The behaviour of gases 2016: Understanding phases

The behaviour of gases 2016
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CHAPTER ONE
MATTER
1.0 Introduction
Long before the science of chemistry was established, materials
were described as existing in one of three physical states. There are
either rigid, solid objects, having a definite volume and a fixed shape,
nonrigid liquids, having no fixed shape other than that of their
containers but having definite volumes or gases, which have neither
fixed shape nor fixed volume.
The techniques used for handling various materials depend on
their physical states as well as their chemical properties. While it is
comparatively easy to handle liquids and solids, it is not as convenient
to measure out a quantity of a gas. Fortunately, except under rather
extreme conditions, all gases have similar physical properties, and the
chemical identity of the substance does not influence those properties.
For example, all gases expand when they are heated in a nonrigid
container and contract when they are cooled or subjected to increased
pressure. They readily diffuse through other gases. Any quantity of gas
will occupy the entire volume of its container, regardless of the size of
the container.
1.1. States of Matter
Matter is anything that has mass and occupies space. All the
material things in the universe are composed of matter, including
anything we can touch as well as the planets in the solar system and all
the stars in the sky. It is composed of tiny particles such as atoms,
molecules, or ions and can exist in three physical states- solid, liquid
and gas.
Solid State
In the solid state, the individual particles of a substance are in
fixed positions with respect to each other because there is not enough
thermal energy to overcome the intermolecular interactions between

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the particles. As a result, solids have a definite shape, volume and are
incompressible. Most solids are hard, but some (like waxes) are
relatively soft. Some large crystals look the way they do because of the
regular arrangement of atoms (ions) in their crystal structure. Solids
usually have their constituent particles arranged in a regular, three-
dimensional array of alternating positive and negative ions called a
crystal. Some solids, especially those composed of large molecules,
cannot easily organize their particles in such regular crystals and exist
as amorphous (literally, ―without form‖) solids. Glass is one example
of an amorphous solid.
Liquid State
A liquid is a nearly incompressible fluid that conforms to the
shape of its container but retains a (nearly) constant volume
independent of pressure. The volume is definite if the temperature and
pressure are constant. The molecules have enough energy to move
relative to each other and the structure is mobile.
Gaseous State
Gases consist of tiny particles widely spaced (Figure 1.1). Under
typical conditions, the average distance between gas particles is about
ten times their diameter. Because of these large distances, the volume
occupied by the particles themselves is very small compared to the
volume of the empty space around them. For a gas at room
temperature and pressure, the gas particles themselves occupy about
0.1% of the total volume. The other 99.9% of the total volume is empty
space (whereas in liquids and solids, about 70% of the volume is
occupied by particles). Because of the large distances between gas
particles, the attractions or repulsions among them are weak.
The particles in a gas are in rapid and continuous motion. For
example, the average velocity of nitrogen molecules, N2, at 20 °C is
about 500 m/s. As the temperature of a gas increases, the particles‘
velocity increases. The average velocity of nitrogen molecules at 100 °C
is about 575 m/s.

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The particles in a gas are constantly colliding with the walls of the
container and with each other. Because of these collisions, the gas
particles are constantly changing their direction of motion and their
velocity. In a typical situation, a gas particle moves a very short
distance between collisions. For example, oxygen, O2, molecules at
normal temperatures and pressures move an average of 10-7 m between
collisions.
Fig.1.1. A Representation of the Solid, Liquid, and Gas States
The various characteristics or properties of the states of matter
discussed above are summarized in table 1.1 below.
Table 1.1. Characteristics of the Three States of Matter
Characteristic Solid Liquid Gas
Shape Definite conforms to the
shape of its
container
Indefinite
Volume Definite Definite Indefinite
Relative
intermolecular
interaction strength
Strong Moderate Weak
Relative particle
positions
in contact and
fixed in place
in contact but not
fixed
not in contact,
random
positions
Compressibility incompressible incompressible Compressible
fluid

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1.2. Phase Transition
Phase transition is a term used to describe a state of change of
matter from one state to another. The state or phase of a given set of
matter can change depending on pressure and temperature conditions,
transitioning to other phases as these conditions change to favour their
existence; for example, solid transitions to liquid with an increase in
temperature. Near absolute zero, a substance exists as a solid. As heat
is added to this substance it melts into a liquid at its melting point,
boils into a gas at its boiling point, and if heated high enough would
enter a plasma state in which the electrons are so energized that they
leave their parent atoms.
1.2.1. Melting point
This is the temperature at which the solid and liquid forms of a
pure substance can exist at equilibrium. As heat is applied to a solid, its
temperature will increase until the melting point is reached. More heat
then will convert the solid into a liquid with no temperature change.
When the entire solid has melted, additional heat will raise the
temperature of the liquid. The melting temperature of crystalline solids
is a characteristic figure and is used to identify pure compounds and
elements. Most mixtures and amorphous solids melt over a range of
temperatures.
The melting temperature of a solid is generally considered to be
the same as the freezing point of the corresponding liquid; because a
liquid may freeze in different crystal systems and because impurities
lower the freezing point, however, the actual freezing point may not be
the same as the melting point. Thus, for characterizing a substance, the
melting point is preferred. A typical example is the change of solid ice
to liquid water as shown below.
H2O(s) → H2O(l) (melting, fusion)
Ice, snow liquid water

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1.2.2. Freezing point
This is the temperature at which a liquid becomes a solid. As
with the melting point, increased pressure usually raises the freezing
point. The freezing point is lower than the melting point in the case of
mixtures and for certain organic compounds such as fats. As a mixture
freezes, the solid that forms first usually has a composition different
from that of the liquid, and formation of the solid changes the
composition of the remaining liquid, usually in a way that steadily
lowers the freezing point. This principle is used in purifying mixtures,
successive melting and freezing gradually separating the components.
The heat of fusion (heat that must be applied to melt a solid), must be
removed from the liquid to freeze it. Some liquids can be supercooled
i.e., cooled below the freezing point without solid crystals forming.
Putting a seed crystal into a supercooled liquid triggers freezing,
whereupon the release of the heat of fusion raises the temperature
rapidly to the freezing point. Freezing of liquid water to ice is a
common example.
H2O(l) → H2O(s) (freezing)
liquid water Ice
1.2.3. Condensation
This is change of a gas to either liquid or solid state, generally
upon a surface that is cooler than the adjacent gas. The change of
vapour to solid is sometimes called deposition. A substance condenses
when the pressure exerted by its vapour exceeds the vapour pressure
of the liquid or solid phase of the substance at the temperature of the
surface where condensation occurs. Heat is released when a vapour
condenses. Unless this heat is removed, the surface temperature will
increase until it is equal to that of the surrounding vapour. In the
atmosphere, however, there is an abundant supply of aerosols, which
serve as nuclei, called condensation nuclei, on which water vapour
may condense. Some are hygroscopic (moisture-attracting), and
condensation begins on them when the relative humidity is less than

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100 percent, but other nuclei require some supersaturation before
condensation begins. Condensation accounts for the formation of dew
(liquid water formed by condensation of water vapour from the
atmosphere), and Frost (solid water formed by direct condensation of
water vapour from the atmosphere without first forming liquid water).
H2O(g) → H2O(l) (condensation)
Water vapour dew
H2O(g) → H2O(s) (condensation, deposition)
Water vapour frost, snow
1.2.4. Vapourization
This refers to the conversion of a substance from the liquid or
solid phase into the gaseous (vapour) phase. Heat must be supplied to
a solid or liquid to effect vaporization. If the surroundings do not
supply enough heat, it may come from the system itself as a reduction
in temperature. The atoms or molecules of a liquid or solid are held
together by cohesive forces, and these forces must be overcome in
separating the atoms or molecules to form the vapour; the heat of
vaporization is a direct measure of these cohesive forces.
H2O(l) → H2O(g)
(vaporization)
Liquid water water vapour
1.2.5. Sublimation
The change of a solid directly to the vapour without its becoming
liquid is specifically referred to as sublimation. Although the vapor
pressure of many solids is quite low, some (usually molecular solids)
have appreciable vapor pressure. Ice, for instance, has a vapour
pressure of 4.7 mmHg at 0oC. For this reason, a pile of snow slowly
disappears in winter even though the temperature is too low for it to
melt. The snow is being changed directly to water vapour.

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H2O(s) → H2O(g) (sublimation)
Ice, snow Water vapour
Sublimation can be used to purify solids such as impure iodine
that readily vaporize. Impure iodine is heated in a beaker so that it
vaporizes, leaving nonvolatile impurities behind. The vapour
crystallizes on the bottom surface of a dish containing ice that rests on
top of the beaker. Freeze-drying of foods is a commercial application of
sublimation. Brewed coffee, for example, is frozen and placed in a
vacuum to remove water vapour. The ice continues to sublime until it
is all gone, leaving freeze-dried coffee. Most freeze-dried foods are
easily reconstituted by adding water. The following diagram
summarizes these phase transitions.
Fig.1.2. Diagram showing the nomenclature for the different phase
transitions.
1.3. Heat of Phase Transition
Any change of state involves the addition or removal of energy
as heat to or from the substance. A simple experiment shows that this
is the case. Suppose you add heat at a constant rate to a beaker
containing ice at -20oC. In Figure 1.3 below, we have plotted the
temperature of the different phases of water as heat is added. The

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temperature of the ice begins to rise from -20oC, as you would expect;
the addition of heat normally raises the temperature of a substance. At
0oC, the ice begins to melt, so that you get a beaker of ice in water. Note
the flat region in the curve, labeled ice and water. Why is this region
flat? It means that heat is being added to the system without a change
in temperature; the temperature remains at 0oC. This temperature, of
course, is the melting point of ice. The heat being added is energy
required to melt ice to water at the same temperature. The
intermolecular forces binding water molecules to specific sites in the
solid phase must be partially broken to allow water molecules the
ability to slide over one another easily, as happens in the liquid state.
Note the flat regions for each of the phase transitions. Because heat is
being added at a constant rate, the length of each flat region is
proportional to the heat of phase transition.
Fig. 1.3. Heating curve for water: Heat is being added at a constant rate to a
system containing water. Note the flat regions of the curve. When heat is
added during a phase transition, the temperature does not change.
The heat needed for the melting of a solid is called the heat of
fusion (or enthalpy of fusion) and is denoted ∆Hfus. For ice, the heat of
fusion is 6.01 kJ per mole.
H2O(s) → H2O(l); ∆Hfus = 6.01 kJ/mol

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The heat needed for the vaporization of a liquid is called the heat of
vaporization (or enthalpy of vaporization) and is denoted ∆Hvap. At
100oC, the heat of vaporization of water is 40.7 kJ per mole.
H2O(l) → H2O(g); ∆Hvap = 40.7 kJ/mol
Note that much more heat is required for vaporization than for
melting. Melting needs only enough energy for the molecules to escape
from their sites in the solid. For vaporization, enough energy must be
supplied to break most of the intermolecular attractions. A refrigerator
relies on the cooling effect accompanying vaporization. The
mechanism contains an enclosed gas that can be liquefied under
pressure, such as ammonia or 1,1,1,2-tetrafluoroethane, CH2FCF3. As
the liquid is allowed to evaporate, it absorbs heat and thus cools its
surroundings (the interior space of the refrigerator). Gas from the
evaporation is recycled to a compressor and then to a condenser,
where it is liquefied again. Heat leaves the condenser, going into the
surrounding air.
1.4. Pressure of Gases
The molecules of a gas, being in continuous motion, frequently
strike the inner walls of their container. As they do so, they
immediately bounce off without loss of kinetic energy, but the reversal
of direction (acceleration) imparts a force to the container walls. This
force, divided by the total surface area on which it acts, is the pressure
of the gas.
The pressure of a gas is observed by measuring the pressure
that must be applied externally in order to keep the gas from
expanding or contracting. To visualize this, imagine some gas trapped
in a cylinder having one end enclosed by a freely moving piston. In
order to keep the gas in the container, a certain amount of weight
(more precisely, a force, f) must be placed on the piston so as to exactly
balance the force exerted by the gas on the bottom of the piston, and
tending to push it up. The pressure of the gas (P) is simply the quotient
f/A, where A is the cross-section area of the piston.

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Example 1.1. If a force of 16N is pressed against an area of 2.44 m2, what
is the pressure in pascals?
Solution
Given force, F = 16N, area, A = 2.44 m2
Apply the relationship,
𝑃 =
𝐹
𝐴
𝑃 =
16𝑁
2.44m2
= 6.57𝑁𝑚−2
1.4.1. Pressure Units
The unit of pressure in the SI system is the pascal (Pa), defined
as a force of one newton per square metre (1 Nm–2 = 1 kg m–1 s–2 ). In
chemistry, it is more common to express pressures in units of
atmospheres or torr:
1 atm = 101325 Pa = 760 torr.
The older unit millimetre of mercury (mm Hg) is almost the
same as the torr; it is defined as one mm of level difference in a
mercury barometer at 0°C. In meteorology, the pressure unit most
commonly used is the bar:
1 bar = 106 N m–2 = 0.987 atm.
For conversion purposes,
1 atm = 760 torr =760 mmHg = 1.01325 × 105 Nm-2
Example 1.2. How many atmospheres are in 1547mmHg
Solution
Use the conversion factor;
1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔
∴ 𝑥 𝑎𝑡𝑚 = 1547 𝑚𝑚𝐻𝑔
Cross multiplying and making 𝑥 the subject gives

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𝑥 =
1 𝑎𝑡𝑚 ×1547 𝑚𝑚𝐻𝑔
760 𝑚𝑚𝐻𝑔
𝑥 = 2.04 𝑎𝑡𝑚
Example 1.3. Write the conversion factor to determine how many
mmHg are in 9.65 atm.
Solution
Use the same conversion factor as in example 1.2 above
∴ 9.65 𝑎𝑡𝑚 = 𝑥 𝑚𝑚𝐻𝑔
Cross multiplying and making 𝑥 the subject give
𝑥 =
9.65 𝑎𝑡𝑚 ×760 𝑚𝑚𝐻𝑔
1 𝑎𝑡𝑚
𝑥 = 7334 𝑚𝑚𝐻𝑔
Example 1.4. How many torr are in 1.56 atm
Solution
Use the conversion factor;
1 𝑎𝑡𝑚 = 760 𝑡𝑜𝑟𝑟
∴ 1.56 𝑎𝑡𝑚 = 𝑥 𝑡𝑜𝑟𝑟
𝑥 =
1.56𝑎𝑡𝑚 ×760 𝑡𝑜𝑟𝑟
1 𝑎𝑡𝑚
𝑥 = 1190 𝑡𝑜𝑟𝑟
Example 1.5. Blood pressures are expressed in mmHg. What would be
the blood pressure in atm if a patient‘s systolic and diastolic blood
pressures are 120 mmHg and 82 mmHg respectively? (In medicine,

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such a blood pressure would be reported as ―120/82‖, spoken as ―one
hundred twenty over eighty-two‖ ).
Solution
Use the same conversion factor as in example one above
∴ 9.65 𝑎𝑡𝑚 = 𝑥 𝑚𝑚𝐻𝑔
Use the same conversion factor as in example one above
∴ 120 𝑚𝑚𝐻𝑔 =
120 𝑚𝑚𝐻𝑔 ×1 𝑎𝑡𝑚
= 0.157 atm
82 𝑚𝑚𝐻𝑔 =
82 𝑚𝑚𝐻𝑔 ×1 𝑎𝑡𝑚
= 0.107 atm
∴
82 𝑚𝑚𝐻𝑔
= 0.157 𝑎𝑡𝑚: 0.107 𝑎𝑡𝑚
1.4.2. Atmospheric Pressure
This is defined as the force per unit area exerted against a
surface by the weight of the air above that surface. In most
circumstances atmospheric pressure is closely approximated by the
hydrostatic pressure caused by the weight of air above the
measurement point. On a given plane, low-pressure areas have less
atmospheric mass above their location, whereas high-pressure areas
have more atmospheric mass above their location. Likewise, as
elevation (altitude) increases, there is less overlying atmospheric mass,
so that atmospheric pressure decreases with increasing elevation.

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1.4.3. Measurement of Gas Pressure
A barometer is piece of lab equipment specifically designed to
measure the atmospheric pressure. Invented in the early 17th century
by the Italian EVANGELISTA TORRICELLI. The barometer consists of
a vertical glass tube closed at the top and evacuated, and open at the
bottom, where it is immersed in a dish of a liquid. The atmospheric
pressure acting on this liquid will force it up into the evacuated tube
until the weight of the liquid column exactly balances the atmospheric
pressure. If the liquid is mercury, the height supported will be about
760 cm; this height corresponds to standard atmospheric pressure.
Fig. 1.4. A simple barometer
The formula for this pressure in the atmosphere is derived as shown
below:
𝒇𝒐𝒓𝒄𝒆 = 𝒎𝒂𝒔𝒔 × 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
or
𝑭 = 𝒎𝒂 or mg

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Earth's acceleration of objects is based on its gravitational field and
equals approximately 9.80665 m s-2. Additionally, since pressure is the
force per the unit area being measured, then
𝑷 =
𝑭
𝑨
=
𝒎𝒈
𝑨
Since mass (m) = density (d) × volume (v)
𝑷 =
𝒈 ×𝒅 ×𝑽
𝑨
Since
𝑉𝑜𝑙𝑢𝑚𝑒 (𝑋3)
𝐴𝑟𝑒𝑎 (𝑋2)
= 𝑕𝑖𝑒𝑔𝑕𝑡 (𝑋)
𝑷 = 𝑔 × 𝑑 × 𝑕
Where d = density, g = gravity and h = height of the liquid or gas.
Example 1.6. Mercury has a density of 13.6 g/cm3 and water has a
density of 1.00 g/cm3. If a column of mercury has a height of 755 mm,
how high would a corresponding column of water be in feet?
Solution:
Let us begin by setting the pressures equal:
Pmercury = Pwater
Since
𝑷 = 𝑔 × 𝑑 × 𝑕
We can write:
𝑕 𝑤𝑎𝑡𝑒𝑟 =
𝑑𝐻 𝑔 × 𝑕𝐻 𝑔
𝑑 𝑤𝑎𝑡𝑒𝑟
=
13.6𝑔𝑐𝑚 −3 ×755 𝑚𝑚
1.00𝑔𝑐𝑚 −3
= 10268 𝑚𝑚 = 33.7 𝑓𝑡
1.4.4. The Manometer
A modification of the barometer, the U-tube manometer,
provides a simple device for measuring the pressure of any gas in a
container. There are a variety of manometer designs. A simple,

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common design is to seal a length of glass tubing and bend the glass
tube into a U-shape. The glass tube is then filled with a liquid, typically
mercury, so that all trapped air is removed from the sealed end of the
tube. The glass tube is then positioned with the curved region at the
bottom. The mercury settles to the bottom.
After the mercury settles to the bottom of the manometer, a
vacuum is produced in the sealed tube. The open tube is connected to
the system whose pressure is being measured. In the sealed tube, there
is no gas to exert a force on the mercury (except for some mercury
vapor). In the tube connected to the system, the gas in the system
exerts a force on the mercury. The net result is that the column of
mercury in the sealed tube is higher than that in the unsealed tube. The
difference in the heights of the columns of mercury is a measure of the
pressure of gas in the system.
In the open-tube manometer, the pressure of the gas is given
by h (the difference in mercury levels) in units of torr or mmHg.
Atmospheric pressure pushes on the mercury from one direction, and
the gas in the container pushes from the other direction. In a
manometer, since the gas in the bulb is pushing more than the
atmospheric pressure, you add the atmospheric pressure to the height
difference:
Pgas > Patm
Gas pressure = atmospheric pressure + h (height of the mercury)
Pgas < Patm
Gas pressure = atmospheric pressure - h (height of the mercury)
The closed-tube manometer look similar to regular manometers except
that the end that is open to the atmospheric pressure in a regular
manometer is sealed and contains a vacuum. In these systems, the
difference in mercury levels (in mmHg) is equal to the pressure in torr.

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Fig.1.5. The Manometer
Example 1.7. Find the pressures using the manometer set up below.
Solution
since Pgas > Patm
Pgas= Patm + h
Pgas= (755 + 24 )mmHg
=779mmHg
since Pgas < Patm
Pgas= Patm ‒ h
Pgas= (763 ‒35)g
Pgas= 728 mmHg

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Example 1.8. Suppose you want to construct a closed-end manometer
to measure gas pressures in the range 0.000–0.200 atm. Because of the
toxicity of mercury, you decide to use water rather than mercury. How
tall a column of water do you need? (The density of water is 1.00
g/cm3; the density of mercury is 13.6 g/cm3).
Solution
Given: pressure range and densities of water and mercury, column
height unknown.
Strategy:
Step 1. Calculate the height of a column of mercury corresponding to
0.200 atm in millimeters of mercury. This is the height needed for a
mercury-filled column.
Step 2. From the given densities, use a proportion to compute the
height needed for a water-filled column.
In millimeters of mercury, a gas pressure of 0.200 atm
1atm = 760mmHg
∴ 0.200 atm will be
0.200 𝑎𝑡𝑚 ×760𝑚𝑚𝐻𝑔
1 𝑎𝑡𝑚
= 152 𝑚𝑚𝐻𝑔
Using a mercury manometer, you would need a mercury column of at
least 152 mm high.
Because water is less dense than mercury, you need a taller
column of water to achieve the same pressure as a given column of
mercury. The height needed for a water-filled column corresponding to
a pressure of 0.200 atm is proportional to the ratio of the density of
mercury to the density of water;
Using 𝑷 = 𝑔 × 𝑑 × 𝑕
Where d = density, g = gravity and h = height of the liquid or gas.
Let us begin by setting the pressures equal:
Pmercury = Pwater
We can then write:
𝑔 × 𝑑𝐻𝑔 × 𝑕𝐻𝑔 = 𝑔 × 𝑑 𝑤𝑎𝑡𝑒𝑟 × 𝑕 𝑤𝑎𝑡𝑒𝑟

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𝑕 𝑤𝑎𝑡𝑒𝑟 =
𝑑𝐻 𝑔 × 𝑕𝐻 𝑔
𝑑 𝑤𝑎𝑡𝑒𝑟
=
13.6𝑔𝑐𝑚 −3 × 152 𝑚𝑚
1.00𝑔𝑐𝑚 −3
= 2070 𝑚𝑚
Comment: it takes a taller column of a less dense liquid to achieve the
same pressure.
1.4.5. Effect of Pressure on the volume of gases
For a gas whose volume is not fixed, increasing the pressure
will cause the gas to contract (reducing the volume), and decreasing
the pressure will cause the gas to expand (increasing the volume). If
the volume is fixed, then increasing the pressure will increase the
temperature, and decreasing the pressure will decrease the
temperature.
1.4.6. Simple Pressure Related Applications
• Drinking straw: A drinking straw is used by creating a
suction with your mouth. Actually this causes a decrease in air
pressure on the inside of the straw. Since the atmospheric pressure is
greater on the outside of the straw, liquid is forced into and up the
straw.
• Siphon: With a siphon water can be made to flow "uphill". A
siphon can be started by filling the tube with water (perhaps by
suction). Once started, atmospheric pressure upon the surface of the
upper container forces water up the short tube to replace water flowing
out of the long tube.
1.5. Density of a Gas
This is defined as mass divided by the volume of a gas
𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑑 =
𝑚𝑎𝑠𝑠 (𝑔)
𝑣𝑜𝑙𝑢𝑚𝑒 (𝐿)

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The density of a gas in grams/L can be obtained from ideal gas
equation as follows:
𝑃𝑉 = 𝑛𝑅𝑇
Number of mole of a gas (n) =
𝑚𝑎𝑠𝑠 (𝑚)
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
Substituting ―n‖ into ideal gas equation above
𝑃𝑉 =
𝑚
𝑀
× 𝑅𝑇
Cross multiplying we have
𝑀 × 𝑃𝑉 = 𝑚 × 𝑅𝑇
Divide both side by V gives
𝑃 × 𝑀 =
𝑚
𝑉
× 𝑅𝑇
Lastly divide both by RT gives density
𝑚
𝑉
=
𝑃 × 𝑀
𝑅𝑇
𝑑 =
𝑃 × 𝑀
𝑅𝑇
Example 1.9. What is the density of oxygen at STP? [R= 0.8206L atm
mol-1K-1]
Solution
Data collection
S.t = 273K
S.p = 1 atm
R= 0.8206L atm mol-1K-1
Molecular weight, M of oxygen = 32.0gmol-1
Using 𝑑 =
𝑚
𝑉
=
𝑃 × 𝑀
𝑅𝑇
=
1 𝑎𝑡𝑚 × 32.0 𝑔𝑚𝑜𝑙−1
0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 273.15𝐾
= 1.428𝑔/𝐿
Example 1.10. A 0.0125g sample of a gas with an empirical formula of
CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at
22.5 °C. What is the molecular formula of the compound? [R= 0.8206L
atm mol-1K-1]

20
Solution
Collect the available data and convert as necessary to agree with the given
unit of R then find the value of density from which the molecular weight of the
gas can be determined using the relation: 𝑑 =
𝑃 × 𝑀
𝑅𝑇
Mass of gas sample = 0.0125g
Volume = 165 mL = 0.156 L
Temperature, T = 22.5°C = 295.7K
Pressure, P = 13.7 mm Hg = 1 atm ×
13.7 𝑚 𝑚𝐻𝑔
760 𝑚 𝑚𝐻𝑔
= 0.0180 𝑎𝑡𝑚 𝑎𝑡 𝑚
Now density, 𝑑=
𝑚
𝑉
𝑑 =
𝑚
𝑉
=
0.0125 𝑔𝑔
0.156 𝐿 𝐿
= 0.0758 𝑔 𝑔/𝐿 𝐿
To find molecular weight of gas, we use
𝑑=
𝑃 × 𝑀
𝑅𝑇
Making molecular weight, M the subject and substituting
M =
𝑑𝑅𝑇
𝑃
=
0.075𝑔𝐿−1
× 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1
𝐾−1
× 273.15𝐾
0.0180 𝑎𝑡𝑚
M = 102𝑔𝑚𝑜𝑙−1
The molecular formula is (CHF2)2 or C2H2F4.
Example 1.11. If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25
oC, what is the pressure of O2 & H2O?[0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1
𝐾−1
, H =
1, 0 = 16, ]
Solution
Step 1: Write the balanced chemical reaction.
Step 2: Calculate the moles of each product.
Step 3: Find the pressure of each via PV = nRT
Equation of reaction : 2H2O2(l) → 2H2O (g) + O2 (g)
From the equation of reaction, 2 mol of 2H2O2 produce 2 mol of H2O
and a mol of O2.

21
Therefore mol of H2O2 =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
=
0.11𝑔
34𝑔/𝑚𝑜𝑙
=0.0032 mol
mol of O2 =
1
2
× 0.0032 mol of H2O2
=0.0016 mol
mol of H2O = 1 × 0.0032 mol of H2O2
=0.0032 mol
Using PV = nRT to calculate the pressure of the gases
P(O2) =
𝑛𝑅𝑇
𝑉
=
0.0016 𝑚𝑜𝑙 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1
𝐾−1
× 298𝐾
2.5 𝐿
= 0.016 atm
P(H2O) =
𝑛𝑅𝑇
𝑉
=
0.0032 𝑚𝑜𝑙 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1
𝐾−1
× 298𝐾
2.5 𝐿
= 0.032 atm
Example 1.12. A chemist has synthesized a greenish-yellow gaseous
compound of chlorine and oxygen and finds that its density is 8.14 g/L
at 47°C and 3.15 atm. Calculate the molar mass of the compound and
determine its molecular formula.
Solution
We can calculate the molar mass of a gas if we know its density,
temperature, and pressure. The molecular formula of the compound
must be consistent with its molar mass. What temperature unit should
we use?
Data provided
density = 8.14 g/L
T = 47°C = 320 K
P = 3.15 atm
Using the relationship, 𝑀 =
𝑑𝑅𝑇
𝑃
to solve for molar mass,

22
=
8.14 𝑔𝐿−1 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾−1 ×320 𝐾
3.15 𝑎𝑡𝑚
= 67.9 𝑔𝑚𝑜𝑙−1
We can determine the molecular formula of the compound by trial and
error, using only the knowledge of the molar masses of chlorine (35.45
g) and oxygen (16.00 g). We know that a compound containing one Cl
atom and one O atom would have a molar mass of 51.45 g, which is too
low, while the molar mass of a compound made up of two Cl atoms
and one O atom is 86.90 g, which is too high. Thus, the compound
must contain one Cl atom and two O atoms and have the formula ClO2
, which has a molar mass of 67.45 g.
Example 1.13. The density of a gaseous organic compound is 3.38 g/L
at 40°C and 1.97 atm. What is its molar mass?
Solution
Data provided
d = 3.38 g/L
T = 40°C = 313 K
P = 1.97 atm
Using the relationship 𝑀 =
𝑑𝑅𝑇
𝑃
=
3.38 𝑔𝐿−1 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾−1 ×313 𝐾
1.97 𝑎𝑡𝑚
= 44.0 𝑔𝑚𝑜𝑙−1
1.5.1. The effects of temperature on density
The density of a gas depends quite strongly on its temperature,
so hot air has a smaller density than does cold air; colder air is more
dense than hot air. From everyday experience, we know that
something is dense if it tries to drop, which is why a stone drops to the
bottom of a pond and a coin sinks to the bottom of a pan of water. This
relative motion occurs because both the stone and the coin have higher
densities than does water, so they drop. Similarly, we are more dense
than air and will drop if we fall off a roof. Just like the coin in water,
cold air sinks because it is denser than warmer air. We sometimes see

23
this situation stated as warm air ‗displaces‘ the cold air, which
subsequently takes its place. Alternatively, we say ‗warm air rises‘,
which explains why we place our clothes above a radiator to dry them,
rather than below it.
Light entering the room above the radiator passes through
these pockets of warm air as they rise through colder air, and therefore
passes through regions of different density. The rays of light bend in
transit as they pass from region to region, much in the same way as
light twists when it passes through a glass of water. We say the light is
refracted. The eye responds to light, and interprets these refractions and
twists as different intensities.
So we see swirling eddy (or ‗convective‘) patterns above a
radiator because the density of air is a function of temperature. If all
the air had the same temperature, then no such difference in density
would exist, and hence we would see no refraction and no eddy
currents – which is the case in the summer when the radiator is
switched off. Then again, we can sometimes see a ‗heat haze‘ above a
hot road, which is caused by exactly the same phenomenon.
1.6. Temperature
` This is the numerical measure of the degree of hotness or
coldness of a body. It is an important property of any gas. If two bodies
are at different temperatures, heat will flow from the warmer to the
cooler one until their temperatures are the same. This is the principle
on which thermometry is based; the temperature of an object is
measured indirectly by placing a calibrated device known as a
thermometer in contact with it. When thermal equilibrium is obtained,
the temperature of the thermometer is the same as the temperature of
the object.
1.6.1. Temperature Scale
A thermometer makes use of some temperature-dependent
quantity, such as the density of a liquid, to allow the temperature to be
found indirectly through some easily measured quantity such as the

24
length of a mercury column. The resulting scale of temperature is
entirely arbitrary; it is defined by locating its zero point, and the size of
the degree unit.
Celsius temperature scale locates the zero point at the freezing
temperature of water; the Celsius degree (C °) is defined as 1/100 of
the difference between the freezing and boiling temperatures of water
at 1 atm pressure.
The older Fahrenheit scale placed the zero point at the coldest
temperature it was possible to obtain at the time (by mixing salt and
ice.) The 100° point was set with body temperature (later found to be
98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The
Fahrenheit scale is a finer one than the Celsius scale; there are 180
Fahrenheit degrees in the same temperature interval that contains 100
Celsius degrees, so 1F° = 9/5 C . Since the zero points are also different
by 32F, conversion between temperatures expressed on the two scales
requires the addition or subtraction of this offset, as well as
multiplication by the ratio of the degree size. These selections allow us
to write the following relations.
t(oF) =
9
5
t(oC) + 32
t(oC) =
9
5
t(oF) – 32
Where
t(oF) is the temperature in degree Fahrenheit and
t(oC) is the temperature in degree Celsius.
1.6.2. Absolute temperature
In 1787 the French mathematician and physicist JACQUES
CHARLES discovered that for each Celsius degree that the
temperature of a gas is lowered, the volume of the gas will diminish by
1/273 of its volume at 0°C. The obvious implication of this is that if the
temperature could be reduced to –273°C, the volume of the gas would
contract to zero. Of course, all real gases condense to liquids before this

25
happens, but at sufficiently low pressures their volumes are linear
functions of the temperature (Charles' Law), and extrapolation of a plot
of volume as a function of temperature predicts zero volume at -273°C.
This temperature, known as absolute zero, corresponds to the total
absence of thermal energy.
Because the Kelvin scale is based on an absolute, rather than on
an arbitrary zero of temperature, it plays a special significance in
scientific calculations; most fundamental physical relations involving
temperature are expressed mathematically in terms of absolute
temperature. The diagram below compares the different temperature
scales with respect to boiling and freezing point of water.
Fig. 1.6. Comparison of Temperature Scales (Schematic)
1.6.3. Conversion between Celsius and Kelvin Scale
In order to covert temperature in degree Celsius to temperature
in Kelvin, the expression below is used.
toC = (273 + t )K = T (K)

26
Where t is the temperature on the Celsius scale, T is the temperature on
the Kelvin scale.
Example 1.14. Covert the following temperatures to Kelvin scale:
a. 27oC; b. -10oC.
Solutions
a. Using the relationship
toC = (273 + t )K = T (K)
27oC = (273 + 27)K = 300K
b. toC = (273 + t )K = T (K)
-10oC = (273- 10)K = 263K
In order to convert absolute temperature T K to degree Celsius,
273 is simply subtracted from the value.
Example 1.15. Covert the following temperatures to degree Celsius:
a. 298K
b. 25K
Solutions
a. Using the relationship
toC = (273 + t )K = T (K)
toC = (298 ‒ 273) oC = 25oC
b. Using the relationship
toC = (273 + t )K = T (K)
toC = (25 ‒ 273) oC = ‒ 248 oC
1.7. The Volume of Gas
The volume of a gas is simply the space in which the molecules
of the gas are free to move. If we have a mixture of gases, such as air,
the various gases will coexist within the same volume. In these

27
respects, gases are very different from liquids and solids, the two
condensed states of matter. The volume of a gas can be measured by
trapping it above mercury in a calibrated tube known as a gas burette
(fig. 1.7). The SI unit of volume is the cubic meter, but in chemistry the
liter and the milliliter (mL) are commonly used.
Fig. 1.7. Gas burette
It is important to bear in mind, however, that the volume of a
gas varies with both the temperature and the pressure, so reporting the
volume alone is not very useful. A common practice is to measure the
volume of the gas under the ambient temperature and atmospheric
pressure, and then to correct the observed volume to what it would be
at standard atmospheric pressure and some fixed temperature, usually
0° C or 25°C. The table below shows some commonly used volume
measurement units and their conversion factor.

28

29
1.8. Effect of Temperature on the volume of gases
If the volume of the container is not fixed, increasing the
temperature will cause a gas to expand (increase the volume), and
contract when cooled (decreasing the volume). This would be the case
for a gas inside a piston, or inside a rubber balloon. If the volume is
fixed, then increasing the temperature will increase the pressure, and
decreasing the temperature will decrease the pressure. This would be
the case for a gas in a closed solid container, like a canister or sealed
metal box.
Why does thunder accompany lightning?
Lightning is one of the most impressive and yet frightening
manifestations of nature. It reminds us just how powerful nature can
be. Lightning is quite a simple phenomenon. Just before a storm
breaks, perhaps following a period of hot, fine weather, we often note
how the air feels ‗tense‘. In fact, we are expressing an experiential
truth: the air contains a great number of ions – charged particles. The
existence of a large charge on the Earth is mirrored by a large charge in
the upper atmosphere. The only difference between these two charges
is that the Earth bears a positive charge and the atmosphere bears a
negative charge.
Accumulation of a charge difference between the Earth and the
upper atmosphere cannot proceed indefinitely. The charges must
eventually equalize somehow: in practice, negative charge in the upper
atmosphere passes through the air to neutralize the positive charge on
the Earth. The way we see this charge conducted between the Earth
and the sky is lightning: in effect, air is ionized to make it a conductor,
allowing electrons in the clouds and upper atmosphere to conduct
through the air to the Earth‘s surface. This movement of electrical
charge is a current, which we see as lightning. Incidentally, ionized air
emits light, which explains why we see lightning. Lightning comprises
a massive amount of energy, so the local air through which it conducts
tends to heat up to as much as a few thousand degrees centigrade. And
we have already seen how air expands when warmed, e.g. as described

30
mathematically by Charles‘s law. In fact, the air through which the
lightning passes increases in volume to an almost unbelievable extent
because of its rise in temperature. And the expansion is very rapid.
1.9. Standard Temperature and Pressure, s.t.p.
Suppose two scientists work on the same research project, but
one resides in the far north of the Arctic Circle and the other lives near
the equator. Even if everything else is the same – such as the air
pressure, the source of the chemicals and the manufacturers of the
equipment – the difference between the temperatures in the two
laboratories will cause their results to differ widely. For example, the
‗room energy‘ RT will differ. One scientist will not be able to repeat the
experiments of the other, which is always bad science.
An experiment should always be performed at known
temperature. Furthermore, the temperature should be constant
throughout the course of the experiment, and should be noted in the
laboratory notebook. But to enable complete consistency, sets of
universally accepted arbitrary standards were devised and are called a
set of standard conditions. ‗Standard pressure‘ was set as 1 atm and
‗Standard temperature‘ has the value of 0oC (273 K). If both the
pressure and the temperature are maintained at these standard
conditions, then we say the measurement was performed at ‗standard
temperature and pressure‘, which is universally abbreviated to ‗s.t.p.‘
If the scientists at the equator and the Arctic Circle perform their work
in thermostatically controlled rooms, both at s.t.p., then the results of
their experiments will be identical. If we know the volume of a
sample of a gas at any condition, we can easily calculate the volume it
would have as an ideal gas at STP by employing the combined gas law.
1.10. Molar volume of a gas
The volume occupied by one mole of a gas under any
conditions of temperature and pressure is called the molar volume, Vm.
The molar volume of an ideal gas depends on the conditions of
temperature and pressure; at s.t.p. it is 22.4 L (or 22400 cm3).

31
How did we arrive at this value?
It is simply the volume of 1.00 mol of gas at STP
At s.t.p, pressure (P) = 1atm, temperature (T) = 27K, for one mole of
gas, n = 1, R= 0.0821 L atm mol-1K-1
Using ideal gas equation to calculate the volume
PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
1.00 𝑚𝑜𝑙 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾−1 ×273 𝐾
1.00 𝑎𝑡𝑚
= 22.4 𝐿
1.11. Molecular weight and density of a gas
The molar volumes of all gases are the same when measured at
the same temperature and pressure. But the molar masses of different
gases will vary. This means that different gases will have different
densities (different masses per unit volume). If we know the molecular
weight of a gas, we can calculate its density.
More importantly, if we can measure the density of an
unknown gas, we have a convenient means of estimating its molecular
weight. This is one of many important examples of how a macroscopic
measurement (one made on bulk matter) can yield microscopic
information (that is, about molecular-scale objects).
Determination of the molecular weight of a gas from its density
is known as the Dumas method, after the French chemist JEAN DUMAS
(1800-1840) who developed it. One simply measures the weight of a
known volume of gas and converts this volume to its STP equivalent,
using Boyle's and Charles' laws. The weight of the gas divided by its
STP volume yields the density of the gas, and the density multiplied by
22.4Lmol–1 gives the molecular weight. Pay careful attention to the
examples of gas density calculations shown below.

32
Example 1.16. Calculate the approximate molar mass of a gas whose
measured density is 3.33 g/L at 30oC and 780 torr.
Solution.
Data provided
Molar mass?
Density = 3.33 g/L
Volume = 1L
Temperature,T = 30oC = (30 +273)K
Pressure, P = 780 torr = (780/760) atm
From the ideal gas equation, the number of moles contained in one litre
of the gas is
𝑛 =
𝑃𝑉
𝑅𝑇
=
780
760
𝑎𝑡𝑚 × (1.00 𝐿)
0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 393𝐾
= 0.413 𝑚𝑜𝑙
Now density, 𝑑 =
𝑉𝑜𝑙𝑢𝑚𝑒 (𝑉)
Therefore, 𝑚 = 𝑑𝑣
But mass (m) = number of mole (n) × molar mass (M)
Therefore 𝑑𝑣 = 𝑛𝑚
M=
𝑑 ×𝑣
𝑛
Substituting gives
M =
33𝑔 𝐿−1 ×1.0 𝐿
0.0413 𝑚𝑜𝑙
= 80.6gmol-1
Example 1.17. The density of air at 15OC and 1.00 atm is 1.23g/L. What
is the molar mass of the air?

33
Solution
First calculate the mole of air from which the molar mass can be gotten.
Data provided
Density = 1.23 g/L
Volume = 1L
Temperature,T = 15oC = (15 +273)K = 288K
Pressure, P = 1 atm
Molar mass?
From the ideal gas equation, the number of moles contained in one litre
of the air is
𝑛 =
𝑃𝑉
𝑅𝑇
=
1 𝑎𝑡𝑚 × (1.00 𝐿)
0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 288𝐾
= 0.0423 𝑚𝑜𝑙
M=
𝑑 ×𝑣
𝑛
Substituting gives
M =
1.23𝑔 𝐿−1 ×1.0 𝐿
0.0423 𝑚𝑜𝑙
= 29.1gmol-1

34
CHAPTER TWO
THE GAS LAWS
2.1. Introduction
Experience has shown that several properties of a gas can be
related to each other under certain conditions. The properties are
pressure (P), volume (V), temperature (T, in kelvins), and amount of
material expressed in moles (n). What we find is that a sample of gas
cannot have any random values for these properties. Instead, only
certain values, dictated by some simple mathematical relationships,
will occur. These properties and other variables such as rate of
diffusion of any gaseous substance bear a simple mathematical
relationship to each other. These are collectively called gas laws.
2.2. Pressure – Volume Relationship
Robert Boyle (1627–1691), an Irish physical scientist, discovered
that the volume of a given sample of a gas at a constant temperature is
inversely proportional to its pressure. This generalization, known as
Boyle’s law, applies approximately to any gas, no matter what its
composition. (It does not apply to liquids or solids.)
Inverse proportionality occurs when one variable gets larger by
the same factor as another gets smaller. For example, average speed and
the time required to travel a certain distance are inversely proportional.
If we double our speed, the time it takes us to complete the trip is
halved. Similarly, if the pressure on a given sample of gas at a given
temperature is doubled (increased by a factor of 2), its volume is
halved (decreased by a factor of 2).
Boyle might have observed the following data on volume and
pressure for a given sample of gas at a given temperature, under four
different sets of conditions:
Volume (L) Pressure (atm)
1 4.00 1.00
2 2.00 2.00
3 1.00 4.00

35
4 0.500 8.00
Note that tabulating data is very helpful when two or more variables
are being considered. The units are usually included in the column
headings in such a table. The data in the table show that the product of
the volume (V) and the pressure (P) is a constant. The table may be
expanded to show this relationship:
Volume (L) Pressure (atm) Volume × Pressure (L. atm)
1 4.00 1.00 4.00
2 2.00 2.00 4.00
3 1.00 4.00 4.00
4 0.500 8.00 4.00
Mathematically expression of the law;
[V α
1
𝑃
]T
[V =
𝐾
𝑃
]T
PV = K
(Where K = constant of proportionality).
A more useful form of the law can be written as:
P1V1= P2V2
Where V1 and P1 refer to the original volume and pressure, V2 and P2
refer to the volume and pressure under the new or changed conditions.
If we place the values of P on the horizontal axis and the values
of V on the vertical axis, plot the preceding tabulated values for P and
V, and smoothly connect the points, we get a curve that can tell us
what the volume will be at any intermediate pressure (Figure 2.1a). We
can also plot 1/V versus P and get a straight line through the origin
(Figure 2.1b).
V (L) 1/V (1/L) P (atm)
1 4.00 0.250 1.00
2 2.00 0.500 2.00
3 1.00 1.00 4.00
4 0.500 2.00 8.00

36
(a) (b)
Fig. 2.1. Graphical illustration of Boyle’s law: (a) Plot of V versus P. (b) Plot
of 1/V versus P.
2.3. Kinetic Theory and Boyle’s Law
The pressure of gas is due to continuous collision of the gaseous
molecules with the walls of the container. At constant temperature, the
average kinetic energy of the gas molecules is constant. If the size of
the container is reduced to a half (volume reduces), the frequency of
collision of the gas molecules with the walls of the container will be
doubled. This is due to the fact that the distance to the walls has been
reduced to a half. Therefore, the gas pressures will double the initial
value.
On the other hand if the volume of the container (size) is
doubled, the frequency of collision of the gas molecules with the walls
of the container will become reduced by a half, since the distance
between the molecules before colliding with the walls has been
doubled. Hence the pressure will be half of the initial value.
Example 2.1. A certain mass of a gas occupies 400cm3 at 1.0 × 105 Nm-2.
Calculate its volume when the pressure is 4.0 × 105 Nm-2 at constant
temperature.

37
Solution
In trying to solve this kind of problem, it is always good to collect the given
information together so as to easily detect the variable you are asked to find.
Data provided;
P1 = 1.0 × 105 Nm-2,
V1 =400cm3,
P2 = 2.0 × 105 Nm-2,
V2 = ?
According to boyle‘s law, P1V1= P2V2
Making V2 the subject,
V2 =
P1V1
P2
On substituting, V2 =
1 × 105 ×400
2 × 105 = 200cm3
Example 2.2. If 4 Liters of methane gas has a pressure of 1.0 atm, what
will be the pressure of the gas if we squish it down so it has a volume
of 2.5 L?
Solution
Data provided;
P1 = 1.0 atm
V1 = 4.0L
P2 = ?
V2 = 2.5 L
According to boyle‘s law, P1V1= P2V2
Making P2 the subject,
P2 =
p1v1
v2
P2 =
1.0 × 4
2.5
= 1.6 𝑎𝑡𝑚
Example 2.3. A 3.50-L sample of gas has a pressure of 0.750 atm.
Calculate the volume after its pressure is increased to 1.50 atm at
constant temperature.
Solution
Alternatively, data collection can be in the form of table as shown
below
Pressure Volume
1 0.750 atm 3.50 L

38
2 1.50 atm ?
Using P1V1= P2V2
V2 =
p1v1
P2
Substitution of the values into the equation yields
V2 =
0.750 ×3.50
1.50
= 1.75 𝐿
Note that multiplying the pressure by 2 causes the volume to be
reduced to half.
Example 2.4. A sample of gas initially occupies 35.0 mL at 1.50 atm.
Calculate the pressure required to reduce its volume to 20.5 mL at
Solution
Data collection
Pressure Volume
1 1.50 atm 35.0 mL
2 ? 20.5 mL
Using P1V1= P2V2
P2 =
p1v1
V2
P2 =
1.50 ×35.0
20.5
= 2.56 𝑎𝑡𝑚
Note that the units of pressure and volume must be the same on each side of
the equation P1V1= P2V2 . If the units given in a problem are not the same,
one or more of the units must be converted.
Example 2.5. A 1.45-L sample of gas has a pressure of 0.950 atm.
Calculate the volume after its pressure is increased to 787 torr at
Solution
Because the pressures are given in two different units, one of them
must be changed.

39
Pressure Volume
1 0.950 atm 1.45 L
2 787 𝑡𝑜𝑟𝑟
1 𝑎𝑡𝑚
760 𝑡𝑜𝑟𝑟
= 1.036 𝑎𝑡𝑚 ?
Using P1V1= P2V2
V2 =
p1v1
P2
V2 =
0.950 ×1.45
1.036
= 1.33 𝐿
Alternatively, we can change 0.950 atm to torr and still arrive at the
same answer.
(722 torr) (1.45 L) = (787 torr)V2
V2 = 1.33 𝐿
Note: 1 atm = 760 torr
Example 2.6. Calculate the initial volume of a sample of gas at 1.20 atm
if its volume is changed to 70.4 mL as its pressure is changed to 744
torr at constant temperature
Solution
Data collection
Pressure Volume
1 1.20 atm ?
2 744 𝑡𝑜𝑟𝑟
1 𝑎𝑡𝑚
760 𝑡𝑜𝑟𝑟
= 0.979 𝑎𝑡𝑚 70.4 L
Using P1V1= P2V2
V1 =
p2v2
P1
V1 =
0.979 ×70.4
1.2
= 57.4 𝐿

40
Example 2.7. Calculate the pressure required to change a 3.38-L sample
of gas initially at 1.15 atm to 925 mL, at constant temperature.
Solution
Collect the data and convert 925 mL to L (mL ≡ cm3, 1000mL = 1L)
Pressure Volume
1 1.15 atm 3.38 L
2 ? 925 mL = 0.925 L
Using P1V1= P2V2
P2 =
p1v1
V2
P2 =
1.15 ×3.38
0.925
= 4.20 𝑎𝑡𝑚
The pressure must be raised to 4.20 atm.
Practice questions
1. State Boyle‘s law (i) in words (ii) mathematically
2. Explain Boyle‘s law in terms of kinetic theory.
3. Fill the following gaps: (Measurements at constant
temperatures).
Initial pressure Initial volume Final pressure Final volume
1.0 × 105 Nm-2 300cm3 1.5 × 105 Nm-2 -
1.0 × 105 Nm-2 225cm3 - 900cm3
- 3.50dm3 760 mmHg 700 cm3
800 mm Hg 300cm3 650 mmHg -
4. 30dm3 of oxygen at 10 atmospheres is placed in a 20dm3
container. Calculate the new pressure if temperature is kept
constant.

41
5. Calculate the initial pressure of a 485-mL sample of gas that has
been changed at constant temperature to 1.16 L and 1.18 atm.
2.4. Temperature – Volume Relationship
In 1787, 125 years after Boyle published the law that bears his
name, J. A. C. Charles (1746–1823) discovered a law relating the
volume of a given sample of gas to its absolute temperature. It took
more than a century to discover this law because of the requirement
that the temperature be absolute.
The volume of a sample of gas varies with the temperature, as
shown in Table 2.1 and plotted in Figure 2.2(a) for a particular sample.
Although the volume changes with the Celsius temperature, the
relationship is not a direct proportionality. That is, when the Celsius
temperature doubles, the volume does not double, all other factors
being held constant. On the graph, the plotted points form a straight
line, but the line does not pass through the origin. For a direct
proportionality to exist, the straight line must pass through the origin.
If the straight line corresponding to the points in Table 12.1 is extended
until the volume reaches 0 L, the Celsius temperature is -273K (Figure
2.2b). Charles defined a new temperature scale in which the lowest
possible temperature is absolute, corresponding to -273K. This
temperature is called absolute zero.
Table 2.1 Temperature and Volume Data for a Particular Sample of Gas at a
Given Pressure
Temperature(°C) Volume(L)
1 0 0.400
2 100 0.548
3 200 0.692
4 300 0.840

42
(a) (b)
Fig. 2.2. Dependence of Volume on Temperature at Constant Pressure (a) Plot
of the data given in Table 2.1. (b) Extension of the line in part (a) to absolute
zero, with the Kelvin scale added to the horizontal axis.
We can state Charles‘ findings in simple terms:
At constant pressure, the volume of a fixed amount of gas is directly
proportional to its absolute temperature. This means an increase in the
temperature of a fixed mass of a gas leads to a corresponding increase
in the volume of the gas by the same proportion, and vice versa, with
the proviso that pressure remains the same.
Mathematically expression of the law;
[ V∝ T ]P
[ V= KT ] P
[ V/𝑇 = K ] P
(Where K = constant of proportionality).
A more useful form of the law can be express as:
𝑉1
𝑇1
=
𝑉2
𝑇2
Where V1 and T1 refer to the original volume and pressure, V2 and T2
refer to the volume and pressure under the new or changed conditions.

43
2.5. Kinetic Theory and Charles’ Law
As the temperature of the gas molecules increase, the average
kinetic energy is equally raised, and hence, the average velocity of gas
molecules. The gas molecules move more rapidly colliding with one
another and more frequently with the walls of the container. For gas
pressure to remain constant, the volume of the container must be
increased with an increase in temperature.
Example 2.8. Assume that the volume of a balloon filled with H2 is 1.00
L at 25°C. Calculate the volume of the balloon when it is cooled to -
78°C in a low-temperature bath made by adding dry ice to acetone.
Solution
Collect the given information and convert as necessary
Data provided;
V1 = 1.00L,
T1 = 250C = (25 + 273)K = 298K
T2 = -780C = (273 - 78)K =195K
V2 = ?
Applying Charles‘ law,
𝑉1
𝑇1
=
𝑉2
𝑇2
V2 =
195 ×1.00
293
= 0.65L
Example 2.9. The volume of a fixed mass of gas measured at
atmospheric pressure and 260C is 3.0 dm3. Calculate the volume at
1270C and at the same pressure.

44
Solution
Data provided;
V1 = 3.0 dm3,
T1 = 260C = (25 + 273)K = 299K
T2 = 1270C = (273 + )K =400 K
V2 = ?
𝑉1
𝑇1
=
𝑉2
𝑇2
V2 =
400 ×3.00
299
= 4.0 dm3
Example 2.10. If 250cm3 of a gas at s.t.p. is heated to 270C at constant
pressure, calculate its new volume.
Solution
Data provided;
V1 = 250 cm3,
T1 = s.t = 273K
T2 = 270C = (273 +27 )K =300 K
V2 = ?
V1/ T1 = V2/ T2
V2 = V1 × T2/ T1
V2 = 250 × 300
273

45
= 274.7 cm3
Example 2.11. Show that the data in Table below prove (a) that the
Celsius temperature is not directly proportional to volume and (b) that
the Kelvin temperature is directly proportional to volume.
Temperature and Volume data for a particular Sample of gas at a given pressure
Solution
As the absolute temperature 273 K is increased to 373 K or 473 K, the
volume increases to 373/273 = 1.37 or 473/273 = 1.37 times the original
volume. The ratio of V to T is constant (see Table above). The volume is
directly proportional to absolute temperature.
Example 2.12. Calculate the Celsius temperature to which a 678-mL
sample of gas at 0oC must be heated at constant pressure for the
volume to change to 0.896 L.
Solution
Data provided
V1 = 678 mL = 0.678 L
T1 = 0oC = 273K
V2 = 0.896 L

46
T2 = ?
Using the relationship
𝑉1
𝑇1
=
𝑉2
𝑇2
T2 =
273 ×0.896
0.678
= 361 𝐾𝐾
= (361 − 273)oC
= 88oC
Note: 1000 mL = 1L
Example 2.13. Calculate the original temperature of a 0.456-mL gas
sample if it is expanded at constant pressure to 1.75 L at 55°C.
Solution
Data provided
V1 = 0.456 mL = 0.000456 L
T1 = ?
V2 = 1.75 L
T2 = 55 OC = (273 + 55)K = 238K
Using the relationship below and making T1 the subject;
𝑉1
𝑇1
=
𝑉2
𝑇2
T1 =
238 ×0.000456
1.75
= 0.1 𝐾K
= (0.1 − 273) oC
= −272.9 oC

47
Example 2.14. A plastic bag of peanuts is laid on a windowsill in the
sun, where its temperature increases from 20OC to 30OC. If the original
volume is 100.0 cm3, what is the final volume after warming?
Solution
Data collection
V1 = 100 cm3
T1 = 20OC = 293 K
V2 = ?
T2 = 30 oC = 303 K
Using the relationship below and making V2 the subject and
substituting;
𝑉1
𝑇1
=
𝑉2
𝑇2
V2=
303 ×100
293
= 103.4 cm3
Example 2.15. The temperature of a 4.00 L sample of gas is changed
from 10.0 °C to 20.0 °C. What will the volume of this gas be at the new
temperature if the pressure is held constant?
Solution
Data collection
V1 = 4.00L
T1 = 10OC = 283 K
V2 = ?
T2 = 20 OC = 293 K

48
substituting;
𝑉1
𝑇1
=
𝑉2
𝑇2
𝑉2 =
𝑇2 𝑉1
𝑇1
=
293 ×4.00
283
= 4.1 𝐿
Example 2.16. Carbon dioxide is usually formed when gasoline is
burned. If 30.0 L of CO2 is produced at a temperature of 1.00 x103 °C
and allowed to reach room temperature (25.0 °C) without any pressure
changes, what is the new volume of the carbon dioxide?
Solution
Data collection
V1 = 30.0L
T1 = 1.00 x103 °C = (273 + 1000)K = 1273 K
V2 = ?
T2 = 25 OC = 298 K
substituting;
𝑉1
𝑇1
=
𝑉2
𝑇2
𝑉2 =
𝑇2 𝑉1
𝑇1
=
298 × 30.00
1273
= 7.0 𝐿

49
Example 2.17. The volume of a gas syringe which contains 56.05
milliliters was raised to 67.7 milliliters at 107.5 oC. Determine the initial
temperature of the gas?
Solution
Data collection
V1 = 56.05 mm = 0.05605L
T1 =
V2 = 67.7 mm = 0.068L
T2 = 107.5 OC = 380.5 K
Using the relationship below and making T1 the subject and
substituting;
𝑉1
𝑇1
=
𝑉2
𝑇2
𝑇1 =
𝑇2 𝑉1
𝑉2
=
380.5 × 0.05605
0.068
= 313.6 𝐾
= (313.6 − 273) = 40.6 oC
Example 2.18. If 15.0 liters of neon at 25.0 °C is allowed to expand to
45.0 liters, what is the new temperature?
Solution
Data provided
V1 = 15.0L

50
T1 = 25 °C = (273 + 25)K = 298 K
V2 = 45.0 L
T2 = ?
Using the relationship below and making T2 the subject and
substituting;
𝑉1
𝑇1
=
𝑉2
𝑇2
𝑇2 =
𝑇1 𝑉2
𝑉1
=
298 × 45.00
15
= 294 𝐾
Example 2.19. A balloon has a volume of 2500.0 mL on a day when the
temperature is 30.0 °C. If the temperature at night falls to 10.0 °C, what
will be the volume of the balloon if the pressure remains constant?
Solution
Data collection
V1 = 2500 mL
T1 = 30OC = 303 K
V2 = ?
T2 = 10 OC = 283 K
substituting;
𝑉1
𝑇1
=
𝑉2
𝑇2

51
𝑉2 =
𝑇2 𝑉1
𝑇1
=
283 ×2500.00
303
= 2335 𝑚𝐿
2.6. Temperature-Pressure Relationship
Boyle‘s Law is the relationship between Pressure and Volume
but does not address temperature. How does temperature change
affect the properties of a sample of gas? Recall that temperature is a
measure of the average kinetic energy of particles. As the particles of a
substance move faster, the substance‘s temperature increases. The
particles bump into each other and the sides of the container more
often.
How would this affect a system where the volume is closed and
constant? This observation was first made by Gay-Lussac. He observed
that pressure has a direct proportional link with temperature of a
sample of gas in a closed container (volume constant). Properly put,
this law states that at constant volume, the pressure of a fixed mass of a
gas is directly proportional to its absolute temperature. The law is
expressed mathematically as follows:
𝑃 ∝ 𝑇 (Constant volume)
𝑃
𝑇
= 𝑘
A more useful form of the law can be express as:
𝑃1
𝑇1
=
𝑃2
𝑇2
Where P1 and T1 refer to the original pressure and temperature, P2 and
T2 refer to the pressure and temperature under the new or changed
conditions.
Note: in solving or addressing mathematical problems with this law,
the temperature must be expressed in Kelvin and the pressure in a
standard uint.

52
Example 2.20. 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What
would be the required temperature (in Celsius) to change the pressure
to standard pressure?
Solution
Data provided
P1 = 97.0 kPa
T1 = 25.0°C = ( 25 + 273)K = 298.0 K
P2 = s.p. = 101.325kPa
T2 = ?
Applying
𝑃1
𝑇1
=
𝑃2
𝑇2
and making T2 the subject
T2 =
𝑇1 𝑃2
𝑃1
=
298.0 𝐾 ×101.325 𝑘𝑃𝑎
97.0 𝑘𝑃𝑎
= 311K
Converting to degree in Celsius;
311K = (311 ‒ 273) °C
= 38°C
Example 2.21. If a gas in a closed container is pressurized from 15.0
atmospheres to 16.0 atmospheres and its original temperature was 25.0
°C, what would the final temperature of the gas be?
Solution
Data provided
P1 = 15 atm

53
T1 = 25.0°C = ( 25 + 273)K = 298.0 K
P2 = 16 atm
T2 = ?
Applying
𝑃1
𝑇1
=
𝑃2
𝑇2
and making T2 the subject
T2 =
𝑇1 𝑃2
𝑃1
=
298.0 𝐾 ×16 𝑎𝑡𝑚
15 𝑎𝑡𝑚
= 317 K
Example 2.22. A 30.0 L sample of nitrogen inside a metal container at
20.0 °C was placed inside an oven whose temperature is 50.0 °C. The
pressure inside the container at 20.0 °C was 3.00 atm. What is the
pressure of the nitrogen after its temperature was increased?
Solution
Collect data and convert temperatures to Kevin
P1 = 3.00 atm
T1 = 25.0°C = ( 20 + 273)K = 293.0 K
P2 = ?
T2 = 50.0°C = ( 50 + 273)K = 323.0 K
Applying
𝑃1
𝑇1
=
𝑃2
𝑇2
and making P2 the subject
P2 =
𝑇2 𝑃1
𝑇1

54
=
323.0 𝐾 ×3.00 𝑎𝑡𝑚
293 𝑎𝑡𝑚
= 3.3 atm
Example 2.23. The temperature of a sample of gas in a steel container at
30.0 kPa is increased from ‒100.0 °C to 1.00 x 103 °C. What is the final
pressure inside the tank?
Solution
Collect data and convert temperatures to Kevin
P1 = 30 kPa
T1 = ‒100.0 °C = (‒100.0 + 273)K = 173.0 K
P2 = ?
T2 = 1.00 x 103 °C = (1.00 x 103 + 273)K = 1273.0 K
Applying
𝑃1
𝑇1
=
𝑃2
𝑇2
and making P2 the subject
P2 =
𝑇2 𝑃1
𝑇1
=
1273 𝐾 ×30 𝑘𝑃𝑎
173 𝐾
= 220 kPa
2.7. The Combined Gas Law
Boyle‘s and Charles‘ laws may be merged into one law, called the
combined gas law, expressed in equation form as derived below:
From Boyle‘ law: V∝ 1/𝑃 (T constant)

55
From Charles‘ law: V ∝ 𝑇 (P constant)
V ∝ 1/𝑃 ∝ T
𝑉 = 𝑘𝑇/𝑃
𝑃𝑉
𝑇
= k
That is, for a given sample of a gas, PV/T remains constant, and
therefore
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
(a given sample of a gas)
This expression is a mathematical statement of the combined (or
general) gas law. In words, the volume of given sample of a gas is
inversely proportional to its pressure and directly proportional to its
absolute temperature.
Note that if the temperature is constant, T1 = T2, then the
expression reduces to the equation for Boyle‘s law, P1V1 = P2V2.
Alternatively, if the pressure is constant, P1 = P2, the expression is
equivalent to Charles‘ law, V1/T1 = V2/T2.
When the initial volume V1 of a gas at temperature T1 and pressure P1
is subjected to changes in temperature to T2 and pressure to P2, its new
volume V2 is obtained from the equation.
To apply this gas law, the amount of gas should remain
constant. As with the other gas laws, the temperature must be
expressed in kelvins, and the units on the similar quantities should be
the same. Because of the dependence on three quantities at the same
time, it is difficult to tell in advance what will happen to one property
of a gas sample as two other properties change. The best way to know
is to work it out mathematically.
Example 2.24. A certain mass of a gas occupies 330 cm3 at 27oC and 9.0
× 104 Nm-2 pressure. Calculate its volume at s.t.p. (s.p = 1.0 × 105 Nm-2).
Solution
Write the given data down, convert as variable to appropriate units
and substitute into the form to find the unknown.

56
Data provided:
V1 = 330 cm3
P1 = 9.0 × 104 Nm-2
T1 = 27oC = (27 + 273)K = 300K
T2 = s.t. = 273K
P2 = s.p. = 1.0 × 105Nm-2
V2 = ?
Using the gas equation:
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
Making V2 the subject of the formula:
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
9.0 × 104 × 330 × 273
1.0 × 105 × 300
= 270 cm3
Example 2.25. Calculate the volume of a sample of gas originally
occupying 908 mL at 717 torr and 20OC after its temperature and
pressure are changed to 72OC and 1.07 atm.
Solution
In attempting this problem, the volume can be stated in millilitres in
both states. The pressure can be stated in atmospheres in both but the
temperature must be in kelvins in both states.
Data provided
V1 = 908 mL

57
P1 =
717
760
𝑎𝑡𝑚 = 0.94 𝑎𝑡𝑚
T1 = 20oC = (20 + 273)K = 293K
T2 = 72 oC = 345K
P2 = 1.07 atm
V2 = ?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
0.94 × 908 × 345
1.07 × 293
= 943 mL
Example 2.27. Calculate the original volume of a sample of gas that is
at 700 torr and 22 oC before its volume, temperature, and pressure are
changed to 998 mL, 82°C, and 2.07 atm
Solution
Data provided
V1 = ?
P1 =
700
760
T1 = 22oC = (22 + 273)K = 295K
T2 = 82 oC = 355K
P2 = 2.07 atm

58
V2 = 998 mL
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V1 =
𝑃2 𝑉2 𝑇1
𝑃1 𝑇2
=
2.07 × 998 × 298
0.92 × 355
= 1884 mL
Example 2.28. 17.3-mL sample of gas originally at standard
temperature and pressure is changed to 10.9 mL at 678 torr. Calculate
its final temperature in degrees
Celsius.
Solution
Data provided
V1 = 17.3 mL
P1 = s.p. = 760 torr
T1 = s.t. = 273 K
T2 = ?
P2 = 678 torr
V2 =10.9 mL
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
Making T2 the subject of the formula:

59
T2 =
𝑃2 𝑉2 𝑇1
𝑃1 𝑉1
=
678 × 10.9 × 273
760 × 17.3
= 153 K
Converting to degree Celsius
= (153 − 273) oC
= 120 oC
Example 2.29. Calculate the volume at standard temperature and
pressure of a sample of gas that has a volume of 49.7 mL at 52°C and
811 torr.
Solution
Data provided
V1 = 49.7 mL
P1 =
811
760
T1 = 52°C = 325 K
T2 = s.t. = 273 K
P2 = s.p. = 1 atm
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1

60
=
1.07 × 49.7 × 273
1.0 × 325
= 45 mL
Example 2.30. Calculate the new volume after a 2.00-L sample of gas at
1.50 atm and 127oC is changed to 27oC at 3.50 atm.
Solution
Data provided
V1 = 2.00 L
P1 =1.50 𝑎𝑡𝑚
T1 = 127°C = 400 K
T2 = 27oC =300 K
P2 = 3.50 atm
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
1.5 × 2.00 × 300
3.50 × 400
= 0.64 L
Example 2.31. 500.0 liters of a gas are prepared at 700.0 mmHg and
200.0 °C. The gas is placed into a tank under high pressure. When the
tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the
volume of the gas?

61
Solution
Data provided
V1 = 500 L
P1 = 700.0 𝑚𝑚𝐻𝑔 = 700
760 𝑎𝑡𝑚 = 0.92 𝑎𝑡𝑚
T1 = 200°C = 473 K
T2 = 20oC =293 K
P2 = 30.0 atm
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
0.92 × 500 × 293
30 × 473
= 9.5 L
Example 2.32. A gas balloon has a volume of 106.0 liters when the
temperature is 45.0 °C and the pressure is 740.0 mm of mercury. What
will its volume be at 20.0 °C and 780 .0 mm of mercury pressure?
Solution
Data provided
V1 = 106 L
P1 = 740.0 𝑚𝑚𝐻𝑔
T1 = 45°C = 318 K

62
T2 = 20oC =293 K
P2 = 780.0 𝑚𝑚𝐻𝑔
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
740 × 106 × 293
780 × 318
= 92.7 L
Example 2.33. The volume of a gas originally at standard temperature
and pressure was recorded as 488.8 mL. What volume would the same
gas occupy when subjected to a pressure of 100.0 atm and temperature
of -245.0 °C?
Solution
Data provided
V1 = 488.8 L
P1 = 𝑠. 𝑝. = 1.0 𝑎𝑡𝑚
T1 = s.t. = 273 K
T2 = ‒245oC =28 K
P2 = 100 𝑎𝑡𝑚
V2 =?

63
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
1.0 × 488.8 × 28
100 × 273
= 0.5 L
Example 2.34. A gas is heated from 263.0 K to 298.0 K and the volume
is increased from 24.0 liters to 35.0 liters by moving a large piston
within a cylinder. If the original pressure was 1.00 atm, what would the
final pressure be?
Solution
Data provided
V1 = 24.0 L
P1 = 1.0 𝑎𝑡𝑚
T1 = 263.0 K
T2 =298.0 K
P2 = ?
V2 = 35.0 L
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
Making P2 the subject of the formula:
P2 =
𝑃1 𝑉1 𝑇2
𝑉2 𝑇1

64
=
1.0 × 24.0 × 298
35 × 263
= 0.78 atm
Example 2.35. The pressure of a gas is reduced from 1200.0 mmHg to
850.0 mmHg as the volume of its container is increased by moving a
piston from 85.0 mL to 350.0 mL. What would the final temperature be
if the original temperature was 90.0 °C?
Solution
Data provided
V1 = 85.0 mL
P1 = 1200 𝑚𝑚𝐻𝑔
T1 = 90.0 °C = 363 K
T2 =?
P2 = 850 𝑚𝑚𝐻𝑔
V2 = 350.0 mL
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
T2 =
𝑃2 𝑉2 𝑇1
𝑉1 𝑃1
=
850 × 350 × 363
85 × 1200
= 1059 K
= (1059 − 273) oC

65
= 786 oC
Example 2.36. If a gas is heated from 298.0 K to 398.0 K and the
pressure is increased from 2.230 x 103 mmHg to 4.560 x 103 mmHg
what final volume would result if the volume is allowed to change
from an initial volume of 60.0 liters?
Solution
Data provided
V1 = 60.0 L
P1 = 2.230 × 103
𝑚𝑚𝐻𝑔
T1 = 298.0 K
T2 =398.0 K
P2 = 4.560 × 103
𝑚𝑚𝐻𝑔
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
2.230 × 103 × 60.0 × 398
4.560 × 103 × 298
= 39.2 L
Example 2.37. A balloon containing a sample of gas has a temperature
of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The
balloon has a volume of 1,070 mL. The balloon is transported by plane

66
to Denver, where the temperature is 11°C and the pressure is 655 torr.
What is the new volume of the balloon?
Solution
Data provided
V1 = 1070 mL
P1 = 1.09 𝑎𝑡𝑚
T1 = 22°C = 295 K
T2 =11°C = 284 K
P2 = 655 𝑡𝑜𝑟𝑟 = 655
760 𝑎𝑡𝑚 = 0.86 𝑎𝑡𝑚
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
1.09 × 1070 × 284
0.86 × 295
= 1306 mL
Example 2.38. A balloon used to lift weather instruments into the
atmosphere contains gas having a volume of 1,150 L on the ground,
where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this
gas has a pressure of 6.88 torr and a temperature of −15°C. What is the
new volume of the gas?

67
Solution
Data provided
V1 = 1150 L
P1 = 0.977 𝑎𝑡𝑚
T1 = 18°C = 291 K
T2 = −15°C = 258 K
P2 = 6.88 𝑡𝑜𝑟𝑟 = 6.88
760 𝑎𝑡𝑚 = 0.0091 𝑎𝑡𝑚
V2 =?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
0.977 × 1150 × 258
0.0091 × 291
= 109465 L
2.8. Relationship between Amount and Volume
2.8.1. Gay-Lussac's Law of Combining Volumes
In the same 1808 article in which Gay-Lussac published his
observations on the thermal expansion of gases, he pointed out that
when two gases react, they do so in volume ratios that can always be
expressed as small whole numbers. This came to be known as the Law
of combining volumes.
Example 2.39. Ammonium carbonate decomposes when heated to
yield carbon dioxide, ammonia, and water vapour. Calculate the ratio

68
of the (separate) volume of ammonia to that of water vapour, each at
450°C and 1.00 atm.
Solution
The mole ratio of the gases, given in the balanced equation, is
2 mol NH3: 1 mol CO2: 1 molH2O
The ammonia and water vapour are separated and measured at the
given temperature and pressure. The ratio of their volumes can be
calculated as follows:
𝑉 𝑁𝐻 3
𝑉 𝐻2
𝑂
=
𝑛 𝑁𝐻 3 𝑅𝑇 /𝑃
𝑛 𝐻2 𝑂 𝑅𝑇 /𝑃
Because R is a constant and both T and P are the same for the two
gases, this equation reduces to
𝑉 𝑁𝐻 3
𝑉 𝐻2
𝑂
=
𝑛 𝑁𝐻 3
𝑛 𝐻2 𝑂
The right side of this equation is the ratio of the numbers of moles—the
ratio given by the balanced chemical equation. The left side of the
equation is the ratio of the volumes, so the ratio given by the balanced
chemical equation is equal to the volume ratio under these conditions.
The ratio is 2: 1.
Example 2.40. If 2.00 L H2 of and 1.00 L of both at standard temperature
and pressure, are allowed to react, will the water vapor they form at
250°C and 1.00 atm occupy 2.00 L?
Solution
2H2 (g) + O2 (g) → 2H2O (g)
The volumes of H2 and O2 that react are in the ratio given in the
balanced equation because the two gases have the same temperature and
pressure. The volume of water vapour formed is not in that ratio,
however, because its temperature is different. Its volume will be much
greater than 2 L.
2.8.2. Avogadro's Law
The work of the Italian scientist Amedeo Avogadro
complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811

69
he published a hypothesis stating that at the same temperature and
pressure, equal volumes of different gases contain the same number of
molecules (or atoms if the gas is monatomic). This law states that equal
volumes of all gases, under the same conditions of temperature and
pressure, contain the same number of molecule.
Mathematically: V ∝ 𝑛 (at constant T and P)
𝑉 = 𝑘𝑛
𝑉
𝑛
= 𝑘
Where V is the volume of gas, n is the number of molecules and 𝑘 is
the proportionality constant.
This law relates the volume of a fixed mass of a gas to the
number of molecules it contains. It shows that the volume occupied by
a gas depends on the number of molecules it contains, at a given
temperature and pressure. An increase in the number of gas molecules
leads to an increase in gas volume, and vice versa.
According to Avogadro‘s law we see that when two gases react
with each other, their reacting volumes have a simple ratio to each
other. If the product is a gas, its volume is related to the volume of the
reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac).
For example, consider the synthesis of ammonia from molecular
hydrogen and molecular nitrogen:
3H2(g) + N2(g) → 2NH3(g)
3 mol 1 mol 2 mol
Because, at the same temperature and pressure, the volumes of gases
are directly proportional to the number of moles of the gases present,
we can now write
3H2(g) + N2(g) → 2NH3(g)
3 volume 1 volume 2 volume
The volume ratio of molecular hydrogen to molecular nitrogen is 3:1,
and that of ammonia (the product) to molecular hydrogen and
molecular nitrogen combined (the reactants) is 2:4, or 1:2.

70
Example 2.41. 50 cm3 of sulphur (IV) oxide were produced at s.t.p.
when some quantity of powdered sulphur was burnt in excess oxygen.
(a) Write a balanced chemical equation for the reaction. (b) Calculate
the volume of oxygen used up during the reaction. (c) Which of the
laws is applicable? State the law.
Solution
(a). S(g) + O2(g) → SO2(g)
(b). From the balanced chemical equation in (a) above;
At s.t.p: 22400 cm3 of SO2 required 22400 cm3 of O2
Hence 1 cm3 of SO2 will require 1 cm3 of O2
∴ 50 cm3 of SO2 will use 50 cm3 of O2
(c). Avogadro‘s law is applicable in (b) above and it state that at the
same temperature and pressure equal volume of gases contain the
same number of molecules.
2.9. The Ideal Gas Law
So far, the gas laws we have used have focused on changing one or
more properties of the gas, such as its volume, pressure, or
temperature. There is one gas law that relates all the independent
properties of a gas under any particular condition, rather than a change
in conditions. This gas law is called the ideal gas law. The general
ideal gas equation is a combination of Boyle‘s, Charles‘ and
Avogadro‘s laws involving the four gas variables: pressure (P), volume
(V), number of mole of gas (n), and temperature (T).
From Boyle‘ law: V∝ 1/𝑃 (T constant)
From Charles‘ law: V ∝ 𝑇 (P constant)
From Avogadro‘s law: V ∝ 𝑛 (P,T constant)
V ∝ 1/𝑃 ∝ T ∝ 𝑛
V = R × 1/𝑃 × T × 𝑛
PV = nRT

71
In this equation, P is pressure, V is volume, n is amount of moles,
and T is temperature. R is called the ideal gas law constant and is a
proportionality constant that relates the values of pressure, volume,
amount, and temperature of a gas sample. The variables in this
equation do not have the subscripts i and f to indicate an initial
condition and a final condition. The ideal gas law relates the four
independent properties of a gas under any conditions.
2.10. Evaluation of the Gas Constant, R
The gas constant can be expressed in various units, all having
the dimension of energy per degree per mol.
From the general equation PV = nRT we get:
𝑅 =
𝑃𝑉
𝑛𝑇
Where P is pressure, V is volume, n is amount, and T is temperature.
R is most easily calculated from the fact that the hypothetical volume
of an ideal gas is 22.4L at STP (273.K and 1 atm).
i. If volume is expressed in liters and pressure in atmospheres,
then the proper value of R is as follows:
𝑅 =
𝑃𝑉
𝑛𝑇
=
1.0 𝑎𝑡𝑚 ×22.414 𝐿
1.0 𝑚𝑜𝑙 ×273.15 𝐾
R= 0.08206 atm L mol-1K-1
ii. if pressure is in Nm-2 and volume in m3 then the proper value of R is
thus:
𝑅 =
𝑃𝑉
𝑛𝑇
Where P = 101325 Nm-2, V = 22.4/1000 = 0.0224 m3
R =
101325 𝑁𝑚−2 ×0.0224 𝑚3
1.0 𝑚𝑜𝑙 ×273.15 𝐾
= 8.314N m mol-1K-1
iii. if pressure is in atm and volume in cm3 then

72
𝑅 =
𝑃𝑉
𝑛𝑇
=
1.0 𝑎𝑡𝑚 ×22414 𝑐𝑚3
1.0 𝑚𝑜𝑙 ×273.15 𝐾
= 82.06 atm cm3 mol-1K-1
[22.414 L =22400 cm3]
iv. if pressure is in Pa and volume in liter
𝑅 =
𝑃𝑉
𝑛𝑇
[1 atm = 1.01325× 105 Pa; 1 L= 10−3m3]
=
1.01325 ×105 𝑃𝑎 ×22.414 × 10−3 𝑚3
1.0 𝑚𝑜𝑙 ×273.15 𝐾
= 8.314 Pa m3 mol-1K-1
v. In JK−1 mol−1,
R = 8.314 kgm2s−2
= 8.314 JK−1 mol−1 [1 Pa = 1 kgm−1 s−2]
vi. In cal K−1 mol−1 (1 cal = 4.184 J),
R = 1.987 calK−1mol−1
Example 2.42. 50.0 g of N2 (M = 28.0 g) occupies a volume of 750mL at
298.15 K. Assuming the gas behaves ideally, calculate the pressure of
the gas in kPa.

73
Solution
Collect the data, convert volume to liter and find the number of mole
of nitrogen before substituting into the ideal gas equation to find
pressure
Data given
Mass of N2 = 50g
Molar mass of N2 = 28g/mol
Volume , V = 750mL =
750
1000
𝐿 = 0.750 𝐿
Temperature, T = 298.15 K
Number of mole of nitrogen gas (n) =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
=
50 𝑔
28𝑔𝑚𝑜𝑙 −1
=1.79 mol
Using PV = nRT
𝑃 =
𝑛𝑅𝑇
𝑉
=
1.79 𝑚𝑜𝑙 ×0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙 −1 𝐾−1 ×298.15 𝐾
0.750 𝐿
= 58.39 atm
Converting to kPa
=
(58.39 × 101325)
1000
= 5.916 × 103
𝑘𝑃𝑎
Example 2.43. Calculate the volume occupied by 2.5 moles of an ideal
gas at -23 oC, and 4.0 atmospheres. [R = 0.082 atm dm3 K-1 mol-1]

74
Solution
Data provided:
P = 4.0 atm
T = -23 + 273 = 250K
n = 2.5 moles
R = 0.082 atm dm3 K-1 mol-1
V = ?
Applying PV = nRT
Making V the subject, and substituting:
V =
𝑛𝑅𝑇
𝑃
=
2.5 ×0.082 ×250
4
=12.8 dm3
Example 2.44. Calculate the volume of 1.63 mol of carbon dioxide gas
at 295 K and 1.14 atm.
Solution
Data provided:
P = 1.14 atm
T = 295K
n = 1.63 moles
R = 0.082 atm L K-1 mol-1
V = ?
Applying PV = nRT

75
V =
𝑛𝑅𝑇
𝑃
=
1.63 ×0.082 ×295
1.14
=34.6 L
Example 2.45. Calculate the volume of 0.898 mol of methane gas, CH4,
at 292 K and 1.06 atm.
Solution
Data provided:
P = 1.06 atm
T = 292 K
n =0 .898 moles
R = 0.082 atm L K-1 mol-1
V = ?
Applying PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
0.898 mol ×0.082 atm L mol −1 K−1 ×292 K
1.06 atm
= 20.3 L
Example 2.46. Calculate the volume of 42.6 g of oxygen gas at 35oC and
792 torr
Solution
1. First convert temperature to Kelvin and pressure to atm.
2. Find the number of mole of oxygen

76
3. Plug data into idea gas equation to find the volume
Data provided:
P =
792
760
𝑎𝑡𝑚 = 1.04 atm
T = 35oC = 308 K
R = 0.082 atm L K-1 mol-1
n = mass/molar mass = 42.6g/32gmol-1 = 1.33 mol
V = ?
Applying PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
1.33 mol ×0.082 atm L mol −1 K−1 ×308 K
1.04 atm
=32.3 L
Students sometimes wonder “How do I decide when to use the combined gas
law and when to use the ideal gas law?” The answer depends on the problem,
naturally. If moles are involved, the combined gas law cannot be used.
Example 2.47. Decide which gas law should be used to solve each of
the following:
(a) Calculate the final volume of a sample of gas that has an initial
volume of 7.10 L at STP if the temperature and pressure are changed to
33oC and 696 torr.
(b) Calculate the volume of 0.977 mol of gas at 33oC and 792 torr.

77
Solution
(a) The combined gas law can be used because it does not involve
number of moles and initial and final conditions are involved.
(b) This problem involves moles and must be solved with the ideal gas
law.
Example 2.48. Calculate the pressure of 0.0789 mol of chlorine gas that
occupies 891 mL at ‒15°C.
Solution
The quantities given are converted to the units generally used with the
ideal gas law equation. Note that the nature of the gas is immaterial as
long as the number of moles is known.
Data provided:
T = ‒15°C = (‒15 +273) K = 258 K
n =0 .0789 moles
R = 0.082 atm L K-1 mol-1
V = 891 mL = (891/1000) L = 0.891 L
P = ?
Applying PV = nRT
Making P the subject, and substituting:
P =
𝑛𝑅𝑇
𝑉
=
0.0789 mol ×0.082 atm L mol −1 K−1 ×258 K
0.891 L
=1.87 atm

78
Example 2.49. Calculate the pressure of 0.0855 mol of neon gas that
occupies 66.1 mL at 25°C.
Solution
Data provided:
T = 25°C = (25 +273) K = 298 K
n =0 .0855 moles
R = 0.082 atm L K-1 mol-1
V = 66.1 mL = (66.1/1000) L = 0.0661 L
P = ?
Applying PV = nRT
P =
𝑛𝑅𝑇
𝑉
=
0.0855 mol × 0.082 atm L mol −1 K−1 ×298 K
0.0661 L
= 31.6 atm
Example 2.50. Calculate the number of moles of oxygen gas in a 2.60-L
container at 19°C and
755 torr.
Solution
Data provided:
T = 19°C = (19 +273) K = 292 K
R = 0.082 atm L K-1 mol-1
V = 2.60 L

79
P = 755 torr =
755
760
𝑎𝑡𝑚 = 0.99 atm
n =?
Applying PV = nRT
Making n the subject, and substituting:
n =
𝑃𝑉
𝑅𝑇
=
0.99 atm ×2.60 L
0.082 atm L mol −1 K−1 ×292 K
= 0.12 mol
5.0g of neon is at 256 mm Hg and at a temperature of 35º C. What is the
volume?
Solution
Step 1: Write down your given information:
P = 256 mmHg
V = ?
m = 5.0 g
R = 0.082 L atm mol-1K-1
T = 35oC
Step 2: Convert as necessary:
T = 35oC = (35 + 273)K = 308 K
P = 256 mmHg = (256/760) atm = 0.34 atm
n = mass/molar mass = 5.0g/20.1797 gmol-1 = 2.5 mol
Applying PV = nRT

80
V =
𝑛𝑅𝑇
𝑃
=
2.5 mol ×0.082 atm L mol −1 K−1 ×308 K
0.34 atm
=186 L
Example 2.51. What is a gas‘s temperature in Celsius when it has a
volume of 25 L, 203 mol, 143.5 atm?
Solution
Data provided:
R = 0.082 atm L K-1 mol-1
V = 25 L
P = 143.5 atm
n = 203 mol
T = ?
Applying PV = nRT
Making T the subject, and substituting:
T =
𝑃𝑉
𝑛𝑅
=
143.5 atm × 25 L
0.082 atm L mol −1 K−1 ×203 mol
= 215.5 K
= (215.5 − 273) oC
= −57.5 oC

81
Example 2.52. Sodium azide (NaN3) is used in some automobile air
bags. The impact of a collision triggers the decomposition of NaN3 as
follows:
2NaN3(s) →2Na(s) + 3N2(g)
The nitrogen gas produced quickly inflates the bag between the driver
and the windshield and dashboard. Calculate the volume of N2
generated at 85°C and 812 mmHg by the decomposition of 50.0 g of
NaN3.
Strategy From the balanced equation we see that 2 mol NaN3 gives 3
mol N2 so the conversion factor between NaN3 and N2 is
3 mol N2
2 𝑚𝑜𝑙 𝑁𝑎𝑁3
Because the mass of NaN3 is given, we can calculate the number of
moles of NaN3 and hence the number of moles of N2 produced. Finally,
we can calculate the volume of N2 using the ideal gas equation.
Solution
The sequence of conversions is as follows:
grams of NaN3 → moles of NaN3 →moles of N2 →volume of N2
First, we calculate the number of moles of N2 produced by 50.0 g of
NaN3:
mole of NaN3 = mass/ molar mass
= 50g/65.02gmol-1
= 0.769 mol

82
Mole of N2 = 0.769 mol ×
3 mol N2
2 𝑚𝑜𝑙 𝑁𝑎𝑁3
= 1.15 mol N2
The volume of 1.15 mol of N2 can be obtained by using the ideal gas
equation:
PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
1.15 mol ×0.082 atm L mol −1 K−1 ×(85+273) K
(
812
760
) atm
=
33.7594 𝐿
1.068
=31.6 L
Example 2.53. The equation for the metabolic breakdown of glucose
(C6H12O6) is the same as the equation for the combustion of glucose in
air:
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
Calculate the volume of CO2 produced at 37°C and 1.00 atm when 5.60
g of glucose is used up in the reaction. [C=12, O=16,H=1]
Solution
Collect given data and convert as necessary
R = 0.082 atm L K-1 mol-1
V = ?
P = 1.00 atm

83
n = ?
T = 37°C = 310 K
Mass of glucose = 5.6g
Strategy
1. First calculate the molar mass of glucose
2. Calculate the mole of C6H12O6 from which the mole of CO2 can
be gotten since 1 mole of C6H12O6 produced 6 moles of CO2
from the equation.
3. Use ideal gas equation to find the volume of CO2 produced.
C6H12O6 =[(12× 6 )+ (1 × 12) + (16× 6)] = 179gmol-1
Mole of C6H12O6 = 5.6g/179gmol-1
= 0.03 mol
From the balanced equation we see that 1 mol C6H12O6 gives 6 mol CO2
so the conversion factor between C6H12O6 and CO2 is
6 mol CO2
1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
Mole of CO2 = 0.03 mol ×
6 mol CO2
1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
=0.18 mol CO2
The volume of 0.18 mol of CO2 can be obtained by using the ideal gas
equation:
PV = nRT
V =
𝑛𝑅𝑇
𝑃

84
=
0.18 mol ×0.082 atm L mol −1 K−1 ×310 K
1.0 atm
= 4.6 𝐿
Example 2.54. Assuming ideal behaviour, which of the following
samples of gases will have the greatest volume at STP? Which of these
gases will have the greatest density at STP? (a) 0.82 mole of He. (b) 24 g
of N2. (c) 5.0 × 1023
molecules of Cl2
Solution
At STP, [T = 273K, P=1.0atm, R= 0.082 atm L mol−1
K−1
]
(a) n = 0.82 mole He
Using PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
0.82 mol ×0.082 atm L mol −1 K−1 × 273 K
1.0 atm
= 18.4 L
Density, d of He =
𝑚
𝑉
But mass, 𝑚 = 𝑛𝑚 = 0.82 × 4.003
= 0.33𝑔
∴ dendity, d =
0.88 𝑔
18.4 𝐿
=0.018g/L
(b) 24 g of N2
No of mole of nitrogen 𝑛 =
𝑚𝑎𝑠𝑠
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

85
=
24g
14𝑔𝑚𝑜𝑙 −1
= 1.71 𝑚𝑜𝑙
Using PV = nRT to calculate the volume of N2 at STP
V =
𝑛𝑅𝑇
𝑃
=
1.71 mol ×0.082 atm L mol −1 K−1 × 273 K
1.0 atm
= 38.3 L
Density, d of N2 =
𝑚
𝑉
=
24 𝑔
38.3 𝐿
= 0.627 g/L
(c) 5.0 × 1023
molecules of Cl2
Using PV = nRT to calculate the volume of chlorine molecule at STP,
Make V the subject, and substite:
V =
𝑛𝑅𝑇
𝑃
=
5.0 × 1023 mol ×0.082 atm L mol −1 K−1 × 273 K
1.0 atm
= 1.1193 × 1025
L
Density, d of Cl2 =
𝑚
𝑉
=
35.5 𝑔
1.1193 ×1025 𝐿
= 3.17 × 10−24
g/L

86
Summary of the results
He gas N2 gas Cl2 gas
Volume at STP (L) 18.4 38.3 1.1193 × 1025
Density at STP (g/L) 0.018 0.627 3.17 × 10−24
Results from the calculations showed chlorine gas has the highest
volume at STP 1.1193 × 1025
L while nitrogen gas has the greatest
density of 0.627 g/L.
Example 2.55. How many moles of O2 are present in a 0.500-L sample
at 25oC and 1.09 atm?
Solution
Collect the given data and convert as necessary
T = 25°C = (25 +273) K = 298 K
R = 0.082 atm L K-1 mol-1
V = 0.500 L
P = 1.09 atm
n =?
Applying PV = nRT to find n of O2;
n =
𝑃𝑉
𝑅𝑇
=
1.09 atm ×0.500 L
0.082 atm L mol −1 K−1 × 298 K

87
=0.022 mol of O2
Example 2.56. What is the volume of 1.00 mol of gas at STP?
Solution
Data provided:
P = s.p. = 1.0 atm
T = s.t. = 273 K
R = 0.082 atm L K-1 mol-1
n = 1.0 mol
V = ?
Applying PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
1.0mol ×0.082 atm L mol −1 K−1 ×273 K
1.0 atm
= 22.4 L
Note that the volume of 1.00 mol of gas at STP is called the molar volume of a
gas. This value should be memorized.
Example 2.57. How many moles of SO2 are present in a 765-mL sample
at 37oC and 775 torr?
Solution
Since R is defined in terms of liters and atmospheres, the pressure and
volume are first converted to those units.

88
Collect the given data
T = 37°C = (37 +273) K = 310 K
R = 0.082 atm L K-1 mol-1
V = 765 mL = (765/1000)L = 0.765 L
P = 775 torr = (775/760) atm = 1.02 atm
n =?
Applying PV = nRT to find n of SO2;
n =
𝑃𝑉
𝑅𝑇
=
1.02 atm ×0.765 L
0.082 atm L mol −1 K−1 × 310 K
= 0.03 mol of SO2
Example 2.57. At what temperature will 0.0750 mol of CO2 occupy 2.75
L at 1.11 atm?
Solution
V = 2.75 L
P = 1.11 atm
n = 0.0750 mol
T = ?
R = 0.082 atm L K-1 mol-1

89
Applying PV = nRT and Making T the subject, and substituting:
T =
𝑃𝑉
𝑛𝑅
=
1.11 atm × 2.75 L
0.0750 mol × 0.082 atm L mol −1 K−1
= 496 K
Example 2.58. What volume will 7.00 g of Cl2 occupy at STP?
Solution
The value of n is not given explicitly in the problem, but the mass is
given, from which we can calculate the number of moles:
Data provided
V = ?
P = s.p. = 1.0 atm
T = s.t. = 273 K
R = 0.082 atm L K-1 mol-1
Number of mole of Cl2 =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑕𝑙𝑜𝑟𝑖𝑛𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
=
7.00 𝑔
35.5 × 2 𝑔/𝑚𝑜𝑙
= 0.0986 mol of Cl2
Applying PV = nRT to find the volume of Cl2
V =
𝑛𝑅𝑇
𝑃
=
0.0986 mol × 0.082 atm L mol −1 K−1 273 K
1.0 atm
= 2.23 L

90
Example 2.59. If 4.58 g of a gas occupies 3.33 L at 27oC and 808 torr,
what is the molar mass of the gas?
Solution
If you do not see at first how to solve this problem to completion, at
least you can recognize that P, V, and T data are given. First calculate
the number of moles of gas present from which you can get the molar
mass.
T = 27°C = (27 +273) K = 300 K
R = 0.082 atm L K-1 mol-1
V = 3.33 L
P = 808 torr = (808/760) atm = 1.06 atm
Mass (m) = 4.58g
n =?
Applying PV = nRT to find n of gas;
n =
𝑃𝑉
𝑅𝑇
=
1.06 atm × 3.33 L
0.082 atm L mol −1 K−1 × 300 K
= 0.143 mol of gas
Recall, number of mole (n) =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
Therefore molar mass of gas (M) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒 (𝑛)
=
4.58 𝑔
0.143 𝑚𝑜𝑙

91
= 32.0g/mol
Example 2.60. What volume is occupied by the oxygen liberated by
heating 0.250 g of KClO3 until it completely decomposes to KCl and
oxygen? The gas is collected at STP.
Solution
From the balanced equation below, we see that 2 mol KClO3 gives 3
mol O2 so the conversion factor between KClO3 and O2 is
3 mol O2
2 𝑚𝑜𝑙 KClO3
2 KClO3 → 2KCl + 3O2
Because the mass of KClO3 is given, we can calculate the number of
moles of KClO3 and hence the number of moles of O2 produced.
Finally, we can calculate the volume of O2 using the ideal gas equation.
T = s.t. = 273 K
R = 0.082 atm L K-1 mol-1
V = ?
P = 𝑠. 𝑝. = 1.0 atm
Mass (m) = 0.250 g of KClO3
n =?
Molar mass of KClO3 [39.10 +35.5 +(16× 3)] = 122.6g/mol
mole of KClO3 = mass/ molar mass
= 0.250g/122.6gmol-1

92
= 0.002 mol KClO3
Mole of O2 = 0.002 mol ×
3 mol O2
2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3
= 0.003 mol O2
The volume of 0.003 mol O2 can be obtained by using the ideal gas
equation:
PV = nRT
V =
𝑛𝑅𝑇
𝑃
=
0.003 mol ×0.082 atm L mol −1 K−1 × 273 K
1.0 atm
= 0.067 L
5.0 moles of oxygen gas are contained in a 1.13 dm3 container at 127oC.
wha is the pressure of the system in Nm-2? [R = 8.314NmK-1mol-1]
Solution
Collect information provided and convert as necessary
T = 127oC = 400 K
n = 0.32 mol
R = 8.314NmK-1mol-1
V = 1.13 dm3 = (1.13/1000) m3 = 1.13 × 10−3
𝑚3
[1000 dm3 = 1 m3]
P =?
Use PV = nRT to find P

93
P=
𝑛𝑅𝑇
𝑉
=
0.32 mol ×8.314 Nm mol −1 K−1 ×400 K
0.00113 m3
= 941762.8 𝑁𝑚−2
= 9.43 × 105
𝑁𝑚−2
Example 2.61. A vessel contains 2.5 dm3 of oxygen gas at 29oC under
2.1 atmospheres. Estimate the amount of the gas at STP.
Solution
We will use the general gas equation to get the volume of oxygen gas
at STP then apply ideal gas law to get the amount of the gas.
Collect information provided and convert as necessary
T1 = 29oC = (29 + 273) K = 302 K
P1 = 2.1 atm
V1 = 2.5 dm3
T2 = s.t. = 273 K
P2 = s.p. = 1.0 atm
V2 = ?
n = ?
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1

94
=
2.1 atm × 2.5 dm 3 × 273 K
1.0 atm × 302 K
= 4.75 dm3
Now we can apply ideal gas equation in order to find the amount of
oxygen gas at s.t.p;
PV = nRT
n =
𝑃𝑉
𝑅𝑇
But P = 1 atm = 101325 𝑁𝑚−2
Volume, V = 4.75 dm3 = (4.75/1000) m3 = 4.75 × 10−3
𝑚3
Temperature, T = 273 K
R = 8.314NmK-1mol-1
n =
101325 Nm −2 × 4.75 × 10−3 m3
8.314 NmK −1mol −1 × 273 K
= 0.212 mol of O2 gas
Example 2.62.Calculate the volume occupied by 40g of carbon dioxide
(CO) at 4.58 × 104
𝑁𝑚−2
and 50oC , assuming ideal gas law is obeyed.
[O = 16, C =12]
Solution
First we calculate the molar mass of CO and find its number f mole
then apply ideal gas equation to get the volume.
Collect the given data and convert as necessary
T = 50°C = (50 +273) K = 323 K

95
R = 8.314NmK-1mol-1
V = ?
P = 4.58 × 104
𝑁𝑚−2
Mass of CO(m) = 40g
n =?
Molar mass of CO = 12 +16 = 28g/mol
Recall, number of mole (n) =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
=
40 𝑔
28𝑔/ 𝑚𝑜𝑙
= 1.43 mol
Applying PV = nRT to find V of CO gas;
V =
𝑛𝑅𝑇
𝑃
=
1.43 mol 8.314 Nm mol −1 K−1 × 323 K
4.58 × 104 𝑁𝑚−2
= 0.082 m3 of CO gas
= 82.0 dm3 of CO gas
[1000dm3 = 1m3]
Example 2.63. How many moles of a gas are contained in 890.0 mL at
21.0 °C and 750.0 mmHg pressure?
Solution
Collect the given data and convert to appropriate units
T = 21°C = (21 +273) K = 294 K
R = 0.082 atm L K-1 mol-1

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