Chapter 11
Gases
Homework
• Assigned Problems (odd numbers only)
• “Questions and Problems” 11.1 to 11.71
(begins on page 338)
• “Additional Questions and Problems” 11.79
to 11.101 (begins on page 371)
• “Challenge Questions” 11.103 and 11.107,
page 372
Gases
• The atmosphere that surrounds us is
composed of a mixture of gases (air)
• The air we breath consists primarily of
– oxygen (about 21 %)
– nitrogen (about 78 %)
•
•
•
•
Industrial and household uses of gases:
chlorine
To purify drinking water
acetylene
Welding
Semiconductors hydrogen chloride, hydrogen bromide
Gases
• A substance that is normally in the gas state
at ordinary temp. and pressure
– oxygen gas
• Vapor: The gaseous form of any substance
that is a liquid or solid at normal temp. and
pressure
– water vapor or steam
• The gas phase is the normal state for eleven
of the elements
– H2, N2, O2, F2, Cl2, and noble gases
Kinetic Molecular Theory
• A mathematical model to explain and predict the
behavior of gases (the properties of an “ideal gas”)
• The theory of moving molecules. It gives an
understanding of pressure and temperature at a
molecular level
– Gases are made up of tiny particles (atoms or
molecules) widely separated from one another and in
constant rapid motion
– The particles have no attraction or repulsion for one
another
– The particles of a gas are extremely small compared to
the total volume which contains them: A gas is mostly
empty space
Kinetic Molecular Theory
– The average kinetic
energy (motion) of the
gas particles is
proportional to the
temperature of the
gas (in Kelvin)
– Particles are in
constant, random
straight-line motion,
colliding with each
other and the
container walls
(causing pressure)
Properties of Gases
• Gases have no definite shape or volume
and completely fill the container that holds
them
• Gases can be compressed to a smaller
volume or expanded to a larger volume
• Gases are mostly empty space and have
low densities compared to solids and
liquids
Properties of Gases
• Four variables are used to define the
physical condition, or state of a gas
• These variables are expressed in equations
that relate the pressure, temperature,
volume, and a specific quantity of a gas
• The behavior of gases can be described by
simple quantitative relationships
• The equations that express these
relationships are called gas laws
Properties of Gases
• Gas laws are generalizations
that describe mathematically
the relationships of these four
properties:
–Pressure (P)
–Volume (V)
–Temperature (T)
–Amount of gas in moles (n)
Properties of Gases
• Only the behavior of gases can be described
by these simple quantitative relationships
• The gas laws are mathematical equations
that predict how gases behave
• Why do they behave in a predictable
manner?
– Why does a gas expand when its
temperature is increased?
– A model has been proposed to understand
the physical properties of gases
Pressure
• A gas exerts pressure as its molecules collide
with the surface of the container that holds it
• The pressure is the result of these collisions
(impacts) divided by the unit area receiving the
force
force
P
area
• The atmosphere exerts pressure on the earth by
the collisions of molecules with every surface it
contacts
Pressure
• Pressure depends on
–The collisions of gas molecules with
the walls of the container
–The number of collisions
–The average force of the collisions
Pressure
• The atmosphere exerts pressure due to
collisions with every surface it contacts
• Changes in atmospheric pressure can affect
the weather
– Low pressure (cloudy and precipitation)
– High pressure (fair, sunny weather)
• Altitude can change pressure
– Atmospheric pressure is lower in Denver than
in Stockton because of “thinner air” due to
less collisions of gas with surface
Pressure Units
• Pressure (defined as force per unit area) is
commonly expressed in these units:
• The atmosphere (atm)
• The pascal (Pa)
• The millimeter of mercury (mm Hg) or Torr
• Pounds per square inch (psi)
Average Air Pressure at Sea Level
1 atm = 760 mm Hg
1 mm Hg = 1 Torr
1 atm = 101, 325 Pa
1 atm = 14.7 psi
The Mercury Barometer
• A barometer is a device used to measure
atmospheric pressure
– Constructed with a glass tube that is sealed
at one end (filled with mercury)
– The open end (initially covered) is submerged
into an open dish of mercury
– The cover is removed from the open end, the
mercury falls a little, then holds
– The length of the column (in mm) measures
the pressure (height of Hg in the glass tube)
– Invented in 1643 by Torricelli
The Mercury Barometer
• The atmosphere exerts
pressure on the mercury
in the dish
• The pressure of the
atmosphere supports the
column of mercury
• The length of mercury in
the column is a measure
of the pressure (mm Hg)
• At sea level, the mercury
would be 760 mm above
the surface of the table
1 atm =
760 mm Hg
Volume
The volume of a gas depends on
1. The pressure of the gas
2. The temperature of the gas
3. The number of moles of the gas
The volume of a gas is expressed in mL and L
Temperature
• Temperature of the gas is related to
the KE of the molecule
• The velocity of gas particles
increases as the temp. increases
When working problems, temperature is expressed
in Kelvin not Celcius
Amount of Gas
• Gases are usually measured in grams
• Gas law calculations require that the
quantity of a gas is expressed moles
• Moles of a gas represents the number
of particles of a gas that are present
When working gas law problems,
convert mass (in grams) to moles
Pressure and Volume
• Boyle, R. discovered the relationship
between gas pressure to gas volume
If the pressure on the gas increases,
then volume decreases proportionally
• As long as T and n are constant
• As the volume decreases, more
collisions with the walls occur
(less space), so pressure increases
Boyle’s Law
• Boyle determined this relationship from
observations, not the theory
• Only pressure and volume are allowed to change
• Inverse relationship between P and V
1
V
P
• Valid only if the temperature and the amount of
gas (n) are held constant
• For calculations
k
V
P
or
PV  k
One gas, two sets of conditions
P1V1  k  P2 V2 P1V1  P2 V2
Temperature and Volume
• Charles, J. found that the volume of
any gas will increase with an increase
in temperature
The volume of a sample of gas is
directly proportional to its temperature (K)
• As long as P and n are held constant
Charles’s Law
• In a closed system (container), what
happens to a gas when the temperature
increases?
– the velocity of molecules increases
– the force with which they hit the walls
increases
force
P
area
– Since force is proportional to area, the area
must increase as much as the force of the
molecules hitting the container
Charles’s Law
• If there is an increase in temperature in a
closed system (constant pressure)
• If the volume can change:
– In a flexible container, the force of the
molecules “stretches” out the container
(increases the volume) to maintain constant
pressure
• If the volume is fixed:
– In a rigid container, the force of the molecules
increases without an increase in area.
– An increase in temperature results in an
increase in pressure
Charles’s Law
• Volume and temperature are proportional
(if pressure and amount of gas is constant)
• Increase temperature, volume increases
• Decrease temperature, volume decreases
• For calculations:
V  bT
or
V
b
T
One gas, two sets of conditions
V1
V2
 b
T2
T2
V1 V2

T1 T2
VT
Charles’s Law
• If we could cool down a gas enough, its volume
would reduce to zero
• The V of a gas never actually reaches zero
• The four dashed lines are theoretical only because at a low
enough temperature, all gases will become a liquid
• Absolute Zero (0 Kelvin) is the point of intersection and is the
starting point for the Kelvin Temp. scale
Volume-temp data for the same gas at
various constant pressures
The Combined Gas Law
• Boyles’ law and Charles’s law show the
relationship of V to P and T (respectively)
T
PV
1
V
C
or
V
VT
P
T
P
• The combination of these two laws gives a
mathematical expression: The combined
gas law
• Can calculate a change in any one of the
three gas law variables brought by changes
in the other two variables
• Valid only if the amount of gas is held
constant
The Combined Gas Law
• An expression which combines
Boyle’s and Charles’s Law
One gas, two sets of conditions
P1V1 P2V2

T1
T2
• Applies to all gases and mixtures of
gases
• If five of the six variables are known,
the sixth can be calculated
Using the Combined Gas Law
•
Allows you to calculate a change in one of
the three variables (P, V, T) caused by a
change in both of the other two variables
1)
2)
3)
Determine the initial values
Determine the final values
Set two equations equal to one another, one for
initial values, one for final values
Solve for the unknown variable
4)
P1V1 P2 V2

T1
T2
P1V1T2
 T1
P2 V2
Volumes and Moles
• Avogadro proposed the relationship
between volume and a quantity of gas
(in moles, n)
• Equal volumes of two different gases at
the same temperature and pressure will
contain the same number of molecules
The volume of a gas varies directly with the number of moles
of gas as long as the pressure and temperature remain constant
Avogadro’s Law
• The volume and the amount of gas are directly proportional:
• If you increase (double) the number of moles, you increase
(double) the volume
• If the number of molecules is cut in half, the volume will be
cut in half
• More molecules push on each other and the walls,
“stretching” the container and making it larger until the
pressure is the same as before
Vn
V
a
n
One gas, two sets of conditions
V1
V2
a
n1
n2
V1 V2

n1 n 2
The Ideal Gas Law
• So far, three independent relationships dealing with the
volume of a gas
1
V
P
V T
V n
Boyle’s Law
n, T constant
Charles’s Law
n, P constant
Avogadro’s Law
P, T constant
Combined into a
single expression
nT
V
P
nT
V R
P
• Alternate, rearranged statement:
PV  nRT
The Ideal Gas Law
• The ideal gas law
PV  nRT
• The ideal gas law describes the
relationships among the four variables:
P, T, V, n (moles)
• P in atmospheres, T in Kelvin, V in liters
• Derived from the three different
relationships concerning the volume of a
gas
Ideal Gas Law
• R symbolizes the ideal gas law
constant
–Make sure that you use the
correct units in the ideal gas law
with the correct version of R
J
L • atm
R  8.314
or 0.0821
mol • K
mol • K
Simple Gas Laws & Ideal Gas Law
• Ideal gas law can be re-arranged to
make the equations from Boyle’s,
Charles’s or Avogadro’s Laws (simple
n, T are
gas laws)
PV  nRT
• Boyle’s Law
constant
–Temperature and number of moles are
constant
P1V1  nRT
P1V1  nRT  P2VP22V2  nRT
P1V1  P2V2
Boyle’s
Law
Simple Gas Laws &
Ideal Gas Law
• Charles’s Law
–Pressure and number of moles are
constant
V nR
n, P are
Ideal gas law

constant
T
P
PPV  nR
nRT
V1 nR

T1
P
V1 nR V2


T1
P T2
V1 V2

T1 T2
V2 nR

T2
P
Charles’s
Law
Simple Gas Laws &
Ideal Gas Law
• Avogadro’s Law
– Pressure and Temperature are constant
Ideal gas law
PPV  nRT
RT
V1 RT

n1
P
V RT

n
P
V1 RT V2


n1
P n2
V1 V2

n1 n2
T, P are
constant
V2 RT

n2
P
Avogadro’s
Law
Using the Ideal Gas Law
• Allows the direct calculation of P,
T, V, or n
• If three of the four variables are
known, (e.g. P, T, n) the unknown
variable, V can be calculated
• Allow the calculation of the molar
mass of a gas
• Can use in chemical reactions
where gases are involved as
reactants or products
Molar Mass of a Gas
(from Ideal Gas Law)
• The ideal gas law is useful for
calculation of the mass, molar
mass, or to determine the density
of a gas
• A modified form of the ideal gas
law (with a mass measurement of a
gas) can be used to calculate the
molar mass of a gas
Molar Mass of a Gas
(from Ideal Gas Law)
• A sample of gaseous compound has
a mass of 3.16 g. Its volume is 2050
L at a temp. of 45 °C and a pressure
of 570 mm Hg. Find its molar mass.
P
570 torr
1atm
760 torr
 0.750 atm
V  2050 mL  2.05 L
T  45 C  273 C  318 K


mass  3.16 g
Molar Mass of a Gas
(from Ideal Gas Law)
ideal gas law
pV
n
RT
0.750 atm  2.05 L
n
L  atm
0.0821
 318 K
mol  K
n = 0.0589 mol
Mass
3.16 g
molar mass 

Moles 0.0589 mol
mm = 53.7 g/mol
Mixtures of Gases: Dalton’s Law
• Not all gas samples are pure
• One gas law is specific for mixtures of
gases
• Dalton, J. (1801) proposed a law by studying
mixtures of gases in the atmosphere
• In a mixture of gases each type of molecule
moves about the container as if the other
gases were not present
– The attractions between particles are insignificant
– A gas in mostly empty space
Mixtures of Gases: Dalton’s Law
• The pressure exerted by a gas in a mixture
is the same as it would be for the individual
gas
• The partial pressure of a gas is the pressure
it would exert if were alone in the container
under the same conditions
• Dalton’s Law: The total pressure
put forth by a mixture of gases is
the sum of the partial pressures of
the individual gases
Mixture of Gases: Dalton’s Law
• Each gas behaves
independently of any
others in a mixture of
ideal gases
• Ptotal is the total
pressure, exerted by
the gases in a
mixture. It is directly
related to ntotal
• ntotal is the sum of the
moles of gases in the
mixture
Ptotal = Pa + Pb + Pc + ● ● ●
Mixtures of Gases: Dalton’s Law
• Each gas behaves independently of any others
in a mixture of ideal gases
• Ptotal is the total pressure, exerted by the gases
in a mixture. It is directly related to ntotal
• ntotal is the sum of the moles of gases in the
mixture
Pa
+
Pb
+
Pc
= Ptotal
Mixtures of Gases: Dalton’s Law
• Using the ideal gas law, for a mixture of
gases, can come up with an equation for Ptotal
Partial pressures
n B RT
n ART
n C RT
pB 
pA 
pC 
V
V
V
Dalton’s Law
ptot = pA  pB  pC
p tot
RT
RT
RT
= nA
 nB
 nC
V
V
V
= ntotal
RT
Ptotal  (n A  n B  nC )
V
p tot = n tot
RT
V
Partial Pressure Example 1
• A halothane-oxygen mixture used for
anesthesia is placed in a 5.0 L tank at 25 °C.
You used 15.0 g of halothane (C2HBrClF3)
vapor and 23.5 g of oxygen.
• What is the total pressure (in torr) of the gas
mixture in the tank?
• (Halothane molar mass: 197.4 g/mol)
Partial Pressure Example 1
To solve:
Calculate the total number of moles
of each component
15.0 g Halo
23.5 g O2
1 mol Halo

 0.0760 mol
197.4 g Halo
Halo
1 mol O2

 0.734 mol O2
32.00 g O2
Partial Pressure Example 1
RT
mol halo + mol oxygen
Ptotal  (n A  n B )
V
ntotal  n A  n B  0.0760 mol  0.734 mol  0.810 mol
T  25  C  273  C  298 K
V  5.0 L
Ptotal  X
Ptotal
(0.810 mol ) (0.08206 L  atm  mol 1 K 1 ) (298 K )

 3.96 atm
5.0 L
p tot
3.96 atm 760 torr

 3010 torr
1 atm
Collecting Gases Over Water
• Small quantities of gases, generated in
a chemical reaction, can be collected
by displacement of water
• Uses an inverted bottle filled with water
• As the gas is bubbled through the
water it collects in the inverted bottle
• The gas collected is impure because of
the water also collected due to
evaporation
Collecting Gases Over Water
• This produces a mixture of gas and water
vapor
• The partial pressure of the water is
temperature dependent and is called its
vapor pressure (see table 11.4, pg 388)
Partial Pressure Example 2
• When magnesium metal reacts with
hydrochloric acid, 355 mL of hydrogen gas
is collected over water at 25 °C at a total
pressure of 752 torr.
– What is the partial pressure of the hydrogen gas?
• How many moles of hydrogen were
collected?
• Water has a vapor pressure of 23.8 torr at
25 °C (See page 388, table 11.4)
Partial Pressure Example 2
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
Vhydrogen= 355 mL
T = 25.0 °C
Ptotal = 752 torr
Mg
Pwater = 23.8 torr
Phydrogen= x
moleshydrogen = y
Partial Pressure Example 2
PH 2  Ptotal  Pwater
PH 2  752 torr  23.8 torr  728 torr
V  355 mL  0.355 L
T  25  C  273  C  298 K
nH 2
(0.958 atm) (0.352 L)

(0.0821 L  atm  mol 1  K 1 ) (298 K )
Convert mL to L
Convert torr to atm
pV
n
RT
= 0.0138 mol H2
Gases in Chemical Reactions
• Gases are involved as reactants and products
in many chemical reactions
• Quantitative relationships (stoichiometry) are
based on balanced chemical reactions
• The coefficients in a balanced equation
correspond to the number of moles of each
reactant and product and are used to make
conversion factors
• Use of the ideal gas law is a convenient way
to relate the moles of a gas to its volume,
pressure, and temperature
Using the Ideal Gas Law
Grams A
Grams B
MM
Moles A
Moles B
pV=nRT
PA, TA, VA
PB, TB, VB
Gas Laws and
Chemical Reactions
• In chapter 8 we used the balanced
equation and mass of one reactant to
calculate the mass of any other
component
• Mass to mass or gram to gram
• Can also do volume to mass and mass
to volume problems where at least one
of the components is a gas
• The ideal gas law will relate the moles
of a gas to its P, T and V
Gas Laws and
Chemical Reactions
• In chapter 8 we used the balanced equation and
mass of one reactant to calculate the mass of any
other component
Mass to mass or gram to gram problems
• In this chapter, where at least one of the
components is a gas
Volume to mass and mass to volume problems
• The ideal gas law will relate the moles of a gas to
its P, T and V
Gases in Chemical Reactions
TB, PB
TA, PA
Use the ideal gas
law and P, V, T
to convert to
moles
pV=nRT
Mol-mol
factor
pV=nRT
Gases in Chemical Reactions
Example I
• Gaseous ammonia is synthesized by the
following reaction.
N2 (g)  3 H2 (g)  2 NH3 (g)
• Suppose you take 355 L of H2 gas at 25 °C and
542 mm Hg and combine it with excess N2 gas.
a. What quantity of NH3 gas, in moles, is
produced?
b. If this amount of NH3 gas is stored in a 125 L
tank at 25 °C, what is the pressure of the gas?
Gases in Chemical Reactions
Example I
Excess N2
H2
Find:
Given:
V = 355 L
T = 25 °C
P = 542 mm Hg
Part a: Mol = ?
Part b: Pressure = ?
NH3
V = 125 L
V = 125 L
T = 25 °C
T = 25 °C
Gases in Chemical Reactions
Example I
3 H2 ( g )  N2 ( g )  2 NH3 ( g )
Given:
V = 355 L,
T = 25 °C,
P = 542 mm Hg
Given:
Excess N2
Find: mol NH3
Equation and Conversion Factors
pV = nRT
volume
grams HLi2
MM= of
Li
PV
nRT
moles H
Li2
3 mol H2 = 2 mol NH3
Solution Map:
grams Li3N
Stoichiometry
MM of Li3N
moles NH3
moles NH3
moles
moles Li
H2
Gases in Chemical Reactions
Example I
1) Find the moles of hydrogen using ideal gas law
N2 (g)  3 H2 (g)  2 NH3 (g)
1 atm
p  542 torr 760
torr  0.713 atm
V  355 L
T
T =2525
°C
+ 273
273 °C=298
298 KK
n?
(0.713 atm)(355 L)
n
L•atm
(0.0821 mol
• K (298 K)
pV
n
RT
n = 10.3 mol H2
2) Determine the moles of ammonia produced using the
mole-mole factor in the balanced equation
N2 (g)  3 H2 (g)  2 NH3 (g)
3 moles H2 = 2 moles NH3
2 mol NH 3
3 mol H 2
and
3 mol H 2
2 mol NH 3
Now, combine steps 1 and 2 to calculate moles
10.3 mol H 2 2 mol NH3
 6.87 mol NH 3
3 mol H 2
Molar Volume at STP
• Gas volumes can only be compared at
the same T and P
• The universally accepted conditions
are T = 0 °C and P = 1.00 atm
Standard Temperature and Pressure = STP
Standard Temperature = 0 °C (273 K)
and Standard Pressure = 1.00 atm (760 mmHg)
Molar Volume at STP
• The volume of one
mole of any gas at
STP is 22.4 L
– Known as the
standard molar
volume
– When used in
calculations at
STP, the
conversion factor(s)
can be used:
1 mol gas = 22.4 L
1 mol gas
22.4 L

22.4 L
1 mol gas
Gases in Chemical Reactions
TB, PB
TA, PA
Use the
equality to
convert V to
moles
at STP
at STP
pV=nRT
1 mol gas =
22.4 L
Mol-mol
factor
1pV=nRT
mol gas =
22.4 L
• end