Chapter 6
Fig. 6-CO, p. 208
p. 242
What do SCUBA and airbags
have to do with gases?
Let’s find out…..
6.1 Properties and Measurements of Gases
Objectives
To describe the characteristics of the three
states of matter – solid, liquid, and gas
 To define the pressure of a gas and the units in
which it is measured

Solid Phase

A solid has fixed shape
and volume.
Solid Br2 at low
temperature
Liquid Phase

A liquid has fixed volume
but no definite shape.
 The density of a solid or a
liquid is given in g/mL.
Liquid Br2
Gas Phase

A gas has no fixed volume or
definite shape.
 The density of a gas is given in
g/L whereas liquids and solids
are in g/mL.
 Can expand or be compressed
depending on the pressure
exerted on a system
Gaseous Br2
6.1 Properties and Measurements of Gases cont’d

Condensed phases
 Includes solids & liquids because they are resistant to
changes in volume caused by changes in pressure
 The density of gases can increase 1000-fold or more
when compressed
6.1 Properties and Measurements of Gases
cont’d

Pressure of a Gas
= the force per unit area exerted on a surface
 Ex: the atmosphere exerts a pressure due to the
weight of gas molecules in the air
○ Air pressure at higher elevations is lower due to
fewer gas molecules above you than at sea level
 A barometer measures the pressure of the
atmosphere
 A manometer measures pressure differences
Pressure of a Gas

The pressure of the
atmosphere is
measured with a
barometer.
Manometers

Both open and closed end manometers
measure pressure differences.
Pgas < Patm
Pgas > Patm
h = Pgas
6.1 Properties and Measurements of Gases
cont’d
 Units
of Pressure Measurement
 SI unit is the pascal (Pa)
 1 Pa = 1N/m2 or 1 kg/m•s2
(N = newton; m = meter, s = seconds)
 This unit of pressure is very small for experiments
typically carried out by chemists
 Typically see other units used as shown in Table
6.1 (p. 212 in TB)
 1 atm of pressure = the normal pressure at sea
level which is defined as the pressure exerted on
a column of mercury 760 mm high
6.1 Properties and Measurements of Gases
cont’d
Relationships Between Pressure Units
1 atm = 760 mm Hg
1 torr = 133.3 Pa
1 atm = 760 torr
1 atm = 14.7 psi
1 atm = 101.325 kPa
1 atm = 1.01325 bar
1 atm = 29.92 in Hg
6.1 Properties and Measurements of Gases
cont’d
Express 433 torr in atmospheres
Answer: 0.570 atm
Express 0.450 atm in kPa
Answer: 45.6 kPa
6.2 Gas Laws
Objectives

Determine how a gas sample responds to
changes in volume, pressure, moles, and
temperature
Why are we interested in determining how
a gas sample responds to changes in one
or more physical properties?

What if you lived near a gas storage tank? …
 Would you want to know how a change in temperature
would affect the pressure in the metal tank that stores
gaseous materials?
 Would you want to know if the tank is designed to
withstand such an increase in pressure?

So let’s learn how to assess how gases behave….
6.2 Gas Laws
The physical properties of all gases behave
in the same general manner, regardless
of the identity of the gas
 Four independent properties define the
physical state of a gas

 Pressure
 Volume
 Temperature
 # of moles
The Effect of Pressure on Gas Volume

Boyle’s Law is the
relationship between
volume & pressure (at a
constant temperature and
amount of gas)

Increasing the pressure
on a gas sample, by
addition of mercury to
an open ended
manometer, causes the
volume of the gas to
decrease.
Boyle’s Law
A plot of volume
versus 1/P is a
straight line.
 V = k1 x 1
P

can be rewritten as
PV = constant
6.2 Gas Laws cont’d

Boyle’s Law cont’d
 Can predict what will happen to the pressure or
volume of a gas if its volume or pressure changes
(when T and n are held constant)
 Since P1V1 = k1 and P2V2 = k1
then P1V1 = P2V2
6.2 Gas Laws, cont’d
Boyle’s Law cont’d
Example: A balloon containing 575 mL
nitrogen gas at a pressure of 1.03 atm is
compressed to a final volume of 355 mL.
What is the resulting pressure of the
nitrogen?
A: 1.67 atm
6.2 Gas Laws, cont’d
• Boyle’s Law cont’d
Example : In the lungs of a
deep-sea diver (V= 6.0 L) the
pressure of the air is 7400 torr.
At a constant temperature of
37°C, to what volume would
the air expand if the diver
were immediately brought to
the surface (1.0 atm)
A: 58 L
The Effect of Temperature on Gas Volume
Fig. 6-7, p. 216
Charles’s Law
A plot of volume versus
temperature is a straight
line.
 Extrapolation to zero
volume yields absolute
zero in temperature:
-273o C.
 V = k2 x T, where T is
given in units of kelvin. ***

6.2 Gas Laws cont’d

Charles’ Law
 V = k2 x T
○ Note temperatures must be in degrees
Kelvin!!!!
○ can be rewritten as V/T = constant
 Can predict what will happen to the
temperature or volume of a gas if its volume
or temperature changes (when P and n are held
constant)
 Use V1/T1 = constant = V2/T2
Problem Solving Using Charles’ Law
The volume of a sample of nitrogen gas
increases from 0.440 L at 27°C to 1.01 L as
it is heated to a new temperature. Calculate
the new temperature of the nitrogen gas
sample.
A: 416°C
6.2 Gas Laws cont’d

Avagadro’s Law
 At a constant temperature and pressure,
equal volumes of gases contain the same
number of particles (independent of the type
of gas)
 Avagadro’s Law:
V = k3 x n
Avogadro’s
Hypothesis
Both containers are
of equal volume,
contain the same #
of moles of gas, are
at the same
pressure, but the
mass of the gases is
different.
Avogadro’s Law
A plot of the volume
of all gas samples, at
constant T and P, vs.
the number of moles
(n) of gas is a
straight line.
 V = k3 x n

6.2 Gas Laws cont’d

The Combined Gas Law
 For a given amount of gas, n, the three
remaining properties of a gas can be related
by the combined gas law
 P1V1 = P2V2
T1
T2
*** Remember, temperatures must be in
degrees Kelvin!
Practice Problem Using the Combined Gas Law
1. The pressure of a sample of gas is 2.6 atm
in a 1.54 L container at a temperature of
0°C. Calculate the pressure exerted by
this sample if the volume changes to 1.00 L
and the temperature changes to 27°C.
A1: 4.40 atm
2. A sample of a gas occupies 4.0 L at 25°C
and 2.0 atm of pressure. Calculate the
volume at STP (T = 0°C, P = 1 atm).
A2: 7.3 L
If you aren’t given gas law relationships,
what can you do…..




If you ever forget any of the gas law
relationships, including the combined gas law,
you can use the ideal gas law to help you out.
Remember P1V1 = n1RT1 & P2V2 = n2RT2
To solve problems, the ideal gas law is
rearranged so variable quantities are on the
left side and constant quantities are listed on
the right. Then the two left hand sides of the
resulting equations are set equal to one
another and you solve for the unknown
quantity
Let’s practice in class……
6.3 The Ideal Gas Law
Objectives
Write the ideal gas law
 Calculate the pressure, volume, amount, or
temperature of a gas given the other three
properties
 Calculate the molar mass and the density of gas
samples by using the ideal gas law

Ideal Gas Law

The ideal gas law combines the three gas
laws into a single equation:
PV = nRT
where: R = 0.08206 L.atm/mol.K
(R determined experimentally)

All common gases follow the ideal gas law at
normal temperatures and pressures

FYI: The volume of one mole of an ideal gas at STP is
22.4 L. This is a commonly used conversion factor that
might come in handy. STP = 1 atm, 0°C
Table 6-2, p. 219
6.3 The Influences of Changing Conditions
on Gases




Q: In A balloon filled with oxygen gas at 25°C
occupies a volume of 2.1 L. Assuming that the
pressure remains constant, what is the volume at
100°C?
A: 2.6 L
Q: The volume of a sample of nitrogen gas at
27°C increases from 0.440 L to 1.101 L when the
sample is heated. What is the second
temperature of the nitrogen sample?
A: 416°C
Practice Problems Using the Ideal Gas Law
Calculate the number of moles of argon gas in a
30.0 L container at a pressure of 10.0 atm and
temperature of 298 K.
 A: 12.3 mol

Using the Ideal Gas Law to
Calculate Molar Mass and Density
The ideal gas law (PV=nRT) can be used to
calculate the density (mass/volume) and molar
mass (mass/moles) of a gas.
 Since the ideal gas law does not contain a
variable for mass, moles must be calculated first
to help determine a molar mass or n/V must be
calculated first to determine a density.
 Let’s see how …..

Calculating Density from the Ideal Gas Law

Calculate the density of CO2 (g) at STP.
 Strategies to consider:
○ Remember that the density of gases is in units of grams/L
○ Given PV = nRT, you can calculate n/V in terms of mol/L.
○ Then you can use the molar mass of the given gas to
calculate density: mol/L x g/mol = g/L

A: 1.96 g/L

Please Note: At constant pressure and temperature the
density of a gas is proportional to its molar mass, so the
higher the molar mass, the greater the density of the gas.
Calculating Molar Mass from the Ideal Gas Law

Calculate the molar mass of a gas if a 1.02 g
sample occupies 220 mL at 95.0°C and a
pressure of 750 torr.
 Strategies to consider:
○ Remember that molar mass = grams/mol
○ You are given grams, so if you calculate moles from
PV=nRT you have everything you need to calculate
molar mass

A: 142 g/mol
Using the Ideal Gas Law to Calculate Molar Mass
A 4.25 g sample of an ideal gas occupies a
volume of 1250 mL at 100.0°C and a
pressure of 707 torr. What is the molar
mass of the gas?
 A: 112 g/mol
 A gas has a density of 0.725 g/L and a
pressure of 1.033 atm. What is the molar
mass of the gas?
 A: 17.0 g/mol

6.4 Stoichiometry Calculations Involving Gases
Objectives
Perform stoichiometric calculations for reactions
in which some or all of the reactants or products
are gases
 Use relative volumes of gases directly in equation
stoichiometry problems

Gases and Chemical Equations
The ideal gas law can be used to determine
the number of moles, n, for use in problems
involving reactions.
 The ideal gas law relates n to the volume of
gas just as molar mass is used with masses
of solids and molarity is used with volumes of
solutions.

Fig. 6-11, p. 224
Example: Gases with Equations

Calculate the volume of O2 gas formed in the
decomposition of 2.21 g of KClO3 at STP.
2KClO3(s)  2KCl(s) + 3O2(g)
6.5 Dalton’s Law of Partial Pressure
Objectives
Use Dalton’s law of partial pressure in
calculations involving mixtures of gases
 Calculate the partial pressure of a gas in a
mixture from its mole fractions

Dalton’s Law of Partial Pressure
The pressure exerted by each gas in a
mixture is called its partial pressure.
 For a mixture of two gases A and B, the total
pressure, PT, is


PT = PA + PB
Pressure of a Mixture of Gases
Solving Partial Pressures Problems
Calculate the pressure in a container that
contains O2 gas at a pressure of 3.22 atm and
N2 gas at a pressure of 1.29 atm.
 A: 4.51 atm

6.5 Dalton’s Law of Partial Pressure
Mole Fraction
Mole fraction (c, chi) is the number of
moles of one component of a mixture
divided by the total number of moles of all
substances present in the mixture.
 cA + cB + cC = 1
 The partial pressure of any gas, A, in a
mixture is given by: PA = cA x PT

Mole Fraction

Mole fraction of
the yellow gas is
3/12 = 0.25 and
the mole fraction
of the red gas is
9/12 = 0.75
Solving Partial Pressure Problem s …..
Calculate the partial pressure of Ar gas in a
container that contains 2.3 mol of Ar and 1.1
mol of Ne and is at a total pressure of 1.4 atm.
 A: 0.95 atm


Now let’s use these skills to solve some
practical problems commonly
encountered in the chemistry lab….
Collecting Gases
Over Water

Water vapor is also
present in a sample
of O2 gas collected
over water.
(Fig 6.14 p. 228)
6.5 Dalton’s Law of Partial Pressure

Collecting Gases by Water Displacement
 Gas samples collected by water displacement are
not pure because some water molecules are also
present in the gas phase
 The total pressure is equal to the gas pressure
plus the pressure of the water vapor
 To calculate pressure due to gas alone you must
subtract the partial pressure of water vapor from
the total pressure
 Pressures due to water vapor at different
temperatures are given in Table 6.3 on p. 228
6.5 Dalton’s Law of Partial Pressure




Q: A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 26°C. The total volume of the collected
gas is 229 mL at a pressure equal to the measured
atmospheric pressure, 754 torr. How many moles of O2
form?
A: 8.95 x 10-3 mol O2
Calculate the number of moles of hydrogen produced by
the reaction of sodium with water. In the reaction, 1.3 L
of gas are collected by water displacement at 26C. The
atmospheric pressure is 756 torr.
A: 0.051 mol H2
Example: Collecting Gases

Sodium metal is added to excess water, and
H2 gas produced in the reaction is collected
over water with the gas volume of 1.2 L. If
the pressure is 745 torr and the temperature
26o C, what was the mass of the sodium?
The vapor pressure of water at 26o C is 25
torr.
2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq)
6.6 The Kinetic Molecular Theory of Gases
Objectives
Show that the predictions of the kinetic
molecular theory are consistent with
experimental observations
 Sketch a Maxwell-Boltzmann distribution curve
for the distribution of speeds of gas molecules
 Perform calculations using the relationships
among molecular speed and the temperature
and molar mass of a gas.


How can a single law (the ideal gas law)
explain the behavior of all gases,
regardless of the nature or size of the
gas particles?
6.6 The Kinetic Molecular Theory of Gases

The four postulates of kinetic molecular
theory are:
1. A gas consists of small particles that are in
constant and random motion
2. Collisions of gas particles with each other and
with the walls of the container are elastic – there
is no loss in the total kinetic energy when the
particles collide – and no attractive or repulsive
forces exist between the gases
6.6 The Kinetic Molecular Theory of Gases

The four postulates of kinetic
molecular theory are cont’d:
3. Gas particles are very small compared to
the average distance that separates them
4. The average kinetic energy of gas particles
is proportional to the temperature on the
Kelvin scale
6.6 The Kinetic Molecular Theory of Gases

The relationship between kinetic energy and
the speed of the particles is given by:
KE = ½ mu2
Average Speed of a Gas
Gas particles move
at different speeds.
 Average speed is
called the root
mean square (rms)
speed, urms, and is
the square root of
the average squared
speed.

Maxwell-Boltzmann
distribution curves
Average Speed of a Gas
urms
3RT

molar mass
R = 8.314 J/mol.K;
molar mass in
kilograms per mole
6.7 Diffusion & Effusion
Objectives
Define diffusion and effusion
 Calculate relative rates of effusion of two
gases and use the data to calculate molar
masses

6.7 Diffusion & Effusion

Diffusion
 the mixing of particles due to motion
 the faster the molecular motion, the faster a gas
diffuses
 Rate of diffusion is less than the root mean speed
(rms) of the gas because collisions prevent
particles from moving in a straight line
6.7 Diffusion & Effusion
Effusion
The passage of a
gas through a small
hole into an
evacuated space
Graham’s Law
states that the rate of
effusion of a gas is
inversely proportional
to the square root of
its molar mass
6.7 Diffusion & Effusion
Graham’s Law can be used to determine
the molar mass of an unknown gas by
measuring the times needed for equal
volumes of a known gas and an unknown
gas to effuse through the same small hole
at constant pressure and temperature
 Note that the rate of effusion is inversely
related to time so Graham’s Law can be
rewritten as:
tB = molar mass of B
tA √ molar mass of A

6.7 Diffusion & Effusion




Q: Calculate the molar mass of a gas if equal volumes
of nitrogen and the unknown gas take 2.2 and 4.1 min,
respectively, to effuse through the same hole under
conditions of constant temperature and pressure
A: 97 g/mol
Q: Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and 8.41
min, respectively, to effuse through the same hole under
conditions of constant temperature and pressure.
A: 214 g/mol
6.8 Deviations from Ideal Behavior
Objectives
Explain why gases deviate from the ideal
gas law under certain conditions
 Use the van der Waals equation to account
for deviations from the ideal gas law

6.8 Deviations from Ideal Behavior

Most gases obey the ideal gas law closely at a
pressure of about 1 atm and a temperature well
above the boiling point of the substance
 in other words at low pressures & temperatures
100°C above the BP of the substance, far from
conditions under which the gases would condense to
a liquid

Gases at high pressures do not behave as
predicted by the ideal gas law
Deviations from Ideal Behavior

Gases
deviate from
the ideal gas
law at high
pressures.
Deviations from Ideal Behavior

A gas deviates
from ideal gas
behavior at low
temperatures
(near the
condensation
point ) and high
pressures.

Behavior of O2 is
shown
6.8 Deviations from Ideal Behavior
Deviations from the ideal gas law occur under
extreme conditions because two assumptions
of the kinetic molecular theory are not
correct when gas molecules are close
together.
 These assumptions are:

(a) that gas particles are small compared to the
distances separating them
(b) that there are no attractive forces between gas
particles
Deviations from Ideal Behavior
The assumption that gas particles are small
compared to the distances separating them fails at
high pressures.
 The observed value of PV/nRT will be greater than
1 under these conditions.

Forces of Attraction in Gases
The forces of attraction
between closely spaced gas
molecules reduce the impact of
wall collisions.
 These attractive forces cause
the observed value of PV/nRT
to decrease below the
expected value of 1 at
moderate pressures.

6.8 Deviations from Ideal Behavior

The van der Waals equation is used to describe
gases under non-ideal conditions :
(P + an2/V2)(V-nb) = nRT
 To correct for the volume occupied by the gas we
subtract the term nb from the volume where n is
the # of moles of a gas and b is a constant that
depends on the gas
6.8 Deviations from Ideal Behavior

The van der Waals equation:
(P + an2/V2)(V-nb) = nRT
 To correct for attractive forces between gas
particles we add the term an2/V2 to the
pressure where a is a constant related to the
strength of the attractive forces, n= # of
moles of the gas, and V = the volume of the
gas sample
 Note that the values of a and b are derived
experimentally and are different for each gas
Gas
6.9
Deviations
from Ideal
Behavior
van der
Waals
constants
H2
He
Ne
H2O
NH3
CH4
N2
O2
Ar
CO2
a
b
(atm•L2/mol2
(L/mol)
0.244
0.034
0.211
5.46
4.17
2.25
1.39
1.36
1.34
3.59
0.0266
0.0237
0.0171
0.0305
0.0371
0.0428
0.0391
0.0318
0.0322
0.0427
6.8 Deviations from Ideal Behavior






Q: Calculate the pressure of 2.01 mol of gaseous
H2O at 400°C in a 2.55 L container, using the ideal
gas law and the van der Waal’s equation.
A: Ideal gas law = 43.6 atm
A: van der Waal’s = 41.2 atm
Q: Calculate the pressure of 0.223 mol of
ammonia gas at 30.0°C in a 3.23 L container,
using the ideal gas law and the van der Waal’s
equation
A: Ideal gas law = 1.72 atm
A: van der Waal’s = 1.70 atm
Fig. 6-16, p. 230