Spencer L. Seager
Michael R. Slabaugh
www.cengage.com/chemistry/seager
Chapter 6:
The States of Matter
Jennifer P. Harris
PHYSICAL PROPERTIES OF MATTER
• All three states of matter have certain properties that help
distinguish between the states. Four of these properties are
density, shape, compressibility, and thermal expansion.
A KINETIC MOLECULAR VIEW OF SOLIDS,
LIQUIDS, AND GASES
DENSITY
• Density is equal to the mass of a sample divided by the volume
of the same sample.
mass
density 
volume
High density
high – med density
low density
OTHER PHYSICAL PROPERTIES
• SHAPE
• The shape matter takes depends upon the
physical state of the matter.
• COMPRESSIBILITY
• Compressibility is the change in volume of a sample of matter
that results from a pressure change acting on the sample.
• THERMAL EXPANSION
• Thermal expansion is the change in volume of a sample of
matter resulting from a change in the temperature of the
sample.
Effusion and Diffusion
• EFFUSION
• Effusion is a process in which a gas escapes from a
container through a small hole in the container.
• DIFFUSION
• Diffusion is a process that causes gases to
spontaneously mix when they are brought together.
CHARACTERISTIC PROPERTIES OF THE
THREE STATES OF MATTER
KINETIC ENERGY
• Kinetic energy is the energy a particle has as a result of being
in motion.
• Kinetic energy (KE) is calculated using the equation:
mv
KE 
2
2
In this equation, m is the mass of a particle and v is its
velocity.
POTENTIAL ENERGY & FORCES
• POTENTIAL ENERGY
• Potential energy is the energy a particle has as a result of
being attracted to or repelled by other particles.
• COHESIVE FORCE
• A cohesive force is an attractive force between particles. It
is associated with potential energy.
• DISRUPTIVE FORCE
• A disruptive force results from particle motion. It is
associated with kinetic energy.
CHANGES IN STATE
• Changes in state are often accomplished by adding or removing
heat from a substance.
• Changes in state caused by adding heat to a substance are
classified as endothermic (heat in) processes.
• Changes in state caused by removing heat are classified as
exothermic (heat out) processes.
Add Heat
Give off Heat
ENDOTHERMIC PROCESSES
HEAT IS ADDED
• EVAPORATION OR VAPORIZATION
• Evaporation or vaporization is an endothermic process in
which a liquid is changed to a gas.
• SUBLIMATION
• Sublimation is an endothermic process in which a solid is
changed to a gas without first melting to a liquid.
• MELTING OR FUSION
• Melting or fusion is an endothermic process in which a solid
is changed to a liquid.
EXOTHERMIC PROCESSES
HEAT IS REMOVED
• CONDENSATION
• Condensation is an exothermic process in which a gas is
changed to a liquid.
• DEPOSITION
• Deposition is an exothermic process in which a gas is
changed into a solid.
• FREEZING
• Freezing is an exothermic process in which a liquid is
changed into a solid.
VAPOR PRESSURE
Vapor pressure is the pressure exerted by a vapor that is in
equilibrium with its liquid.
High vapor pressure = low boiling point
ENERGY AND THE STATES OF MATTER
• At 760 torr, constant heat is applied until a 1 g sample of ice at
-20°C is converted to steam at 120°C.
• This is a five step process: (AB) heating ice to melting point,
(BC) melting ice, (CD) heating liquid to boiling point, (DE)
boiling water, and (EF) heating steam.
HEATS OF FUSION & VAPORIZATION
• SPECIFIC HEAT
• The specific heat of a substance is the amount of heat
required to raise the temperature of exactly 1 g of a
substance exactly 1°C.
• HEAT OF FUSION
• The heat of fusion of a substance is the amount of heat
required to melt exactly 1g of a solid substance at constant
temperature.
• HEAT OF VAPORIZATION
• The heat of vaporization of a substance is the amount of
heat required to vaporize exactly 1g of a liquid substance at
constant temperature.
GASES
• Intermolecular forces are minimal
• Move independent of each other
• High velocity – move in random motion
• Higher kinetic energy than liquids or solids
• Volume occupied is negligible compared to
volume of container
PRESSURE
• PRESSURE
• Pressure is the force pushing on a unit
area of surface on which the force acts.
• STANDARD ATMOSPHERE OF PRESSURE
• A pressure of one standard atmosphere is
the pressure needed to support a 760-mm
(76.0-cm) column of mercury in a
barometer.
• ONE TORR OF PRESSURE
• One torr of pressure is the pressure
needed to support a 1-mm column of
mercury in a barometer. A pressure of 760
torr is equal to one standard atmosphere.
OFTEN-USED UNITS OF PRESSURE
TEMPERATURE
• The temperature of a gas sample is a measurement of the
average kinetic energy of the gas molecules in the sample.
• The Kelvin temperature scale is used in all gas law
calculations.
• ABSOLUTE ZERO
• A temperature of 0 K
is called absolute zero.
It is the temperature at
which gas molecules
have no kinetic energy
because all motion
stops. On the Celsius
scale, absolute zero is
equal to -273°C.
THE IDEAL GAS LAW
• The ideal gas law allows calculations to be done in which the amount
of gas varies as well as the temperature, pressure, and volume.
• Mathematically, the ideal gas law is written as follows:
PV= nRT
P
Pressure
atm
V
Volume
Liters
n
Number of moles
moles
T
Temperature
Kelvin
R
Universal gas constant
L atm
R = 0.0821
mol K
IDEAL GASES vs. REAL GASES
• No ideal gases actually exist.
• If they did exist, they would behave exactly as predicted by
the gas laws at all temperatures and pressures.
• Real gases deviate from the behavior predicted by the gas
laws, but under normally encountered temperatures and
pressures, the deviations are small.
• Consequently, the gas laws can be used for real gases.
• Interparticle attractions make gases behave less ideally.
• The gas laws work best for gases made up of single atoms
or nonpolar molecules.
AVOGADRO’S LAW
• Avogadro’s law states that equal volumes of gases
measured at the same temperature and pressure contain
equal numbers of molecules.
• STANDARD CONDITIONS
• STP = standard temperature and pressure
• 0°C (273 K)
• 1.00 atm
• MOLAR VOLUME AT STP
• 1 mole of any gas molecules
has a volume of 22.4 L at STP.
IDEAL GAS LAW CALCULATIONS
• Example 1: A 2.50 mole sample of gas is confined in a 6.17 liter
tank at a temperature of 28.0°C. What is the pressure of the gas
in atm?
• Solution: The ideal gas equation is first solved for P:
nRT
P
V
The known quantities are then substituted into the equation,
making sure the units cancel properly to give units of atm in the
answer:
L atm 

2.50 mol  0.0821
301K 
mol K 

P
 10.0 atm
6.17 L 
IDEAL GAS LAW CALCULATIONS
(continued)
• Example 2: A 4.00 g sample of gas is found to exert a pressure of
1.71 atm when confined in a 3.60 L container at a temperature of
27°C. What is the molecular weight of the gas in grams per mole?
• Solution:
• The molecular weight is equal to the sample mass in grams
divided by the number of moles in the sample.
• Because the sample mass is known, the molecular weight can
be determined by calculating the number of moles in the
sample.
• The ideal gas equation is first solved for n:
PV
n
RT
• The known quantities are then substituted into the equation,
making sure units cancel properly to give the units of mol for the
answer.
IDEAL GAS LAW CALCULATIONS
(continued)

1.71 atm 3.60 L 
n
 0.250 mol
L atm 

 0.0821
300 K 

mol K 
• The units are seen to cancel properly to give the number of
moles as the answer. The molecular weight is calculated by
dividing the number of grams in the sample by the number of
moles in the sample:
4.00 g
g
mw 
 16.0
0.250 mol
mol
PRESSURE, TEMPERATURE, & VOLUME
RELATIONSHIPS FOR GASES
• Mathematical equations relating the pressure, temperature, and
volume of gases are called gas laws.
• All of the gas laws are named after the scientists who first
discovered them.
VARIATION OF WATER
BOILING POINT WITH CHANGE IN PRESSURE
PV = nRT
(Assuming V and n are equal)
• at high elevation  P decreases, T decreases
• Pressure cooker  P increases, T increases (cooks faster)
THE COMBINED GAS LAW
• Boyle's law and Charles's law can be combined to give the
combined gas law that is written mathematically as follows:
PV
 k' '
T
In this equation, P, V and T have the same meaning as
before and k'' is another experimentally determined constant.
• The combined gas law can be
expressed in another useful form
where the subscript i refers to an
initial set of conditions and the
subscript f refers to a final set of
conditions for the same gas sample.
Pi Vi Pf Vf

Ti
Tf
GAS LAW EXAMPLE
• A gas sample has a volume of 2.50 liters when it is at a temperature of
30.0°C and a pressure of 1.80 atm. What volume in liters will the sample
have if the pressure is increased to 3.00 atm, and the temperature is
increased to 100°C?
• Solution: The problem can be solved:
• using the combined gas law.
• by identifying the initial and final conditions.
• making sure all like quantities are in the same units.
• expressing the temperatures in Kelvin.
• Thus, we see that the combined gas law must be solved for Vf.
GAS LAW EXAMPLE (continued)
• The result is:
Pi Vi Tf
Vf 
TiPf
• Substitution of appropriate values gives:
Vf

1.80 atm 2.50 liters 373 K 

 1.85 liters
303 K 3.00 atm 
PARTIAL PRESSURE
• The partial pressure of an individual gas of a mixture is the
pressure the gas would exert if it were alone in the container at
the same temperature as the mixture as shown in the following
illustration:
DALTON'S LAW OF PARTIAL PRESSURES
• According to Dalton's law, the total pressure exerted by a
mixture of gases is equal to the sum of the partial pressures of
the gases in the mixture.
Ptotal   Pindiv idual gases
Zn(s) + NH4NO3(s) → N2(g) + 2 H2O(g) + ZnO(s)
P total = PN2 + PH2O