Chapter 13 – Gases
13.1 The Gas Laws
13.2 The Ideal Gas Law (Part I – Equation
Only)
13.3 Gas Stoichiometry
Section 13.2 The Ideal Gas Law
The ideal gas law relates the number of
particles to pressure, temperature, and
volume.
• Relate the amount of gas present to its pressure,
temperature, and volume using the ideal gas law.
Section 13.2 The Ideal Gas Law
Key Concept
• The ideal gas law relates the amount of a gas present
to its pressure, temperature, and volume
PV = nRT
Gas Laws – Mechanistic and Empirical
“Named” laws described in section 13.1
were empirically determined from
experiments – stated relationships “fit the
data”
Example: Boyle experimentally determined
that P & V are inversely related but there
was no underlying understanding of why
this occurs – this is an empirical model
Gas Laws – Mechanistic and Empirical
Ideal gas law is mechanistic – it can be
derived by using the basic theory (kineticmolecular) of how gases behave; the
underlying mechanism that leads to the
equation is known and understood
Ideal gas law was formulated after the
empirical (named) laws were discovered
It predicted the same behavior as the
empirical laws but provided a way to
understand why they were true
Kinetic-Molecular Theory (KMT) of
Gases (see section 12.1)
Mostly empty space
No attractions/repulsions between
particles
Constant straight-line motion until
collisions with other particles or walls
Collisions perfectly elastic: KE constant
• T constant, KE constant
All gases have same KEavg at a given T
Ideal Gas Law
Relationship between pressure P,
volume V, Kelvin temperature T, and
number of moles n
PV = nRT
T must be expressed in units of kelvin
Equation derived using KMT
assumptions
R = Gas Constant – value depends
upon units used for P
R (Gas Constant) – Table 13.2
Value
Units
0.0821
L atm
mol K
L kPa
mol K
L mm Hg
mol K
8.314
62.4
Ideal Gas Law – Dependence on n
PV = nRT
Fixed V, T then
Pn
Fixed P, T
Vn
Ideal Gas Law: Example Problem 13.6
Calculate number of moles of NH3 for
V = 3.0 L, T = 3.00x102 K, P = 1.50 atm
PV = nRT  n = PV / RT
n = (1.50 atm  3.0 L) /
(8.21x10-2 L atm/mol K  3.00x102 K)
n = 0.18 mol
Note that value of R was chosen that
has atm as the pressure unit
Practice (Ideal Gas Law)
Problems 26 – 30, page 455
Problems 68, 69, 74, 75, 77 – 78,
pages 468 – 9
Problems 15, 16, 18, page 985
Ideal Gas Law and the
Combined Gas Law
PV = nRT
If the # of moles of gas (n) is fixed, then
PV/T = nR = constant
If P,V,T conditions change (from 1 to 2)
P1V1/T1 = P2V2/T2
(have derived the Combined Gas Law)
Combined & Named Gas Laws
(Subject of Section 13.1)
For given mass (fixed # moles) of gas
P1V1/T1 = P2V2/T2
3 laws can be derived from this:
Fix T: P1V1 = P2V2 (Boyle’s)
Fix P: V1/T1 = V2/T2 (Charles’)
Fix V: P1/T1 = P2/T2 (Gay-Lussac’s)
Chapter 13 – Gases
13.1 The Gas Laws
13.2 The Ideal Gas Law (Part I)
13.3 Gas Stoichiometry
Section 13.1 The Gas Laws
For a fixed amount of gas, a change in
one variable—pressure, temperature, or
volume—affects the other two.
• State the relationships among pressure, temperature,
and volume of a constant amount of gas.
• Apply the gas laws to problems involving the pressure,
temperature, and volume of a constant amount of gas.
Section 13.1 The Gas Laws
Key Concepts
• Boyle’s law states that the volume of a fixed amount of
gas is inversely proportional to its pressure at constant
temperature.
P1V1 = P2V2
• Charles’s law states that the volume of a fixed amount
of gas is directly proportional to its kelvin temperature
at constant pressure.
Section 13.1 The Gas Laws
Key Concepts
• Gay-Lussac’s law states that the pressure of a fixed
amount of gas is directly proportional to its kelvin
temperature at constant volume.
• The combined gas law relates pressure, temperature,
and volume in a single statement.
Boyle’s Law
Volume of given mass (fixed # of
moles) of gas held at constant T varies
inversely with pressure
P1V1 = P2V2
PV = constant
V2 = V1 (P1/P2)
10 L
1 atm
5L
2 atm
2.5 L
4 atm
Constant = PV = 10 atm L
Boyle’s Law – Example Problem 13.1
Diver blows bubble of volume = 0.75 L at 10
m depth & pressure = 2.25 atm. Volume of
bubble at surface when pressure = 1.03 atm?
P1V1 = P2V2
V2 = V1 (P1/P2)
V2 = 0.75 L (2.25 atm / 1.03 atm)
V2 = 1.6 L
Practice (Boyle’s Law)
Problems 1- 3 page 443
Problems 59 – 60, page 468
Problems 1 – 2, page 984
Charles’s Law
Volume of given mass (fixed # of
moles) of gas held at constant P is
directly proportional to its kelvin
temperature
V1/T1 = V2/T2
V2 = V1(T2/T1)
TK = 273.15 + TC
V change with
T at fixed P
Volume vs Celsius Temperature
Volume vs Kelvin Temperature
Volume vs T (Kelvin) – Different Gases
Gases follow Charles’s Law but liquefy at different T–
dashed lines are extrapolations of gas phase behavior
Ideal gases
occupy no
volume at
absolute
zero
Charles’s Law: Example Problem 13.2
Balloon has volume of 2.32 L at 40.0 C.
Volume at 75.0 C if P remains constant?
V1/T1 = V2/T2
T1 = 40.0 C + 273.15 = 313.2 K
T2 = 75.0 C + 273.15 = 348.2 K
V2 = V1 (T2/T1)
V2 = 2.32 L (348.2 K/ 313.2 K)
V2 = 2.58 L
Practice (Charles’s Law)
Problems 4 – 7, page 446
Problems 55, 58, 59 page 448
Problems 3 – 4, page 984
Joseph Louis Gay-Lussac
(1778 – 1850)
http://en.wikipedia.org/wiki/Gay-Lussac
French chemist & physicist. Known
mostly for 2 laws related to gases, &
for his work on alcohol-water mixtures,
which led to the degrees Gay-Lussac
used to measure alcoholic beverages
in many countries.
Prof. of physics at Sorbonne (1808 to
1832), a post which he only resigned
for the chair of chemistry at the Jardin
des Plantes. In 1802, first formulated law stating that
if mass and pressure of gas are held constant, then
gas volume increases linearly as temperature rises.
Gay-Lussac’s Law
P change with T at
fixed V
Gay-Lussac’s Law
Pressure of given mass (fixed # of
moles) of gas in a constant volume is
directly proportional to its kelvin
temperature
P1/T1 = P2/T2
P2 = P1(T2/T1)
TK = 273.15 + TC
Gay-Lussac’s Law
At 0 K, ideal
gas exerts zero
pressure
Gay-Lussac’s Law: Example Problem 13.3
Pressure of O2 in canister is 5.00 atm at
25.0 C. New pressure at T of -10.0 C?
P1/T1 = P2/T2
T1 = 25.0 C + 273.15 = 298.2 K
T2 = -10.0 C + 273.15 = 263.2 K
P2 = P1(T2/T1)
P2 = 5.00 atm (263.2 K / 298.2 K)
P2 = 4.41 atm
Practice (Gay-Lussac’s Law)
Problems 8 – 10, page 448
Problem 18, page 451
Problem 59 page 468
Problems 5 – 7, page 984
Combined Gas Law
If P,V,T conditions change (from 1 to 2)
P1V1/T1 = P2V2/T2
Combined Gas Law: Practice Problem 13.4
Gas at 110 kPa and 30.0 C fills flexible
container with initial V of 2.00 L. If T and P
raised to 80.0 C and 440 kPa, new V?
T1 = 30.0 C + 273.15 = 303.2 K
T2 = 80.0 C + 273.15 = 353.2 K
P1V1/T1 = P2V2/T2
V2 = V1 (P2 /P1) (T1/T2)
V2 = 2.00 L (440 kPa/ 110 kPa)
(303.2 K/ 353.2 K)
V2 = 0.58 L
Practice (Combined Gas Law)
Problems 11 – 13, page 450
Problem 60 page 468
Problems 8 – 9, page 984
Gas Laws Summary – Table 13.1
All require fixed n (moles of gas)
Chapter 13 – Gases
13.1 The Gas Laws (includes Combined)
13.2 The Ideal Gas Law (Part II Avogadro’s Principle, Gas Density,
Real Gases)
13.3 Gas Stoichiometry
Section 13.2 The Ideal Gas Law
The ideal gas law relates the number of
particles to pressure, temperature, and
volume.
• Relate number of particles and volume using
Avogadro’s principle.
• Compare the properties of real and ideal gases.
Section 13.2 The Ideal Gas Law
Key Concepts
• Avogadro’s principle states that equal volumes of
gases at the same pressure and temperature contain
equal numbers of particles.
• The ideal gas law can be used to find molar mass if
the mass of the gas is known, or the density of the gas
if its molar mass is known.
• At very high pressures and very low temperatures,
real gases behave differently than ideal gases.
Avogadro’s Principle
Equal volumes of gases at same T and P
contain equal numbers of particles
Size of gas particles negligible compared to
volume occupied, so it doesn’t matter which
gas you are using
Molar Volume of Gas
Volume occupied by one mole of gas at
0.00 C and 1.00 atmosphere pressure
T, P conditions known as STP –
Standard Temperature and Pressure
At STP, 1 mole of any gas occupies
22.4 L
Key to lots of problems involving gases
at STP
Molar Volume of Gas:
Example Problem 13.5
V of 2.00 kg of methane at STP?
2.00 kg CH4  1.00x103 g/kg  mol
CH4/16.05 g CH4 = 125 mol CH4
125 mol CH4  22.4 L CH4 / mol CH4 (at
STP)
V = 2.80x103 L
Practice
(Avogadro’s Principle & Molar Volume)
Problems 20 – 25, page 453
Problems 67, 71, 73 pages 468-9
Problems 10 - 14, page 984
Ideal Gas Law: Calculate M or Density
PV = nRT
n = # moles = mass /molar mass = m / M
PV = mRT / M  MP = mRT / V 
M = mRT / PV
M = (m/V) RT / P
(m/V) = density = D
M = D RT / P or
D = MP / RT
For a given P, T , if know:
1. mass m & V, can compute molar mass M
2. density D, can compute molar mass M
3. molar mass M, can compute density D
Practice
(Molar Mass, Density from Ideal Gas Law)
Problems 70, 72, 76 page 469
Problems 17 – 18, page 985
Ideal Gas Law - Deviations
No gas truly ideal, but most behave as
ideal gases over wide range of T & P
Deviations happen at high P and low T
where assumptions of KMT break down
• Volume of gas no longer negligible
• An ideal gas can’t be liquefied
• Intermolecular forces more significant

Polar molecules especially affected
Ideal Gas Law - Deviations
Polar molecules have larger attractive
forces between particles and generally
do not behave as ideal gases at low T
where slower-moving particles can be
affected by these forces
Large gas particles occupy more space
and deviate more from ideal gases at
high density (high P, low T) where there
is less empty space
Real Gases - PV/RT vs P for 1 Mole of Gas
Approach
ideal gas at
low P
2.0
PV
RT
1.0
0
200 400 600 800 1000 1200
P (atm)
Chapter 13 – Gases
13.1 The Gas Laws
13.2 The Ideal Gas Law
13.3 Gas Stoichiometry
Section 13.3 Gas Stoichiometry
When gases react, the coefficients in
the balanced chemical equation
represent both molar amounts and
relative volumes.
• Determine volume ratios for gaseous reactants and
products by using coefficients from chemical equations.
• Apply gas laws to calculate amounts of gaseous
reactants and products in a chemical reaction.
Section 13.3 Gas Stoichiometry
Key Concepts
• The coefficients in a balanced chemical equation
specify volume ratios for gaseous reactants and
products.
• The gas laws can be used along with balanced
chemical equations to calculate the amount of a
gaseous reactant or product in a reaction.
Balanced Equation & Gas Volumes
Gas laws can be applied to calculate
stoichiometry of reactions in which gases
are reactants or products
2H2(g) + O2(g) → 2H2O(g)
2 mol H2 reacts with 1 mol O2 to produce 2
mol water vapor
Coefficients in balanced equation represent
volume ratios for gases if P and T are fixed
during reaction
Balanced Equation & Gas Volumes
CH4(g) +
2O2(g)  CO2(g) +
2H2O(g)
There is a one-to-one relationship between
number of moles and gas volume
Balanced Equation & Gas Volumes
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
Interpreted as moles:
2 mol butane + 13 mol oxygen 
8 mol carbon dioxide + 10 mol water
Avogadro’s principle  can also be volume
2 L butane + 13 L oxygen 
8 L carbon dioxide + 10 L water
Gas Stoichiometry Map
Practice – Gas Stoichiometry
How many mol of hydrogen gas are
required to react with 1.50 mol oxygen
gas in the following reaction?
2H2(g) + O2(g) → 2H2O(g)
A.
B.
C.
D.
1.00
2.00
3.00
4.00
?
V toV Stoichiometry:
Practice Problem 13.7
Volume of oxygen gas needed for complete
combustion of 4.00 L of propane (C3H8)
assuming P and T remain constant?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
4.00 L C3H8  5 L O2 / L C3H8 = 20.0 L O2
Practice – Gas Stoich. (all V to V)
Problems 38 – 41, page 461
Problems 84 – 86, page 469
Problems 19 – 20, page 985
Combo Mass/Volume Problems
Need:
a) Balanced equation
b) At least one mass or V for product
or reactant
c) Conditions under which volumes
measured
Combo Mass/Volume – Prob 13.8
N2(g) + 3H2(g)  2NH3(g)
5.00 L N2 reacts completely
P = 3.00 atm T = 298 K
Mass of ammonia produced?
5.00 L N2  [2 L NH3/1 L N2] =
10.0 L NH3
volume ratio from equation
Combo Mass/Volume – Prob 13.8
N2(g) + 3H2(g)  2NH3(g)
10.0 L NH3 P = 3.00 atm T = 298 K
n = PV/RT = (
3.00 atm  10.0 L
)
(298 K  0.0821 L atm/mol K)
= 1.23 mol NH3
Molar mass NH3 = 17.04 g/mol
1.23 mol NH3  17.04 g NH3/mol NH3
= 21.0 g NH3
Practice (Stoich. m & V combo)
Problems 42 – 45, page 463
Problems 88 – 90, page 469
Problems 21 – 22, page 985