Solids, Liquids, and Gases
5th International Junior
Science Olympiad (IJSO)
Dr. Yu-San Cheung
yscheung@cuhk.edu.hk
Department of Chemistry
The Chinese University of Hong Kong
1
Basic Properties of Solids, Liquids,
and Gases
2
Characteristics of Gases
No definite volume or shape:
A gas fills whatever volume is available to it and is easy to
compress.
Low densities: (density = mass  volume)
Compared with those of liquids and solids: one mole of
liquid water at 20°C (298 K) and 1 atm pressure occupies a
volume of 18.8 cm3, whereas the same quantity of water
vapor at the same temperature and pressure has a volume
of 30200 cm3, more than 1000 times greater.
3
Properties of Gases
What can we study about a gas (e.g. in a balloon)?
Pressure: the gas makes the balloon expand (against ambient
atmospheric pressure and tension of the balloon).
Temperature: if the balloon is left in a room long enough, the
temperatures of the balloon and the gas are the same as that of the
room.
Volume: the gas fills out the whole space inside a container. (The
volume of gas is taken as the container capacity, which is usually
assumed to be the container volume if the wall of a container is
thin.)
What is the relationship between these properties?
4
The Pressure of a Gas
The molecules of a gas, being in continuous motion,
frequently strike the inner walls of their container. As they
do so, they immediately bounce off without loss of kinetic
energy(動能) , but the reversal of direction (acceleration)
imparts a force to the container walls. This force, divided by
the total surface area on which it acts, is the pressure of
the gas.
5
Pressure
When a force (F) is acting on a surface with area A, a pressure (p)
exerts on the surface:
Downward force due to
weight of mass acting
on piston of area A,
creating a pressure p
(assuming no air outside)
p=F/A
The pressure of a gas is
observed by measuring the
pressure that must be
applied externally in order
to keep the gas from
expanding or contracting.
Moveable
piston
(frictionless
and
weightless)
When the piston is
stationary,
the
gas
pressure
is
exactly
equal to p.
That
means the gas creates
an opposing force F =
pA which is equal but
opposite in direction to
the force created by
gravity acting on the
weight.
6
How is pressure measured?
Barometer(氣壓計)
Atmospheric
Pressure
is
measured by an instrument
called barometer ( 氣 壓 計 ) ,
invented in the early 17th
century.
The barometer (氣壓計) consists
of a vertical glass tube closed at
the top and evacuated, and
open at the bottom.
liquid
http://en.wikipedia.org/wiki/Barometer
7
How is pressure measured?
8
p = p0 + p
p = pressure exerted by the height difference
9
Pressure Exerted by a Liquid Column
Column height = h
Cross-section area = A
Volume = h A
Mass = volume  density = h A
( = density)
Force produced by the column
= Weight = mg = h A g
A
h
(g = gravity acceleration
= 9.80665 m ·s2)
Pressure = F / A =  g h
Note that the cross-section
must be uniform, but not
necessarily circular.
10
Choice of Medium
p =  g h
For the same p, higher density, smaller height difference.
For mercury: 1 standard atmospheric pressure  760 mm
For water: water density = 1/13.6 of mercury density
So, if water is used, 1 standard atmospheric pressure
 760 mm  13.6 = 10.34 m
In practice, mercury is used.
Other advantages?
But any disadvantages?
11
Mercury as the Medium
Advantages:
- High density
- Chemically inert
(e.g., water dissolves some gases such as CO2, NH3, HCl)
- Not sticking to the glass wall (compared with water)
Disadvantages:
- Vapor is toxic
- Difficult to deal with spillage
(not absorbed by paper towel, tending to form small
droplets, difficult to remove with dropper)
12
Mercury Barometer
One end open to atmosphere
Accuracy: depending on the
ruler readout (usually 1 mm)
The other end connected to closed
container containing the gas
13
Mercury Barometer (Closed-end)
sealed end
vacuum
inside
p = p
14
Pressure Gauge
Also called barometer, but the term “Pressure Gauge”
(壓強計) is more commonly use in the market
For measuring both “positive pressure” and “negative
pressure” (relative to atmospheric pressure)
“Negative pressure”: also called vacuum
“Positive pressure”: can be measured up to 1380
atmospheric pressure
http://en.wikipedia.org/wiki/Pressure_measurement
15
Principle of Pressure Gauge
A typical Bourdon tube contains a curved tube. It is open to
external pressure input on one end and is connected mechanically
to an indicating needle on the other end.
The external pressure is guided into the tube. Change in pressure
causes the tube to flex, resulting in a change in curvature of the
tube. This curvature change is linked to the dial indicator for a
number readout.
Alternatively, a strain gauge
circuit can be used to produce
output electronically. If you are
interested, see:
http://www.efunda.com/DesignS
tandards/sensors/strain_gages/s
train_gage_theory.cfm
http://www.efunda.com/DesignStandards/
sensors/bourdon_tubes/bourdon_intro.cfm
http://en.wikipedia.
org/wiki/Pressure_
measurement
16
Principle of Thermo-conductivity Gauge
Principle:
- A filament is heated by running electrical current
through it.
- A thermocouple thermometer measures the filament
temperature.
- High pressure: more gas molecules  higher heat loss
(by convection)  lower temperature  electrical signal
High pressure vs. electrical signal: may be complicated
Different gases: different calibration
Range: 10 – 0.001 mmHg
http://www.varianinc.com.cn/
products/vacuum/measure/
transducers/531gauge/shared/
531gauge-180.jpg
17
Principle of Ionization Gauge
Principle:
- A filament is heated to emit electrons.
- These electrons ionized gaseous molecules.
- Ions (gaseous molecule ions) are attracted to the
cathode of a electrical circuit and an electrical current is
resulted.
High pressure vs. electrical signal: may be complicated
Different gases: different calibration
Range: 10–3 – 10–10 mmHg
http://www.varianinc.com.cn/products/vacuum/measure/shared/uhv24a-180.jpg
18
Units of Pressure
The unit of pressure in the SI system is Pascal
(Pa)(帕斯卡)
1 Pa  1 N·m2  1 J·m3  1 kg·m1·s2
p=F/A
Energy = F ·d
F = m· a
 1 Pa = 1 N·m2
 1 J = 1 N·m = 1 (Pa·m2) · m
 1 N = 1 kg·m·s2
19
Units of Pressure
Other commonly used units:
(i) psi = pound per square inch
1 pound = 453.59237 g = 0.45359237 kg
1 inch = 2.54 cm = 0.0254 m
Exercise: 1 psi = ?? Pa
(You also need: gravity acceleration g = 9.80665 m ·s2)
20
Units of Pressure
(ii) 1 torr = 1 mm Hg
(implying the pressure produced by a Hg column of 1 mm height)
Pressure = F / A =  g h (p. 11)
1 torr = 133.32 Pa
( = 13595.1 kg·m3 for mercury)
(iii) 1 atm = 760 torr
1 atm = 101325 Pa = 14.696 psi
(iv) 1 bar = 105 Pa
Important: 1 bar is close to 1 atm, but not exactly equal!
21
Determination of Volume
By calculations:
Rectangular shape: width × depth × height
4
 r3
3
1
Pyramid:
× base area × height
3
Sphere:
(including cone)
More complicated shape but well-defined by
mathematical formulas: by calculus
Irregular shape: by empirical measurements
(1) Immersing the object in liquid
(2) For container only: check how much liquid is needed
to fill out the container (thinner wall, better result)
22
Units of Volume
Unit of volume = (Unit of length)3
SI unit for volume : m3
Units used in daily life: liter, dm3, ml (or mL), cm3
1 m3 = (1 dm)3 = 1 dm3, also called liter (L)
1 ml = 1 milli-liter = 10-3 L = 10-6 m3 = 10-6 (100 cm)3 = 1 cm3
Smaller unit: L (micro-liter, 10-6 L) , nL (nano-liter, 10-9 L)
Summary: fill out the blanks below with appropriate units
m3
 1000
L
dm3
 1000
mL
ml
cm3
 1000
L
 1000
nL
23
Types of thermometer
(1) Glass thermometer(玻管液體溫度計)
Glass tube containing a liquid, e.g., _________, ________
High temperature: _______________ of liquid
Low temperature: _______________ of liquid
(2) Resistance thermometer(電阻溫
度計)
Consisting of a probe (e.g.,
platinum) and a controller (with battery,
read-out, and electronic circuit). The
resistance of the probe changes with
temperature.
By measuring the
resistance, the temperature can be
determined.
http://en.wikipedia.org/wiki/Thermometer
24
Types of thermometer
(3) Infra-red (IR) thermometer(紅外
線溫度計)
All objects emit IR light. The hotter
an object is, the stronger the IR light
emitted. By measuring the IR light
intensity, the temperature can be
determined.
(What is its main advantage over the
other types?)
(4) Liquid crystal thermometer
Containing dots of heat-sensitive liquid crystals on a plastic strip.
The dots change color at different temperatures.
http://en.wikipedia.org/wiki/Thermometer
25
Unit of Temperature
Celsius scale:
0 C = freezing temperature of water
100 C = boiling temperature of water
at 1 atm pressure for both
26
Pressure-volume relations:
Boyle‘s law (波義耳定律)
Robert Boyle (1627 – 91)
Volume
Pressure
Modified from:
http://www.grc.nasa.gov/WW
W/K-12/airplane/aboyle.html
27
Boyle's law & J-tube
Each time Hg is added, h1 and
h2 changes, but p0 and h0
remain unchanged.
p0
Trapped air
h0
Pressure of trapped air = p0 + h1 – h2
Volume = (h0 – h2) 
cross-sectional area
Exercise:
p0 = 0.987 atm
h1 = 45.9 cm, h2 = 10.7 cm
h1
h2
p = ?? cm Hg
28
Boyle's Law
p  1/V
(i.e., p is inversely
proportional to V )
1 / Volume
We can also say that:
p is proportional to 1/V




Pressure
p = “constant” × 1/V
Plot of p vs. 1/V yields a straight line
passing through the origin
29
Boyle's Law
p = “constant” × 1/V
1 / Volume
leads to:
pV = “constant”




Pressure
“Constant” here means that:
It does not change with p and V.
If p changes, V also changes. V changes
in such a way that pV remains the same.
30
Boyle's Law
Alternative form (more useful):
Initial: p1, V1
Final: p2, V2
p1V1 = constant = p2V2
i.e.,
p1V1 = p2V2
31
Boyle's Law
Example:
p1V1 = p2V2
If p1 = 700 torr, V1 = 1 L, p2 = 100 torr, V2 = ?
Ans.: V2 = p1V1 / p2 = (700 × 1) / (100) = 7 L
Exercise:
If p1 = 700 torr, V1 = 1 L, V2 = 100 L, p2 = ?
Ans.:
32
Effect of Temperature on p-V Curve
Modified from:
http://www.che
m1.com/acad/w
ebtext/gas/gas_
2.html
33
Effect of Temperature on p-V Curve
Try finding out
the “constant” for
each curve:
500K: pV =
400K: pV =
300K: pV =
200K: pV =
100K: pV =
50K: pV =
34
Effect of Temperature on p-1/V Curve
1/V
Temperature:
increasing or
decreasing?
p
35
Exercise:
In an industrial process, a gas confined to a volume of 1 L at
a pressure of 20 atm is allowed to flow into a 4 L container
by opening the valve that connects the two containers.
What will be the final pressure of the gas? (Hint: at the end,
the gas fills up both containers.)
20 atm at
the
beginning
Vacuum at
the
beginning
4L
1L
36
Temperature – Volume
Relationship:
Charles‘ Law (查理定律)
With the pressure held constant, the volume of a gas
changes by the same amount for each C change in
temperature.
37
Put it another way, V varies linearly with T, i.e., the
T-V graph is a straight line.
V, liter
p1
When the straight lines are
extrapolated, they reaches
the x-axis at –273.15 °C.
p2
p3
0
T, °C
http://www.chem1.com/acad/webtext/gas/gas_2.html
38
Kelvin Scale (K): adding 273.15 to the C value
e.g., -273.15 C
0.00 C
25.00 C
100.00 C
 0.00 K (Note: the symbol “” is not used)
 273.15 K

K

K
If talking about temperature difference or temperature change,
change in 1 C = change in 1 K.
Diff =
°C
Diff =
°C
-273.15 C
0.00 C
25.00 C
100.00 C

0.00 K
 273.15 K
 298.15 K
 373.15 K
Diff =
K
Diff =
K
39
Temperature-Volume Relationship
V
Using the Kelvin scale, we have
TV
(T is proportional to V )
or
T / V = constant
-273.15 °C
0 °C
T
V
or
T 1 / V1 = T 2 / V2
T / V = constant 
(i) V = 0 if T =0
(ii) V is –ve if T is –ve
0K
273.15 K
T
40
Kelvin Scale: what a big deal?
Fahrenheit set zero degree (0 F = –17.78 C) for the lowest
temperature reachable at that time (1724), so as to avoid negative
temperature in practical life.
Celsius: more open-mined, not minding negative temperature,
choosing convenient scale: freezing point of water for 0 C.
But there is no clue what the lowest temperature is in nature.
Charles’ law: 0 K is the lowest temperature in nature, otherwise
we have negative volume of a gas, which is physically unacceptable.
0 K: also called absolute zero
Kelvin Scale: also called the absolute temperature scale
Kelvin Scale is assumed in scientific formulas
(unless otherwise specified).
41
Charles’ Law
Example:
T 1 / V1 = T 2 / V2
If T1 = 200 K, V1 = 1 L, T2 = 100 K, V2 = ?
Ans.: V2 = T2V1 / T1 = (200 × 1) / (100) = 2 L
Exercise:
If T1 = 200 C , V1 = 1 L, V2 = 100 cm3, T2 = ?
Ans.:
42
Boyle’s Law + Charles’ Law = What?
Boyle’s Law
p  1/ V
or
V  1/p
+
Charles’ Law
VT
VT/p
pV / T = constant
or
p1V1 / T1 = p2V2 / T2
43
Exercise:
The tires of a car were filled with air to a pressure
of 30 psi at 25 C. After the car running for several
hours, the temperature raised to 100 C and the
tire volume increased by 10%. What was the
pressure of the air in the tires?
44
Amadeo Avogadro (1776-1856): "E.V.E.N principle"
Equal volumes of gases, measured at the same
temperature and pressure, contain equal numbers of
molecules.
i.e., V  N
(N = no. of gas molecules)
Combined with pV / T = constant, we have
pV / NT = constant = k
(Boltzmann constant)
or
pV = NkT
(Ideal Gas Law)
45
Mole Concept
32 g of oxygen gas
= 6.022  1023 oxygen gas molecules
= 1 mole of oxygen gas molecules
i.e., 1 mole = 6.022 x 1023
c.f.: 1 dozen of pencils = 12 pencils
1 kB = 1024 bytes
1 catty of nails = ??? pieces of nails
Avogadro Number: NA = 6.022  1023 mol1
Mole number (n) = numbers of moles = N / NA
e.g., 0.01 mol of O2 gas molecules
= 0.01 x 6.022 x 1023 = 6.022 x 1021 O2 gas molecules
46
Mass and Mole
?? grams of a gas = 1 mole of the gas molecules
You need to find out the atomic mass units of the
atoms (from books or internet)
e.g. H:
1.0 amu (amu = atomic mass unit)
C: 12.0 amu
Cl: 35.5 amu
H2: 2 x 1.0 = 2.0

2.0 g of H2 = 1 mole of H2
CH3Cl: 12.0 + 3 x 1.0 + 35.5 = 50.5

50.5 g of CH3Cl = 1 mole of CH3Cl
47
Molecular Mass
50.5 g of CH3Cl = 1 mole of CH3Cl
We define:
molar mass / molecular mass / molecular weight (Mw)
e.g. Mw for CH3Cl = 50.5 g mol1
e.g. Mass of 0.1 mol of CH3Cl = 50.5 g mol1 x 0.1 mol
= 5.05 g
Mw
NA
Mass  Mole  Number of molecules
48
Ideal Gas Law / Perfect Gas Law
pV = NkT = (NAn)kT = n (NAk)T = n (NAk)T = nRT
(R = NAk)
R: gas constant or universal gas constant
Exercise:
(i) What is the S.I. unit for R ?
(ii) What is the S.I. unit for k ?
(iii) If the experimental value for R is 8.314 in S.I. unit.
What is the value for k in S.I. unit?
49
Ideal Gas Law / Perfect Gas Law
n : no. of moles
R : gas constant = 8.314
Unit to be filled out
Must be
memorized!!
pV = nRT
pV = NkT
N : no. of
molecules
More commonly used
k : Boltzmann constant =
Value to be filled out
50
Ideal Gas Law / Perfect Gas Law
(a)
pV = nRT
There are five symbols in the equation. If four of them
are given, the remaining one can be determined.
If three of p, V, n, and T are known, the remaining one
can be determined because R can be found out from
literature.
Exercise: how much gas is needed to fill an 1-L box to 700
torr at 25 C?
51
Ideal Gas Law / Perfect Gas Law
(b)
pV = nRT  pV / nT = R = constant
 p1V1 / n1T1 = p2V2 / n2T2
This equation is more useful if there are changes for the
variables (p, V, n, and T ).
Exercise:
A box contained a gas of 1700 torr at 25 C. A hole was
punched and gas leaked out. After the hole was sealed, it
was found that half of the gas remained and the temperature
dropped to 0 C. What was the final pressure of the gas?
52
Ideal Gas Law / Perfect Gas Law
p1V1 / n1T1 = p2V2 / n2T2
This form includes the Boyle’s Law, Charles’ Law, and EVEN
Principle. You can show that by keeping some variables
constant.
Boyle’s Law:
p1V1 = p2V2
Charles’ Law: V1 / T1 = V2 / T2
EVEN Principle: V1 / n1= V2 / n2
Constants:______
Constants:______
Constants:______
53
Gas Mixture
P : total pressure
V : volume of the container
N : total number of gaseous molecules
T : temperature of the mixture
Idea: The two gases act independently.
p = pA + pB
n = nA + nB
pAV = nART
pBV = nBRT
V & T : same for both gases
Adding: (pA + pB)V = (nA + nB)RT

pV = nRT
54
Air Composition
Air is described as a gas containing 20% of O2 and 80% of N2 by
volume. How to understand this?
O2
1 atm
+
N2
1 atm
O2
1 atm
N2
1 atm
Air
1 atm
O2
+ N2
0.2 atm
0.8 atm
55
When dealing with gas mixing, the “volume %”
concept is more convenient,e.g.,
N2
Volume:
4
(Same pressure)
Air
O2
:
1
Air
56
Partial Pressure
Partial Pressure:
the pressure of each gas in a gas mixture
If gases A and B with the same pressure are mixed
in volume ratio VA:VB, then the partial pressures ratio
in the mixture are
pA:pB = VA:VB
(according to the figure on p. 55)
and the mole ratio nA:nB is also VA:VB
(because p  n according to the ideal gas law)
57
Exercise:
The following gases are mixed and put into a 2L box:
3 L of O2 at 0.5 atm, 2 L of N2 at 1 atm, and
1 L of CO2 at 1 atm
(i) Calculate the pressure for each of the three gases
before mixing if the volume of each gas is changed to
2L box.
(ii) Determine the partial pressures for the three gases
after mixing.
(iii) Calculate the total pressure of the mixture.
58
Molar Volume of a Gas:
Standard Temperature and Pressure
For convenient comparison it is customary to use the
conditions of STP, standard temperature and
pressure where T = 273.15 K and p = 1 atm.
Molar volume is the volume of one mole of a substance.
Standard molar volume is the molar volume at STP.
Exercise:
Calculate the standard molar volume of a gas.
59
Density of a Gas
(i) Number density:
no. of gas particles per unit volume
= N / V = p / kT
(ii) Mole density:
no. of moles of gas particles per unit volume
= n / V = p / RT
(iii) Mass density:
mass of gas particles per unit volume
= mass of gas / V
= n  molecular weight / V
= molecular weight  p / RT
60
Mass Density
(iii) Mass density:
molecular weight  p / RT
We can conclude that:
(a) hot air is lighter than cold air.
(Where should we put air heater and
conditioner in a room? Near Ceiling or on the
floor?)
(b) gas with smaller molar mass is lighter than gas
with larger molar mass.
61
Change of States
Liquid
Solid
Sublimation
Deposition
Gas
62
Vapor Pressure
When the temperature of a liquid reaches its boiling point, it becomes a gas. Why?
At the boiling point, the molecules have enough energy to move into the space
above the liquid.
But even if the boiling point is not reached, some molecules still have enough
energy to leave the liquid (surface). They become gas (though we call them vapor
in our daily life). Like a pure gas, this vapor also exerts a pressure. The pressure is
called vapor pressure.
63
64
As long as the liquid and vapor co-exist, the vapor pressure
depends on temperature only.
~
v.p. = p0·exp(Hvap/RT )
~
Hvap = amount of heat needed to evaporate the liquid
(similar to the latent heat of vaporization)
p0 is a constant
Some observations from the equation:
(i)
T   v.p. 
(ii)
If we put the setup into a water-bath to measure v.p.
~
at different temperature, we can obtain Hvap.
65
Condensation
Exercise:
Why, when a person walks out a building in summer,
his/her glasses get moist?
66
How does the vapor pressure change in the following cases?
(Temperature remains unchanged)
?

 
 
Liquid
A closed container
filled with 1/3 of
liquid.
?
?
?
Liquid
?
(i) Same volume of liquid and
space above the liquid, but the
surface area of the liquid is
doubled.
? ?
?
?
?
Liquid
(ii) Same
container,
but the
volume of
the liquid is
is doubled.
67
Answer:
The same in both cases because the temperature is the
same.
(i) Puzzle: the surface area of the liquid is increased, so
more molecules escape from liquid?
Answer: there are also more molecules returning to the
liquid.
(ii) The vapor also obeys the ideal gas law. Though p and
T are the same as before, V is halved and so is n, i.e., the
amount of vapor is halved.
68
Comparisons of Gas and Vapor
(In many cases, the term “vapor” implying that it is in
contact with liquid.)
(i) Gas obeys the ideal gas equation. If T is fixed, two of
the variables, p, V, and n are free to be varied. (The
remaining one is fixed through the law.)
(ii) Vapor obeys the ideal gas equation and the formula on
p. 65. If T is fixed, p is also fixed. Only one of the
variables, V and n can be varied. In addition, the mole
density (n / V ) is constant. That is, once T is fixed, so is
the mole density (and other types of density such as
number density and mass density.)
69
Vapor Pressure of Some Liquids
In general, the weaker the intermolecular
interaction, the higher vapor pressure is.
70
From Vapor to Gas
An 1-m3 box contains 3 mol of ethanol at 273 K. At the beginning, liquid and
vapor co-exist. How does the v.p. change with increasing temperature?
p-T curve for gas
density of 3
mole/m3
(iii) After all
liquid evaporates,
the vapor
becomes a “pure
gas” and the v. p.
follows the p-T
curve.
(ii) At this point, all
liquid evaporates.
(i) At the beginning, the v. p.
follow this curve, more and
more liquid evaporates.
71
Boiling
A liquid boils when its vapor pressure is equal to the
external pressure acting on the liquid surface.
72
Boiling at Different External Pressures
If the external pressure increases, higher vapor pressure is
required for boiling, so the boiling point is higher. And vice versa.
Question: from the graph
on the left, what are the
boiling points for ethanol
at external pressure =
600 and 800 torr?
Pressure cooker:
boiling point  100 C
At hill top,
boiling point  100 C
73
Evaporation vs. Boiling
Ordinary evaporation is a surface phenomenon. Below the boiling point,
the vapor pressure is lower than the outside pressure and bubbles of
water vapor cannot form. But at the boiling point, the vapor pressure is
equal to atmospheric pressure. Bubbles form and the vaporization
becomes a volume phenomena.
74
Properties of Solid
(i) Having a definite shape
(ii) Difficult to compress (as the particles are already
packed closely together)
(iii) Usually denser than the liquid (as the particles are
packed more closely together than in liquid), except:
water (according to http://en.wikipedia.org/wiki/Ice, it
is the only non-metallic substance having this property)
(iv) With particles vibrating around fixed locations, not
moving around
75
Why Ice Floating on Water?
Ice floats on water because of its lower density.
At 0 C,
density of ice = 0.917 g/cm3
density of water = 0.9988 g/cm3
There are specific orientations
between
different
water
molecules, leaving many holes
inside ice and a hollow
structure is resulted.
http://www.uwgb.edu/dutchs/PETROLGY/Ice%20Structure.HTM
76
“Heavy Ice” Floating on Water
However, heavy water ice (ice of the heavy water, D2O) sinks in
water!
The structure of a solid mainly depends of the chemical
properties of the particles. We expect the two types of ices to
have the same structure and the numbers of molecules per unit
volume (“number densities”) are the same.
But
molar mass of H2O = 1x2+16 = 18 g/mol
molar mass of D2O = 2x2+16 = 20 g/mol
Density of D2O ice = (20/18) x density of D2O ice = 1.02 g/cm3
i.e., 2% higher than that of water!
Would you expect that heavy water ice floats or sinks in heavy
water?
77
Phase Diagram
Whether a substance exists in this phase or that phase depends
on the pressure and temperature. A phase diagram indicates the
phase of a substance at various combinations of pressure and
temperature.
p
Common phases for most of the
substances are: solid, liquid, and
gas.
Hence, there are usually
three regions in a phase diagram.
The following is a typical one.
(Note the locations of the regions.)
Question: can you tell which region is for solid,
which one is for liquid, and so on?
T
78
Phase Diagram
Answer: from left to right, temperature increases, so
you go through the three phases in the order of:
solid  liquid  gas
p
Liquid
Solid
Gas
T
79
Phase Diagram
Line segments: places where two phases co-exist
p
solid & liquid
Liquid
Solid
Gas
liquid & gas
T
solid & gas
Triple point: three phases co-existing
80
Phase Diagram
What does it tell us?
Melting point & boiling point
p
Liquid
p1
Solid
Gas
Melting point at p1
T
Boiling point at p1
81
Phase Diagram
How melting point & boiling point change with external pressure
p2
p1
p
Liquid
Solid
Gas
T
Melting point at p1
Melting point at p2
Boiling point at p1
Boiling point at p2
82
Phase Diagram
The steepness of a line segment tells us the following:
(i) If the slope is positive, the melting point/boiling point
increases with increasing temperature.
 solid-gas and liquid-gas segments: positive slope
 solid-liquid segment: usually positive slope
(one exception: water)
 solid-liquid segment: usually very steep
(whether +ve or –ve)
(ii) If the line is steep, the melting point/boiling point is
insensitive to pressure change.
 solid-liquid segment: usually very steep
(whether +ve or –ve)
 solid-gas and liquid-gas segments: usually not as steep
83
Decreasing Melting Point of Water With
Increasing Temperature
Hollow structure of solid
 Volume decreasing on melting
 M. P. decreases with increasing pressure
 Negative slope of the solid-liquid segment
Note: the phase diagram indicates (or shows)
that the m.p. of water decreases with
increasing pressure. It does not prove or
explain it!
84
From Solid to Gas: Sublimation
There are not many
substances subliming
at 1 atm. Examples
are CO2 and iodine.
But under high-enough
external
pressure,
these substances can
go through the stages
from solid to liquid to
gas.
http://userpages.umbc.edu/~neumann/Chem102/Notes/ch11/c1306d.html
85
From Solid to Gas: Sublimation
Substances not subliming
at 1 atm can sublime if
pressure is lowered.
http://userpages.umbc.edu/~neumann/Chem102/Notes/ch11/c1306d.html
86
Significance of Triple-point
External pressure > triple-point pressure: solid  liquid  gas
External pressure < triple-point pressure: solid

gas
87
Critical Point
In a phase diagram, the liquid-gas boundary stops somewhere.
88
Critical Point
A gas can usually
be liquefied by
increasing pressure
and/or decreasing
temperature. As
the path crosses
the liquid-gas
boundary, the gas
changes to liquid.
89
Critical Point
But, if the temperature
is too high (above the
“critical temperature”
(臨界溫度)), the gas
cannot be liquefied,
however high the
pressure is applied.
In order to achieve
liquefaction, the
temperature must be
lowered to a value
below the critical
temperature.
90
Critical Point
How to rationalize the existence of critical temperature(臨界
溫度)?
If the temperature is too high, the motions of particles are
too vigorous. Attraction between the particles is not strong
enough to hold the particles together, even though they are
brought very closely together (by high pressure).
Therefore, the gas must be cooled down to slow down the
particle motions. The molecular attraction may overcome
the particle motions.
91
Supercritical Fluid
The substance at the
rectangular region with T
 Tc and p  pc is called
supercritical fluid.
The fluid is not a liquid,
but more like a gas. It
may have high density
(e.g., comparable to that
of liquid).
pc:
Critical
pressure
Tc:
Critical temperature
92
Formation of Supercritical Fluids
http://www.chem.leeds.ac.uk/People/CMR/criticalpics.html
93
Properties of Supercritical Fluid
http://sunny.vemt.bme.hu/sfe/angol/supercritical.html
94
Green Chemistry
in Electronic Industry
Silicon wafer
coating
etching
washing
Chips
soldering
Circuit board
95
Cleansing of Electronic Circuit Boards
Using Chlorofluorocarbons (CFCs)
Reasons for using CFCs as solvent in electronics cleaning:
1.
2.
3.
4.
5.
Inert
Volatile
Low surface tension
Non-flammable
Non-corrosive
Cl
Cl
Cl
C
F
C
F
F
1,1,2-trichloro-1,2,2-trifluoroethane
(CFC-113)
96
Ozone Depletion Caused By CFCs
Enormous ozone “hole” over Antarctica
Increased damages to
•
•
•
Human immune system
Skin
Ecology
97
Supercritical CO2 Cleaning System
http://www1.boc.com/eco-snow/index.htm
98
Supercritical CO2 in Extraction
Also used in extraction (of components in herbs)
Advantages over boiling water:
(1)
(2)
(3)
Commercial product:
http://elchem.kaist.ac.kr/vt
/chem-ed/sep/sf/sfe.htm
Home-made apparatus at CUHK 
99
Graph Plotting – A tool for data analysis
• Data: collected observations and facts
• Regularities in observations can be found through careful
analysis of data.
• Graphing is a way to present data that shows relationships
among data analyzed.
100
Graph Plotting
101
What Should Be Included in a Graph
•
Title
•
Axis, legend, number, tick, unit
•
Data point
•
If you fit the data to an equation, the best-fit line (or “trendline”
in some computer softwares), the equation and the square of
correlation coefficient (R2) obtained should also be included.
•
If your purpose is to present a formula with a graph, connect the
data points with lines.
102
Choosing x and y for Plotting
A straight line has the form y = mx + c (m: slope, c: y-intercept)
Sometimes we have to rearrange the equation
Constants
(known/unknown)
y = m x + c
Known
(variables/constants)
103
Choosing x and y for Plotting
A straight line has the form y = mx + c (m: slope, c: y-intercept)
Sometimes we have to rearrange the equation
e.g., A particle falls freely. The distance at different time is
measured. We have a series of data (t, x). According to the
classical mechanics, t and x obeys the equation: x = x0 – (1/2)gt2.
x = –(1/2)g t 2 + x 0
cf. y = m x + c
To obtain a straight line, we plot x against t2.
So, slope = –(1/2)g, y-intercept = x0
Line-fitting gives the value of the slope and yintercept.
We get the value of g (gravity acceleration) from the
slope.
x
104
Example
The mass, A, of a radioactive substance is:
A = A0ekt
where A0 is the amount at t = 0 and k is an unknown constant
Take ln on both sides:
ln(A) = ln(A0ekt) = ln(A0) + ln(ekt) = ln(A0)  kt
ln(A) = kt + ln(A0)
y=
,x=
,m=
,c=
.
105
Line-fitting
Some computer software (such as Excel) provide fitting other
than straight line: polynomial, logarithmic, exponential, and
power.
106
Reading a Graph
Experiment measuring the density of water:
A beaker containing different volumes of water is weighed. The mass of
the beaker, M, is related to the volume of the water, V, in the following
manner:
M = V + M0
Plot: M against V
Slope =  = 0.9933
y-intercept = M0 = 30.283
Unit:
(i) Unit for y-intercept
= (unit for y) = ________
(ii) Slope = y / x
Unit for slope
= (unit for y)  (unit for x)
= ________
So,  =
and M0 =
107
Reading a Graph
Mass of Beaker vs. Volume for Water (273 K)
•
The axes do not have to start
from zero. In the previous
graph, if both axes start at
zero, the graph is squeezed
and space is wasted.
The R2-value indicates how
good the data fit the
trendline. 0 < R2 < 1. For a
perfect fit (e.g., fitting a
straight line to two data
points), R2 = 1. The closer to
1 it is, the better is the fitting.
60
50
40
30
20
10
0
y = 0.9933x + 30.283
2
R = 0.9787
0
5
10
15
20
25
Volume (ml)
Mass of Beaker vs. Volume for Water (273 K)
47
Mass of Beaker (g)
•
Mass of Beaker (g)
Also note the followings:
46.5
46
45.5
R2 = 1.00000000000
45
15
15.5
16
16.5
Volume (ml)
108
Exercise
In a second-order reaction, two molecules of A react to form
molecule B. The concentration of A, a, changes with time as:
a=
a0
1 + kta0
where a = concentration of A (variable)
a0 = initial concentration of A (i.e., a at t = 0, a constant)
k = “rate constant” (constant)
t = time (variable)
The following is a set of experimental data:
t (s)
199
246
367
a (mol·dm–3)
0.0094 0.0079 0.056
686
1200
0.0027 0.0015
Rearrange the formula in the form of y = mx + c, make a plot,
and determine k and a0 (with appropriate units).
Ans:
k = 0.42 dm3·mol–1·s–1
a0 = 0.041 mol·dm–3
109
Linear Scale vs. Log-Scale
The following graph shows the pH of a weak acid at different
concentration, x0. We can tell the pH at x0 = 0.01 M, 0.02 M, … But it
is hard to tell the pH below 0.01 M, where pH  3 – 5.
pH
x0 (M)
110
Linear Scale vs. Log-Scale
The part for small x0 can be expanded with “log-scale” as shown below.
We can tell easily that pH = 5.0 at x0 = 0.00001 M and pH = 3.45 at x0
= 0.001 M.
pH
x0 (M)
111
Comparison
Linear scale: equal distance, equal increment.
Linear scale: equal distance, equal multiplication factor.
0.00009
0.00002
0.00003
0.09
0.02
0.03
112
Reading Log-Scale
= 105
= 104
= 103
1/3
0.00001
= 105
= 102
= 101
= 100
1/10
= 104.666667
= 0.000022
0.0001
= 104
= 104.1
= 0.000079
113
Exercise
http://electrochem.usask.ca/Chem112/Notes/L22.pdf
Shown on the right is
the phase diagram for
carbon. Find out:
(a)Transition pressure
for: graphite 
diamond at 1000 °C
(b)Minimum
temperature and
pressure for melting
graphite into liquid
(c)Boiling point of
carbon liquid at 500
atm
Ans. (a) 25000 atm (b)
3600 °C, 52 atm (c)
3700 °C
114