Chapter 5
Gases and the Kinetic-Molecular Theory
Macroscopic vs. Microscopic
Representation
Kinetic Molecular Theory of Gases
1. Gas molecules are in constant motion in
random directions. Collisions among
molecules are perfectly elastic.
Kinetic Molecular Theory of Gases
2. The average kinetic energy of the molecules is proportional
to the temperature of the gas in kelvins. Any two gases at
the same temperature will have the same average kinetic
energy
3. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be points;
that is, they possess mass but have negligible volume.
4. Gas molecules exert neither attractive nor repulsive forces
on one another.
5. Each gas molecule “behaves” as if it were alone in
container (due to #3 and #4)
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(Kk) of the particles is constant.
Distribution of molecular speeds at three temperatures.
Relationship between molar mass and
molecular speed.
Physical Characteristics of Gases
•
Gases assume the volume and shape of their containers.
•
Gases are the most compressible state of matter.
•
Gases will mix evenly and completely when confined to
the same container.
•
Gases have much lower densities than liquids and solids.
Pressure – KMT Viewpoint
• Origin of Pressure –
Gas molecules hitting
container walls
–  Temp, KE,  #
collisions,  P
–  Volume,  #
collisions,  P
Pressure – Macroscopic
Viewpoint
Pressure = Force
Area
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
•  Temp, KE, 
Force,  P
• Volume,  Area,
P
Common Units of Pressure
Unit
Atmospheric Pressure
Scientific Field
pascal(Pa);
kilopascal(kPa)
1.01325x105Pa;
101.325 kPa
SI unit; physics, chemistry
atmosphere(atm)
1 atm
chemistry
millimeters of
mercury(Hg)
760 mmHg
chemistry, medicine, biology
torr
760 torr
chemistry
14.7lb/in2
engineering
0.01325 bar
meteorology, chemistry,
physics
pounds per square
inch (psi or lb/in2)
bar
Atmospheric Pressure
Barometer
Two types of manometer
Converting Units of Pressure
PROBLEM:
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 = 291.4mmHg. Calculate the CO2 pressure in torrs,
atmospheres, and kilopascals.
PLAN: Construct conversion factors to find the other units of pressure.
291.4mmHg
1torr
= 291.4torr
1mmHg
291.4torr
1atm
= 0.3834atm
760torr
0.3834atm
101.325kPa
1atm
= 38.85kPa
Effect of Pressure on Volume
Boyle’s Law
1 atm
2 atm
5 atm
5
5
5
3
3
3
1
1
1
The relationship between volume and the pressure of a gas.
Boyle’s Law
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
Boyle’s Law
PV = constant
V a
1
P
n and T are fixed
V = constant/P
Kinetic
Molecular theory of gases and …
• Boyle’s Law
P a collision rate with wall
Collision rate Increases with decreased volume
P a 1/V
Increase P, decrease volume
A sample of chlorine gas occupies a volume of 946 mL
at a pressure of 726 mmHg. What is the pressure of
the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL?
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
Applying the Volume-Pressure Relationship
PROBLEM:
Boyle’s apprentice finds that the air trapped in a J tube occupies
24.8cm3 at 1.12atm. By adding mercury to the tube, he increases
the pressure on the trapped air to 2.64atm. Assuming constant
temperature, what is the new volume of air (inL)?
PLAN:
SOLUTION:
V1 in cm3
1cm3=1mL
V1 in mL
unit
conversion
103mL=1L
V1 in L
xP1/P2
P1 = 1.12atm
P2 = 2.64atm
V1 = 24.8cm3
V2 = unknown
24.8cm3
gas law
calculation
P1V1
V2 in L
n1T1
V2 =
P1V1
P2
P and T are constant
=
1mL
L
1cm3
103mL
P2V2
= 0.0248L
P1V1 = P2V2
n2T2
1.12atm
= 0.0248L
2.46atm
= 0.0105L
Effect of Temperature on Volume
(Charles’ Law)
Low Temperature
High Temperature
Va T
V = kT
V/T = k
V1/T1 = V2/T2
As T increases, V Increases
Charles Law Animation
Applying the Temperature-Pressure Relationship
A 1-L steel tank is fitted with a safety valve that opens if the
internal pressure exceeds 1.00x103 torr. It is filled with helium at
230C and 0.991atm and placed in boiling water at exactly 1000C.
Will the safety valve open?
T1 and T2(0C)
P1(atm)
1atm=760torr
P1(torr)
K=0C+273.15
T1 and T2(K)
x T2/T1
P2(torr)
P2 = P1
P1 = 0.991atm
P2 = unknown
T1 = 230C
T2 = 100 oC
P1V1
=
P2V2
n1T1
n2T2
0.991atm 760torr
1atm
= 753torr
T2
T1
= 753torr
373K
296K
= 949torr
P1
T1
=
P2
T2
Kinetic theory of gases and …
• Charles’ Law
-Average kinetic energy a T
-Increase T, Gas Molecules hit walls with greater Force,
this Increases the Pressure
BUT since pressure must remain constant, and only
volume can change
-Volume Increase to reduce Pressure
-Increase Temperature, Increase Volume
Determination of Absolute Zero
A sample of carbon monoxide gas occupies 3.20 L at
125 0C. At what temperature will the gas occupy a
volume of 1.54 L if the pressure remains constant?
V1/T1 = V2/T2
T2 =
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
V2 x T1
V1
=
1.54 L x 398.15 K
3.20 L
= 192 K
Avogadro’s Law
V a number of moles (n)
V = constant x n
V1/n1 = V2/n2
5.3
Applying the Volume-Amount Relationship
A scale model of a blimp rises when it is filled with helium to a
volume of 55dm3. When 1.10mol of He is added to the blimp, the
volume is 26.2dm3. How many more grams of He must be added
to make it rise? Assume constant T and P.
We are given initial n1 and V1 as well as the final V2. We have to find
n2 and convert it from moles to grams.
n1(mol) of He
P and T are constant
x V2/V1
n1 = 1.10mol
n2 = unknown
n2(mol) of He
V1 = 26.2dm3
V2 = 55.0dm3
subtract n1
mol to be added
V1
n1
=
V2
n2
n2 = n1
xM
g to be added
n2 = 1.10mol
55.0dm3
26.2dm3
P1V1
n1T1
=
P2V2
n2T2
V2
V1
= 2.31mol
4.003g He
mol He
= 4.84g He
Kinetic theory of gases and …
• Avogadro’s Law
More moles of gas, more collisions with walls of
container
More collisions, higher pressure
BUT since pressure must remain constant and
only volume can change
Volume increases to decrease pressure to original
value
V a
Boyle’s Law
V
= constant
T
P a T
Amonton’s Law
T
combined gas law
P
V a T
Charles’s Law
P
1
= constant
V a
n and T are fixed
P and n are fixed
V = constant x T
V and n are fixed
P = constant x T
T
P
V = constant x
T
PV
P
T
= constant
Gas Law Animation
Ideal Gas Equation
Boyle’s law: V a 1 (at constant n and T)
P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Va
nT
P
V = constant x
nT
=R
nT
R is the gas constant
P
P
R = 0.082057 L • atm / (mol • K)
PV = nRT
Obtaining Other Gas Law
Relationship
• PV = nRT
PV
R
nT
P1V1 P2V2

n1T1
n2T2
THE IDEAL GAS LAW
PV = nRT
R=
PV
nT
=
1atm x 22.414L
1mol x 273.15K
=
0.0821atm*L
mol*K
IDEAL GAS LAW
nRT
PV = nRT or V =
fixed n and T
Boyle’s Law
V=
constant
P
P
fixed n and P
fixed P and T
Charles’s Law
Avogadro’s Law
V=
constant X T
V=
constant X n
Argon is an inert gas used in lightbulbs to retard the
vaporization of the filament. A certain lightbulb
containing argon at 1.20 atm and 18 0C is heated to
85 0C at constant volume. What is the final pressure of
argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant
nR
P
=
= constant
T
V
P1
P2
=
T1
T2
P1 = 1.20 atm
P2 = ?
T1 = 291 K
T2 = 358 K
T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1
Types of Problems
P1V1 P2V2

n1T1
n2T2
Given initial conditions,
determine final conditions;
Cancel out what is constant
• Make Substitution into
PV = nRT
moles (n) 
mass, g
MolarMass, g / mole
mass
Density 
Volume
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 765torr and 2250C is needed to form 35.5g of Cu from
copper (II) oxide?
PLAN: Since this problem requires stoichiometry and the gas laws, we have
to write a balanced equation, use the moles of Cu to calculate mols
and then volume of H2 gas.
mass (g) of Cu
SOLUTION:
divide by M
mol of Cu
35.5g Cu
CuO(s) + H2(g)
mol Cu
0.559mol H2 x 0.0821
use known P and T to find V
L of H2
1mol H2
63.55g Cu 1 mol Cu
molar ratio
mol of H2
Cu(s) + H2O(g)
atm*L
mol*K
1.01atm
x
= 0.559mol H2
498K
= 22.6L
Solving for an Unknown Gas Variable at Fixed Conditions
A steel tank has a volume of 438L and is filled with 0.885kg of
O2. Calculate the pressure of O2 at 210C.
V, T and mass, which can be converted to moles (n), are given. We
use the ideal gas law to find P.
0.885kg
V = 438L
T = 210C (convert to K)
n = 0.885kg (convert to mol)
P = unknown
103g
mol O2
kg
32.00g O2
= 27.7mol O2
24.7mol x 0.0821
P=
nRT
V
atm*L
mol*K
=
210C + 273.15 = 294K
x 294K
= 1.53atm
438L
Using the Ideal Gas Law in a Limiting-Reactant Problem
PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)]
to form ionic metal halides. What mass of potassium chloride forms
when 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g of
potassium?
PLAN: After writing the balanced equation, we use the ideal gas law to find the
number of moles of reactants, the limiting reactant and moles of product.
SOLUTION:
2K(s) + Cl2(g)
2KCl(s)
P = 0.950atm
V = 5.25L
PV
0.950atm x 5.25L
T = 293K
n = unknown
n =
=
= 0.207mol
Cl2
RT
atm*L
0.0821
x 293K
mol*K
2mol KCl
mol
K
0.207mol
Cl
= 0.414mol
2
17.0g
= 0.435mol K
1mol Cl2
KCl formed
39.10g K
2mol KCl
Cl2 is the limiting reactant.
0.435mol K
= 0.435mol
2mol K
KCl formed
74.55g KCl
0.414mol KCl
= 30.9 g KCl
mol KCl
Summary of the stoichiometric relationships
among 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
ideal
gas
law
Standard Molar Volume
Density (d) Calculations
PM
m
d=
=
V
RT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT
M=
P
d is the density of the gas in g/L
Finding the Molar Mass of a Volatile Liquid
A chemist isolates from a petroleum sample a colorless liquid
with the properties of cyclohexane (C6H12). The Dumas method
is used to obtain the following data to determine its molar mass:
Volume of flask = 213mL
T = 100.00C
Mass of flask + gas = 78.416g
Mass of flask = 77.834g
P = 754 torr
Is the calculated molar mass consistent with the liquid being cyclohexane?
Use unit conversions, mass of gas and density-M relationship.
m = (78.416 - 77.834)g = 0.582g
M=
m RT
VP
=
0.582g x 0.0821
atm*L
mol*K
x 373K
= 84.4g/mol
0.213L x 0.992atm
M of C6H12 is 84.16g/mol and the calculated value is within experimental error.
Calculating Gas Density
Calculate the density (in g/L) of carbon dioxide and the number
of molecules per liter (a) at STP (00C and 1 atm) and (b) at
ordinary room conditions (20.0C and 1.00atm).
Density is mass/unit volume; substitute for volume in the ideal gas
equation. Since the identity of the gas is known, we can find the molar
mass. Convert mass/L to molecules/L with Avogardro’s number.
MxP
d = mass/volume PV = nRT
V = nRT/P
d =
RT
44.01g/mol
(a)
mol CO2
L
44.01g CO2
= 1.96g/L
d=
0.0821
1.96g
x 1atm
atm*L
x 273K
mol*K
6.022x1023molecules
mol
= 2.68x1022molecules CO2/L
Calculating Gas Density
continued
(b)
44.01g/mol
x 1atm
d=
0.0821
1.83g
mol CO2
L
44.01g CO2
= 1.83g/L
atm*L x 293K
mol*K
6.022x1023molecules
mol
= 2.50x1022molecules CO2/L
The Molar Mass of a Gas
n=
mass
M
=
PV
RT
M=
m RT
VP
d=
m
V
d RT
M=
P
Dalton’s Law of Partial Pressures
V and T
are
constant
P1
P2
Ptotal = P1 + P2
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
P1= c1 x Ptotal
where c1 is the mole fraction
c1 =
n1
n1 + n2 + n3 +...
=
n1
ntotal
Applying Dalton’s Law of Partial Pressures
PROBLEM:
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
0.75atm to simulate high altitude. Calculate the mole fraction
and partial pressure of 18O2 in the mixture.
PLAN: Find the c 18 and P18 from Ptotal and mol% 18O2.
O2
O2
4.0mol% 18O2
18
SOLUTION:
mol% O2
=
c 18
O2
100
divide by 100
c
18O
P18
2
multiply by Ptotal
partial pressure P
18O
2
O2
= 0.040
= c18 x Ptotal = 0.040 x 0.75atm = 0.030atm
O2
Kinetic theory of gases and …
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the
presence of another gas
Ptotal = SPi
Calculating the Amount of Gas Collected Over Water
PROBLEM:
Acetylene (C2H2), an important fuel in welding, is produced in
the laboratory when calcium carbide (CaC2) reaction 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 738torr and
the volume is 523mL. At the temperature of the gas (230C), the
vapor pressure of water is 21torr. How many grams of
acetylene are collected?
PLAN: The difference in pressures will give us the P for the C2H2. The ideal
gas law will allow us to find n. Converting n to grams requires the
molar mass, M.
P
C2H2 = (738-21)torr = 717torr
Ptotal
P
C2H2
atm
P
= 0.943atm
PV
717torr
H2O
n=
760torr
RT
n
g
C2H2
C2H2
xM
Calculating the Amount of Gas Collected Over Water
continued
n
C2H2
0.943atm x
=
0.0821
atm*L
0.523L
= 0.203mol
x 296K
mol*K
0.203mol
26.04g C2H2
mol C2H2
= 0.529 g C2H2
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
V a
Avogadro’s Law
n
Ek = 1/2 mass x u 2
Ek = 1/2 mass x speed2
u 2 is the root-mean-square speed
urms =
√3RT
R = 8.314Joule/mol*K
M
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
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 for each gas. Find the molar mass of both gases and find
the inverse square root of their masses.
SOLUTION:
M of CH4 = 16.04g/mol
rate
He
rate
CH4
=
√
16.04
4.003
= 2.002
M of He = 4.003g/mol
Real Gases
• Nonideal Conditions when gas gets close to
conditions where it
will liquify
– Lower Temperature
– Higher Pressure

an 2 
 P  2 V  nb   nRT
V 

Real Gases
Effect of Intermolecular
Forces
– Corrected Pressure, a
– Corrected Volume, b