Transcript chapter6

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.
DENSITY
• Density is equal to the mass of a sample divided by the volume
of the same sample.
mass
density =
volume
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.
CHARACTERISTIC PROPERTIES OF THE
THREE STATES OF MATTER
KINETIC MOLECULAR THEORY OF MATTER
• The kinetic molecular theory of matter is a useful tool for
explaining the observed properties of matter in the three
different states of solid, liquid and gas.
• Postulate 1: Matter is made up of tiny particles called
molecules.
• Postulate 2: The particles of matter are in constant motion
and therefore possess kinetic energy.
• Postulate 3: The particles possess potential energy as a
result of repelling or attracting each other.
• Postulate 4: The average particle speed increases as the
temperature increases.
• Postulate 5: The particles transfer energy from one to
another during collisions in which no net energy is lost from
the system.
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.
THE SOLID STATE
• The solid state is characterized by a high density, a definite
shape that is independent of its container, a small
compressibility, and a very small thermal expansion.
THE LIQUID STATE
• The liquid state is characterized by a high density, an indefinite
shape that depends on the shape of its container, a small
compressibility, and a small thermal expansion.
THE GASEOUS STATE
• The gaseous state is characterized by a low density, an
indefinite shape that depends on the shape of its container, a
large compressibility, and a moderate thermal expansion.
A KINETIC MOLECULAR VIEW OF SOLIDS,
LIQUIDS, AND GASES
THE GAS LAWS
• The gas laws are
mathematical equations
that describe the behavior
of gases as they are
mixed, subjected to
pressure or temperature
changes, or allowed to
diffuse.
• The pressure exerted on
or by a gas sample and the
temperature of the sample
are important quantities in
gas law calculations.
PRESSURE
• PRESSURE
• Pressure is defined as a force pushing on a unit area of
surface on which the force acts.
• In gas law calculations, pressure is often expressed in units
related to the measurement of atmospheric pressure.
• 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 1mm 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.
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.
BOYLE'S LAW
• Boyle's law is a gas law that describes the pressure and volume
behavior of a gas sample that is maintained at constant
temperature.
• Mathematically, Boyle's law is written as follows:
k
or
P=
PV = k
v
In these equations, P is the pressure, V is the volume, and k
is an experimentally determined constant.
CHARLES'S LAW
• Charles's law is a gas law that describes the temperature and
volume behavior of a gas sample that is maintained at constant
pressure.
• Mathematically, Charles's law is written as follows:
V = k' T
or
V
= k'
T
In these equations, V is the volume, T is the temperature in
Kelvin, and k' is an experimentally determined constant.
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
THE COMBINED GAS LAW
• Crushing Soda Can Demonstration
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)
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.
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
In this equation, P is the pressure of a gas sample, V is the
sample volume, T is the sample temperature in Kelvin, n is
the number of moles of gas in the sample, and R is a
constant called the universal gas constant. A commonlyused value for R is:
L atm
0.0821
mol K
• In calculations, the quantities V, P, and T must be expressed in units
that match the units of R, liters (L), atm, and Kelvin, respectively.
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 ö
æ
(2.50 mol)ç 0.0821
÷(301K )
mol K ø
è
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)
n=
(1.71 atm)(3.60 L )
L atm ö
æ
ç 0.0821
÷(300 K )
mol K ø
è
= 0.250 mol
• 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
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.
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 = å Pindividual gases
Zn(s) + NH4NO3(s) → N2(g) + 2 H2O(g) + ZnO(s)
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:
GRAHAM'S LAW
• Graham's law is a mathematical
expression that relates the rates of
effusion or diffusion of two gases to
the masses of the molecules of the two
gases.
• 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.
MATHEMATICAL EXPRESSION OF
GRAHAM'S LAW
effusion rate of A
molecular mass of B
=
effusion rate of B
molecular mass of A
GRAHAM'S LAW CALCULATION EXAMPLE
• Compare the rates of effusion or diffusion for neon and krypton
gases.
• Solution: The molecular masses of neon and krypton are 20.18 u
and 83.80 u, respectively. These are the molecular weights of
the gases from the periodic table. Substitution into the Graham's
law equation gives the following:
rate Ne
83.80 u
=
= 4.153 = 2.038
rate Kr
20.18 u
Thus, the rate of Ne = (2.038) rate of Kr. Stated another way,
neon gas effuses or diffuses about twice as fast as krypton gas.
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.
ENDOTHERMIC PROCESSES
• 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
• LIQUEFACTION OR CONDENSATION
• Liquefaction or condensation is an
exothermic process in which a gas is
changed to a liquid.
• DEPOSITION OR CONDENSATION
• Deposition or condensation is an
exothermic process in which a gas is
changed into a solid.
• FREEZING OR CRYSTALLIZATION
• Freezing or crystallization 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.
BOILING POINT
• The boiling point of a liquid is the temperature at which the
vapor pressure of the liquid is equal to the prevailing
atmospheric pressure.
• The normal or standard boiling point is the temperature at
which the vapor pressure of a liquid is equal to 1 standard
atmosphere (760 torr).
VARIATION OF WATER
BOILING POINT WITH ELEVATION
SUBLIMATION AND MELTING
• Sublimation is the endothermic process in which a solid is
changed directly to a gas without first becoming a liquid.
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.
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.
HEATS OF FUSION & VAPORIZATION
• 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.