Chapter 5. Gases
Properties of Gases (Sections 5ol - 5.2)
Gas Laws (Section 5.3)
Ideal Gases (Sections 5.4 & 5.6)
Gas Stoichiometry (Section 5.5)
Kinetic Molecular Theory of Gases (Section 5.7)
Nonideal Gases (Section 5.8)
SUMMARY
Properties of Gases (Sections 5.1 - 5.2)
Properties of Gases. All matter exists as either a solid, a liquid, or a gas, depending on the
temperature and the pressure. The elements that are gases at T=25 °C and 1 atm, except
CI2, are at the top and right of the periodic chart, in rows 1 and 2 and column 18. Gaseous
molecules include CO, CO2, HCI, NH3, CH4, NO2, N204, SO2 (See Table 5.1, pg 155). What
all these species have in common are small intermolecular (or interatomic) forces (Chapter
11), which keep them from condensing to liquids or solids under atmospheric conditions.
Consequently, very different gases can have very similar physical properties:
1. Expandability. Gases expand indefinitely to fill the space available to them.
2. Indefinite shape. Gases fill all parts of a container evenly and so have no definite shape
of their own.
3. Compressibility. Gases are the most compressible of the states of matter.
4. Miscibility. Two or more gases w!ll mix evenly and completely when confined to the
same container.
5. Low density. Gases have much lower densities than liquids and solids. Typically,
densities of gases are about 1/1000 those of liquids and solids.
Gas Pressure. Gas particles are in constant motion. Their collisions with the container
exert pressure on the container. Pressure is force per unit area. Therefore, the SI unit for
pressure (force/area) is the pascal (Pa = N/m2). The pascal is not the most convenient
pressure unit; the pressure of a small puff of air is > 100,000 Pa. (The most useful units
produce values near 1 for ordinary amounts.) The atmosphere is a more convenient unit:
1 atm = 101,325 Pa.
The torr is based on using the barometer to measure atmospheric pressure. The
simplest barometer consists of a tube of mercury inverted in a small pool of
mercury. The height of the mercury column rises and fails in response to the
pressure the atmosphere exerts on the mercury in the pool. Atmospheric pressure
of 1 atm will support a mercury column that is 760 mm high.
Examples
5.1 -5.2
Exercises
5-1, 5-2
89
90
Chemistry, Ch. 5: Gases
Gas Laws (Sections 5.3 - 5.4)
The Pressure-Volume Relationship. Of the three states, the gaseous state is the simplest
to describe mathematically. In the seventeenth century, Robert Boyle made many
observations of the relationship between gas pressure and gas volume. He discovered that
the volume occupied by molecules in a container decreases as the pressure increases at
fixed temperature. This conclusion is summarized in Boyle’s Law:
p = ...v or PV = kpv for constant amounts of gas at constant temperature
V
Example
5.3
One of the most useful applications of Boyle’s law is the prediction of pressure or volumeExercise
changes of fixed amounts of gases at constant pressure, as might occur when gas is
released from a container. The form of Boyle’s law for this application is ~
often abbreviated P~V~ = P2V2.
The Temperature-Volume Relationship. In the nineteenth century Jacques Charles and
Joseph Gay-Lussac observed that the volume occupied by molecules in a container
increases as the temperature increases at fixed pressure. In other words, the volume
occupied by a fixed amount of gas is proportional to the temperature at fixed pressure.
Their observations are summarized by Charles’ Law:
V = kvTT or -~ = kvT for constant amounts of gas at constant pressure
Applications of Charles’ Law (Gay-Lussac often gets short schrift because his
name is long) include the prediction of temperature changes with volume, such as
the cooling that occurs when gases expand. The form of Charles’ law for this
Exercises
application is
5-4 - 5-7
The Volume-Amount Relationship. Around the same time, Amadeo Avogadro observed
that equal volumes of different gases contained the same number of molecules. In other
words, that the volume occupied by molecules in a container increases as the number of
gas molecules in the vessel increases at fixed pressure and temperature. These
observations are summarized as Avogadro’s Law
V = knvn at constant temperature and pressure.
One consequence of this relationship is the fact that, ideally, one mole quantities of different
gases have the same volumes when the temperature and pressure are constant. The molar
volume is the volume of one mole of gas at 0 °C and 1 arm (called standard temperature
and pressure, STP): 22.41 L. This is a very important observation, because a reaction like
2 H2(g)+O2(g)~ 2 H20(g)
Chemistry, Ch. 5: Gases 91
reduces the volume by a factor of 2/3. Avogadro’s law was useful to scientists attempting to
understand the stoichiometry of chemical reactions. This subject is discussed in more detail
in Section 5.5.
ideal Gases (Sections 5.4 & 5.6)
The Ideal Gas Law. All the gas relationships can be combined into a single equation called
the Ideal Gas Law (IGL):
PV = nRT
where R is the ideal gas constant, R = 0.0821 L-atm/K.mol. The dots in the units are to
remind us that both L and atm are in the numerator, and K and mol are in the denominator.
The units of R may also be written, R = 0.0821 L.atm.K-l.mo1-1. The molecules of an ideal
gas
¯ have no intermolecular interactions (attraction or repulsion)
¯ take up no volume (compared to container volume)
The properties of many real gases can be described by the IGL at high temperatures (>0 °C)
and low pressures (<10 atm). Given any three of the four properties related by the
IGL, the value of the fourth can be calculated. Property changes can be predicted Example
by rearranging
5.4, 5.5
Exercises
5-8 - 5-13
Another IGL rearrangement relates gas properties to the gas density. Remembering that
the number of moles is equal to the ratio of the sample mass to its molar mass, n = m
M’
m PM
dV RT
One of the applications of this relationship is that gas density, which can be
measured experimentally, can be combined with pressure and temperature to
determine the molar mass.
Exercises
5-14-5-17
Dalton’s Law of Partial Pressures. Another consequence of the IGL is that the pressure
of a mixture of gases is the sum of the pressures of the individual components. The IGL
reduces to P = kpn because the volume and temperature are the same for all the
n
components. For example, the pressure of air is the sum of the pressures of several gases,
mostly nitrogen, air and carbon dioxide:
Pair = + Po2 + Poo2
compounds
92
Chemistry, Ch. 5: Gases
Another way to state this relationship is based on the mole fraction, X~, the ratio of the number of moles of a component to the total number of moles of gas, n~,
nT
Example 5.6
Exercise 518
Gas Stoichiometry (Section 5.5)
Chapter 3 presented the idea that balanced chemical reactions describe the
relationships between moles (or molecules) of reactants and products. The IGL
relates the number of moles to gas volumes or pressures, making it possible to
establish relationships between the amounts of reactants and products without
converting volumes to moles first. For exarfiple, we can predict that at STP,
22.41 L of H2 (1 mole) will react with 11.21 L of Oz (0.5 mole) to produce 22.41 L
of H20 (1 mole).
Example 5.8
Exercises
5-19 - 5-21
Kinetic Molecular Theory of Gases (Section 5,7)
The gas laws let us predict gas behavior, but they don’t explain gas behavior. The kinetic
molecular theory of gases provides explanations of gas behavior at the molecular level. The
principal hypotheses of the kinetic theory of gases are
H-0. Gas particles obey Newton’s laws of motion
H-l. Gas molecules are separated by distances much larger than their dimensions and
can be considered as points (mass, but no volume). H-1 explains compressibility.
H-2. Gas molecules are in constant motion in random directions and often collide. Gas
collisions are perfectly elastic, i.e., total amount of energy in the colliding molecules
is the same after collision. H-2 explains pressure-volume relationship.
H-3. There are no intermolecular forces (attraction or repulsion) between gas
molecules. H-3 explains Dalton’s Law of Partial Pressures.
H-4. The energy of gas molecules is primarily kinetic (internal E is small) and depends
on mass and speed of molecules: ~-~ =l-m~, where m is the mass and ~ is the
2
average of the squares of the speeds of all the molecules. The kinetic energy is
also proportional to the absolute temperature (temperature in Kelvins):
KE = kBT, where k~ is the Boltzmann constant (the real gas constant for
individual molecules rather than a mole of them). H-4 explains Charles’ Law.
The kinetic molecular theory has several other applications. The molecules in a container of
gas move randomly, but it is possible to determine the distribution of molecular speeds from
kinetic molecular theory.. As the temperature increases, the range of speed narrows. The
Chemistry, Ch. 5: Gases 93
average speed of the molecules in a gas can be calculated by combining the two features of
H-4:
1M~= 3 RT
2
2
Example
5.9
Exercises
5-22 - 5-24
Nonideal Gases (Section 5.8)
The kinetic theory and ideal gas law describe gas molecules as independent of one another,
not exerting attractive or repulsive forces on one another. A plot of PV/RT versus P is a
horizontal line for an ideal gas, but this relationship is valid for real gases only at low
pressures (less than about 10 atm) as Figure 5.19 of the text shows. The condensation of
real gases to liquids at low temperatures is a second indication of the non-ideality of real
gases. At low temperatures, the gas molecules lack the energy they need to break away
from intermolecular forces.
van der Waals Equation. In the late nineteenth century J.D. van der Waals modified the
ideal gas law to account for two molecular properties that produce deviations from ideal gas
behavior: intermolecular forces and finite molecular volume. According to van der Waals,
the pressure of an ideal gas would be larger than that of a real gas because attraction to
neighboring molecules reduces the impact real molecules make with the container wall.
Similarly, the volume occupied by an ideal gas will be reduced by the small, but finite,
volume of the molecules. Incorporating these corrections into the IGL, the formula becomes
n2
Po~s + a ,~- I ( Vobs - nb ) = n R T
where Pobs and Vobs are
Example
5.10
the observed pressure and volume, respectively and the
constants a and b are specific for a particular gas. The larger the values of a and Exercises
5-25 - 5-26
b, the less ideal the behavior of the gas.
Chemistry, Ch. 5: Gases 95
GLOSSARY LIST
pressure
Pascal
atmospheric pressure
atmosphere
torr
barometer
manometer
Boyle’s law
Charles’ law
Avogadro’s law
ideal gas law
gas constant
standard temperature
and pressure
Dalton’s law
mole fraction
partial pressure
kinetic molecular theory
absolute temperature
root-mean square speed
Boltzmann constant
van der Waals equation
EQUATIONS
Algebraic Equation
PV = nRT
RT
Pi = XiPT
~-K = 1--2 mug = ~ kBT
Urms = ~/~ =13~T
Pobs + a ,~-2 ](Vo~,s - nb ) = nRT
English Translation
The energy of a gas (PV) is constant at
constant temperature
WORKED EXAMPLES
EXAMPLE 5.1 Pressure Measurement
What is the pressure of the gas trapped in the J-tube shown below if the atmospheric pressure
is 0.803 atm?
open to
the air
Hg
¯ Solution
The pressure of the gas is sufficient to support a column of mercury 42 mm high and hold
back the pressure of the atmosphere as well. Therefore the pressure of the gas is:
Pgas = Ph + Patm
The pressure from the mercury column is just the difference ~n heights of the two mercury
surfaces, 42 mmHg. The atmospheric pressure is 610 mmHg.
Chemistry, Ch. 5: Gases 99
Pgas =
42 mmHg + 610 mmHg
=
P~as 652 mmHg
EXAMPLE 5.2 Converting Pressure Units
The atmospheric pressure on Mars is about 0.22 inches of Hg as compared to about 30
inches of Hg on Earth. Express the Martian pressure in mmHg, atmospheres, and pascals.
¯ Solution
Treat this like any other factor label problem. First, state the problem:
? mmHg = 0.22 in Hg
Recall that there are 2.54 cm/in, therefore there are 25.4 mm/in.
The unit factor needed is
25.4 mm
1 in
Converting inches to millimeters:
? mmHg = 0.22 in Hg x 25.4 mm = 5.6 mmHg
1 in
The number of atmospheres is
? atm = 5.6 mmHg x
1 atm
=7.4x10 3atm
760 mmHg
The pressure in pascals is
? pascals = 7.4 x 10-3 atm x 1.013x105 Pa
1 atm
- 750
Pa
100 Chemistry, Ch, 5: Gases
EXAMPLE 5.3 Charles’ Law Calculation
When 2,0 L of chlorine gas (CI2) at STP is warmed at constant pressure to 100 °C, what is the
new volume?
Solution
The volume of a fixed amount of a gas at constant pressure is proportional to the absolute
temperature according to Charles’s law, where T1 = 273 K and T2 = (100 °C + 273 °C) K =
373 K.
Rearranging gives:
V2 = V~ xT_2, where V~,the initial volume, is 2.0 L.
Substituting, we get:
2.0Lx
373 K
- 2.7L
273 K
¯ Comment
Note that we expect V2 > V~ because of the temperature increase. This provides us with a
simple check of our answer. If we calculated that V2 < Vl, we would know our answer was
wrong.
EXAMPLE 5.4 Combined Gas Law Calculation
Given 10.0 L of neon gas at 5 °C and 630 mmHg, calculate the new volume at 400 °C and
2.5 atm.
¯ Solution
First note that both the pressure and the temperature of the gas are changed, but that the
number of moles is constant. Write the ideal gas equation with all the constant terms on one
side.
PV
--= nR = a constant
T
We see that PV/T is a constant. Therefore
where the subscripts 2 refer to the final state, and the subscripts 1 refer to the initial state.
Chemistry, Ch. 5: Gases 101
Rearranging to solve for V2 gives
1 atm
= 0.829 atm
where PI = 630 mmHg ×
760 mmHg
P2 = 2.5 atm
T1 = (273 °C + 5 °C) K = 278 K
T2 = (273 °C + 400 °C) K = 673 K
Substituting
V2=
10.0 L x
0.829 atmx673 K
=8.0 L
2.5 atm 278 K
¯ Comment
Note that P1 and P2 must be expressed in the same units and both temperatures in kelvin.
--_XAMPLE 5.5 Ideal Gas Law
What volume will be occupied by 0.833 mole of fluorine (F2) at 645 mmHg and 15.0 °C?
¯ Solution
The ideal gas law relates the gas volume to temperature, pressure and number of moles.
V=--
nRT
P
First, convert the pressure into atmospheres:
P = 645 mmHg x
1 atm
= 0.849 atm
760 mmHg
= (0.833 mol)(0.0821 L.atm/K.mol)(288 K) = 23.2 L
0.849 atm
¯ Comment
Even though it takes longer to write out the units for each term in the equation, doing so is
helpful because when the units cancel and leave you with the desired units (in this case liters)
yeu are probably closer to the right answer.
EXAMPLE 5.6 Partial Pressures
When oxygen gas is collected over water at 30 °C and the total pressure is 645 mmHg:
a. What is the partial pressure of oxygen? Given the vapor pressure of water at 30 °C is
31.8 mmHg.
b. What are the m~le fractions of oxygen and water vapor?
102 Chemistry, Ch, 5: Gases
¯ Solution for (a)
Mixtures of gases obey Dalton’s law of partial pressures which says that the total pressure is
the sum of the partial pressures of oxygen and water vapor.
Pt = Po2 + P~2o
Po2 = Pt - PH~o = 645 mmHg - 31.8 mmHg
Po~ = 613 mmHg
Solution for (b)
Recall that the partial pressures of 02 and H20 are related to their mole fractions,
P02 = Xo2Pt
PH20 = XH20Pt
Therefore, on rearranging, we get:
_
XOz P-r
X.2o
=
Xo2 = 613/645 = 0.950, and XH2o = 31.8/645 = 0.049
¯ Comment
Note also that the sum of the mole fractions is 1.0, within the number of significant figures,
given:
Xo2 + XH~O = 0.950 + 0.049 = 0.999
EXAMPLE 5.7 Molar Mass of a Gas
A gaseous compound has a density of 1.69 g/L at 25 °C and 714 tom What is its molar mass?
¯ Solution
The density of an ideal gas is directly proportional to (ts molecular mass, We use the equation
d = P_~
RT
Convert pressureinto units of atmospheres.
P = 714 tort x
1 atm = 0.939 atm
760 tort
Chemistry, Ch. 5: Gases 103
Rearranging and substituting, we get:
dRT (1.69 g/L)(0.0821 L. atm/K, mol)(298 K) = 44.0 g/mol
# = ~-- =
0.939 atm
EXAMPLE 5.8 Gas Stoichiometry
Calculate the volume of ammonia gas, measured at 645 torr and 21 °C, that is produced by
the complete reaction of 25.0 g of quicklime, CaO, with excess ammonium chloride, NH4CI,
solution.
¯ Solution
First write the balanced chemical equation.
CaO(s) + 2NH4CI(aq) -~ 2NH3(g) + CaCl2(aq) + H20([)
Here we must determine the number of moles of NH3 formed in the reaction and convert this
into the volume of an ideal gas. Three steps are required: (1) convert g CaO to moles CaO,
(2) convert moles CaO to moles NH3 produced, and (3) use the ideal gas equation to calculate
the volume of NH3.
1
2
3
grams CaO --> moles CaO -~ moles NH3 ~,~ volume NH3
We already know how to find conversion factors 1 and 2; therefore bet’s first find the number of
moles of NH3 formed.
Stating the problem:
? mol NH3 = 25.0 g CaO
"~
¯ mol NH3 = 25.0 g CaO x 1 mol CaO x 2 mol NH3- = 0.891 mol NH3
56.1 g CaO
1 mol CaO
The volume of 0.891 mol NH3 can be calculated from the ideal gas equation:
V-
nRT
P
Convert pressure into units of atm before substituting into the ideal gas equation.
P = 645 torr x 1 atm = 0.849 atm
760 tort
V= 0.891 mo1(0.0821 L.atm/K.mol)(294 K) = 25.3 L
0.849 atm
104 Chemistry, Ch. 5: Gases
EXAMPLE 5,9 Root-Mean-Square Speed
Calculate the root-mean-square speed of gaseous argon atoms at STP,
¯ Solution
The root-mean-square speed is given by the equation
where R = 8.314 J/K.moL T is the absolute temperature 273 K; and/: is the molar mass of At
in kilograms, 0.0399 kg/mol. Note that standard pressure is irrelevant here because the
average kinetic energy depends only on the temperature. Before substituting into the
equation, we recall that 1 J = 1 kg m2/s2.
3(8.314 J/K .mol)(273 K)x 1 kg m2/s2
Urms = (17.1 x 104 m2/s2)~/2 : 413 m/s
EXAMPLE 5.10 van der Waals Equation
The molar volume of isopentane, CsH12, is 1.0 L at 503 K and 30.0 atm:
a. Does isopentane behave as an ideal gas?
b. Given that a = 17.0 L2 atm/mol2 and b = 0.136 L/tooL calculate the pressure of isopentane
as predicted by the van tier Waals equation.
Solution for (a)
According to the ideal gas equation the pressure of 1 mole of a gas at 503 K that occupies 1.0
L is:
p= nRT (1.0 mo0(0.082 L-atm/K.mol)(503 K)
V
1.0 L
P = 41 arm
This calculated result differs considerably from the observed pressure of 30 atm. In
fact, the percent error which is the difference between the two values divided by the actual
pressure is:
41 arm-30.0 atm
% error =
x 100%
30.0 atm
% error = 38%
We conclude that under these conditions C5H12 behaves in a nonideal manner.
Chemistry, Ch. 5." Gases
¯ Solution for (b)
In this case, write the van der Waals equation
Pre~, + ~22/(V -rib) = nRT
and substitute into it, but first calculate the correction terms.
an_~~ = (17.0 L2 .arm/tool2)(1.0 mol)~ = 17 atm
V~
1.0 L~
nb= 1.0 mol (0.136 L/tool) = 0.14 L
nRT = (1.0 mol)(0.0821 L.atm/K.mol)(503 K) = 41 L arm
Now substitute using the van der Waals equation.
(P + 17.0 arm)(1.0 L- 0.14 L) = 41 L atm
(P + 17.0 atm)(0.9 L) = 41 atm
P + 17.0 arm = 50 atm
P = 33 arm
Thus, the pressure calculated by the van tier Waals equation is much closer to the actual
value of 30.0 atm. The percent error is only 10 percent.