Phases of Matter
Intro
Solid – State of matter with definite shape and
definite volume
Liquid – State of matter with definite volume,
but not definite shape
Gas – State of matter with no definite shape
and no definite volume
Intro
Plasma – State of matter in which atoms have
been stripped of all of their electrons
Plasmas can only occur at very high energies
It is estimated that 99% of matter in the
universe is plasma
Kinetic Theory of Matter
All particles of matter are in
constant motion
There are forces of
attraction between all atoms
Behavior of Gases
Gas particles behave like billiard balls
They will continue in a straight line until they
hit the side of their container, or another gas
particle
Air particles move at approximately 1000mph at
room temperature
Weak force of attraction between particles
Behavior of Gases
Behavior of Liquids
Water also has kinetic energy at room temperature
So, why aren’t water molecules flying around and
colliding with each other and the walls?
Water molecules have much less kinetic energy and
much stronger forces of attraction
The same is true for all liquids at room
temperature
Behavior of Liquids
Behavior of Liquids
Behaviors of Solids
Solids have low velocity and strong
forces of attraction at room
temperature
This is what gives them their
definite shape and definite volume
Behaviors of Solids
All Three Phases
Phase Changes
Freezing – Liquid to Solid
Melting – Solid to Liquid
Phase Changes
Vaporization – Liquid to Gas
Condensation – Gas to Liquid
Phase Changes
Boiling and Evaporation are both types of
Vaporization
Boiling – When a liquid is heated to its boiling point
the atoms at the bottom of the liquid are moving
fast enough to overcome the forces of attraction
Evaporation – Occurs at temperatures below the
boiling point and occurs at the surface of a liquid
Phase Changes
Phase Changes
Sublimation – Solid to Gas
Deposition – Gas to Solid
Phase Changes
Ionization – Gas to Plasma
Recombination – Plasma to Gas
Phases Changes
Plasma
Gas
Forces of Attraction
Kinetic Energy
and
Temperature
Deposition
Liquid
Sublimation
Solid
Heating Curves
Heating Curve – Graph that shows how much heat it takes
a substance to increase in temperature and change phases
How we calculate heat
q=mcΔT
q-heat (Joules or J)
m-mass (grams)
c-specific heat
ΔT-change in temperature (C)
Heating Curves
Review Problem
For this unit we will always use water. 5g of water is
heated up from 30°C to 50°C. The specific heat of water
is 4.186. How much heat did the water gain?
m=5g
c=4.186
ΔT=50-30=20°C
q=mcΔT
q=(5)(4.186)(20)=418.6J
Heating Curves
When a substance goes through a phase
change, the temperature of the
substance does not increase until every
molecule has changed phases
This is what heating curves show us
Heating Curves
Heating Curves
To calculate how much heat a phase change
needs we use the latent heat of fusion equation
q=mLf
Latent heat of fusion for water is 334J/g
Remember: you DO NOT need to memorize any
equations or constants
Heating Curves
Practice
How much heat would it take to convert 5g of
ice into 5g of water?
m=5g
Lf=334J/g
q=mLf
q=(5)(334)=1670J
Heating Curves
Water freezes at 0°C
Water vaporizes at 100°C
You do need to memorize these, but
they’re not hard to remember
Heating Curves
Practice
How much heat does it take to heat 10g of ice
from -20°C to 50°C?
BREAK THIS PROBLEM INTO 3 STEPS
Step 1: How much heat does it take to get 10g of
ice from -20°C to 0°C?
m=10g
c=4.186
ΔT=20°C
q=mcΔT
q=(10)(4.186)(20)
q=837.2J
Heating Curves
Step 2:How much heat does it take to change
10g of ice into 10g of water?
m=10g
Lf=334J/g
q=mLf
q=(10)(334)
q=3340J
Heating Curves
Step 3: How much heat does it take to heat up
10g of water from 0°C to 50°C?
m=10g
c=4.186
ΔT=50°C
q=mcΔT
q=(10)(4.186)(50)=2093J
Heating Curves
Add the heat from each step to get your
final answer
837.2J + 3340J + 2093J
It takes 6720.2J to heat 10g of ice from
-20°C to 50°C
Gas Laws
Pressure – Result of a Force distributed over an area
SI units – Pascals (Pa)
Equation – P=F/A
Force is measured in Newtons (N)
Area is measured in meters squared (m2)
Gas Laws
Practice
A force of 20N is applied to the top of a
container that has an area of 2m2. What is the
pressure applied to that container?
F=20N
A=2m2
P=F/A
P=20/2=10Pa
Gas Laws
Gas Pressure – The force exerted by a gas on the walls of its
container
Factors that affect gas pressure
1)Number of molecules
2)Temperature
3)Volume
Gas Laws
The more gas molecules in a container, the
higher gas pressure there will be
You would say that the number of gas
molecules in a container is directly
related to the gas pressure
As the number of molecules increases
the gas pressure also increases
Gas Laws
Changing the volume of a container will also change the gas
pressure in a container
As you decrease the volume the
gas pressure will increase
Volume and gas pressure are
inversely related
As one goes up the other goes down
Gas Laws
Changing the temperature in a container will change the
gas pressure of the container
When temperature increases the molecules move faster
and collide with the walls of the container more
How are temperature and gas pressure related
Directly Related
As one goes up, the other also goes up
Gas Laws
Charle’s Law shows the relationship between volume and
temperature
𝑉1 𝑉2
=
𝑇1 𝑇2
If the volume of a container increases, the temperature of
the container must also increase
How are temperature and volume related?
Directly Related
Gas Laws
Boyle’s Law shows the relationship between pressure and
volume
𝑃1 𝑉1 = 𝑃2 𝑉2
If the pressure of the container increases, the volume of
the container must decrease and vice versa
How are pressure and volume related?
Inversely Related
Gas Laws
SI Units
Pressure – Pascals (Pa)
Volume – Liters (L)
Temperature – Kelvin (K)
Gas Laws
Practice
A container of gas initially as a volume of 5L and a pressure of
10Pa. The volume of the container then increases to 10L.
What is the final pressure of the container?
V1=5L
P1=10Pa
V2=10L
P2=?
P1V1=P2V2
(5)(10)=P2(10)
50=P2(10)
5=P2
P2=5Pa
Gas Laws
Practice
A container of gas initially has a volume of 5L and a temperature
of 100K. Then, the temperature is increased to 300K. What is
the final volume of the gas?
V1=5L
T1=100K
T2=300K
V2=?
V1/T1=V2/T2
5/100=V2/300
0.05=V2/300
V2=15
V2=15L
Gas Laws
𝑉1 𝑉2
=
𝑇1 𝑇2
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
𝑃1 𝑉1 = 𝑃2 𝑉2
Combined Gas Law
Gas Laws
Practice
A container of gas initially has a volume of 10L, a temperature of
150K, and a pressure of 50Pa. Then the volume is increased to
20L and the temperature is increased to 300K. What is the final
gas pressure of the container?
V1=10L
T1=150K
P1=50Pa
V2=20L
T2=300K
P2=?
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2
50 10
𝑃2 20
=
150
300
3.33 = 0.06𝑃2
𝑃2 = 55.5𝑃𝑎