Gases and the Kinetic Molecular Theory
5.1 An Overview of the Physical States of Matter
5.2 Gas Pressure and Its Measurement
5.3 The Gas Laws and Their Experimental Foundations
5.4 Further Applications of the Ideal Gas Law
5.5 The Ideal Gas Law and Reaction Stoichiometry
5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior
5.7 Real Gases: Deviations from Ideal Behavior
1

The three 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.
2

Common Units of Pressure
Unit pascal(Pa); kilopascal(kPa) atmosphere(atm)
Atmospheric Pressure
1.01325x10
5 Pa;
101.325 kPa
1 atm*
Scientific Field
SI unit; physics, chemistry chemistry millimeters of mercury
(Hg)
760 mm Hg* torr 760 torr* chemistry, medicine, biology chemistry pounds per square inch
(psi or lb/in 2 )
14.7lb/in 2 engineering
Bar 1.01325 bar meteorology, chemistry, physics
*This is an exact quantity; in calculations, we use as many significant figures as necessary.
Boyle’s Law
The relationship between the volume and pressure of a gas.
Boyle’s Law – the compressibility of gases
PV = C
3

The relationship between the volume and temperature of a gas.
Charles’s Law – The effect of temperature on gas volume
Boyle’s Law V
α
V x P = constant
1
P n and T are fixed
V = constant / P
Charles’s Law V
α
T
V
T
= constant
Amontons’s Law P
α
T
P
T
= constant
P and n are fixed
V = constant x T
V and n are fixed
P = constant x T combined gas law V
α
T
P
V = constant x
T
P
PV
T
= constant
4

Standard molar volume.
bar bar bar
R is the universal gas constant
3 significant figures
fixed n and T
Boyle’s Law
V = constant
P
IDEAL GAS LAW
PV = nRT or V = nRT
P fixed n and P
Charles’s Law
V = constant X T fixed P and T
Avogadro’s Law
V = constant X n
5

Applying the Volume-Pressure Relationship
PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies
24.8 cm 3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?
PLAN:
SOLUTION: P and T are constant
V
1 in cm 3
P
1
= 1.12 atm P
2
= 2.64 atm
V
1
1cm 3 =1mL in mL unit conversion
10 3 mL=1L
V
V
2
1 in L xP
1
/P in L
2
V
1
= 24.8 cm
24.8 cm 3
3
1 mL
V
L
2
= unknown
= 0.0248 L
1 cm 3 10 3 mL gas law calculation
V
2
=
P
1
V
1
P
2
P
1
V
1
P
2
V
2
= n
1
T
1 n
2
T
2
= 0.0248 L
1.12 atm
2.46 atm
P
1
V
1
= P
2
V
2
= 0.0105 L
Applying the Temperature-Pressure Relationship
PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x10
3 torr. It is filled with methane at 23 0 C and 0.991 atm and placed in boiling water at exactly 100 0 C. Will the safety valve open?
PLAN: SOLUTION:
P
1
(atm) T
1 and T
2
( 0 C)
P
1
= 0.991atm
P
1
1atm=760torr
(torr) x T
2
/T
1
K= 0 C+273.15
T
1 and T
2
(K)
T
P
1
V
1 n
1
T
1
1
= 23 0 C
=
P
2
(torr)
P
2
V
2 n
2
T
2
0.991 atm
760 torr = 753 torr
1 atm
T
2
P
2
= P
1
T
1
373K
= 753 torr
296K
= 949 torr
P
2
= unknown
T
2
= 100 0 C
P
T
1
1
=
P
T
2
2
6

Applying the Volume-Amount Relationship
PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm 3 . When 1.10 mol of He is added to the blimp, the volume is 26.2 dm 3 . How many grams of He must be added to make it rise? Assume constant T and P.
PLAN:
We are given initial n
1 n
2 and V
1 as well as the final V and convert it from moles to grams.
2
. We have to find n
1
(mol) of He x V
2
/V
1 n
2
(mol) of He subtract n
1 mol to be added x M g to be added
SOLUTION: n
V
1
1
= 1.10 mol
= 26.2 dm 3 n
V
2
2
P and T are constant
= unknown
= 55.0 dm 3
P
1
V
1 n
1
T
1
=
P
2
V
2 n
2
T
2
V n
1
1
=
V n
2
2 n
2
= n
1
V
2
V
1 n
2
= 1.10 mol
55.0 dm 3
26.2 dm 3
4.003 g He
= 2.31 mol mol He
= 9.24 g He
Lecture 1
Solving for an Unknown Gas Variable at Fixed Conditions
PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O
2
.
Calculate the pressure of O
2 at 21 0 C.
PLAN: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P.
SOLUTION: V = 438 L n = 0.885 kg (convert to mol)
T = 21 0 C ( convert to K )
P = unknown
0.885kg
10 3 g kg mol O
2
32.00 g O
2
= 27.7 mol O
2
21 0 C + 273.15 = 294.15K
P = nRT
=
V
24.7 mol x 0.0821
atm*L mol*K x 294.15K
438 L
= 1.53 atm
7

density = m/V n = m / M
PV = nRT PV = (m/ M )RT m/V = M x 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.
Calculating Gas Density
PROBLEM:
To apply a green chemistry approach, a chemical engineer uses waste
CO
2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO
2 and the number of molecules
(0 0 C and 1 atm) and (b) at room conditions (20.
(a) at STP
0 C and 1.00 atm).
PLAN:
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 Avogadro’s number.
d = mass/volume
PV = nRT d =
M x P
RT
V = nRT/P
8

Calculating Gas Density
SOLUTION:
1.96 g
L
(a) mol CO
2
44.01 g CO
2
44.01 g/mol x 1atm d = atm*L
0.0821
mol*K
6.022x10
23 molecules x 273.15K
= 1.96 g/L
= 2.68x10
22 molecules CO
2
/L mol
(b) d =
44.01 g/mol x 1 atm
0.0821
atm*L x 293K mol*K
1.83g
L mol CO
2
44.01g CO
2
6.022x10
23 molecules mol
= 1.83 g/L
= 2.50x10
22 molecules CO
2
/L
Using Gas Variables to Find Amount of Reactants and Products
PROBLEM: Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H
2 reduces the metal oxide, forming the pure metal and H
2
O. On a laboratory scale, what volume of H
2 at 765 torr and 225 0 C is needed to reduce 35.5 g of 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 H
2 gas.
mass (g) of Cu
SOLUTION: CuO( s ) + H
2
( g ) Cu( s ) + H
2
O( g ) divide by M mol of Cu molar ratio mol of H
2 use P and T to find V
L of H
2
35.5 g Cu mol Cu
63.55 g Cu
0.559 mol H
2 x 0.0821
1 mol H
2
1 mol Cu atm*L mol*K x
1.01 atm
= 0.559 mol H
498K
2
= 22.6 L
9

Take-home message
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 5.25
L of chlorine gas at 0.950 atm and 293K reacts with 17.0 g 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 ) + Cl
2
( g ) 2KCl( s )
P = 0.950 atm V = 5.25 L
PV
=
0.950 atm x 5.25L
= 0.207 mol
T = 293K n = unknown n =
Cl
2
RT atm*L
0.0821
x 293K
17.0g
mol K mol*K
= 0.435 mol K
0.207 mol Cl
2
2 mol KCl
1 mol Cl
2
= 0.414 mol
KCl formed
39.10 g K
Cl
2 is the limiting reactant.
0.435 mol K
2 mol KCl
2 mol K
0.414 mol KCl
74.55 g KCl mol KCl
= 30.9 g KCl
= 0.435 mol
KCl formed
• 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
P total
= P
1
+ P
2
+ P
3
+ ...
P
1
= c
1 x P total where c
1 is the mole fraction c
1
= n
1 n
1
+ n
2
+ n
3
+...
= n
1 n total
10

Applying Dalton’s Law of Partial Pressures
PROBLEM: In a study of O
2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol % N
2
, 17 mol % 16 O
2, and
4.0 mol % 18 O
2
. (The isotope 18 O will be measured to determine the O
2 uptake.)
The pressure of the mixture is 0.75atm to simulate high altitude.
Calculate the mole fraction and partial pressure of 18 O
2 in the mixture.
PLAN: Find the
χ
18 O
2 mol % 18 O
2 and P from P
O
2 total and mol % 18 O
2
.
SOLUTION: divide by 100 %
Mole fraction : c 18 O
2
χ
18 O
2
=
4.0 mol% 18 O
2
100%
= 0.040
multiply by P total
P =
18 O
2
χ
18 O
2 x P total
= 0.040 x 0.75 atm partial pressure P
18 O
2
= 0.030 atm
Lecture 2
Avogadro’s Law V
α n
E k
= 1/2 mass x speed 2 u rms
=
3RT
M
E k
= 1/2 mass x u 2 u 2 is the root-mean-square speed
R = 8.314Joule/mol*K
Graham’s Law of Effusion
Effusion is the flow of gas molecule at low pressure through the tiny holes, depending on the pressure and the relative speed of the particles.
The rate of effusion of a gas is inversely related to the square root of its molar mass.
rate of effusion a =
√ M
1
11

Applying Graham’s Law of Effusion
PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH
4
).
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.
rate of effusion a =
1
√
M
SOLUTION:
M of CH
4
= 16.04g/mol M of He = 4.003g/mol rate
He rate
CH
4
=
16.04
4.003
= 2.002
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 (E k
) of the particles is constant.
12

Distribution of molecular speeds at three temperatures.
The average kinetic energy of a gas molecule is directly proportional to the temperature.
The average speed of a gas molecule is directly proportional to the square root of absolute temperature.
The average speed of a gas molecule is reversely proportional to the square root of its molar mass.
Average speed can be expressed by the equation: u = (3RT/MM) 1/2 .
13