Chapter 5:
Gases
5.1 Measurements on Gases
• Volume- amount of space the gas occupies:
1 L = 1000 mL = 1000 cm3 = 1 x10-3 m3
• Amount – most commonly expressed in terms of
moles (n):
m = MM x n
• Temperature – measured in degrees Celsius but
commonly must convert to Kelvin:
TK = t*C
+
273.15
• Pressure – gas molecules are constantly
colliding & because of this they exert a force
over an area:
1.013 bar = 1 atm = 760 mmHg = 1 x 105 Pa = 14.7 psi
Barometer
Manometer
Example 5.1
• A balloon with a volume of 2.06 L contains
0.368 g of helium at 22 degrees Celsius
and 1.08 atm. Express the volume of the
balloon in m3, the temperature in K, and
the pressure in mmHg.
– V = 2.06 x 10-3 m3
– nHe = 0.0919 mole
– T = 22 + 273.15 = 295 K
– P = 821 mmHg
Gas Laws
• Boyle’s Law:
P1V1 = P2V2
• Charles’ Law:
V1 = V2
T1
T2
• Gay-Lusaac’s Law: P1 = P2
T1 T2
• Combined Gas Law:
P1V1 = P2V2
T1
T2
Example
• A tank is filled with a gas to a pressure of
977 mmHg at 25*C. When the tank is
heated, the pressure increases to 1.50
atm. To what temperature was the gas
heated?
– 75*C
5.2 The Ideal Gas Law &
5.3 Gas Law Calculations
– The Ideal Gas Law Constant (R):
0.0821 L atm/mol K
- ideal gas law problems
8.31 J/ mol K
- equations involving energy
8.31 x 103 g m2/s2 mol k- molecular speed problems
Molar Volume
Initial & Final State Problems
• Starting with a sample of gas at 25*C and 1.00
atm you might be asked to calculate the
pressure developed when the sample is heated
to 95*C at a constant volume. Determine a twopoint equation and solve for the final pressure.
– Initial State: P1V = nRT1
– Final State: P2V = nRT2
– Divide the 2 equations to derive a “two-point” equation:
• P1 = T1
P2 = T2
– Rearrange to solve for the variable you want: P2 = P1 T2
T1
Ans: 1.23 atm
Example 5.2
• A 250.0 mL flask, open to the atmosphere,
contains 0.0110 mol of air at 0 *C. On
heating, part of the air escapes: how much
remains in the flask at 100 *C?
– 0.00805 mol of air
Example 5.3
• If 2.50 g of sulfur hexafluoride is
introduced into an evacuated 500.0 mL
container at 83*C, what pressure (atm) is
developed?
– Ans: 1.00 atm
Density & The Ideal Gas Law
The ideal gas law offers a simple approach to the
experimental determination of the molar mass of a gas.
– Remember that m = MM x n and
n = PV and d = m
RT
V
– So you can substitute these equations
into the ideal gas law to solve fro
density (d) or molar mass (M)
–Gas Density and Human Disasters: Many gases that are denser
than air have been involved in natural and human-caused
disasters. The dense gases in smog that blanket urban centers,
such as Mexico City (see photo), contribute greatly to respiratory
illness. In World War I, poisonous phosgene gas (COCl2) was used
against ground troops as they lay in trenches. In 1984, the
unintentional release of methylisocyanate from a Union Carbide
India Ltd. chemical plant in Bhopal, India, killed thousands of
people as vapors spread from the outskirts into the city. In 1986 in
Cameroon, CO2 released naturally from Lake Nyos suffocated
thousands as it flowed down valleys into villages. Some
paleontologists suggest that a similar process in volcanic lakes
may have contributed to dinosaur kills.
Example 5.4
Acetone is widely used in nail polish remover. A
sample of liquid acetone is placed in a 3.00 L flask
and vaporized by heating to 95*C at 1.02 atm. The
vapor filling the flask at this temperature and
pressure weighs 5.87 g:
(a) What is the density of acetone vapor under
these conditions?
Ans: 1.96 g/L
(b) Calculate the molar mass of acetone.
Ans: 58.1 g/mol
(c) Acetone contains three elements C, H, and O.
When 1.00 g of acetone is burned 2.27 g of CO2
and 0.932 g of H2O are formed. What is the
molecular formula of acetone?
Ans: C3H6O
5.3 Stoichiometry of Gaseous Reactions
• A molar ratio from a balanced chemical
reaction is also used in reactions involving
gases however, the ideal gas law can now be
applied.
• Example 5.5: A nickel smelter in Sudbury,
Ontario produces 1% of the world’s supply of
sulfur dioxide by the reaction of nickel II sulfide
with oxygen another product of the reaction is
nickel II oxide:
What volume of sulfur dioxide at 25*C and a
pressure of one bar is produced from a metric
ton of nickel II sulfide?
Ans: 2.73 x 105 L
Gas A to Gas B
Example 5.6
• Octane, C8H18, is one of the hydrocarbons
in gasoline. On combustion octane
produces carbon dioxide and water. How
many liters of oxygen, measured at 0.974
atm and 24*C, are required to burn 1.00 g
of octane?
– Ans: 2.73 L
Law of Combining Volumes
• The volume of any 2 gases in a reaction at
constant temperature and pressure is the
same as the reacting molar ratio:
2 H2O (l)  2H2 (g) + O2 (g)
4 L H2 x 1 L O2 = 2 L O2
2 L H2
Example 5.7
• Consider the reaction for the formation of water
from its elemental units.
– (a) What volume of hydrogen gas at room
temperature and 1.00 atm is required to react with
1.00 L of oxygen at the same temperature and
pressure?
• Ans: 2.00 L hydrogen gas
– (b) What volume of water at 25*C and 1.00 atm
(d=0.997 /mL) is formed from the reaction in (a)?
• Ans: 1.48 mL of water
– (c) What mass of water is formed from the reaction
assuming a yield of 85.2%?
• Ans: 1.26 g of water
Limiting Reactant Problems
The alkali metals react with the halogens to form
ionic metal halides. What mass of potassium
chloride forms when 5.25 L of chlorine gas at
0.950 atm and 293 K reacts with 17.0 g of
potassium?
Ans: 30.9 g KCl
5.5 Dalton’s Law of Partial Pressures
• The total pressure of a gas mixture is the
sum of the partial pressures of the
components of the mixture.
Ptot = PA + PB
+…..
PH2 = 2.46 atm
PHe = 3.69 atm
then Ptot = 6.15 atm
Wet Gases
• When a gas is collected by bubbling through
water then it picks up water vapor. Then the
total pressure is the sum of the pressure of the
water vapor and the gas collected. So Dalton’s
Law can be applied by:
Ptot = PH2O
+
PA
*The partial pressure of water is equal to the vapor
pressure of water. This has a fixed value at a given
temperature (PH2O @ 25*C = 23.76 mmHg)
Gas collection by water displacement.
Example 5.8
• A student prepares a sample of hydrogen
gas by electrolyzing water at 25*C. She
collects 152 mL of hydrogen gas at a total
pressure of 758 mmHg. Calculate:
– (a) the partial pressure of hydrogen gas.
• Ans: 734 mmHg
– (b) the number of moles of hydrogen gas
collected.
• Ans: 0.00600 mol of hydrogen gas
Partial Pressures & Mole Fraction
• The partial pressure of a gas (PA) divided
by the total pressure (Ptot) is equal to the
number of moles of that gas divided by the
total moles of gases:
• PA = nA
Ptot ntot
• Mole fraction: XA = nA
ntot
• Partial Pressures: PA = XA Ptot
Example 5.9
• Methane burns in air. When one mole of
methane and four moles of oxygen are heated:
(a) What are the mole fractions of oxygen,
carbon dioxide, and water vapor in the resulting
mixture (assume all the methane is converted)?
XCH4 = 0, XCO2 = 0.200, XH2O =0.400, XO2 = 0.400
(b) If the total pressure of the mixture is 1.26
atm, what are the partial pressures of each gas?
PCO2 = 0.252 atm, PH2O =0.504 atm, PO2 = 0.504 atm
5.6 Kinetic Theory of Gases
The Molecular Model of Gases (pg 115):
• Gases are mostly empty space (assumes that
gases do not have their own volume).
• Gas molecules are in constant and chaotic
motion. Their velocities are constantly changing
because of this.
• Collisions of gases are elastic (assumes no
attractive forces).
• Gas pressure is caused by collisions of
molecules with the walls of the container. As a
result, pressure increases with the energy and
frequency of these collisions.
Average Speed
• The equation below is derived from the average
translational kinetic energy of a gas molecule:
• It follows that at a given temperature, molecules of
different gases have the same average kinetic energy of
translational motion
and
• the average translational kinetic energy of a gas
molecule is directly proportional to the Kelvin
temperature so that:
u = (3RT) ½
(MM)
* An R value of 8.31 x 103 g m2/(s2 mol K) is used for average speed
calculations.
Graham’s Law of Effusion
• The average speed is inversely
proportional to the square root of the molar
mass (MM). So for two different gases A
and B at the same temperature then we
can write:
rate of effusion B = (MMA) 1/2
rate of effusion A = (MMB)
Example: Using Graham’s Law
A mixture of helium (He) and methane (CH4) is
placed in an effusion apparatus. Calculate the ratio of
their effusion rates:
Ans: He effuses 2.002 times faster than
methane
Example 5.11
In an effusion experiment, argon gas is
allowed to expand through a tiny opening into an
evacuated flask of volume 120.0 mL for 32.0 s, at
which point the pressure in the flask is found to be
12.5 mmHg. This experiment is repeated with a gas X
of unknown molar mass at the same T and P. It is
found that the pressure in the flask builds up to 12.5
mmHg after 48.0 s. Calculate the molar mass of X.
Ans: 89.9 g/mol
Real Gases
• The ideal gas law has been used with the
assumption that it applies exactly. However, all
real gases deviate at least slightly from the ideal
gas law.
• These deviations arise because the ideal gas
law neglects two factors:
– 1. attractive forces between gas particles
– 2. the finite volume of gas particles
*In general, the closer a gas is to the liquid state, the
more it will deviate from the ideal gas law.
Correction for Real Gas Behavior