States of Matter This unit mostly contains material from Chapter 10

1

• States of Matter (This unit mostly contains
  material from Chapter 10)
• Kinetic Molecular Theory pg. 329-332
• 1. An ideal gas particle occupies ____ volume.
• 2. An ideal gas has perfectly ___________collision
  s (no losses in energy).
• 3. Ideal gas particles are in a ___________ state
  of motion.
• 4. Ideal gas particles exhibit _____
  intermolecular forces for each other.
• 5. Temperature is a measure of the _____________
  kinetic energy of the particles of the gas.
• The larger and more polar the gas particle, the
  further it will deviate from an ideal gas.
• The gas pressure inside of a container is a
  result of molecules being in motion, this will
  cause collisions with the walls of the container
  to put a _______ on the walls of the container.
  P F/A
• Diffusion and Effusion
• 1. Diffusion the ___________of gas particles
  caused by their random motion (see fig. 2 pg.
  331).
• 2. Effusion the process by which gas particles
  pass through a ____________ opening.
• The heavier the gas, the slower it will travel.
  
• ______
• v1 / v2 vM2 / M1 pg. 387
• Sample problem J pg. 388 practice problems 2,3
  pg. 388

no
elastic
constant
no
average
force
mixing
narrow
_______
VH2
?
VH2
3.98
MO2


VO2
VO2
MH2
___________
OR
VH2
?
2(16.00 g/mol)

VH2 3.98 VO2
VO2
2(1.01 g/mol)
This means H2 effuses 3.98 times faster than O2
2

• 2. VH2 9.0
• Vu

Homework pg. 353 1-6a pg 392-3 55 57 58
Bonus pg. 393 67
3. VB VNe
VB VNe
_______
?
20.18 g/mol
VB 400. m/s
58.14 g/mol
VB 236 m/s
3

• Liquids pg. 333-336
• 1. A liquid has a definite volume and assumes the
  _________ of its container.
• 2. Liquid particles are _________ together than
  gaseous particles.
• 3. Particles within a liquid have mobility and
  thus have a flow. A substance which can _______
  and assume the shape of the container they are
  placed in are called fluids (gases are also
  fluids, solids are not).
• 4. Liquids are relatively incompressible and are
  nearly as ________ as solids.
• Changes of State pg. 333-336 342-344
• 1. Solid particles have relatively _______
  positions relative to one another and therefore
  do not assume the shape of their container.
• 2. Liquids have a definite freezing point
  (temperature where the liquid converts to
  _________).
• 3. Freezing point is the same as the melting
  point. The temperature at which the solid and
  liquid are at equilibrium at standard temperature
  and pressure.
• 4. Sublimation is the conversion of a solid
  directly to a __________ (dry ice, iodine
  crystals).
• 5. Evaporation-process where particles at the
  surface of a nonboiling liquid enter the gas
  phase. The particles which escape have a ______
  kinetic energy (energy of motion). The greater
  the temperature, the ________ particles there are
  with enough KE to escape into the gaseous phase.
  Fig. 15 pg. 345. Substances which readily
  evaporate are said to be _________.
• The evaporation of a liquid will cause the liquid
  to __________ (examples sweating to cool
  yourself)
• Vapor pressure is the pressure of the vapor above
  a liquid when the rates of evaporation and
  condensation are __________(called equilibrium).
  Fig 13. pg. 343.

shape
closer
flow
dense
fixed
solid
0oC, 1 atm
gas
high
more
volatile
cool
equal
4

• 6. Boiling the conversion of a liquid to a vapor
  within the liquid as well as at the surface.
  Boiling will occur when the vapor pressure of the
  liquid is ______ to atmospheric pressure. The
  lower the atmospheric pressure the lower the
  boiling point (or the higher the atmospheric
  pressure, the higher the boiling point). So
  boiling will occur at __________ temperatures at
  higher altitudes.
• 7. Condensation is the process by which a ______
  converts to a liquid.
• Solid energy ? liquid (melting) Liquid ?
  solid energy (freezing )
• liquid energy ? gas (boiling) gas? liquid
  energy (condensation)
• Solid energy ? gas (sublimation) gas? solid
  energy (deposition)
• Gas pressure ? liquid (liquefaction) solid
  pressure ? liquid (regelation)
• Phase Diagrams a graph of P vs. T which shows the
  different phases that a certain material exists.
  (See Resource Disc)
• See Fig. 16. pg. 347 Note the difference between
  this figure and the one on page 354.
• Notice that if increase the pressure on ice, you
  can liquefy it. This can not be done with
  substances like CO2. This is because water has
  more volume as a solid than a liquid at
  temperatures near freezing (examples ice skating
  and making snowballs).
• Phase Diagram Features
• Normal Freezing and Boiling Points are measured
  at ___ atm of pressure.
• Triple Point Where a substance exists in __
  states (solid, liquid and gas)
• Critical Temperature (Tc) the temperature
  _________ which, no matter how great the pressure
  applied, cannot liquefy a gas. Note the
  ___________ the intermolecular forces the lower
  the critical temperature.
• Critical pressure (Pc) the pressure required to
  ____________ a gas at the critical temperature.

equal
lower
gas
1
3
above
weaker
liquefy
5
higher

• The stronger the intermolecular force the _______
  the freezing and boiling points and critical
  temperature.
• The stronger the intermolecular force the
  ___________ the vapor pressure.
• Enthalpy of Fusion and Vaporization
• Enthalpy of fusion (?H fus) energy to ________1
  gram of a substance at its normal freezing point
  (334 J/g for water given as 6.009 kJ/mol in
  book). This is also the energy released in
  freezing 1 gram of a substance.
• Enthalpy of vaporization (?Hvap ) energy to
  ____________1 gram of a substance at its normal
  boiling point (2260 J/g for water given as 40.79
  kJ/mol in book). This is also the energy released
  in condensing 1 gram of a substance.
• Sample Problem A pg. 351 practice problem 1, 2
  pg. 351

lower
melt
vaporize
334 J g
/ /
47.0 g
15,700 J
H2O (s) ?H fus ? H2O (l)
?J
2260 J g
?J
47.0 g
/ /
106,000 J
H2O (l) ?H vap ? H2O (g)
334 J g
/ /
1.
?J
506 g
169,000 J
H2O (l) ? H2O (s) ?H fus
_g_ 2260 J
1000J kJ
/ /
2.
?g
4.97.105 kJ
/ /
219,000 g
H2O (g) ? H2O (l) ?Hvap
6

• Homework pg. 353 13a 14-17 pg. 354 28 30
  35 Also Looking at the CO2 phase diagram fig. on
  pg. 354. State the phases that exists for the
  following
• a. -50oC , 2 atm b. -50oC , 8 atm c. -50oC ,
  12 atm

7
Ch 16 Thermochemistry Calorimetry
Measurements of ___________ absorbed or released
by chemical or physical change. Temperature A
measure of the avg. ______ of particles in a
sample of matter. Heat the energy transferred
between samples of matter because of a difference
in ________________. Specific Heat amount of
energy required to raise the temperature of __ g
of substance ___oC (or 1 K). a. For water this
is ________ J/goC (1 calorie/goC). b. Metals
(good conductors) have ________ specific heats
than water as seen that they heat up easier than
water. See table 1 pg. 533. c. Solving for
heats during a chemical or physical change q
mCp ?t q heat (J) m mass (g) Cp specific
heat (J/goC) ?t change in temperature ?t (tf
ti) When a substance is cooled ?t is negative so
heat will be negative. Here we are interested in
the magnitude (size) of the heat with no regard
to sign so we will treat ?t as a value.
energy or heat
KE
temperature
1 1
4.184
lower
Sample A pg. 533
b. q m Cp ?t
a. Cp q / m ?t
q (4.0 g)(0.2000 J/goC )(344K 314K)
Cp 32 J / (4.0g(40.K))
Cp 0.20 J/gK
q (4.0 g)(0.2000 J/goC )(30. K)
q 24 J
8

• A 41.65 g sample of a metal at 100.0oC is placed
  in 125 g of water at 25.0oC. The final
  temperature of the water is 30.0oC. Determine
  the specific heat and identity of the metal.

Here the metal will lose heat and the water will
gain heat.
The quantity lost will be equal to the quantity
gained (opposite sign) or qlost - qgained
This negative sign will make ?t a positive value
for the cooling metal
41.65 g, 100.0oC
(mCp?t)metal (mCp?t)water
(mCp?t)metal (mCp?t)water
Cpmetal (mCp?t)water
(m?t)metal
Cpmetal 125 g(4.184 J/ goC)(30.0oC
25.0oC) 41.65 g (100.0oC 30.0oC)
125 g 25.0oC
Cpmetal 125 g(4.184 J/ goC)(5.0oC)
41.65 g (70.0oC)
30.0oC
Cpmetal 0.90 J/goC

The metal is Aluminum see table 1 pg. 533
9

• Homework pg. 552 s 7 9 pg. 557 3
• A-1 A 25 g sample of metal (initially at
  800.00oC) is dropped into 200. g of water
  (initially at 30.00oC). The final temperature of
  the system is 40.22oC. Find the specific heat
  and identify the metal.
• A-2 A piece of iron at 75.0oC is dropped into
  50.0 g of water at 20.0oC. The final temperature
  of
• the water is 28.3oC. Determine the mass
  of the piece of iron.
• Bonus A 1.0 kg sample of metal with a specific
  heat of 0.50 J/goC is heated to 100.0 oC and then
  placed in a  50.0 g sample of water at 20.0 oC. 
  What is the final temperature of the metal and
  the water?

10

• Solving Problems involving Changes of State
• Specific heat (Cp) - the energy required to raise
  the temperature of 1 gram of a substance 1 oC.
• Cp (ice) 2.06 J/goC Cp (water) 4.184
  J/goC Cp (steam) 2.02 J/goC
• ?Hfus 334 J/g ?Hvap 2260 J/g
• Equation to use when there is no change of state
  (for phase changes use factor-label)
• q mCP?t
• q heat (J) m mass (g) Cp specific heat
  (J/goC) ?t change in temperature

11

• Sample Problem 1 How much heat is required to
  raise the temperature of 25.0 g of ice at 0.0oC
  to water at 25.0oC?

\
q2
q1
q125.0 g
334 J g
/ /
qtot q1 q2
8350 J
qtot 8350 J 2615 J
q2mCp?t
qtot 10965J
q2(25.0g)(4.184 J/goC)(25.0oC- 0oC )
qtot 10970J or 1.097.104 J
q2(25.0g)(4.184 J/goC)(25.0oC)
q22615 J
12
q1
q3
q2

• Problem 2 How much heat is required to raise
  the temperature of 11.5 g of H2O at -25oC to
  water at 15oC?

q1mCp?t
q1(11.5g)(2.06J/goC)(0 (-25oC))
qtot q1 q2 q3
q1(11.5g)(2.06J/goC)(25oC)
qtot 592.25 J 3841 J 721.74 J
q1592.25 J
/ /
q2 11.5 g
334 J g
qtot 5154.99 J
3841 J
qtot 5150 J or 5.15.103 J
q3mCp?t
q3(11.5g)(4.184J/goC)(15oC- 0oC)
q3(11.5g)(4.184J/goC)(15oC)
q3 721.74 J
13
q1
q3
q2
Problem 3 How much energy is required to
vaporize 225 g of water at 20.0oC to steam at
150.0oC?
q1mCp?t
qtot q1 q2 q3
q1(225g)(4.184J/goC)(100.0oC-20.0oC )
qtot 75312 J 508500 J 22725 J
q1(225g)(4.184J/goC)(80.0oC)
q1 75312 J
qtot 606537J
2260 J g
/ /
q2 225 g
508500 J
qtot 607,000J or 6.07.105 J
q3mCp?t
q3(225g)(2.02J/goC)(150.0oC- 100.0oC)
q3(225g)(2.02J/goC)(50.0oC)
q3 22725 J
14
q1
q3
q2
Problem 3 with a twist 805,000J of energy is
required to vaporize an unknown mass of water at
20.0oC to steam at 150.0oC, what is the mass?
q1mCp?t
qtot q1 q2 q3
q1(m)(4.184J/goC)(100.0oC-20.0oC )
805,000 J 334.72 m J/g 2260 m J/g 101.00 m
J/g
q1(m)(4.184J/goC)(80.0oC)
q1 334.72 m J/g
805,000 J 2695.72 m J/g
2260 J g
q2 m
2260 m J/g
m 805,000 J/ 2695.72 J/g
q3mCp?t
m 299 g
q3m(2.02J/goC)(150.0oC- 100.0oC)
q3 m(2.02J/goC)(50.0oC)
q3 101.00 m J/g
15

• Homework Day 1 pg. 351 section review 4, 5,
  7 (read pg. 350 for help)
• 1. How much heat is required to heat 15.0 g of
  ice from -15o C to water at 0.oC?
• 2. How much heat is required to heat 15.0 g of
  water at 0oC to steam at 120.oC?
• 3. How much heat is required to heat 15.0 g of
  ice at -15oC to steam at 120. oC?
• (hint see homework problems 1 and 2)
• Bonus
• If 15 kJ of heat is added to ice at -10.oC and
  the ice melts to water at 25oC. What is the mass
  of the ice?
• Day 2 1. What is the name given to liquefying a
  gas with application of pressure? What is the
  name of liquefying a solid with application of
  pressure?
• 2. Substances with small IM forces have ____
  vapor pressures and are said to be ________.
• 3a. What does temperature measure? b. How does
  the diffusion rate depend upon mass?
• 4. Utilize fig. 16 pg. 347 to estimate the
  critical temperature and pressure of water.
• 5. Determine the heat required to raise the
  temperature of 50.0 g of ice at 0.0oC to water at
  45.0oC.
• Pg. 357 2 6 7 pg. 348 7 pg. 393 70 (also if
  CO diffuses at 0.50 m/s, what will be the
  diffusion rate of SO3?)