AP Chemistry 4: States of Matter
A.
Gas State (10.2 to 10.9)
1. tend to be small, non-polar molecules
2. form homogenous mixtures
3. distributed throughout the entire container
(molecules typically occupy 0.1 % of the volume)
4. kinetic theory for gases (ideal gas)
a. molecules in continuous, chaotic motion, which is
proportional to temperature
1. kinetic energy, Kmole = 1/2(MM)v2 = 3/2RT
a. R = 8.31 J/mol•K
b. MM (molar mass) in kg
c. T in kelvin (TK = ToC + 273)
2. root-mean-square speed, v = (3RT/MM)½
3. Graham’s law
a. rate r of effusion (leaking) or diffusion
(spreading out) is proportional to speed
b. rA/rB = (MMB/MMA)½
c. time is inversely proportional to rate 
TB/TA = rA/rB
b. molecular volume is insignificant compared to
container volume (approximation—see real gas)
c. collisions produce pressure w/o loss of total
kinetic energy
d. Bonding between molecules is insignificant
(approximation—see real gas)
5. gas laws
a. Ideal gas Law equation: PV = nRT
1. molecules generate pressure via collisions
a. pressure = force/area
b. 1 atm = 101 kPa = 760 mm Hg (torr)
c. measuring tools
1. barometer: atmospheric pressure
2. manometer: enclosed gas pressure
Pgas = Patm ± h (in mm Hg)
2. gas pressure is affected by:
a. n: each molecule exerts pressure  more
molecules exert more pressure: P  n
b. T: hotter molecules move faster and
collide with greater force  generate
more pressure: P  T
c. V: molecules spread out which reduces
collision per surface area  generate
less pressure: P  1/V
3. ideal gas law constant R (V in L, T in K)
a. 8.31 J/mol•K (P in kPa)
b. 0.0821 atm•L/mol•K (P in atm)
4. molar volume at STP = 22.4 L/mol
(standard T = 0oC, standard P = 1 atm)
5. derived equations
a. P1V1/T1 = P2V2/T2
b. MM = mRT/PV = dRT/P
b. Dalton’s law (P  n)
1. Ptot = PA + PB
2. PA = XAPtot , where XA = molA/(molA + molB)
6. real gases
a. Van der Waals:(Preal + n2a/V2)(Vreal – nb) = nRT
b. "a" corrects for molecular bonding
1. "low" temperature (close to boiling point)
molecules clump and collide less often, which
generates less pressure  Preal < Pideal
2. a is proportional to molecular polarity
c. "b" corrects for molecular volume
1. high pressure is generated by crowded
molecules where the volume of empty space
(Videal) is significantly less than 100 % of the
total volume (Vreal) Vreal > Videal
2. b is proportional to molar mass
Name __________________________
B.
Phase Change (11.1 to 11.2, 11.4 to 11.6)
1. molecular-level comparison of gas, liquid, and solid
states of a substance
Characteristic
Gas
Liquid
Solid
Energy
Highest
Middle
Lowest
Disorder
Greatest
Middle
Least
Occupied space
Whole
Bottom
Own
Compressibility
Yes
No
No
Flow rate
Fast
Middle
No
Diffusion
Rapid
Slow
No
2. cohesive forces (van der Waals forces)
a. attraction between molecules
(covalent bonds hold atoms together in molecule)
b. dipole-dipole forces
1. polar molecules
2.  of one molecule attracts  of a neighbor
3. strength  to polarity, if all else is even
c. London dispersion forces
1. attraction between nuclei of one molecule's
atoms for the electrons in a neighboring
molecule causes temporary polarization
throughout the liquid or solid (polarizability)
2. generalization
a. operates between all molecules (stronger
than dipole-dipole for large molecules, i.e.
large nonpolar > small polar)
b. only force for nonpolar (strength  to
mass: Xe > He, I2 > F2, C3H8 > CH4)
d. hydrogen bonding
1. super strong dipole-dipole force (stronger than
dispersion forces)
2. H bonded to N, O or F
a. H  +1 charge and N, O or F  –1 charge
because of extreme electronegative
difference and small radius
b. bonding is ionic like (E  Q1Q2/d)
3. explains unusual properties of water
a. each water molecule bonds to 4, which
makes a 3-d structure with open cavities
b. high melting and boiling temperatures
c. low vapor pressure (low volatility)
3. cooling profile for water from 110oC to -10oC
A
B
C
100oC
D
E
0 oC
a.
b.
F
Heat Removed  (J)
slope C-D < slope E-F  more heat is
removed when one mole H2O(l) is cooled 1oC
compared to one mole H2O(s)
length B-C > length D-E  more heat is
removed when one mole H2O(g)  H2O(l)
compared to one mole H2O(l)  H2O(s)
c. calculations
Process
Formula
Condensation
B-C
Q = nHvap
(100oC)
Step
C-D Cooling liquid
Freezing
(0oC)
D-E
E  Cooling solid
Hvap = 40.7 kJ/mol
Q = nClT
Cl = 75.3 J/mol•K
Q = nHfus
Hfus = 6.01 kJ/mol
Q = nCsT
Cs = 37.8 J/mol•K
phase diagram
Pressure
4.
Constants
Temperature
point A: triple point (three phases at equilibrium)
1. below triple point: sublimation
2. above triple point: melting and vaporization
b. line A-B: equilibrium vapor-pressure curve for liquid
(normal boiling point occurs at 1 atm pressure)
c. B: critical point, where there is no distinction
between liquid and vapor (no liquid-vapor surface)
d. line A-C: equilibrium vapor-pressure curve for solid
e. line A-D: melting point of solid at various pressures
(normal freezing point occurs at 1 atm)
1. positive slope when solid is the densest
phase (melting point increases with pressure)
2. negative slope when liquid is the densest
phase (melting point decrease with pressure)
vapor
a. some surface molecules in the condensed phase
have enough kinetic energy (speed) to escape
surface (evaporate) below boiling point
b. as temperature increases more molecules have
sufficient kinetic energy  more vapor molecules
a.
5.
c.
cooling process (hottest evaporate first, leaving
cooler molecules behind)
d.
C.
equilibrium between liquid and vapor
1. evaporation rate = condensation rate in a
closed container
2. concentration of vapor measured as Pvap
3. independent of container size until no liquid
4. Pvap increases at higher temperature because
a. more molecules are in vapor phase
b. vapor molecules exert greater pressure
e. boiling occurs when Pvap = Patm  boiling point
decreases with elevation (lower air pressure, Patm)
f. high Pvap indicates volatility—tendency to evaporate
Crystalline Solids (11.8)
1. ions, atoms or molecules fit into a regular geometric
pattern (crystal lattice)
2. minimum energy state—maximum bond energy
3. intermolecular forces (attraction between + and -) or
bonds (covalent, ionic or metallic) hold particles together
4. 4 types of solids
a. metallic—metals only
1. attraction between cations and delocalized
valence electrons (electron sea model)
b.
c.
d.
2. melting point: variable ( bond strength)
3. conductivity: free electrons  yes
4. malleable: non-directional bond  yes
5. water solubility: no molecular interactions  no
6. examples: Cu, Ag, etc.
covalent network—nonmetals w/o H or halogen
1. atoms covalently bond throughout w/o size
limit (different than large molecule)
2. melting point: strong bonds  high
3. conductivity: no free electrons  no
4. malleability: bond highly directional  brittle
5. water solubility: no molecular interactions  no
6. three examples
a. diamond and graphite—C
1. allotropes (2 forms in the same state)
2. diamond: covalently bonded 3-d
structure—good abrasive
3. graphite: covalent bonded planar
sheets linked by dispersion forces
a. separate easily—good lubricant
b. electron flow—good conductivity
b. quartz—SiO2
1. 3-d structure similar to diamond
2. softens when heated until liquid
3. fast cooling = non-crystalline glass
molecular—nonmetals often with H and/or halogen
1. attraction between + of one molecule with
– of another
2. melting point: weak bonds  low
3. conductivity: no free electrons  no
4. malleability: non-directional  yes
(H-bonding in H2O(s) is somewhat directional)
5. water solubility ("like dissolves like")  yes/no
6. examples: H2O, C6H12O6, etc.
Ionic—metal plus nonmetals
1. attraction between cations and anions
2. melting point: strong bonds 
3. conductivity: no free electrons  no
(fused or dissolved state is conducting)
4. malleability: bond highly directional  brittle
5. water solubility: ion-dipole interaction  yes
6. examples: NaCl, CaCO3, etc.
1.
2.
3.
Gas State
What features of the kinetic theory of gases
a. describe all gas molecules?
Molecules are in continuous, chaotic motion.
Collisions produce pressure w/o loss of total energy.
b. describe ideal gas molecules only?
Molecules have zero volume.
Molecules don't interact with each other.
Consider one mole of Ne gas at 274 K. Determine
a. the total kinetic energy.
Kmole = 3/2RT
Kmole = 3/2(8.31 J/mol•K)274 K = 3420 J
b. the average speed.
u = (3RT/MM)½
u = [(3)(8.31)(274)/(20.2/1000)]½  u = 582 m/s
Consider the drawing below.
7.
a.
8.
25oC,
a.
4.
5.
6.
A and B are He and O2, at
which is which?
Explain
A is O2 and B is He. O2 is heavier and would have
slower speeds compared to He. u = (3RT/MM)½
b. A and B are at 100 K and 200K, which is which?
Explain
B is the higher temperature. Hotter gas has greater
speed than cooler gas. u = (3RT/MM)½
a. What alkane effuses at 1/5 the rate of He?
rA/rB = (MMB/MMA)½
5/1 = (MMB/4)½  MMB = 100 g mol-1  (C7H16)
b. How many times faster does C2H2 diffuse compared to
the alkane?
TB/TA = (MMB/MMA)½
TB/TA = (100/26)½ = 2 times faster
Determine the pressure of 1.22 atm in the following units.
mm Hg
kPa
torr
1.22 atm x 760 = 1.22 atm x 101 = 1.22 atm x 760 =
927 mm Hg
123 kPa
927 torr
If the atmospheric pressure is 749 mm Hg, what is the
pressure of the enclosed gas in each case below?
749 mm Hg
–520 mm Hg
229 mm Hg
749 mm Hg
+67 mm Hg
816 mm Hg
103 mm Hg
A gas is confined inside a container with a movable piston
held down by a fixed pressure.
9.
What affect would doubling the number of gas
molecules at the same temperature have on the
system? Explain.
The piston would move upward to double the volume
because the greater number of molecules exerts twice
the pressure.
b. What affect would doubling the Kelvin temperature of
the gas have on the system? Explain.
The piston would move upward to double the volume
because the faster moving molecules exert twice the
pressure.
c. What affect would doubling pressure by the piston at
the same temperature have on the system? Explain.
The piston would move downward to halve the volume
so that molecules are more crowded in order to
produce enough collisions to match the pressure.
Complete the following table for an ideal gas:
P
V
n
T
2.00 atm
1.00 L
1.500 mol
16.2 K
30.3 kPa
1.250 L
0.0152 mol
27oC
650 torr
11.2 L
0.333 mol
350 K
10.4 atm
585 mL
0.250 mol
295 K
Oxygen gas in a 10.0-L container has a pressure of 94.6 kPa
and temperature of 25oC.
a. How many moles of oxygen gas are in the container?
PV = nRT
(94.6)(10.0) = n(8.31)(25 + 273)  n = 0.382 mol
b. How many grams of oxygen gas are in the container?
0.382 mol O2 x 32.0 g/1 mol = 12.2 g
10. A sample of gas occupies 350 mL at 15oC and 750 torr.
What temperature will the gas have at the same pressure if
its volume increases to 450 mL?
P1V1/T1 = P2V2/T2
(350)/(15 + 273) = (450)/T2  T2 = 370 K (97oC)
11. Determine the molar mass of an unknown gas given the data.
Mass
Volume
Temperature
Pressure
4.93 g
1.00-L
400. K
1.05 atm
MM = mRT/PV
MM = (4.93)(0.0821)(400)/(1.05)(1.00) = 154 g/mol
12. Calculate the density of ammonia, NH3, at STP.
MM = dRT/P
(17.0) = d(0.0821)(0 + 273)/(1)  d = 0.758 g/L
13. Consider the samples of gases.
I
II
III
The samples are at the same temperature. Rank them
with respect to the following (1 is highest).
I
II
1
3
Total Pressure
1
2
Partial Pressure of He
2
3
Density
1
1
Average Kinetic energy per molecule
1
3
Total Kinetic energy
III
1
3
1
1
1
14. Each bulb contains a gas at the pressure and volume
shown and temperature of 25oC. Determine
a.
b.
Record the collected data.
m1
m2
V
15.
16.
17.
18.
19.
the number of moles of each gas.
n = PV/RT = (265/760)(1.0)/(0.0821)(298)
N2
n = 0.014 mol
n = PV/RT = (800/760)(1.0)/(0.0821)(298)
Ne
n = 0.043 mol
n = PV/RT = (532/760)(0.5)/(0.0821)(298)
H2
n = 0.014 mol
b. the total pressure after all stopcocks are opened.
P = nRT/V = (0.071)(0.0821)(298)/(2.5)
P = 0.695 atm (528 torr)
c. the partial pressure of each gas.
PN2 = XN2Ptot = (0.014/0.071)(528 torr)
N2
PN2 = 104 torr
PNe = XNePtot = (0.043/0.071)(528 torr)
Ne
PNe = 320 torr
PH2 = XH2Ptot = (0.014/0.071)(528 torr)
H2
PH2 = 104 torr
2.00 L of Hydrogen gas is collected over water at 30.0oC.
The total pressure is 740 torr (PH2O = 32 torr).
a. What is the partial pressure of the hydrogen gas?
Ptot = PH2 + PH2O
740 = PH2 + 32  PH2 = 708 torr
b. How many moles of hydrogen gas are collected?
PV = nRT
(708/760)(2.00) = n(0.0821)(30 + 273)  n = 0.0747 mol
A 20-L flask holds 0.20 mol O2 and 0.40 mol NO2 at 27oC.
a. What is the pressure of the mixture in kPa?
PV = nRT
(P)(20) = (0.60)(8.31)(27 + 273)  P = 75 kPa
b. What is the partial pressure of oxygen in kPa?
PA = XAPtot
PO2 = (0.020/0.060)(75 kPa) = 25 kPa
Which gas, SO2 or CO2, should be least ideal at STP?
Explain
SO2, it is both heavier and more polar than CO2.
a. Why do gases under high pressure deviate from ideal
behavior?
high pressure is generated by crowded molecules
where the volume of empty space (Videal) is significantly
less than 100 % of the total volume (Vreal)  Vreal > Videal
b. Why do gases at temperatures near their boiling point
deviate from ideal behavior?
close to boiling point, molecules clump and collide less
often, which generates less pressure  Preal < Pideal
Molar Mass of a Gas Lab
Measure the mass and volume of butane released from a
lighter, use the data to determine the molar mass and
compare the experimental value with known molar mass.
Mass a butane lighter (m1). Fill a 50 mL graduated cylinder
with water and place it upside down in a trough filled with
water. Release butane into the graduated cylinder until the
water level in the cylinder equals the water level in the trough.
Record the volume (V). Dry the butane lighter thoroughly and
then mass it (m2). Record the room temperature (T), room
pressure (Plab) and vapor pressure (PH2O).
Plab
PH2O
Complete the following calculations to determine the
molar mass and percent difference from known value.
Pgas
T
V
m
(atm)
(K)
(L)
(g)
Experiment MM
(g/mol)
a.
T
Known MM
(g/mol)
%
c.
How would the following affect the molar mass value?
(1) The butane lighter was not thoroughly dried.
m2 used in the calculation would be greater than it
should be. This would make the calculated mass of
gas too smaller  the molar mass would be less.
(2) The vapor pressure of water was not subtracted
from the room pressure.
P used in the calculation would be greater than it
should be. This would make the molar mass too small
since P is in the formula's denominator.
(3) The water level in the inverted graduated cylinder
was higher than the level in the trough.
P used in the calculation would be smaller than it
should be. This would make the molar mass too large
since P is in the formula's denominator.
Phase Change
20. Which letter illustrates the types of molecular forces?
b
a
Dipole-Dipole d
Dispersion
H-bond
21. For each pair, highlight the molecule with the higher boiling
point and then justify your choice.
Pair
Justification
H2O & H2S H-bonding > dipole-dipole forces
Ne & Kr
Greater atomic mass
More polar
Cl2 & SO2
22. Explain the boiling points for the two isomers.
Propanol molecules H-bond, which is a stronger
attraction then ethyl methyl ether's dipole forces.
23. In addition to dispersion, what type of force would you
expect between the following molecules?
H2
H2S
CHF3
NH3
none
dipole
dipole
H-bond
24. Consider the heating profile for water in your notes.
a. What can you conclude about the value of Cl compared
to Cs based on the slope of line C-D compared to E-F?
The steeper slope for ice, E-F, means Cs < Cl.
What can you conclude about Hfus compared to Hvap
based on the length of line B-C compared to D-E?
The longer boiling line, B-C, means that Hvap > Hfus.
b.
25. Calculate the amount of heat needed to warm 2 mole of
30. Explain why baking takes longer at high elevations.
H2O between the given temperatures or phase change.
Water boils at a lower temperature at high altitudes,
which means that water based foods cook slower
Q
=
nC

T
=
(2
mol)(37.8
J/mol•K)(10
K)
s
-10oC  0oC
because they cook at the boiling point.
Q = 756 J (0.756 kJ)
31.
0.010 moles of water is added to a 5.0-L container filled
Q = nHfus
melt
with dry air at 20oC (vapor pressure = 20 torr). The
Q = (2 mol)(6.01 kJ/mol) = 12.02 kJ
container is then sealed and equilibrium is established.
Q = nClT = (2 mol)(75.3 J/mol•K)(100 K)
0oC  100oC
a. How many moles of water evaporate?
Q = 15,060 J (15.1 kJ)
PV = nRT
Q = nHvap
boil
(20/760)(5.00) = n(0.0821)(20 + 273)  n = 0.0055 mol
Q = (2 mol)(40.7 kJ/mol) = 81.4 kJ
b. What percentage of the water evaporates?
Q = 0.756 + 12.02 + 15.1 + 81.4
Total
Q = 109.3 kJ
0.0054/0.010 x 100 = 54 %
26. Consider the phase diagram in your notes.
a. How does melting point change when pressure increases?32. Explain why water droplets form on a cold water bottle.
Water vapor in the air in contact with the cold surface of
The melting point increases in pressure increases.
the bottle condenses because the low temperature
b. How would the diagram differ for water?
vapor pressure is lower than the pressure exerted by
The solid/liquid line would have a negative slope.
the water vapor in the room.
Crystalline Solids
27. Answer the questions based on the phase diagram.
33. Use the electron-sea model of metals to explain
a. malleability.
Cations are not in a fixed position. Thus, metal will
bend when stress is placed on it.
b. conductivity.
Valence electrons are free to move throughout the
solid.
34. What are allotropes?
Allotropes are different structural forms of the same
substance in the same physical state.
35. In what way is SiO2 like diamond; unlike diamond?
SiO2 is like diamond because it has covalent bonding
throughout its 3-d structure.
270 K
Normal melting point
SiO2 is unlike diamond because it does not melt at a
370 K
Normal boiling point
constant temperature.
36. What two factors affect ionic bond strength (lattice energy)?
Solid
Most dense phase
Phase at 150 K, 0.2 atm
Gas
Phase at 100 K, 0.8 atm
Solid
Phase at 300 K, 1.0 atm
Liquid
Phase change: 50 K  150 K at 0.2 atm
Sublimation
Phase change: 1.0 atm  0.2 atm at 300 K
Vaporization
Melting
Phase change: 200 K  300 K at 1.0 atm
28. Answer the questions based on the vapor pressure curve.
Vapor pressure at 30oC
Temperature where pressure = 300 mm Hg
400 mm Hg
27oC
48oC
Normal boiling point
29. Explain how pure water can boil at room temperature when
placed in an evacuated bell jar.
Water boils when vapor pressure equals atmospheric
pressure. As air is pumped out of the bell jar, its
pressure decreases until it reaches the vapor pressure.
Ionic charge and distance between ions.
37. Complete the chart for each type of solid.
Covalent
Metallic
Molecular
Ionic
Network
ion
atom
molecule
ion
Structural Unit
metallic covalent molecular
ionic
Bond name
strong
weak
strong
Bond strength variable
high
low
high
Melting point variable
low
low
variable
high
Solubility
high
low
low
low
Conductivity
high
low
variable
low
Malleability
copper diamond
water
salt
Example
38. Explain the following observations. You must discuss both
of the substances in your explanation.
a. SO2 melts at 201 K and SiO2 melts at 1,883 K.
SO2 forms London dispersion and dipole forces
between distinct molecules whereas, SiO2, a covalent
network solid, forms covalent bonds throughout. The
much stronger covalent bonds, which are broken
during melting of SiO2, require much more energy
(higher temperature) to break.
b. Cl2 boils at 238 K and HCl boils at 188 K.
Liquid Cl2 is held together by London dispersion
forces, which although weak increase in strength as
the number of electrons increases. Liquid HCl is held
together by dipole forces in addition to London
dispersion forces, but the addition of dipole forces
between HCl molecules must not make up for the
fewer electrons around the HCl molecule compared to
Cl2.
c. KCl melts at is 776oC and NaCl melts at 801oC.
The stronger ionic bond in NaCl is due to the smaller
Na+ ion compared to K+, which allows the Cl- ion to get
closer and strengthens the attraction between ions,
making the NaCl bond stronger and the melting point
higher than KCl.
d. Si melts at 1,410oC and Cl2 melts at -101oC.
Si is a covalent network solid with strong covalent
bonds between atoms. Cl2 has discrete molecules
with weak London dispersion forces between
molecules. Therefore, melting Si requires a higher
temperature than Cl2.
6.
Which occurs when the pressure increases from 0.5 to 1.5
atm at a constant temperature of 60°C?
(A) Sublimation
(B) Condensation
(C) Freezing
(D) Fusion
As the pressure increases at 60oC, the substance
crosses the vapor/liquid line  condensation.
7. The normal boiling point of the substance is closest to
(A) 20oC
(B) 40oC
(C) 70oC
(D) 100oC
Normal boiling occurs at the intersection of the vapor/
liquid line and 1 atm pressure  70oC.
Questions 8-9The graph shows the temperature of a pure solid
substance as it is heated at a constant rate to a gas.
Practice Quiz
Multiple Choice (no calculator)
Briefly explain why the answer is correct in the space provided.
1
2
3
4
5
6
7
8
9 10 11 12 13
D
A
D
A
A
B
C
C
A
C
B
C
D
14 15 16 17 18 19 20 21 22 23 24 25
C
D
B
C
A
C
C
D
D
D
B
C
Questions 1-2 The molecules have the normal boiling points.
Molecule
HF
HCl
HBr
HI
Boiling Point, oC
+19
-85
-67
-35
1. The relatively high boiling point of HF can be correctly
explained by which of the following?
(A) HF gas is more ideal.
(B) HF molecules have a smaller dipole moment.
(C) HF is much less soluble in water.
(D) HF molecules tend to form hydrogen bonds.
HF forms H-bonds (F = small, high electronegativity),
which are stronger than dipole forces  high BP.
2. The increasing boiling points for HCl, HBr and HI can be
best explained because of the increase in
(A) dispersion force
(B) dipole moment
(C) valence electrons
(D) hydrogen bonding
Bonding is similar, except I has more electrons =
greater dispersion forces  higher boiling temp.
3. A sample of an ideal gas is cooled from 50oC to 25oC in a
sealed container of constant volume. Which of the
following values for the gas will decrease?
I. The average kinetic energy of the molecules
II. The average distance between the molecules
III. The average speed of the molecules
(A) I only
(B) II only (C) III only (D) I and III
Kinetic energy (3/2RT) & speed (3RT/MM)½ decrease with
temperature, but spacing is unchanged.
Questions 4-7 refer to the phase diagram of a pure substance.
4.
Which phase is most dense?
(A) solid
(B) liquid
(C) gas
(D) can't determine
The solid/liquid line has a positive slope. Increasing
pressure favors densest phase  solid is densest.
5. Which occurs when the temperature increases from 0°C to
40°C at a constant pressure of 0.5 atm?
(A) Sublimation
(B) Condensation
(C) Freezing
(D) Fusion
0.5 atm is below the triple point, so the solid sublimates
rather than melts.
8.
Pure liquid exists at time
(A) t1
(B) t2
(C) t3
(D) t4
Pure state exists during the temperature increase
phases. The first rise is solid, second rise is liquid.
9. Which of the following best describes what happens to the
substance between t4 and t5?
(A) The molecules are leaving the liquid phase.
(B) The solid and liquid phases coexist in equilibrium.
(C) The vapor pressure of the substance is decreasing.
(D) The average intermolecular distance is decreasing.
Phase change occurs along the plateaus. The first
plateau is solid to liquid and second is liquid to gas.
10. Which actions would be likely to change the boiling point of
a sample of a pure liquid in an open container?
(A) Placing it in a smaller container
(B) Increasing moles of the liquid in the container
(C) Moving the container to a higher altitude
(D) Increase the setting on the hot plate
Boiling occurs when vapor pressure = air pressure.
Higher altitude = lower air pressure  lowers BP.
11. Gas in a closed rigid container is heated until its absolute
temperature is doubled, which is also doubled?
(A) The density of the gas
(B) The pressure of the gas
(C) The average speed of the gas molecules
(D) The number of molecules per liter
Rigid container = constant V  P  T (PV = nRT).
Density, molecules/liter are unchanged. Speed x 2.
12. A 2.00-L of gas at 27oC is heated until its volume is 5.00 L.
If the pressure is constant, the final temperature is
(A) 68oC
(B) 120oC (C) 477oC (D) 677oC
P1V1/T1 = P2V2/T2
(2.00 L)/(300 K) = (5.00 L)/T2  T2 = 750 – 273 = 477oC
13. Under the same conditions, which of the following gases
effuse at approximately half the rate of NH3?
(A) O2
(B) He
(C) CO2
(D) Cl2
rateNH3/rateA = (MMA/17)½
22 = MMA/17  MMA = 4(17) = 68 MMCl2 = 71
14. What is the partial pressure (in atm) of N2 in a gaseous
mixture, which contains 7.0 moles N2, 2.5 moles O2, and
0.50 mole He at a total pressure of 0.90 atm.
(A) 0.13
(B) 0.27
(C) 0.63
(D) 0.90
PN2 = XN2Ptot = 7.0/(7.0 + 2.5 + 0.50)(0.90 atm)
PN2 = (0.70)(0.90 atm) = 0.63 atm
Questions 15-16 refer to the following gases at 0°C and 1 atm.
(A) Ne
(B) Xe
(C) O2
(D) CO
15. Has an average atomic or molecular speed closest to that
of N2 molecules at 0°C and 1 atm
Speed is related to MM (u = (3RT/MM)½)  MM similar to
N2 will have the same speed. MMN2 = 28 = MMCO
16. Has the greatest density
at STP density is proportional to MM (MM = dRT/P). The
molecule/atom with the greatest MM is Xe.
17. A 2-L container will hold about 4 g of which of the following
gases at 0oC and 1 atm?
(A) SO2
(B) N2
(C) CO2
(D) C4H8
At STP one mole of gas = 22.4 L.
MM/4 g = 22.4 L/2 L MM = 44.8 g/mol  CO2 (MM = 44)
18. Which is the same for the structural isomers C2H5OH and
CH3OCH3? (Assume ideal behavior.)
(A) Gaseous densities at STP
(B) Vapor pressures at the same temperature
(C) Boiling points
(D) Melting points
C2H5OH (H-bond) is more polar than CH3OCH3 (dipole) 
(B), (C), (D) would be different. density = STP.
19. As the temperature is raised from 20oC to 40oC, the
average kinetic energy of Ne atoms changes by a factor of
(A) ½
(B) (313/293)½
(C) 313/293
(D) 2
k = 3/2RT, 40oC + 273 = 313 K, 20oC + 273 = 293 K
 E313/E293 = T313/T293 = 313/293
20. The system shown above is at equilibrium at 28°C. At this
temperature, the vapor pressure of water is 28 mm Hg.
1.
2.
5.00 g CO2 x 1 mol/44.0 g CO2 = 0.114 mol CO2
3.
PO2 + 28 mm Hg = 161 mm Hg  PO2 = 133 mm Hg
21. The partial pressure of toluene is 22 mm Hg and that of
benzene is 75 mm Hg in a mixture of these two gases.
What is the mole fraction of benzene in the gas mixture?
(A) 0.23
(B) 0.29
(C) 0.50
(D) 0.77
Pbenzene = XbenzenePtot
Xbenzene = 75 mm Hg/(75 mm Hg + 22 mm Hg)
22. In which of the processes are covalent bonds broken?
(A) I2(s)  I2(g)
(B) CO2(s)  CO2(g)
(C) NaCl(s)  NaCl(l)
(D) C(diamond)  C(g)
(D), covalent network = covalent bonds throughout. (A),
(B) = molecular, (C) = ionic.
23. Of the following compounds, which is the most ionic?
(A) SiCl4
(B) BrCl
(C) PCl3
(D) CaCl2
Ionic compounds are cation + anion. Metals form
cations, not nonmetals. Ca is the only metal  CaCl2.
24. Which of the following oxides is a gas at 25°C and 1 atm?
(A) Rb2O
(B) N2O
(C) Na2O2 (D) SiO2
Ionic (A), (C) and covalent network (D) are solids,
Molecular (B) can be gaseous.
25. Which of the following has the highest melting point?
(A) S8
(B) I2
(C) SiO2
(D) SO2
Covalent network (C) has high melting points, but
molecular (A), (B), (D) tend to have low melting points.
b. What is the volume of CO2 at 1.04 atm pressure?
PV = nRT
(1.04)V = (0.114)(0.0821)(22.0 + 273)  V = 2.65 L
The vapor pressure of solid iodine, I2, at 30oC is 0.47 torr.
a. How many moles of iodine will sublime into an
evacuated 1.0-L flask?
PV = nRT
(0.47/760)(1.0) = n(0.0821)(30 + 273)  n = 2.5 x 10-5 mol
b. If 2.0 x 10-5 mol of I2 is used, what is the final pressure?
PV = nRT
P(1.0) = (2.0 x 10-5)(0.0821)(30 + 273) = 5.0 x 10-4 atm
c. If 3.0 x 10-5 mol of I2 is used, what will the final
pressure be? Explain your answer.
0.47 torr (maximum vapor pressure at 30oC).
3.327 g of an unknown gas occupies 1.00-L at 25oC and
103 kPa. What is the molar mass of the gas?
MM = mRT/PV
MM = (3.327)(8.31)(298)/(103)(1.00) = 80.0 g/mol
b. What is the density of this gas at STP (standard
temperature—0oC, and pressure—1 atm)?
MM = dRT/P
80.0 g/mol = d(0.0821)(273 K)/1 atm  d = 3.56 g/L
c. Which noble gas would have twice the effusion rate?
rA/rB = (MMB/MMA)½
2/1 = (80/MMA)½  MMA = 20 g/mol  Ne
5. N2 with a volume of 200. mL, pressure of 99.7 kPa, and
temperature of 27.0oC is mixed with O2 and transferred to
a 750.-mL container at 27.0oC. The total pressure of the
mixture is found to be 90.4 kPa, at 27.0oC.
a. Calculate the moles of N2.
PV = nRT
(99.7)(0.200) = (n)(8.31)(300)  n = 0.00800 mol
b. Calculate the total moles of gas.
PV = nRT
(90.4)(0.750) = (n)(8.31)(300)  n = 0.0272 mol
c. Calculate the partial pressure of each gas.
PN2 = XAPtot = (0.00800/0.0272)(90.4) = 26.6 kPa
Ptot = PO2 + PN2
90.4 kPa = PO2 + 26.6 kPa  PO2 = 63.8 kPa
6. Explain why methane gas does not behave as an ideal gas
at low temperatures and high pressures.
Low temperature molecules form clusters, which
produce fewer collisions and less pressure than ideal.
High pressure molecules occupy a significant amount of
volume, which is unaccounted for in an ideal gas.
7. In addition to dispersion, what type of forces would you
expect in the following molecules?
C2H2
NO2
NH3
CF4
none
dipole
H-bond
none
4.
The partial pressure (in mm Hg) of O2(g) in the system is
(A) 28
(B) 56
(C) 133
(D) 161
Free Response (calculator)
Explain why a helium filled balloon expands in each case.
a. Additional helium is added.
Additional atoms strike the walls more frequently,
increasing the internal pressure, which causes the
balloon to expand until pressure is equalized.
b. The temperature is increased.
Hotter atoms move faster and strike the walls with
more force and more often. This increases the internal
pressure, which causes the balloon to expand until
pressure is equalized.
5.00 g of dry ice (solid CO2) is placed in a flexible balloon
and allowed to warm to room temperature (22oC).
a. How many moles of CO2 are in the balloon?
a.
8.
9.
Rank butane, ethane, methane and propane from lowest
boiling point to highest.
methane < ethane < propane < butane
Consider the following heating profile for C5H12.
D
36oC
-130oC
B
E
C
Heat Added (J)
a. Highlight the correct answer.
Which is greater?
Cs
Cl
Which is greater?
Hfus
Hvap
Which interval is melting?
B-C
D-E
o
Which is the boiling point?
-130 C
36oC
b. How much heat is needed to vaporize 5.0 g of liquid
C5H12 from 20.oC to 36oC?
(Cl = 120. J/mol•K, Hvap = 26.2 kJ/mol)
Q = nClT + nHvap
Q = (5.0/72)[(120)(16) + 26200] = 2.0 x 103 J
10. 10 g of water is added to a 10.0-L container filled with dry air
at 20oC (PH2O = 20 torr). The container is sealed.
a. How many grams of the water will evaporate?
PV = nRT
(20 x 0.133)(10.0) = n(8.31)(20 + 273)
n = 0.0109 moles x 18.0 g/1 mole H2O = 0.197 g
b. Would the amount of water that evaporates increase
(), remain the same (=) or decrease () for the
following changes?
=


x
Use a 5.0 L container
x
Use humid air
Raise the temperature to 25oC
x
Add 20.0 g of water
11. Consider the following phase diagram.
Answer the following questions.
What is the normal melting point?
Which state is the densest?
What phase change occurs when
going from 40oC to 0oC at 0.5 atm
pressure?
What phase change occurs when
going from 0.50 atm to 1.0 atm at
60oC?
x
35oC
solid
deposition
condensation
12. Consider the following solids.
a. Rank the solids from highest melting point (1) to
lowest.
CH4
H2O
MgO
Na
NaCl
SiO2
6
5
2
4
3
1
b. Justify your relative ranking of CH4 and H2O.
H2O is higher because its bonding is stronger ( H-bond
+ dispersion) compared to CH4 (dispersion).
c. Justify your relative ranking of MgO and NaCl.
MgO is higher because of higher ionic charges (2+ and
2-) compared to NaCl (1+ and 1-), and smaller ionic
radii compared to NaCl's.
13. Explain the following observations.
a. NH3 boils at 240 K, whereas NF3 boils at 144 K.
NH3 has dispersion forces and hydrogen-bonding. NF3
has dispersion and dipole forces. The higher boiling
point for NH3 is due to the greater strength of the
hydrogen-bonding between NH3 molecules.
b. At 25°C and 1 atm, F2 is a gas, whereas I2 is a solid.
The intermolecular force for both F2 and I2 is London
dispersion. The strength of London forces is related
to the number of electrons surrounding the atom.
Thus, I2 with more electrons forms stronger bonds and
is a solid.
14. A rigid 5.00 L cylinder contains 24.5 g of N2(g) and 28.0 g
of O2(g) at 298 K.
a. Calculate the total pressure, in atm, of the gas mixture
in the cylinder at 298 K.
24.5 g N2 x 1 mol/28.0 g = 0.875 mol N2
28.0 g O2 x 1 mol/32.0 g = 0.875 mol O2
PV = nRT
P(5.00) = (1.75)(0.0821)(298)  P = 8.56 atm
b. Calculate the following for N2(g).
(1) mole fraction
XN2 = mol N2/total moles = 0.875/1.75 = 0.500
(2) partial pressure
PN2 = XN2Ptot = (0.500)(8.56 atm) = 4.28 atm
c.
If the cylinder develops a pinhole-sized leak and some
of the gaseous mixture escapes, would the ratio moles
of N2(g) to moles O2(g) in the cylinder increase,
decrease, or remain the same? Justify your answer.
N2 will effuse faster than O2 because N2 has a lower
molar mass  the moles of N2 will decrease faster than
O2.
