior of Gases
• Factors that Affect the Properties of
Gases
• The Ideal Gas Law
• Kinetic-Molecular Theory of Gases
• Gases and Chemical Reactions
9-1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Properties of Gas Molecules
• Gases consist of particles that are
relatively far apart (as compared to
liquids and solids).
– This results in observed properties:
• Lower densities
• Compressible by applying external pressure
• Gas particles move about rapidly.
• Gas particles have little effect on one
another unless they collide.
– When they collide, gas particles simply
bounce off one another.
• Gases expand to fill containers.
– Take volume and shape of their containers
9-2
Effect of Temperature and Density
• All gases expand if heated and contract if
cooled.
– Heat increases the kinetic energy of gas particles,
making them move faster and farther apart.
• Since the gas particles move farther apart, there
are fewer particles within a given volume.
– Therefore, warm gases have lower densities.
9-3
1
Effect of Temperature and Density
9-4
Pressure
• Is amount of force
applied per unit
area
P = force
area
• For a gas in a
container:
P = force of gas particles
area of container
9-5
Pressure
Figure 9.9
9-6
2
Pressure
9-7
Pressure
• Pressure is
measured using a
barometer.
• Units of pressure
1 atm = 760 mm Hg
(mm Hg and torr are
the same)
1 atm = 14.7 lb/in2 (psi)
1 atm = 101, 325 Pa
Figure 9.11
9-8
Practice – Pressure Conversions
• The pressure of a gas is 8.25 x 104 Pa.
What is this pressure expressed in units
of atm and torr?
9-9
3
Practice Solutions – Pressure
Conversions
• The pressure of a gas is 8.25 x 104 Pa.
What is this pressure expressed in units
of atm and torr?
• Knowing the conversion between atm
and Pa is: 1 atm = 101, 325 Pa
8.25 x 104 Pa x 1 atm
= 8.14 x 10-1 atm
101,325 Pa
• Knowing the conversion between atm
and torr is: 1 atm = 760 torr
8.14 x 10-1 atm x 760 torr = 6.19 x 102 torr
1 atm
9 - 10
Factors That Affect Gases
•
•
•
•
•
Volume
– Measured in liters (L)
Pressure
– Measured in atmospheres
(atm)
Temperature
– Measured in Kelvin (K)
Amount of particles
– Measured in moles
An ideal gas is a gas that
behaves according to predicted
linear relationships
9 - 11
Volume vs. Pressure
Figure 9.14
9 - 12
4
Volume vs. Pressure
•
Boyle’s law
– For a given mass of
gas at constant
temperature, volume
varies inversely with
pressure.
•
Figure 9.15
V∞1
P
P1V1 = P2V2
As volume
increases, pressure
decreases.
9 - 13
Volume vs. Temperature
9 - 14
Volume vs. Temperature
•
Charles’s law
– For a given mass of gas at
constant pressure, volume
is directly proportional to
temperature on an absolute
(kelvin) scale.
•
TK = T°C + 273.15
V ∞ T (in K)
V1 = V2
T1 T2
As volume increases,
temperature increases.
9 - 15
5
Practice – Boyle’s and
Charles’s Law
• What pressure is needed to
compress 455 mL of oxygen gas at
2.50 atm to a volume of 282 mL?
• A sample of carbon monoxide gas
occupies 150.0 mL at 25.0°C. It is
then cooled at constant pressure
until it occupies 100.0 mL. What is
the new temperature in degrees
Celsius?
9 - 16
Practice Solutions – Boyle’s
and Charles’s Law
•
What pressure is needed to compress 455 mL of
oxygen gas at 2.50 atm to a volume of 282 mL?
P1 = 2.50 atm
V1 = 455 mL
Using Boyle’s Law:
P2 = ?
V2 = 282 mL
P1V1 = P2V2
P1V1 = P2
V2
P2 = (2.50 atm x 455 mL) = 4.03 atm
282 mL
2. Solving for P2:
9 - 17
Practice Solutions – Boyle’s
and Charles’s Law
A sample of carbon monoxide gas occupies
150.0 mL at 25.0°C. It is then cooled at
constant pressure until it occupies 100.0
mL. What is the new temperature in degrees
Celsius?
1. You MUST convert temperatures to Kelvin
when dealing with gases – ALWAYS!
•
T1 = 25.0°C + 273.15 = 298.2 K
T2 = ?
9 - 18
6
Practice Solutions – Charles’s Law
V1 = 150.0 mL
V2 = 100.0 mL
T1 = 298.2 K
T2 = ?
2. Using Charles’s Law:
V1 = V2
T1 T 2
3. Solving for T2: T2 = V2T1
V1
T2 = (100.0 mL x 298.2 K) = 198.8 K
150.0 mL
T2 (in °C) = 198.8 K – 273.15 = -74.4°C
9 - 19
Volume, Pressure and
Temperature
• Combined gas law
– For a constant amount of gas, volume
is proportional to absolute temperature
divided by pressure.
V ∞ (T/P)
P1V1 = P2V2
T1
T2
9 - 20
Combining Volumes
• Gay-Lussac’s law of combining volumes
– Gases combine in simple whole-number
volume proportions at constant temperature
and pressure.
Figure 9.19
9 - 21
7
Volume vs. Amount
•
Avogadro’s hypothesis
– The volume occupied by a
gas at a given temperature
and pressure is directly
proportional to the number
of gas particles and thus to
the moles of gas.
V∞n
V1 = V2
n1 n2
•
As volume increases, the
moles of gas increase.
9 - 22
Volume vs. Amount
• Avogadro’s hypothesis Cont’d
– At a given pressure and temperature,
equal volumes of all gases contain
equal numbers of moles (or particles).
• This hypothesis was measured at
standard temperature and pressure
(STP):
0°C and 1 atm
• The volume of an ideal gas at STP is
called its molar volume:
1 mole of gas = 22.414 L
9 - 23
Practice – The Combined Gas Law
and Avogadro’s Hypothesis
• A sample of hydrogen gas occupies
1.25 L at 80.0°C and 2.75 atm. What
volume will it occupy at 185°C and
5.00 atm?
• If a balloon initially contains 0.35 mol
of gas and occupies 8.2 L, what is the
volume of the balloon when 1.2 mol of
gas is present if temperature and
pressure remain constant?
9 - 24
8
Practice Solutions – The Combined
Gas Law and Avogadro’s Law
A sample of hydrogen gas occupies 1.25 L at
80.0°C and 2.75 atm. What volume will it
occupy at 185°C and 5.00 atm?
1. First, convert the temperatures into Kelvin:
T1 = 80.0°C + 273.15 K = 353.2 K
T2 = 185°C + 273.15 K = 458 K
•
V1 = 1.25 L
T1 = 353.2 K
P1 = 2.75 atm
V2 = ?
T2 = 458 K
P2 = 5.00 atm
9 - 25
Practice Solutions – The Combined Gas Law
V1 = 1.25 L
T1 = 353.2 K
P1 = 2.75 atm
V2 = ?
T2 = 458 K
P2 = 5.00 atm
2. The equation we want to use is the Combined
Gas Law:
P1V1 = P2V2
T1
T2
3. Solving for V2:
V2 = P1V1T2
T1P2
V2 = (2.75 atm x 1.25 L x 458 K) = 0.891 L
(353.2 K x 5.00 atm)
9 - 26
Practice Solutions – The Combined
Gas Law and Avogadro’s Law
If a balloon initially contains 0.35 mol of gas
and occupies 8.2 L, what is the volume of the
balloon when 1.2 mol of gas is present if
temperature and pressure remain constant?
n1 = 0.35 mol
n2 = 1.2 mol
V1 = 8.2 L
V2 = ?
1. Using Avogadro’s Hypothesis equation:
V1 = V2
n1 n2
2. Solving for V2:
V2 = V1n2 = (8.2 L x 1.2 mol) = 28 L
n1
0.35 mol
•
9 - 27
9
•
•
•
•
•
Summary – The Gas Laws
Boyle’s law
V ∞ 1 at constant T, n
P
Charles’s law
V ∞ T at constant P, n
Avogadro’s hypothesis
V ∞ n at constant T, P
Combining them all into one proportionality:
V ∞ nT
P
To express this as an equality, we use a
constant:
V = constant x nT
9 - 28
P
The Ideal Gas Law
V = constant x nT
P
•
•
This constant is called the ideal gas constant,
R:
V = RnT
P
This gives us the Ideal Gas Law:
PV = nRT
•
R is calculated at STP using the molar volume
of a gas:
R = (1.000 atm x 22.414 L) = 0.08206 L atm
(1.000 mol x 273.15 K)
K mol
9 - 29
Practice - Calculations with the
Ideal Gas Law
• The volume of an oxygen cylinder used
as a portable breathing supply is 2.025 L.
When the cylinder is empty at 29.2°C, it
has a pressure of 723 torr. How many
moles of oxygen gas remain in the
cylinder?
• The volume of an oxygen cylinder used
as a portable breathing supply is 1.85 L.
What mass of oxygen gas remains in the
cylinder when it is empty if the pressure
is 755 torr and the temperature is 18.1°C?
9 - 30
10
Practice Solutions - Calculations
with the Ideal Gas Law
The volume of an oxygen cylinder used as a
portable breathing supply is 2.025 L. When the
cylinder is empty at 29.2°C, it has a pressure of
723 torr. How many moles of oxygen gas
remain in the cylinder?
1. When using the Ideal Gas Law, you want to
make sure the units of each variable (V, P, n,
and T) match those of R:
R = 0.08206 L atm/(mol K)
T = 29.2°C + 273.15 = 302.4 K
V = 2.025 L
P = 723 torr x 1 atm = 0.951 atm
n=?
760 torr
•
9 - 31
Practice Solutions - Calculations
with the Ideal Gas Law
R = 0.08206 L atm/(mol K)
T = 29.2°C + 273.15 = 302.4 K
V = 2.025 L
P = 723 torr x 1 atm = 0.951 atm
n=?
760 torr
2. Using the Ideal Gas Law:
PV = nRT
3. Solving for n:
n = PV =
(0.951 atm x 2.025 L)
= 0.0776 mol
RT ((0.08206 L atm) x 302.4 K)
mol K
9 - 32
Practice Solutions - Calculations
with the Ideal Gas Law
The volume of an oxygen cylinder used as a
portable breathing supply is 1.85 L. What mass of
oxygen gas remains in the cylinder when it is
empty if the pressure is 755 torr and the
temperature is 18.1°C?
1. When using the Ideal Gas Law, you want to make
sure the units of each variable (V, P, n, and T)
match those of R:
R = 0.08206 L atm/(mol K)
MM = 32.0 g/mol
T = 18.1°C + 273.15 = 291.3 K
V = 1.85 L
P = 755 torr x 1 atm = 0.993 atm mass = ?
760 torr
•
9 - 33
11
Practice Solutions - Calculations
with the Ideal Gas Law
R = 0.08206 L atm/(mol K)
MM = 32.0 g/mol
T = 18.1°C + 273.15 = 291.3 K
V = 1.85 L
P = 755 torr x 1 atm = 0.993 atm
mass = ?
760 torr
2. Using the Ideal Gas Law:
PV = nRT
3. Solving for n (n is the only variable we don’t
have and we will use it to get m):
(0.993 atm x 1.85 L)
= 0.0769 mol
n = PV =
RT ((0.08206 L atm) x 291.3 K)
mol K
9 - 34
Practice Solutions - Calculations
with the Ideal Gas Law
4. Now we need to convert from moles (n)
to mass (m):
MM = 32.0 g/mol
n = 0.0769 mol
n = mass
MM
5. Solving for mass:
mass = n x MM
mass = 0.0769 mol x 32.0 g = 2.46 g
1 mol
9 - 35
Density of a Gas
Density (d) = mass (m)
Volume (V)
•
We can solve for density in the Ideal Gas Law
using substitution:
PV = nRT
n = m/MM
PV = mRT
MM
d = m = P(MM)
V
RT
9 - 36
12
Dalton’s Law of Partial Pressures
• Gases in a mixture
behave
independently and
exert the same
pressure they
would exert if they
were in a container
alone.
Ptotal = PA + PB + PC + …
Ptotal = Pdry air + Pwater
9 - 37
Table 9.2 Vapor Pressure of Water at
Various Temperatures
9 - 38
Practice – Partial Pressures
• Suppose 2.25 L of H2 gas is
collected over water at 18.0°C
and 722.8 torr. How many
moles of H2 are produced in
this reaction?
9 - 39
13
Practice Solutions – Partial
Pressures
• Suppose 2.25 L of H2 gas is collected
over water at 18.0°C and 722.8 torr. How
many moles of H2 are produced in this
reaction?
V = 2.25 L
n=?
T = 18.0°C + 273.15 = 291.2 K
P = 722.8 torr x 1 atm = 0.951 atm
760 torr
9 - 40
Practice Solutions – Partial
Pressures
V = 2.25 L
R = 0.08206 L atm/(mol K)
T = 18.0°C + 273.15 = 291.2 K
n=?
P = 722.8 torr x 1 atm = 0.951 atm
760 torr
1. Using the Ideal Gas Law:
PV = nRT
2. Solving for n: n = PV
RT
n = (0.951 atm x 2.25 L) = 0.0895 mol
(0.08206 L atm x 291.2 K)
mol K
9 - 41
Kinetic-Molecular Theory of Gases
• A model that explains experimental
observations about gases under normal
temperature and pressure conditions
that we encounter in our environment
• Has 5 postulates:
1. Gases are composed of small and
widely separated particles (molecules
or atoms).
– Low volumes of particles
– Low densities of gas
– High compressibility
9 - 42
14
Kinetic-Molecular Theory of Gases
•
Postulates continued:
2. Particles of a gas behave independently of
one another.
– Independent movement, unless two
particles collide.
– No forces of attraction or repulsion
operate between and among gas particles.
3. Each particle in a gas is in rapid, straight-line
motion, until it collides with another
molecule or with its container.
– When collisions occur, the collisions are
perfectly elastic.
– Energy transferred from one particle to
another with no net loss of energy.
9 - 43
Kinetic-Molecular Theory of Gases
• Postulates continued:
4. The pressure of a gas arises from the
sum of the collisions of the particles
with the walls of the container.
– The smaller the container, the more
collisions between the gas particles and
the walls of the container, resulting in
higher pressures.
– Predicts pressure should be proportional
to the number of gas particles.
9 - 44
Kinetic-Molecular Theory of Gases
• Postulates continued:
5. The average kinetic energy of gas
particles depends on the absolute
temperature.
– Relationship between kinetic energy (KE)
and velocity (v) of the gas particles:
KEav = ½m(vav)2
– The average velocity for gas particles is
greater at higher temperatures.
– Thus, if the particles move faster, they hit
the walls of the container more often,
resulting in higher pressure.
9 - 45
15
Kinetic-Molecular Theory of Gases
• 5th Postulate continued:
Figure 9.22
9 - 46
Kinetic-Molecular Theory of Gases
Figure 9.23
9 - 47
Diffusion
• The movement of gas particles from regions of
high concentration to regions of low
concentration.
KEav = ½m(vav)2
• Gases have different average velocities even
though they all have the same average kinetic
energy.
• Light molecules or atoms move faster than
heavy molecules or atoms.
9 - 48
16
Effusion
• The passage of a
gas through a
small opening.
• Smaller particles
effuse faster than
larger ones.
Figure 9.25
9 - 49
Gases and Chemical Reactions
• Product volume from reactant
volume
– We convert from volume to moles or
moles to volume using the Ideal Gas
Law as our conversion factor.
PV=nRT Volume
Mole
Volume PV=nRT
Mole A
Mole B
A
B
ratio
9 - 50
Gases and Chemical Reactions
• Moles and mass from volume
Remember the stoichiometry pattern we used before:
Mass
A
Molar
Mass
Mole A
Mole
ratio
Mole B
Molar
Mass
Mass
B
Modifying the pattern to include volume:
Volume
A
PV=nRT
Mole
Mole A ratio
Molar
Mole B Mass
Mass
B
9 - 51
17
Practice – Gas Reactions
•
•
A sample of hydrogen gas has a volume of 7.49
L at a pressure of 22.0 atm and a temperature of
32.0°C. What volume of gaseous water is
produced by the following reaction at 125.0°C
and 0.975 atm, if all the hydrogen gas reacts
with iron(III) oxide?
Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(l)
How many moles of H2SO4 form if 2.38 L of SO3
gas, measured at 65.0°C and 1.05 atm, react
with sufficient water? How many grams? The
balanced equation is as follows:
SO3(g) + H2O(l) H2SO4(l)
9 - 52
Practice Solutions – Gas Reactions
A sample of hydrogen gas has a volume of 7.49
L at a pressure of 22.0 atm and a temperature of
32.0°C. What volume of gaseous water is
produced by the following reaction at 125.0°C
and 0.975 atm, if all the hydrogen gas reacts
with iron(III) oxide?
Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(l)
Vhydrogen = 7.49 L
Vwater = ?
P2 = 0.975 atm
P1 = 22.0 atm
T1 = 32.0°C + 273.15 = 305.2 K
T2 = 125.0°C + 273.15 = 398.2 K
•
9 - 53
Practice Solutions – Gas Reactions
Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(l)
Vhydrogen = 7.49 L
Vwater = ?
P1 = 22.0 atm
P2 = 0.975 atm
T1 = 305.2 K
T2 = 398.2 K
1. First, we need to find moles of hydrogen gas.
The way to do that is to use the Ideal Gas Law.
PV = nRT
nhydrogen = PV = (22.0 atm x 7.49 L) = 6.58 mol H2
RT (0.08206 L atm x 305.2 K)
mol K
9 - 54
18
Practice Solutions – Gas Reactions
Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(l)
Vhydrogen = 7.49 L
Vwater = ?
P1 = 22.0 atm
P2 = 0.975 atm
T1 = 305.2 K
T2 = 398.2 K
2. Then, we need to find moles of water from
moles of hydrogen gas. We use a mole ratio
from the balanced chemical equation to do this.
6.58 mol H2 x 3 mol H2O = 6.58 mol H2O
3 mol H2
9 - 55
Practice Solutions – Gas Reactions
Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(l)
Vhydrogen = 7.49 L
Vwater = ?
P2 = 0.975 atm
P1 = 22.0 atm
T1 = 305.2 K
T2 = 398.2 K
3. Finally, we need to find the volume of water. The
way to do that is to use the Ideal Gas Law.
PV = nRT
Vwater = nwaterRT = (6.58 mol H2O x 0.08206 (L atm/mol K) x 398.2 K)
P
(0.975 atm)
Vwater = 221 L
9 - 56
Practice Solutions – Gas Reactions
How many moles of H2SO4 form if 2.38 L of SO3 gas,
measured at 65.0°C and 1.05 atm, react with sufficient
water? How many grams? The balanced equation is as
follows:
SO3(g) + H2O(l) H2SO4(l)
VSO3 = 2.38 L
nH2SO4 = ?
T = 65.0°C + 273.15 = 338.2 K
P = 1.05 atm
mH SO = ?
2 4
1. First, we need to find moles of SO3.
PV = nRT
nSO = PV =
(1.05 atm x 2.38 L)
= 0.0900 mol SO3
3
RT (0.08206 L atm x 338.2 K)
mol K
•
9 - 57
19
Practice Solutions – Gas Reactions
SO3(g) + H2O(l) H2SO4(l)
VSO3 = 2.38 L
nH2SO4 = ?
T = 65.0°C + 273.15 = 338.2 K
P = 1.05 atm
mH SO = ?
2 4
2. Next, we will find moles of H2SO4. We use a mole ratio
to convert from SO3 to H2SO4.
nH SO = 0.0900 mol SO3 x 1 mole H2SO4 = 0.0900 mol H2SO4
2 4
1 mole SO3
3. Finally, we need to find the mass of H2SO4. The
conversion factor between moles and mass is Molar
Mass.
mH SO = 0.0900 mol H2SO4 x 98.09 g H2SO4 = 8.83 g H2SO4
2 4
1 mol H2SO4
9 - 58
Solving Simple Algebraic
Equations Math Toolbox 9.1
• We can manipulate equations in any way
that does not destroy the equality.
• Operations that maintain equalities are:
– adding the same number to both sides of
the equation.
– subtracting the same number from both
sides of the equation.
– multiply or dividing both sides of the
equation by the same number.
– raising both sides of the equation to the
same power.
9 - 59
Solving Simple Algebraic
Equations Math Toolbox 9.1
•
•
•
For example, let’s say we need to solve for volume
(V) in the Ideal Gas Law.
PV = nRT
When we solve for a variable (this time V), we
want it alone by itself on one side of the equality.
To rearrange the Ideal Gas Law, we need to divide
both sides by P:
PV = nRT
P
P
On the left side of the equation, P is both in the
numerator and the denominator; it therefore
cancels out.
V = nRT
P
9 - 60
20
Graphing Math Toolbox 9.2
• Steps for graphing
(by hand):
1. Examine and
organize the data
to be plotted.
• Placing the data in
a table may be
easiest to transfer
to the graph.
9 - 61
Graphing Math Toolbox 9.2
•
Steps for graphing (by hand):
2. Identify and label the axes.
– The dependent variable is the variable measured
experimentally. Plot it on the vertical axes (the y
axis).
– The independent variable is the variable controlled
and varied by the experimenter. Plot it on the
horizontal axis (the x axis).
– Clearly label each axis with the name of the variable
and the units. Put the units either in parenthesis
after the name of the variable or separated from the
variable name by a comma: volume (L) or volume, L.
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display
9 - 62
Graphing Math Toolbox 9.2
•
Steps for graphing (by hand):
3. Decide on the scales and limits of values to be
plotted along each axis.
– The graph should nearly fill the graph paper.
– Adjust the scale so the data can be entered easily on
the graph and read off the graph with about the same
precision as they were measured.
– You don’t need to begin each axis scale with zero if it
is not convenient.
– Increments on the x axis can differ from those on the
y axis.
4. After selecting the scales for each axis, clearly mark
the value of each major division along each axis.
9 - 63
21
Graphing Math Toolbox 9.2
Steps for graphing (by hand):
5. Place dots on the graph at the intersection of x and y
values to represent a data point.
– Place a circle 1-2 mm in diameter around each data
point to emphasize its location.
– Use squares or triangles to represent different sets
of data.
6. Draw a smooth line through the data set.
– Do NOT connect the dots with separate straight-line
segments.
– Not all data points may actually fall on the line.
– Label the completed graph with an appropriately
identifying title in a clear space at the top.
9 - 64
Graphing Math Toolbox 9.2
Volume
(L)
Temperature
(K)
2.20
300
2.37
325
2.59
350
2.95
400
3.30
450
3.71
500
4.37
600
5.17
700
9 - 65
Graphing Math Toolbox 9.2
Gas volume versus absolute temperature
6.00
5.00
Volume (L)
•
4.00
3.00
2.00
200
300
400
500
600
700
800
Temperature (K)
9 - 66
22