South Pasadena • AP Chemistry
Name
10 ▪ States of Matter
Period
10.2 LESSON
Date
– LIQUIDS AND GASES
Phase Diagrams
A phase diagram shows the states of matter of a substance at various pressures and temperatures.
Critical Point
Boiling Point
Melting Point
Triple Point
Example 1: Use the Phase Diagram for H2O.
 Define and identify normal melting point and normal boiling point for H2O. Write a chemical equation for
what is taking place at those points.
Normal melting point – the temperature at 1 atm at which solid and liquid states are at equilibrium.
Heat + H2O(s)  H2O(ℓ)
Normal boiling point – the temperature at 1 atm at which liquid and gas states are at equilibrium.
Heat + H2O(ℓ)  H2O(g)

Describe what happens when a sample of H2O(s) is heated from −10°C to 200°C at 1.00 atm.
The solid warms up from −10°C to 0°C, until the solid melts to liquid. Then the liquid increases in
temperature until 100°C, when the liquid boils. The vapor increase in temperature until 200°C.

Describe what happens when a sample of H2O(s) is heated from −10°C to 200°C at 0.0010 atm.
The solid warms up from −10°C to its sublimation point, and sublimes to gas. The vapor then warms up
until 200°C.

Define triple point and critical point.
Triple point: The temperature and pressure at which solid, liquid, and gas states exist in equilibrium.
Critical point: The temperature and pressure beyond which the substance exists as supercritical fluid, or
plasma.

Compare the densities of the states of H2O and CO2 using their phase
diagrams. Explain the difference.
Substances at high pressure have the greatest density. For water,
Pliquid > Psolid > Pgas, so Dliquid > Dsolid > Dgas. However, for most
substances and CO2 in particular, Dsolid > Dliquid > Dgas. The density
of H2O(s) is less than that of H2O(ℓ) because the hexagonal lattice
of H2O molecules in the solid occupies a greater volume than the
disorganized molecules in liquid state.
Properties of Liquids
Vapor Pressure or Equilibrium Vapor Pressure
(VP or Pvap)is the pressure of vapor above a liquid
when the liquid and gas states are at equilibrium:
heat + A(ℓ)  A(g)
Kp = Pvap,A
Substances with high vapor pressures (vaporize
easily) are said to be volatile.
 How does the vapor pressure change at higher temperatures?
As temperature increases, more molecules have energy to escape the liquid phase, increasing the Pvap.
 How does the vapor pressure change with a substance with weaker IMF?
It is easier for molecules of substances with weaker IMF to vaporize, so such substances have higher Pvap.
 What is expected to happen if PA > Pvap,A? What is expected to happen if PA < Pvap,A?
If PA > Pvap,A, condensation takes place; if PA < Pvap,A, evaporation takes place.
Vaporization is the general term that describes the liquid−gas transition. There are two types:
 Evaporation is the vaporization that takes place below the boiling point, when molecules on the liquid
surface escapes to become gas. This happens when the atmospheric pressure is less than the vapor pressure.
 Boiling is the vaporization that that takes
place at the boiling point, when vapor is
forming below the surface of the liquid. This
happens when the atmospheric pressure is
equal to the vapor pressure. The boiling
point is the temperature at which the vapor
pressure is equal to the atmospheric pressure.
Example 2: For the liquids diethylether (C4H10O), ethanol
(C2H5OH), and water (H2O):
(a) What is the boiling point of each liquid when the
atmospheric pressure is 400 mmHg?
Diethyl ether: 15°C; ethanol: 63°C; water: 85°C.
(b) Which liquid(s) will be boiling at 50°C and 500 torr?
Only diethylether will be boiling.
(c) Samples of the liquids at 20°C are placed in a container at
800 torr, and the pressure is reduced using a vacuum
pump. In which order will the liquids boil?
1st: Diethylether, 2nd: ethanol, 3rd: water
(d) Compare the strength of IMFs among the liquids using
their vapor pressures.
Pvap,diethylether > Pvap,ethanol > Pvap,water
So diethylether has the weakest IMFs, and water has the strongest IMFs.
Example 3: For the liquids benzene (C6H6), acetone (C3H6O), and water (H2O), rank in increasing (and explain):
(a) Surface tension
(c) ∆H°vap
C6H6 < C3H6O < H2O
C6H6 < C3H6O < H2O
(b) Boiling point temperature
(d) Equilibrium vapor pressure
C6H6 < C3H6O < H2O
H2O < C3H6O < C6H6
Properties of Gases
Kinetic Molecular Theory – states that gas particles move in constant, rapid, random, straight-line motion. We
assume that there are no attractive forces between gas particles, collisions between particles are completely
elastic, and that these particles occupy negligible volume.
Gas samples that behave according to these assumptions are called ideal gases. While no real gases are
completely ideal, these assumptions allow us to make very good approximations when describing gases.
Properties of ideal gases relate to each other by the Ideal Gas Law:
P·V = n·R·T
Example 4: Consider the following samples of gas:
 Container A: He(g) at 600 torr and 40°C
 Container B: SO2(g) at 600 torr and 40°C
(a) Find the density of the gas in each container.
m
m
m·R·T
P·V = n·R·T
MM = n
so
P·V = MM ·R·T
and
MM = P·V
m
D·R·T
MM·P
atm·L
Since D = V
MM = P
and
D = R·T
R = 0.0821 mol·K
4.026 g 600 torr 
64.07 g 600 torr 
 mol 760 torr/atm
 mol 760 torr/atm
DA =
= 0.124 g/L
DB =
= 1.97 g/L
atm·L
0.0821
(273 + 40 K)
0.0821 atm·L (273 + 40 K)
mol·K
mol·K


(b) Find the average (rms) velocity of the gases in each container.
3
At the same temperature, molecules have the same average kinetic energy (KE = 2 R·T), but the lighter
1
molecules have a faster speed (KE = 2 m·v2). For a molar sample, m = MM.
1
3
KE = 2 (MM)(v)2 = 2 R·T
so v =
3·R·T
MM
J
R = 8.314 mol·K and MM is in kg/mol.
vA =
3·R·T
MMA =
3·(8.314 J/mol·K)(313 K)
(4.026 × 10−3 kg/mol) = 1390 m/s
vB =
3·R·T
=
MMB
3·(8.314 J/mol·K)(313 K)
= 349 m/s
(64.07 × 10−3 kg/mol)
MMA 4.026 g/mol 1 vA 1390 m/s 4
vA
MMB
Note that MM = 64.07 g/mol = 16 , v = 349 m/s = 1 since v =
MMA .
B
B
B
(c) Each container has 0.50 moles of the gas. The volume of Container A was 16.07 L, while the volume of
Container B was 15.96 L. Explain the difference.
Both containers have the same moles of gas at the same temperature and pressure, so they are expected
n·R·T (0.50 mol)(0.0821 atm·L/mol·K)(313 K)
to occupy the same volume:
V= P =
= 16.27 L.
(600/760 atm)
However, because real gases have small IMFs and take up volume and deviate from ideal behavior.
Substances with stronger IMFs and are larger deviate more from ideal behavior.
The van der Waals Equation for real gases is:
P + a n 2[V − n·b] = n·R·T

V 
P is the observed pressure of the gas sample, V is the
volume of the container. a is a parameter related to the
attraction between particles, and b is a parameter related to the volume per
particle.
 Under what conditions (temperature and pressure) will a gas behave like an
ideal gas?
Low pressure and high temperature
 What kind of gases are most ideal?
Small size with weak IMFs