Gases

(P383)

EARTH’S ATMOSPHERE, which is mostly made of

nitrogen and oxygen gases, extends about 350 miles up
from Earth’s surface. Over 95% of the mass of the
atmosphere, however, is in the 30 miles closest to the
surface. Most of what we perceive as “weather” occurs in
the first 10 miles above the surface.
Substances That Exist as Gases

Gases under normal atmospheric conditions, which

are defined as 25°C and 1 atmosphere (atm)
pressure.
Substances That Exist as Gases
Ionic compounds do not exist as gases at 25°C
and 1 atm .
NaCl : m.p. 801°C
Molecular compounds is more varied.
Some are gases (e.g. CO, CO2,
HCl, NH3, and CH4 (methane)), but
the majority of molecular
compounds are liquids or solids at
room temperature.
Substances That Exist as Gases
Substances That Exist as Gases

All gases have the following physical characteristics:

• Gases assume the volume and shape of their

containers.
• Gases are the most compressible of the states of
matter.
• Gases will mix evenly and completely when
confined to the same container.
• Gases have much lower densities than liquids and
solids.
Gases Pressure
Pressure, P, is defined in science as the force, F,
that acts on a given area, A

SI Units of Pressure
(m/s, cm/s)

(m/s2 , cm/s2)
Gases Pressure
SI unit of force is the newton (N):
1 N = 1 kg m/s2
We define pressure as force applied per unit area:

The SI unit of pressure is the pascal (Pa),

defined as one newton per square meter:
1 Pa = 1 N/m2
Gases Pressure
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by Earth’s
atmosphere. The actual value of atmospheric pressure
depends on location, temperature, and weather conditions.

The force experienced by any

area exposed to Earth’s
atmosphere is equal to the
weight of the column of air
above it.
Gases Pressure
Barometer The standard atmosphere
equals a pressure of 760
Evangelista Torricelli (1608–1647) mmHg, where mmHg
represents the pressure
exerted by a column of
mercury 1 mm high. The
mmHg unit is also called
the torr.
1 torr = 1 mmHg
1 atm = 760 mmHg
1 atm = 101,325 Pa = 1.01325 × 105 Pa
1 atm = 1.01325 × 102 kPa
 Gases Pressure
Example
The pressure outside a jet plane flying at high altitude falls
considerably below standard atmospheric pressure. Therefore,
the air inside the cabin must be pressurized to protect the
passengers. What is the pressure in atmospheres in the cabin
if the barometer reading is 688 mmHg?
Solution The pressure in the cabin is given by
Gases Pressure
EXAMPLE
The atmospheric pressure in San Francisco on a certain
day was 732 mmHg. What was the pressure in kPa?

Solution: The pressure in kPa is

 Example：
Converting Pressure Units
(a) Convert 0.357 atm to torr. (b) Convert 6.6 × 10-2 torr to
atmospheres. (c) Convert 147.2 kPa to torr.
Gases Pressure
manometer

(a) (b)
＝(764.7 torr)
The Gas Law
PV = k1 P = k1 × 1/V
 Boyle’s law- The pressure-Volume Relationship
Boyle’s law, which summarizes these observations, states that the
volume of a fixed quantity of gas maintained at constant temperature is
inversely proportional to the pressure.
k1 is a constant called the
“proportionality constant”.

PV = k1 = nRT
The Temperature-Volume Relationship:
Charles’s and Gay-Lussac’s
Law

Jacques Charles (1746–1823)

French scientist

Volume increases as temperature increases and

decreases as temperature decreases. (1787)
 Absolute
zero

Figure: Variation of the volume of a gas sample with temperature, at

constant pressure. Each line represents the variation at a certain pressure.
The pressures increase from P1 to P4.
Kelvin temperature scale

Absolute zero = 0 K = 273.15°C

≈ 273 °C
In terms of the Kelvin scale, Charles’s law states: The volume of a
fixed amount of gas maintained at constant pressure is directly
proportional to its absolute temperature.
 Charles’s and Gay-Lussac’s law
(Charles’s law )
where k2 is the proportionality constant

k2 = nR/P
Heating or cooling a gas at constant pressure
The Volume-Amount Relationship:
Avogadro’s Law

where n represents the number of moles and k4

is the proportionality constant

k4 = RT/P
Avogadro’s hypothesis: Equal
volumes of gases at the same
temperature and pressure contain
equal numbers of molecules.
Avogadro’s law.
The Ideal Gas Law

“ideal gas equation”

R :the gas constant

An ideal gas is a hypothetical gas whose pressure-volume-

temperature behavior can be completely accounted for by the ideal
gas equation.

At 0ºC (273.15 K) and 1 atm pressure, many real gases

behave like an ideal gas.
The Ideal Gas Law
The conditions 0ºC and 1 atm are called standard
temperature and pressure, often abbreviated STP.

A comparison of the
molar volume at STP
(which is approximately
22.4 L) with a basketball.
Comparison of molar volumes at STP:

§Suggest an explanation for the “ideal” nature of helium compared to

the other gases.
§What are the characteristics of ideal gas?
The Ideal Gas Law
Example: Sulfur hexafluoride (SF6) is a colorless, odorless,
very unreactive gas. Calculate the pressure (in atm) exerted
by 1.82 moles of the gas in a steel vessel of volume 5.43 L
at 69.5oC.

Solution：
The Ideal Gas Law
Example: Calculate the volume (in liters) occupied by 7.40
g of NH3 at STP.
Solution:
grams of NH3  moles of NH3  liters of NH3 at STP

So the volume of NH3 is given by

The Ideal Gas Law

If n1 = n2
The Ideal Gas Law
An inflated helium balloon with a volume of 0.55 L at sea
level (1.0 atm) is allowed to rise to a height of 6.5 km, where
the pressure is about 0.40 atm. Assuming that the
temperature remains constant, what is the final volume of
the balloon?
Solution: P1V1 P2V2
We start with the equation n T = n T
1 1 2 2

Because n1 = n2 and T1 = T2, P1V1 = P2V2

(The Boyle’s law)
Therefore,
The Ideal Gas Law
Argon is an inert gas used in lightbulbs to retard the
vaporization of the tungsten filament. A certain lightbulb
containing argon at 1.20 atm and 18°C is heated to 85°C
at constant volume. Calculate its final pressure (in atm).
Solution:
Because n1 = n2 and V1 = V2, P1 P2
=
T1 T2
The final pressure is given by
The Ideal Gas Law
A small bubble rises from the bottom of a lake, where the
temperature and pressure are 8°C and 6.4 atm, to the
water’s surface, where the temperature is 25°C and the
pressure is 1.0 atm. Calculate the final volume (in mL) of the
bubble if its initial volume was 2.1 mL.
Solution:
According to the equation
P1V1 P2V2
=
n1T1 n2T2

We assume that the amount of air in the bubble remains constant,

that is, n1 = n2 so that
The Ideal Gas Law
Density Calculations
Carbon dioxide gas
flows downhill
because it is denser
than air. The CO2
“fog” is not the gas
made visible but
rather is made up of
drops of water that
have condensed from
water vapor in the air.
The Ideal Gas Law
Calculate the density of carbon dioxide (CO2) in grams per
liter (g/L) at 0.990 atm and 55°C.
Solution:
we convert temperature to kelvins (T = 273 + 55 = 328 K) and
use 44.01 g for the molar mass of CO2:
The Ideal Gas Law
Alternatively, we can solve for
the density by writing

Assuming that we have 1 mole of CO2, the mass is 44.01

g. The volume of the gas can be obtained from the ideal
gas equation

Therefore, the density

of CO2 is given by
The Ideal Gas Law
The Molar Mass of a Gaseous Substance

Example:
A chemist has synthesized a greenish-yellow
gaseous compound of chlorine and oxygen and
finds that its density is 7.71 g/L at 36°C and
2.88 atm. Calculate the molar mass of the
compound and determine its molecular formula.
The Ideal Gas Law
Solution:

Alternatively, we can solve for the molar mass by writing

The Ideal Gas Law
From the given density we know there are 7.71 g of the
gas in 1 L. The number of moles of the gas in this volume
can be obtained from the ideal gas equation

Therefore, the molar mass is given by

The Ideal Gas Law
Example: Chemical analysis of a gaseous compound
showed that it contained 33.0 percent silicon (Si) and 67.0
percent fluorine (F) by mass. At 35°C, 0.210 L of the
compound exerted a pressure of 1.70 atm. If the mass of
0.210 L of the compound was 2.38 g, calculate the
molecular formula of the compound.
Solution:
The number of moles of Si and F are given by
Gas Stoichiometry
Example: Calculate the volume of O2 (in liters) required
for the complete combustion of 7.64 L of acetylene (C2H2)
measured at the same temperature and pressure.

Solution:
5mol O2 ~ 2 mol C2H2  5L O2 ~ 2L C2H2.
The volume of O2 that will react with 7.64 L
C2H2 is given by
Gas Stoichiometry
E xam p l e: S o d i u m a z i d e ( N a N 3 ) i s u s e d i n s o m e
automobile air bags. The impact of a collision triggers the
decomposition of NaN3 as follows:

The nitrogen gas produced quickly

inflates the bag between the driver
and the windshield and dashboard.
Calculate the volume of N2
generated at 80°C and 823 mmHg
by the decomposition of 60.0 g of
NaN3.
Gas Stoichiometry
Gas Stoichiometry
Example:
Aqueous lithium hydroxide solution is used to purify air in
spacecrafts and submarines because it absorbs carbon
dioxide, which is an end product of metabolism, according
to the equation
2LiOH(aq) + CO2(g)  Li2CO3(aq) + H2O(l)
The pressure of carbon dioxide inside the cabin of a
submarine having a volume of 2.4 × 105 L is 7.9 × 10-3
atm at 312 K. A solution of lithium hydroxide (LiOH) of
negligible volume is introduced into the cabin. Eventually
the pressure of CO2 falls to 1.2 × 10-4 atm. How many
grams of lithium carbonate are formed by this process?
Gas Stoichiometry
Dalton’s Law of Partial Pressures
Dalton’s law of partial pressures states that the total
pressure of a mixture of gases is just the sum of the
pressures that each gas would exert if it were present alone.
Dalton’s Law of Partial Pressures
Two gases, A and B are in a container of volume V.
The pressure exerted by gas A, according to the ideal gas
equation, is

The pressure exerted by gas B is

Thus, according to Dalton’s law

n = nA + nB
Dalton’s Law of Partial Pressures
The relationship between each partial pressure and the total
pressure:

(XA is called the

mole fraction of A)

In general, the mole fraction of component i in a mixture

is given by
Dalton’s Law of Partial Pressures
Thus, the expression of the partial pressure of A and B are

If only two components are present, then

Dalton’s Law of Partial Pressures
Example: A mixture of gases contains 4.46 moles of
neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of
xenon (Xe). Calculate the partial pressures of the gases if
the total pressure is 2.00 atm at a certain temperature.
Solution:
According to the equation Pi = XiPT
Dalton’s Law of Partial Pressures
We calculate the mole fraction of Ne as follows
Air is approximately 78%
nitrogen, 21% oxygen, plus a
mixture of many other gases.
Dalton’s Law of Partial Pressures
For example,
Dalton’s Law of Partial Pressures
Example: Oxygen gas generated by the decomposition
of potassium chlorate is collected. The volume of oxygen
collected at 24°C and atmospheric pressure of 762 mmHg
is 128 mL. Calculate the mass (in grams) of oxygen gas
obtained. The pressure of the water vapor at 24°C is 22.4
mmHg.
Dalton’s Law of Partial Pressures
From the ideal gas equation we write

where m and M are the mass of O2 collected and the

molar mass of O2, respectively. Rearranging the
equation we obtain
The Kinetic Molecular Theory of Gases
Ludwig Boltzmann and James Clerk Maxwell, found that the physical properties
of gases can be explained in terms of the motion of individual molecules.

Kinetic energy (KE) is the type of energy

expended by a moving object, or energy of motion.

kinetic molecular theory of gases, or simply

the kinetic theory of gases.
The Kinetic Molecular Theory of Gases
Assumptions:
1. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be
“points”; that is, they possess mass but have negligible
volume.

2.Gas molecules are in constant motion in random

directions, and they frequently collide with one another.
Collisions among molecules are perfectly elastic. In other
words, energy can be transferred from one molecule to
another as a result of a collision.
Nevertheless, the total energy of all the molecules in a
system remains the same.
The Kinetic Molecular Theory of Gases

The pressure of a gas is caused

by collisions of the molecules
with the walls of the container
The Kinetic Molecular Theory of Gases
3. Gas molecules exert neither attractive nor repulsive
forces on one another.

4.The average kinetic energy of the molecules is

proportional to the temperature of the gas in kelvins.
Any two gases at the same temperature will have the
same average kinetic energy. The average kinetic energy
of a molecule is given by
Mean square
speed
The Kinetic Molecular Theory of Gases
Application to the Gas Laws
• Compressibility of Gases (assumption 1)
• Boyle’s Law
• Charles’s Law
• Avogadro’s Law

• Dalton’s Law of Partial Pressures (assumption 3)

The Kinetic Molecular Theory of Gases
Distribution of Molecular Speeds

The distribution
of speeds for
nitrogen gas at
three different
temperatures. At
the higher
temperatures,
more molecules
are moving at
faster speeds.
The Kinetic Molecular Theory of Gases

The distribution of
speeds for three
gases at 300 K. At
a given
temperature, the
lighter molecules
are moving faster,
on the average.
The Kinetic Molecular Theory of Gases
Root-Mean-Square Speed
root-mean-square (rms) speed (urms), is an average
molecular speed.
One of the results of the kinetic theory of gases is that
the total kinetic energy of a mole of any gas equals
3/2RT.
The Kinetic Molecular Theory of Gases
Example: Calculate the root-mean-square speeds of
helium atoms and nitrogen molecules in m/s at 25°C.
Solution:
The Kinetic Molecular Theory of Gases
Check
Because He is a lighter gas, we expect it to move faster,
on average, than N2. A quick way to check the answers
is to note that the ratio of the two urms values (1.36 ×
103/515 ≈ 2.6) should be equal to the square root of the
ratios of the molar masses of N2 to He, that is, √28/4 ≈
2.6.
Gas Diffusion and Effusion
Diffusion is the gradual mixing of molecules of one gas with
molecules of another by virtue of their kinetic properties

Graham’s law of diffusion: under the same conditions of

temperature and pressure, rates of diffusion for gases are
inversely proportional to the square roots of their molar
masses. It is expressed mathematically as

where r1 and r2 are the

diffusion rates of gases 1 and
2, and M1 and m2 are their
molar masses, respectively.
The Kinetic Molecular Theory of Gases
Gas Effusion
Effusion is the process by which a gas under pressure
escapes from one compartment of a container to another
by passing through a small opening.

Effusion Diffusion
 The Kinetic Molecular Theory of Gases
Example： A flammable gas made up only of carbon and hydrogen is found to
effuse through a porous barrier in 1.50 min. Under the same conditions of
temperature and pressure, it takes an equal volume of bromine vapor 4.73 min
to effuse through the same barrier. Calculate the molar mass of the unknown
gas, and suggest what this gas might be.

methane (CH4)
Real Gases: Deviation from Ideal Behavior
True or false: Nitrogen gas behaves more like an
ideal gas as the temperature increases.
Real Gases: Deviation from Ideal Behavior

To study real gases accurately, then, we need to modify the

ideal gas equation, taking into account intermolecular
forces and finite molecular volumes.

Effect of intermolecular forces on

the pressure exerted by a gas.
Real Gases: Deviation from Ideal Behavior

where a is a constant and n

and V are the number of
moles and volume of the
container, respectively

van der Waals equation

Real Gases: Deviation from
Ideal Behavior

Example: Given that 3.50

moles of NH3 occupy 5.20 L at
47°C, calculate the pressure of
the gas (in atm) using (a) the
ideal gas equation and (b) the
van der Waals equation.
(b) It is convenient to first calculate the correction terms in
separately. From Table , we have