Gases and the Kinetic-Molecular Theory

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
5.6

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 • ↑ Temp, ↑KE, ↑
Area Force, ↑ P

Units of Pressure • ↓Volume, ↓ Area,
1 pascal (Pa) = 1 N/m2 ↑P
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa

Atmospheric Pressure

Barometer

Effect of Pressure on Volume
Boyle’s Law

1 atm 2 atm 5 atm

55 5
33 3
11 1

The relationship between volume and the pressure of a gas.

Boyle’s Law

P α 1/V Constant temperature
P x V = constant Constant amount of gas
P1 x V1 = P2 x V2

Boyle’s Law 1 n and T are fixed
Vα P

PV = constant V = constant/P

Kinetic
Molecular theory of gases and …

• Boyle’s Law

P α collision rate with wall
Collision rate Increases with decreased volume
P α 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

P1 = 726 mmHg P2 = ?

V1 = 946 mL V2 = 154 mL

P2 = P1 x V1 = 726 mmHg x 946 mL = 4460 mmHg
V2 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: P and T are constant

V1 in cm3 P1 = 1.12atm P2 = 2.64atm
V1 = 24.8cm3 V2 = unknown
1cm3=1mL unit 24.8cm3 1mL L = 0.0248L
conversion
V1 in mL

103mL=1L

V1 in L 1cm3 103mL
xP1/P2
gas law P1V1 = P2V2 P1V1 = P2V2
V2 in L calculation n1T1 n2T2

V2 = P1V1 = 0.0248L 1.12atm = 0.0105L
P2 2.46atm

Vα 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?

P1(atm) T1 and T2(0C) P1 = 0.991atm P2 = unknown
T2 = 100 oC
1atm=760torr K=0C+273.15 T1 = 230C

P1(torr) T1 and T2(K) P1V1 P2V2 P1 = P2
x T2/T1 n1T1 n2T2 T1 T2
=
P2(torr)
0.991atm 760torr = 753torr

1atm

T2 = 753torr 373K = 949torr
P2 = P1 296K

T1

Kinetic theory of gases and …

• Charles’ Law

-Average kinetic energy α 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

V1 = 3.20 L V2 = 1.54 L
T1 = 398.15 K T2 = ?

T2 = V2 x T1 = 1.54 L x 398.15 K = 192 K
V1 3.20 L

Avogadro’s Law

V α 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 P1V1 P2V2
n2(mol) of He n1T1 n2T2
subtract n1 V1 = 26.2dm3 V2 = 55.0dm3 =

mol to be added V1 V2 n2 = n1 V2
xM n1 = n2 V1

g to be added n2 = 1.10mol 55.0dm3 4.003g He = 4.84g He
26.2dm3 = 2.31mol

mol 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

Boyle’s Law 1 n and T are fixed
Vα P

Charles’s Law Vα T P and n are fixed

V V = constant x T
= constant

T

Amonton’s Law P α T V and n are fixed

P P = constant x T
= constant
V = constant x T PV = constant
T P T

T
combined gas law V α P



Ideal Gas Equation

Boyle’s law: V α 1 (at constant n and T)
P

Charles’ law: V α T (at constant n and P)

Avogadro’s law: V α n (at constant P and T)

V α nT R is the gas constant
P

V = constant x nT = R nT
PP

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

PV 1atm x 22.414L 0.0821atm*L
R= = =
nT 1mol x 273.15K mol*K

fixed n and T IDEAL GAS LAW fixed P and T
nRT

PV = nRT or V = P

fixed n and P

Boyle’s Law Charles’s Law Avogadro’s Law
V = constant X T V = constant X n
V = constant
P

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 P1 = 1.20 atm P2 = ?
V T

P1 P2 T1 = 291 K T2 = 358 K
T1 T2
=

P2 = P1 x T2 = 1.20 atm x 358 K = 1.48 atm
T1 291 K

Types of Problems

P1V1 = P2V2 • Make Substitution into
n1T1 n2T2 PV = nRT

moles(n) = mass, g

MolarMass, g / mole

Given initial conditions, Density = mass
determine final conditions; Volume
Cancel out what is constant

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: CuO(s) + H2(g) Cu(s) + H2O(g)

divide by M 35.5g Cu mol Cu 1mol H2 = 0.559mol H2
mol of Cu
63.55g Cu 1 mol Cu
molar ratio
mol of H2 0.559mol H2 x 0.0821 atm*L x 498K = 22.6L
mol*K

use known P and T to find V 1.01atm
L of H2

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.

V = 438L T = 210C (convert to K)
n = 0.885kg (convert to mol) P = unknown

103g mol O2 = 27.7mol O2 210C + 273.15 = 294K
0.885kg 32.00g O2
x 294K
kg = 1.53atm

24.7mol x 0.0821 atm*L

nRT mol*K
P= =
V 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 = 0.207mol T = 293K n = unknown
nCl2 = RT =

0.0821 atm*L x 293K

mol*K 0.207mol Cl2 2mol KCl = 0.414mol
1mol Cl2
17.0g mol K = 0.435mol K KCl formed
39.10g K

Cl2 is the limiting reactant. 0.435mol K 2mol KCl = 0.435mol

0.414mol KCl 74.55g KCl = 30.9 g KCl 2mol K KCl formed
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 amount amount P,V,T
of gas A (mol) (mol) of gas B

ideal of gas A of gas B ideal
gas gas
law molar ratio from law
balanced equation

Standard Molar Volume

Density (d) Calculations

d= m = PM m is the mass of the gas in g
V
RT M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

M = dRT d is the density of the gas in g/L
P

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 P = 754 torr

Mass of flask + gas = 78.416g Mass of flask = 77.834g

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 = 0.582g x 0.0821 atm*L x 373K = 84.4g/mol
VP mol*K

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.

d = mass/volume PV = nRT V = nRT/P d= MxP
RT

(a) 44.01g/mol x 1atm = 1.96g/L
d=

0.0821 atm*L x 273K

mol*K

1.96g mol CO2 6.022x1023molecules = 2.68x1022molecules CO2/L
L 44.01g CO2 mol

Calculating Gas Density

continued

(b) 44.01g/mol x 1atm = 1.83g/L
d=

0.0821 atm*L x 293K
mol*K

1.83g mol CO2 6.022x1023molecules = 2.50x1022molecules CO2/L
L 44.01g CO2 mol

The Molar Mass of a Gas

mass PV
n= =
M RT

m RT m
M= d=

VP 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= χ1 x Ptotal where χ1 is the mole fraction

χ1 = n1 = n1
n1 + n2 + n3 +... 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%
16O2, and 4.0 mol% 18O2. (The isotope 18O will be measured to
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 χ 18Oa2 nd P18O2 from Ptotal and mol% 18O2.

mol% 18O2 SOLUTION: χ 18O2 = 4.0mol% 18O2 = 0.040
divide by 100 100

χ 18O2 P18O2 = χ18Ox2 Ptotal = 0.040 x 0.75atm = 0.030atm
multiply by Ptotal

partial pressure P
18O2

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 = ΣPi