8.1 States of Matter and Their
Changes
Outline
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8.1 States of Matter and Their Changes
8.2 Gases and the Kinetic–Molecular Theory
8.3 Pressure
8.4 Boyle’s Law: The Relation Between Volume and Pressure
8.5 Charles’s Law: The Relation Between Volume and Temperature
8.6 Gay-Lussac’s Law: The Relation Between Pressure and Temperature
8.7 The Combined Gas Law
8.8 Avogadro’s Law: The Relation Between Volume and Molar Amount
8.9 The Ideal Gas Law
8.10 Partial Pressure and Dalton’s Law
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Chapter Eight
► Matter exists in any of three phases, or states—solid,
liquid, and gas, depending on the attractive forces
between particles, temperature, and pressure.
► In a gas, the attractive forces between particles are
very weak compared to their kinetic energy, so the
particles move about freely, are far apart, and have
almost no influence on one another.
► In a liquid, the attractive forces between particles are
stronger, pulling the particles close together but still
allowing them considerable freedom to move about.
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►Phase change or change of state: The transformation
of a substance from one state to another.
►Melting point (mp): The temperature at which solid
and liquid are in equilibrium.
►Boiling point (bp): The temperature at which liquid
and gas are in equilibrium.
►Sublimation: A process in which a solid changes
directly to a gas.
►Melting, boiling, and sublimation all have ∆H > 0,
and ∆S > 0. This means they are nonspontaneous
below and spontaneous above a certain temperature.
In a solid, the attractive forces are much stronger than
the kinetic energy of the particles, so the atoms,
molecules, or ions are held in a specific arrangement
and can only wiggle around in place.
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Kinetic Theory of Gases
A gas consists of small particles that
• move rapidly in straight lines.
• have essentially no attractive (or
repulsive) forces.
• are very far apart.
• have very small volumes compared to
the volume of the container they
occupy.
• have kinetic energies that increase
with an increase in temperature.
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8.2 Gases and the Kinetic-Molecular
Theory
► The behavior of gases can be explained by a group
of assumptions known as the kinetic–molecular
theory of gases. The following assumptions account
for the observable properties of gases:
► A gas consists of many particles, either atoms or
molecules, moving about at random with no
attractive forces between them. Because of this
random motion, different gases mix together quickly.
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Chapter Eight
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8.3 Pressure
► Collisions of gas particles, either with other
particles or with the wall of their container, are
elastic; that is, the total kinetic energy of the
particles is constant. The pressure of a gas against
the walls of its container is the result of collisions of
the gas particles with the walls. The number and
force of collisions determines the pressure.
► A gas that obeys all the assumptions of the kinetic–
molecular theory is called an ideal gas. All gases
behave somewhat differently than predicted by the
kinetic–molecular theory at very high pressures or
very low temperatures. Most real gases display
nearly ideal behavior under normal conditions.
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► The amount of space occupied by the gas particles
themselves is much smaller than the amount of
space between particles. Most of the volume taken
up by gases is empty space, accounting for the ease
of compression and low densities of gases.
► The average kinetic energy of gas particles is
proportional to the Kelvin temperature. Thus, gas
particles have more kinetic energy and move faster
as the temperature increases. (In fact, gas particles
move much faster than you might suspect. The
average speed of a helium atom at room temperature
and atmospheric pressure is approximately 1.36
km/s, or 3000 mi/hr, nearly that of a rifle bullet.)
► Pressure (P) is defined as a force (F) per unit area
(A) pushing against a surface; P = F/A.
► A barometer measures pressure as the height of a
mercury column. Atmospheric pressure presses
down on mercury in a dish and pushes it up a tube.
► Pressure units:
1 atm = 760 mm Hg = 14.7 psi = 101,325 Pa
1 mm Hg = 1 torr = 133.32 Pa
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Learning Check
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Solution
A. What is 475 mm Hg expressed in atm?
1) 475 atm
2) 0.625 atm
3) 3.61 x 105 atm
A. What is 475 mm Hg expressed in atm?
2) 0.625 atm
475 mm Hg x 1 atm
= 0.625 atm
760 mm Hg
B. The pressure in a tire is 2.00 atm. What is this
pressure in mm Hg?
1) 2.00 mm Hg
2) 1520 mm Hg
3) 22 300 mm Hg
B. The pressure of a tire is measured as 2.00 atm. What is
this pressure in mm Hg?
2) 1520 mm Hg
2.00 atm x 760 mm Hg = 1520 mm Hg
1 atm
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Atmospheric Pressure
Gas pressure inside a container is often measured
using an open-end manometer, a simple instrument
similar in principle to the mercury barometer.
Atmospheric pressure
is the pressure exerted by a
column of air from the top of
the atmosphere to the surface
of the Earth.
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Altitude and Atmospheric Pressure
Learning Check
Atmospheric pressure
A. The downward pressure of the Hg in a barometer is
_____ than (as) the pressure of the atmosphere.
• is about 1 atmosphere at sea
level.
• depends on the altitude and
the weather.
• is lower at high altitudes
where the density of air is
less.
• is higher on a rainy day
than on a sunny day.
1) greater
2) less 3) the same
B. A water barometer is 13.6 times taller than a Hg
barometer (DHg = 13.6 g/mL) because
1) H2O is less dense than mercury.
2) H2O is heavier than mercury.
3) air is more dense than H2O.
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8.4 Boyle’s Law: The Relation Between
Volume and Pressure
Solution
► Boyle’s law: The volume of a gas is inversely
proportional to its pressure for a fixed amount of gas
at a constant temperature. That is, P times V is
constant when the amount of gas n and the
temperature T are kept constant.
► V ∝ 1/P or PV = k if n and T are constant
► If: P1V1 = k and P2V2 = k
► Then: P1V1 = P2V2
A.The downward pressure of the Hg in a barometer is 3)
the same (as) the pressure of the atmosphere.
B. A water barometer is 13.6 times taller than a Hg
barometer (DHg = 13.6 g/mL) because
1) H2O is less dense than mercury.
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3
The volume of a gas decreases proportionately as its
pressure increases. If the pressure of a gas sample is
doubled, the volume is halved.
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Chapter Eight
Graph (a) demonstrates the decrease in volume as
pressure increases, whereas graph (b) shows the
linear relationship between V and 1/P.
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Solving for a Gas Law Factor
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Calculations with Boyle’s Law
The equation for Boyle’s Law can be rearranged to
solve for any factor.
P1V1 = P2V2
Boyle’s Law
To solve for V2 , divide both sides by P2.
P1V1 = P2V2
P2
P2
V1 x
P1
P2
=
V2
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Calculation with Boyle’s Law
(Continued)
Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems. What is
the new volume (L) of a 8.0 L sample of Freon gas initially at
550 mm Hg after its pressure is changed to 2200 mm Hg at
constant T?
2. When pressure increases, volume decreases.
Solve Boyle’s Law for V2:
P1V1 = P2V2
1. Set up a data table:
Conditions 1
P1 = 550 mm Hg
V1 = 8.0 L
V2
Conditions 2
P2 = 2200 mm Hg
V2 = -?
V2
= V1 x P1
P2
= 8.0 L x 550 mm Hg =
2200 mm Hg
2.0 L
pressure ratio
decreases volume
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Learning Check
Solution
For a cylinder containing helium gas indicate if cylinder A
or cylinder B represents the new volume for the following
changes (n and T are constant).
For a cylinder containing helium gas indicate if cylinder A
or cylinder B represents the new volume for the following
changes (n and T are constant):
1) Pressure decreases B
1) pressure decreases
2) pressure increases
2) Pressure increases A
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8.5 Charles’ Law: The Relation Between
Volume and Temperature
► Charles’s law: The volume of a gas is directly
proportional to its Kelvin temperature for a fixed
amount of gas at a constant pressure. That is, V
divided by T is constant when n and P are held
constant.
► V ∝ T or V/T = k if n and P are constant
► If: V1/T1 = k and V2/T2 = k
► Then: V1/T1 = V2/T2
Charles’ Law: V and T
• For two conditions, Charles’ Law is written
V1 = V2 (P and n constant)
T2
T1
• Rearranging Charles’ Law to solve for V2
T2 x V1
= V2 x T1
T1
T1
V2
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V1 x T2
T1
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If the Kelvin temperature of a gas is doubled, its
volume doubles.
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=
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As the temperature goes up, the volume also goes up.
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Calculations Using Charles’ Law
(continued)
Calculations Using Charles’ Law
A balloon has a volume of 785 mL at 21°C. If the
temperature drops to 0°C, what is the new volume of
the balloon (P constant)?
2. Solve Charles’ law for V2:
V1 = V2
T1
T2
1. Set up data table:
Conditions 1
Conditions 2
V1 = 785 mL
V2 = ?
T1 = 21°C = 294 K T2 = 0°C = 273 K
V2 = V1 x T2
T1
V2 = 785 mL x 273 K = 729 mL
294 K
Be sure to use the Kelvin (K) temperature in
gas calculations.
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Learning Check
Solution
Use the gas laws to complete sentence with
Use the gas laws to complete with
1) increases
1) increases
2) decreases.
2) decreases.
A. Pressure _______, when V decreases.
A. Pressure 1) Increases, when V decreases.
B. When T decreases, V _______.
B. When T decreases, V 2) Decreases.
C. Pressure _______ when V changes from 12 L to 24 L
C. Pressure 2) Decreases when V changes from 12 L to 24 L
D. Volume _______when T changes from 15 °C to 45°C
D. Volume 1) Increases when T changes from 15°C to 45°C
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8.6 Gay-Lussac’s Law: The Relation Between
Pressure and Temperature
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As the temperature goes up, the pressure also goes up.
► Gay-Lussac’s law: The pressure of a gas is directly
proportional to its Kelvin temperature for a fixed
amount of gas at a constant volume. That is, P
divided by T is constant when n and V are held
constant.
► P ∝ T or P/T = k if n and V are constant
► If: P1/T1 = k and P2/T2 = k
► Then: P1/T1 = P2/T2
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Learning Check
Solution
Solve Gay-Lussac’s Law for P2.
P1
T1
=
Solve Gay-Lussac’s Law for P2.
P1 =
P2
T1
T2
P2
T2
Multiply both sides by T2 and cancel
=
P2 x T 1
P1 x T 2
T1
T1
P2 = P1 x T 2
T1
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8.7 The Combined Gas Law
Learning Check
► Since PV, V/T, and P/T all have constant values for
a fixed amount of gas, these relationships can be
merged into a combined gas law for a fixed amount
of gas.
► Combined gas law: PV/T = k if n constant
► P1V1/T1 = P2V2/T2
► If any five of the six quantities in this equation are
known, the sixth can be calculated.
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A gas has a volume of 675 mL at 35°C and 0.850 atm
pressure. What is the volume(mL) of the gas at -95°C and a
pressure of 802 mm Hg (n constant)?
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Chapter Eight
8.8 Avogadro’s Law: The Relation
Between Volume and Molar Amount
Solution
Conditions 1
T1 = 308 K
► Avogadro’s law: The volume of a gas is directly
proportional to its molar amount at a constant
pressure and temperature. That is, V divided by n is
constant when P and T are held constant.
► V ∝ n or V/n = k if P and T are constant
► If: V1/n1 = k and V2/n2 = k
► Then: V1/n1 = V2/n2
Data Table
Conditions 2
T2 = -95°C + 273 = 178K
V1 = 675 mL
V2 = ???
P1 = 646 mm Hg
P2 = 802 mm Hg
Solve for V2
V2 = V1 x P1 x T2
P2
T1
V2 = 675 mL x 646 mm Hg x 178K
802 mm Hg x 308K
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= 314 mL
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Learning Check
Solution
If 0.75 mole helium gas occupies a
volume of 1.5 L, what volume will
1.2 moles helium occupy at the
same temperature and pressure?
3) 2.4 L
STEP 1 Conditions 1
Conditions 2
V1 = 1.5 L
V2 = ???
n1 = 0.75 mole Hen2 = 1.2 moles He
1) 0.94 L
2) 1.8 L
3) 2.4 L
STEP 2 Solve for unknown V2
V2 = V1 x n2
n1
STEP 3 Substitute values and solve for V2.
V2 = 1.5 L x 1.2 moles He = 2.4 L
0.75 mole He
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► The molar amounts of any two gases with the same
volume are the same at a given T and P.
► Standard temperature and pressure:
(STP) = 0°C (273.15 K) and 1 atm (760 mm Hg)
► Standard molar volume of a gas at STP = 22.4 L/mol
Molar Volume as a Conversion
Factor
The molar volume at STP can be used to
form conversion factors.
22.4 L
1 mole
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Chapter Eight
1 mole
22.4 L
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Using Molar Volume
8.9 The Ideal Gas Law
► Ideal gas law: The relationships among the four
variables P, V, T, and n for gases can be combined
into a single expression called the ideal gas law.
► PV/nT = R (A constant value) or PV = nRT
► If the values of three of the four variables in the
ideal gas law are known, the fourth can be
calculated.
► Values of the gas constant R:
For P in atm:
R = 0.0821 L·atm/mol·K
For P in mm Hg: R = 62.4 L·mm Hg/mol·K
What is the volume occupied by 2.75 moles N2 gas
at STP?
The molar volume is used to convert moles to liters.
2.75 moles N2 x
and
22.4 L = 61.6 L
1 mole
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8.10 Partial Pressure and Dalton’s law
► Dalton’s law: The total pressure exerted by a gas
mixture of (Ptotal) is the sum of the partial pressures
of the components in the mixture.
► Dalton’s law Ptotal = Pgas1 + Pgas2 + Pgas3 + …
► Partial pressure: The contribution of a given gas in
a mixture to the total pressure.
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Chapter Eight
Illustrating Partial Pressures
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Learning Check
For a deep dive, a scuba diver uses a mixture of helium
and oxygen with a pressure of 8.00 atm. If the oxygen
has a partial pressure of 1280 mm Hg, what is the
partial pressure of the helium?
1) 520 mm Hg
2) 2040 mm Hg
3) 4800 mm Hg
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Gases We Breathe
Solution
The air we breathe
• is a gas mixture.
• contains mostly N2
and O2 and small
amounts of other
gases.
3) 4800 mm Hg
PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg
1 atm
PTotal = PO + PHe
2
PHe
-TABL
E 6.4
= PTotal - PO2
PHe = 6080 mm Hg - 1280 mm Hg
= 4800 mm Hg
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Chapter Summary Cont.
Chapter Summary
►According to the kinetic–molecular theory of gases,
the behavior of gases can be explained by assuming
that they consist of particles moving rapidly at
random, separated from other particles by great
distances, and colliding without loss of energy.
►Boyle’s law says that the volume of a fixed amount
of gas at constant temperature is inversely
proportional to its pressure.
►Charles’s law says that the volume of a fixed amount
of gas at constant pressure is directly proportional to
its Kelvin temperature.
►Gay-Lussac’s law says that the pressure of a fixed
amount of gas at constant volume is directly
proportional to its Kelvin temperature.
►Avogadro’s law says that equal volumes of gases at
the same temperature and pressure contain the same
number of moles.
►The four gas laws together give the ideal gas law,
PV = nRT, which relates the effects of temperature,
pressure, volume, and molar amount.
►At 0°C and 1 atm pressure, called standard
temperature and pressure (STP), 1 mol of any gas
occupies a volume of 22.4 L.
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Chapter Summary Cont.
►The pressure exerted by an individual gas in a
mixture is called the partial pressure. Dalton’s law:
the total pressure exerted by a mixture is equal to the
sum of the partial pressures of the individual gases.
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