Chapter 5
Gases and the Kinetic-Molecular Theory
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practice assignments for discussion time.
Why do we need to learn about
gases? To protect our earth!
Table 5.1 from 4th ed. Some Important Industrial Gases
Name
Methane
(CH4)
Origin and Use
natural deposits; domestic fuel
Ammonia
(NH3)
from N2+H2; fertilizers, explosives
Chlorine (Cl2)
electrolysis of seawater; bleaching and
disinfecting
Oxygen (O2)
liquefied air; steelmaking
Ethylene
(C2H4)
high-temperature decomposition of
natural gas; plastics
Atmosphere-Biosphere Redox Interconnections
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.
Figure 5.1
The three states of matter.
Bromine as a gas, a liquid and a solid.
Four measurementss define the state
of a gas
1.
2.
3.
4.
Quantity in moles, n
Volume in Liters
Temperature in Kelvin
Pressure in Atm
Gas Pressure
Pressure = Force/unit area
Force causes something to move a distance
D in work
Gravity is a weak force, F = ma
On earth F = mg, where g is the
acceleration due to earth’s gravity
A BAROMETER
Measures pressure of atmosphere above
Essential principle of the barometer –
force of air on surface of water keeps
water in the tube
If we had a long enough tube, about 34 ft.,
we'd reach the limit of air pressure to
hold the water in
Figure 5.3
A mercury barometer.
Patm is due to weight of
atmosphere above the
earth
Barometers
Why do our barometers contain mercury? Because of
its density. Hg is 13.6 g/mL vs. water at ~1.0
g/mL.
P = Force/Area
= mass of Hg column * grav. constant/area of
column
= vol of Hg column * density of Hg/area of column
= height of Hg*area of column* mass/area of
column
= height of Hg column * density of Hg
Normal atmospheric pressures is measured at sea level
on a calm day: it is 760 mm Hg = one standard
atmosphere pressure (both are exact quantities)
Figure 5.2
Effect of atmospheric pressure on objects
at the Earth’s surface.
A metal can filled with
air has equal pressure
on the inside & outside.
When the air inside the
can is removed,
atmospheric pressure
crushes the can.
Gas Pressure
MEMORIZE:
1 atm = 760 mm Hg exactly = 760 torr
exactly = 101.325 kPa = 14.7 psi
Practice: convert 29.5 inches of Hg to mm
Hg, torr, atm and psi. (Hint – do you
remember how to convert inches to cm?)
MANOMETERS

Definition: A manometer is a
scientific instrument used to
measure gas pressures. Open
manometers measure gas pressure
relative to atmospheric pressure. A
mercury or oil manometer
measures gas pressure as the
height of a fluid column of mercury
or oil that the gas sample supports.
Figure 5.4 from 4th ed.
closed-end
Pgas = ht of Hg, Dh
Two types of
manometers
open-end
The Simple Gas Laws
Memorize Boyle’s, Charles’, Avogadro’s, and
Amonton’s (Gay-Lussac’s) laws
Figure 5.4
The relationship between the volume
and pressure of a gas.
Boyle’s Law
See notes below or look in textbook.
Boyle’s Law
As shown in Fig 5.4, as Pressure  the Volume ,
which means V ~ 1/P (inversely proportional)
P*V is a constant
P1V1 = Cb and P2V2 = Cb
Therefore, P1V1 = P2V2
Example: gaseous CO2 at 55 torr occupies 125mL.
It is transferred to new flask and P increases to
78 mm of Hg. What is the new volume?
First think about whether V inc or dec?
Dec. Why? P Inc.
V2 = P1V1/P2 = 55 mm*125 mL/78 mm = 88 mL
Figure 5.5
The relationship between the
volume and temperature of a
gas.
Notice all lines extrapolate to
-273.15 oC! What is that?
Charles’ Law
See notes below or
in textbook.
Charles’ Law
As shown in Fig 5.5, as Temperature  the Volume , i.e.,
directly proportional. Therefore:
V/T = Cc
BUT Temperature has to be in positive increments, never
negative.
This is why the absolute scale, Kelvin, was developed.
V1/T1 = Cc. V2/T2 = Cc. Therefore, V1/T1 = V2/T2
Example: A balloon is filled with He at 4.5 L and 25.0°C. Take it
outside in winter in Chicago, -10.0°C. What is new volume?
First: is new V larger or smaller?
Smaller, because T dec.
V2 = V1*T2/T1 = 4.5 L*263 K/298 K = 4.0 L
Figure 5.6
An experiment to study the relationship between the volume and
amount of a gas comparing 0.1 mol and 0.2 mol.
Avogadro’s Law
At a given external P & T, a given amt of CO2 is placed in tube.
When it changes from solid to gas, it pushes the piston up until
Pgas = Patm. When twice as much CO2 is used, twice the volume
is occupied, showing V is proportional to amt of gas.
Avogadro’s Law
As shown in Fig 5.6, as the quantity
of a gas increases, the volume
increases.
V1/n1 = Ca = V2/n2
therefore, V1/n1 = V2/n2
n = moles, which we know represents
#'s of molecules
Amonton’s Law
(also Gay-Lussac)
P & T are directly related
P1/T1 = P2/T2 if n and V are constant
Does T have to be in Kelvin? Why?
Practice with the Simple Gas Laws
Work on problems 8a, 16a and 18.
8a: Convert 76.8 cm Hg to atm
16a: What is the effect on the volume
of 1 mole of an ideal gas if the
pressure is reduced by a factor of
four at constant temperature?
18: A 93 L sample of dry air is cooled
from 145oC to -22oC at constant
pressure. What is its final volume?
Video Lecture on Ideal Gas Law

Caveat: you will use R = 0.082057
L-atm/mol-K, because we want 5
sig figs for many of our calcs. You
will also use more sig figs for
temperatures, so add 273.15 to
convert to Kelvin. (Also her final
answer’s sig figs should be at least
3.)
IDEAL GAS LAW:
Combining the work of Boyle, Charles,
Amonton, and Avogadro:
V=Cb/P
V=Cc*T
V=Ca*n
V=(Cb*Cc*Ca)*(n*T/P)
V = R*nT/P which rearranges to PV = nRT
R = 0.082057 L-atm/mol-K (memorize)
EXAMPLE of Ideal Gas Law
A big balloon is filled with 1300.
moles H2, T is 20.0°C, P = 750.
torr. What is Volume of balloon?
P = 750. torr*(1 atm/760 torr)
=0.98684 atm
V = nRT/P =
1300. mol* 0.082057 * 293.15K
0.98684 atm
= 3.17 x 104 L
Figure 5.7
Standard molar volume.
One mole of any ideal gas occupies 22.414 L at STP. Notice these three gases behave
ideally. Masses are different, therefore density will be different.
STP: standard temperature & pressure
Defined as 273.15 K and 1.0000 atm
At STP, one mole of any gas occupies
22.414 L, called the standard molar
volume of a gas
Try V = nRT/P using this STP for one
mole of gas
Simple combined gas law:
P1V1/T1 = P2V2/T2
Works if moles of a gas are constant.
Example
If you have 20.0 L of He at 150. atm and 30.0°C,
then how many balloons can be filled with 5.00 L
each at 755 mm and 22.0°C?
(Really asking for V2 at new T & P, note that n is
constant.)
Convert P2 to atm: 755 mm (1 atm/760 mm)
= 0.99342 atm
V2 = P1V1T2/P2T1 = 150atm*20.0L*295.15K
0.99342atm*303.15K
= 2940 L
2940L/5.00L = 588 balloons
MY SUPER-COMBINED GAS LAW:
Rearrange R = P1V1 = P2V2
n1T1 n2T2
Then "uncombine" these to get the simple gas laws:
@ const T & n : cross out all n & T to get Boyle's
law
@ const P & n : cross out all n & P to get Charles'
law
@ const P & T : cross out all P & T to get
Avogadro's law
@ const V & n : cross out all V & n to get
Amonton’s Law
IDEAL AND COMBINED GAS LAWS:
Practice: A sample of O2 gas occupies
255.5 mL at 25.5oC and 757.7 torr. (First
determine the number of moles present.)
If the temperature drops to 15.5oC and
the pressure drops to 745.0 torr, and
you add 0.00100 moles of gas, what
will the new volume read?
(You should get 275.3 mL)
Practice with Combined Gas Laws
and Ideal Gas Law
Work on problems 20, 24 and 83.
20: Calculate V of a sample of CO
that is at -14oC and 367 torr if it
first occupies 3.65 L at 298K and
745 torr.
24: A 75.0 g sample of N2O is held in
a 3.1 L vessel at 115oC. What is its
pressure?
83: (look in book)
The Density of a Gas
Recall that Density = m/V
Recall we can find moles: n = m/M
PV = nRT; substitute PV = (m/M)RT
Rearrange: m/V = M P/ RT
or D = M 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.
Rearrange again to find molar mass: M = DiRT/P
Example:
An unknown gas has a density of 1.429 g/L
at STP. Find its molar mass. (Hard way
& easy way if at STP.)
MA = DRT/P
= 1.429 g/L*0.082057*273.15K/1 atm
= 32.03 g/mol
or:
MA = 1.429 g/L * 22.414 L/mol
= 32.03 g/mol
Mixtures of Gases and Dalton’s Law
of Partial Pressures


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 + …
Mixtures of Gases and Mole Fraction:
cA = nA/nT = the ratio of the number
of moles of a given component in a
mixture to the total number of
moles in the mixture
PO2= nO2RT/V = nO2 = cO2
PT
nTRT/V
nT
Therefore, PO2 = cO2 * PT
Example:
We react N2 and H2 to make NH3. We start with 10.3 moles
of H2 and 3.71 moles of N2. We use up the H2 to make
6.87 mol of NH3. Find the total pressure in the vessel
and the partial pressure of each gas, assuming that the
product is collected in a 125 L tank at 25.0°C.
N2(g) + 3 H2(g)  2 NH3(g)
Step 1: Determine moles NH3 produced and also find how
much excess N2 remains:
10.3 mol H2 * 2NH3/3H2 = 6.87 mol NH3 produced
10.3 mol H2 * 1N2/3H2 = 3.43 moles N2 used
3.71moli - 3.43 molused = 0.28 moles N2 remaining
Add that to moles of ammonia produced for total moles:
nT = mol NH3 + mol N2 =6.87 + 0.28
= 7.15 moles total in tank
Example continued
Step 2: Determine total pressure, then pressure of
each gas.
PT=nTRT/V = 7.15 mol * 0.082057 * 298K/125L
= 1. 399 atm
Partial pressure for each gas could be determined using
moles of each in the ideal gas law, or by using mole
fraction, as shown below.
cN2 = nN2/nT = 0.28/7.15 = 0.0392,
cNH3 = 0.961 (from 1.000-.0392)
PN2 = cN2*PT = 0.0392*1.399 = 0.0548 atm
PNH3 = cNH3*PT 0.961*1.399 = 1.344 atm
Figure 5.10
Collecting a water-insoluble gaseous reaction
product and determining its pressure.
Example:
We collected 0.500 L of hydrogen gas over water at
a room temperature of 26.0°C and atmospheric
pressure of 745 torr. How many moles of
hydrogen gas were collected?
Look up PH2O at 26.0oC and find it is 25.2 mm Hg.
Find PH2: Pbar = PH2 + PH2O
745 torr = PH2 + 25.2 torr
PH2 = 720. torr
Convert: 720. torr (1 atm/760 torr) = 0.94737 atm
Rearrange Ideal Gas Law: n = PV/RT
n = (0.94737 atm) * 0.500 L
0.082057 L.atm/mol.K*299.15K
n = 0.0193 mol of H2
Practice Problems
Work on problems 32, 34, 38, and
72ab.
32: What is the density of Freon gas
(CFCl3) at 120.oC and 1.50 atm?
34: The density of a noble gas is 2.71
g/L at 3.00 atm and 0.00oC. Name
the gas.
38: A 355 mL container holds 0.146 g
of Ne and some amt of Ar, at 35.0oC
and total pressure of 626 torr.
Calculate moles of Ar present.
STOICHIOMETRY OF GASES:
Stoichiometry problems that you will love to
do!
At STP, one mole of any gas occupies
22.414 L.
WE CAN USE THIS A LOT!
This means that equal volumes of gases at
STP have the same # of moles!!!!
11.2 L of N2 = 0.5 moles = 11.2 L of O2 =
0.5 moles = 11.2 L of He, etc.
We can relate volumes through stoich coeff
just like moles
Stoichiometry of Gases
Example:
N2(g) + 3 H2(g)  2 NH3(g) (over Fe
catalyst at 500°C)
Given 355 L of hydrogen, how many L
of nitrogen required?
355L H2 * 1 N2/3H2 = 118 L
How many L of ammonia produced?
355L H2 * 2NH3/3H2 = 237 L
Stoichiometry at STP: regular?
Example 2: 12.0 g Zn react with excess sulfuric
acid, how many liters of gas will you have at STP?
Zn(s) + H2SO4(aq)  ZnSO4(aq) + H2(g)
12.0 g/(65.4g/mol) = 0.183 mol Zn * 1 H2/1 Zn
= 0.183 mol H2(g)
0.183 mol gas * 22.414 L/mol = 4.11 L
Note: if at other conditions, must use Ideal Gas Law
to find volume, etc.
You Try “Regular” Stoichiometry:
During ammonia production, with 355 L of hydrogen @
25.0°C & 542 mm of Hg plus 105 L of nitrogen @ 20.0°C
& 645 torr.
Find volume and mass of ammonia produced at STP.
(154 L, 117 g)
Postulates of the Kinetic-Molecular
Theory
Postulate 1: Particle Volume
The volume of an individual gas particle is so small
compared to the volume of its container, that the gas
particles are considered to have mass, but essentially
no volume.
Postulate 2: Particle Motion
Gas particles are in constant, rapid, random, straightline 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 (although
energy can be transferred).
Figure 5.12
Distribution of molecular speeds at three temperatures.
See note below or in textbook.
Figure at
end of
section 5.6
Diffusion of a gas particle through a
space filled with other particles.
distribution of molecular speeds
mean free path
collision frequency
A bizarre and dangerous molecular highway!
GRAHAM'S LAWS OF
DIFFUSION AND EFFUSION:
Diffusion is the gradual mixing of molecules
of 2 or more gases in a container, owing
to the continual molecular motion.
Effusion is the escape of molecules through
a tiny hole in a pressurized container into
a vacuum or lower pressure system.
Rate of effusion is related inversely to the
molar mass of the gas, because of its
relation to average speed of the gas
molecules.
Rategas1/Rategas2 = (MAgas2/MAgas1)½
GRAHAM'S LAWS OF
DIFFUSION AND EFFUSION:
Gas molecules are moving with various velocities; we
calculate the average, called root-mean-square
speed as shown, where R = 8.314 J/mol.K and M is
molar mass in kg/mol:
urms = (3RT/M)1/2
For example, calculate the different velocities of H2 and
O2 at 25.0oC.
1921 m/s for H2 vs. 481 m/s for O2
Velocity depends inversely on molar mass! Lighter
molecules move faster!
Figure 5.17
Relationship between molar mass and molecular speed.
Ek = 3/2 (R/NA) T
At a given temperature, gases with lower molar mass (in
parenthesis) have higher most probable speeds (peak of each
curve)
Example of Graham’s Law:
Methane and an unknown gas are observed to
escape. When counted, 1.00x104 of molecules
take 1.50 minutes for methane and the same
number take 4.73 minutes for the unknown. Find
MA for the unknown.
RateCH4/Rateunk = (1.00x104/1.50min)
(1.00x104/4.73min)
= (Munk/16g/mol)½
Square both sides: 9.986 = Munk/16g/mol
Munk = 159.8g/mol
NOTE: these ratios are also related to urms1/urms2
REAL GASES DEVIATE FROM
IDEAL GAS BEHAVIOR


When gases are at low temperature
or high pressure, they approach a
phase change to a liquid and their
behavior deviates from ideal gas
behavior.
When a gas molecule exhibits a
force of attraction for another gas
molecule, the behavior will deviate
from ideal gas behavior. (Review
KMT)
Figure 5.18
The behavior of several
real gases with increasing
external pressure.
Practice




Work on problems 57, 61 and 63a.
57: What is the ratio of effusion
rates for O2 and Kr?
61: An unknown gas effuses in 11.1
min. An equal volume of H2 effuses
in 2.42 min. What is the molar
mass of the unknown gas?
63a: Calc urms for He at 0.00oC and
at 30.0oC.