1版权所有：华东理工大学物理化学教研室
Physical Chemistry
Peter Atkins
(Sixth edition)
Bilingual Program
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Part 1,Equilibrium
Part 3,Change
Part 2,Statistics
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
An Outline
Part 1,Equilibrium
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Part 1,Equilibrium
Part 3,Change
Part 2,Statistics
An Outline
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Part 2,Statistics
11,Quantum theory,introduction and principles
12,Quantum theory,techniques and applications
13,Atomic structure and atomic spectra
14,Molecular structure
15,Molecular symmetry
16,Spectroscopy 1
17,Spectroscopy 2
18,Spectroscopy 3
19,Statistical thermodynamics,the concepts
20,Statistical thermodynamics,the machinery
21,Diffraction techniques
22,The electric and magnetic properties of molecules
23,Macromolecules and colloids
An Outline
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Part 1,Equilibrium
Part 3,Change
Part 2,Statistics
An Outline
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24,Molecules in motion
25,The rates of chemical reactions
26,The kinetics of complex reactions
27,Molecular reaction dynamics
28,Processes at solid surfaces
29,Dynamic electrochemistry
Part 3,Change
An Outline
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Following each chapter,there are Exercise and
Problem sections,An Exercise is a straightforward,
direct application of an item in the text,A problem
is more complex and may draw on the literature.
The further problems are summarized in Micro-
Projects in the end of each Part,These Micro-
Projccts are designed to draw on knowledge from
all the chapters in each part,The MicroProjccts are
intended to be helpful when reviewing the material
of each part of the text,and also provide some
interesting applications,
An Outline
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Part 1,Equilibrium
Bilingual Program
0,Introduction
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Introduction
Physical chemistry is the branch of chemistry that
develops principles of the subject,Its concepts are used
to explain observations on the physical and chemical
properties of matter,Physical chemistry is also essen-
tial for developing and explaining the modern techni-
ques used to determine the structure and properties of
matter,such as new synthetic materials and biological
macromolecules.
In this section,we will concentrate on concepts of:
Matter and Energy
System and its surroundings
State and Equations of states
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0,Basic concepts
0-1 Fundamentals
1),Law,a summary of experience.
2),Hypothesis,a guess at an explanation in
terms of more fundamental concepts.
3),Model,a simplified version of the system
that focuses on the essentials of the
problem.
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0-2 Matter
A substance is a distinct,pure form of matter,The amount
of substance,n,in a sample is reported in terms of a unit
called a mole (mol),The formal definition of 1 mol is that it
is the amount of substance that contains as many objects as
there are atoms in exactly 12 g of carbon-12,This number
is found experimentally to be around 6.02× 1023,If a
sample contains N entities,the amount of substance it
contains is n = N/NA,where NA is the Avogadro constant:
NA = 6.02× 1023 mol-1,Note that NA is not a pure number.
1),Substance:
0,Basic concepts
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0-2 Matter
2),Intensive property:
a property that is independent of the amount of
the substance,
3),Extensive property:
a property that depends on the amount of the
substance.
1),Substance:
0,Basic concepts
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0-3 Energy
The central concept of all explanations in physical
chemistry is that of energy,Briefly,energy is the
capacity to do work.
There are two contributions to the total energy of
a system from the matter it contains,The kinetic
energy and the potential energy.
0,Basic concepts
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0-3 Energy
1),Kinetic energy:
the energy it possesses as a result of its motion.
2),Potential energy:
the energy it possesses as a result of its position.
3),The unit of energy,
Joule (J),SI unit; 1 J = 1 kg m2 s-2,
0,Basic concepts
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0- 4 Systems and surroundings
For the purposes of physical chemistry,the universe is
divided into two parts,the system and its surroundings.
The system is the part of the world in which we have a
special interest,The surroundings are where we make
our measurements,The type of system depends on the
characteristics of the boundary that divides it from the
surroundings.
0,Basic concepts
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(a),An open system:
can exchange matter and
energy with its surroundings.
(b),A closed system:
can exchange energy with its
surroundings,but it cannot
exchange matter.
(c),An isolated system:
can exchange neither energy nor
matter with its surroundings.
Systems
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0-5 The equation of States
1),State:
the entire property of a system,
2),Function of states:
the property determined by sole value of states.
3),Equation of states:
a mathematical relation that interrelates
variables of function of states,
0,Basic concepts
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure
substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
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Part 1,Equilibrium
Bilingual Program
1,The properties of gases
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This chapter establishes the properties of gases that
will be used throughout the text,It begins with an
account of an idealized version of a gas,a perfect gas,
and shows how its equation of state may be assembled
experimentally,We then see how this relation between
the properties of the gas can be explained in terms of
the kinetic model,in which the gas is represented by a
collection of point masses in continuous random
motion,Finally,we see how the properties of real
gases differ from those of a perfect gas,and construct
an equation of state that describes their properties,
1,The properties of gases
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The perfect gas
1.1 The states of gases
1.2 The gas laws
1.3 The kinetic model of gases
Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
1,The properties of gases
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The perfect gas
The simplest state of matter is a gas,a form of matter
that fills any container it occupies,A gas may be
pictured as a collection of molecules in continuous
random motion,with speeds that increase as the
temperature is raised,A gas differs from a liquid in
that the molecules of a gas are widely separated from
one another and move in paths that are largely
unaffected by intermolecular forces.
1,The properties of gases
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1.1 The states of gases
The physical state of a sample of a substance is defined by
its physical properties,and two samples of a substance that
have the same physical properties are in the same state.The
state of a pure gas is specified by giving the values of its
volume,V,amount of substance,n,pressure,p,and temp-
erature,T,However,it has been established experimentally
that it is sufficient to specify only three of these variables,
for then the fourth variable is fixed,That is,it is an experi-
mental fact that each substance is described by an equation
of state,an equation that interrelates these four variables.
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1),Pressure,p
Definition,Pressure is defined as force divided
by the area to which the force is applied.
Unit,Pascal (Pa),1 Pa= 1 Nm-2; the SI unit,
It is defined as 1 newton per meter squared;
Standard pressure:,?p Pa10
5p
1.1 The states of gases
The general form of an equation of state is,p= f (T,V,n)
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Mechanical equilibrium
Two gases are in separate
containers that share a
common movable wall
The stage when the two pressures
are equal and the wall has no
further tendency to move,This
condition of equality of pressure
on either side of a movable wall is
a state of mechanical equilibrium
between the two gases,
p1= p2
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2),Temperature,T
Temperature scales:
The Celsius scale of temperature,?,℃
The thermodynamic temperature scale,T,K
T / K=? / ℃ +273.15
The temperature is the property that tells us the
direction of the flow of energy,
1.1 The states of gases
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Thermal equilibrium
Energy flows as heat from a region
at a higher T to one at a lower T if
the two are in contact through a
diathermic wall,as in (a) and (c).
However,if the two regions have
identical temperatures,there is no
net transfer of energy as heat even
t h ou gh t h e t wo regi on s are
separated by a diathermic wall (b).
The latter condition corresponds to
t h e t w o r e g i o n s b e i n g at
thermal equilibrium,T1= T2
1.1 The states of gases
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The Zeroth Law of
thermodynamics
Zeroth Law of
thermodynamics
If an object A is in
thermal equilibrium
with B and if B is in
thermal equilibrium
with C,then C is in
thermal equilibrium
with A.
1.1 The states of gases
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3),Volume,V,m3
4),The amount of substance
n,mole (mol)
1.1 The states of gases
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1),Three important gas Laws
3),Mixtures of gases
2),The combined gas Law
1.2 The gas laws
4),Mole fractions and partial pressures
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a),Boyle’s Law- at constant T
At constant T,p and V of a fixed
amount of gas are related by
pV = constant
Each of the curves in the graph
corresponds to a single T and is
called an isotherm,
According to Boyle's law,each of
the isotherms is hyperbola.
1.2 The gas laws — 1).Three important gas Laws
The p-V dependence of a fixed
amount of perfect gas at different
T.
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Vp
1?
pV
1?
At constant T,the pressure
of a sample of gas is inversely
proportional to its volume,
and the volume it occupies is
inversely proportional to its
pressure.p against 1/V
1.2 The gas laws — 1).Three important gas Laws
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The limiting law
Boyle's law is valid only at low pressures;and that
real gases obey it only in the limit of the pressure
approaching zero ( p?0).
Equations that are valid in this limiting sense will
be signaled by a oon the equation number in our
text book.
1.2 The gas laws — 1).Three important gas Laws
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b),Charls’s Law- at constant p
The variation of V with T
c o n s t a n t?TV
The volume of any gas
should extrapolate to zero
at? = -273 ℃ or T= 0 K.
1.2 The gas laws — 1).Three important gas Laws
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At a given pressure and temperature,the molar volume
of a gas is approximately the same; the volume of a
sample of gas is proportional to the amount of molecules
present,
Avogadro's principle,The equal volumes of gases at the
same pressure and temperature contain the same
numbers of molecules.
c),Avogadro’s Principle
nV c o n s t a n t
at constant T and p:
nVV m?
1.2 The gas laws — 1).Three important gas Laws
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b),Charls’s Law
c o n s t a n t?Tp
The pressure varies
linearly with the
temperature,and
extrapolates to zero
at T= 0K
The variation of p with T
1.2 The gas laws — 1).Three important gas Laws
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When combine the Boyle’s Law,Charle’s Law and
Avogadro’s Principle,then:
2),The combined gas Law
the perfect gas equation
nTpV c o n s t a n t
n R TpV?
1.2 The gas laws
RT
M
mpV? RTpV?m
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2),The combined gas Law
the perfect gas equation
A gas that obeys this Equation exactly under
all conditions is called a perfect gas,A real gas
behaves more like a perfect gas the lower the
pressure,and is described exactly by this
equation in the limit p? 0,
n R TpV?
1.2 The gas laws
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When combine the Boyle’s Law,Charle’s Law and
Avogadro’s Principle,then:
2),The combined gas Law
the perfect gas equation
The constant,R,is called gas constant,and is the
same for all gases.
1-1-
1-1-2
K m o l J314.8
K m o l a t m L1020578.8
K15.273m o l1
L414.22a t m1
R
1.2 The gas laws
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2),The combined gas Law
R,The gas constant
8.31451 JK-1mol-1
8.20578 10-2LatmK-1mol-1
8.31451 10-2LbarK-1mol-1
1.98722 cal K-1mol-1
The perfect gas equation
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2),The combined gas Law
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The pressure exerted by a mixture of perfect gases
is the sum of the partial pressures of the gases.
Dalton’s Law
BA ppp
V
RTnp J
J?
:Js u b s t a n c ee a c hf o r
1.2 The gas laws — 3),Mixtures of gases
Generally?
ipp
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The partial pressure of a perfect gas is the pressure that
it would exert if it occupied the container alone..
a),Dalton’s Law
gas mixture
n=nA+ nB
P V
gas A
nA
P V
gas B
nA
P V
p= pA+ pB
V
RTnnp )(
BA BABA ppVRTnVRTn
1.2 The gas laws — 3),Mixtures of gases
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A container of volume 10.0 L holds 1.00 mol N2 and
3.00 mol H2 at 298 K,What is the total pressure in
atmospheres if each component behaves as a perfect
gas?
Example;BA ppp
V
RTn
p JJ?
Since
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a t m879
10.0L
298K)m olL a t m K10( 8,20 6
3,00 m ol )( 1,00 m ol
)(
1-1-2-
BABA
.?
V
RT
nnpppWe have
Example
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The mole fraction,xJ,is the amount of
component J expressed as a fraction of the
total amount of molecules,n,in the sample:
4),Mole fractions and partial pressures
BAJJ nnnnnx
1.2 The gas laws
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1AA xx
xJ = 0,no J molecules are present;
xJ = 1,only J molecules are present
For example,a mixture of 1.0 mol N2 and 3.0
mol H2,and therefore of 4.0 mol molecules in
all,consists of mole fractions 0.25 of N2 and
0.75 of H2,It follows from the definition xJ,
that,whatever the composition of the mixture
1.2 The gas laws
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It follows that:
4),Mole fractions and partial pressures
p
pxx
pp
)( BA
BA
pxp JJ?
The partial pressure of a gas J in a
mixture is defined as:
The partial pressures pA and
pB of a binary mixture of
gases of total pressure p as
the composition changes
from pure A to pure B,
1.2 The gas laws
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The perfect gas
1.1 The states of gases
1.2 The gas laws
1.3 The kinetic model of gases
Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
1,The properties of gases
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1),Molecular speeds
3),The mean free path
2),The collision frequency
1.3 The kinetic model of gases
Three assumptions for the kinetic model,
1),The size of the molecules is negligible,in the sense
that their diameters are much smaller than the
average distance travelled between collisions.
2),The molecules do not interact,except that they
make perfectly elastic collisions,
3),The gas consists of molecules of mass m in ceaseless
random motion.
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Maxwell distribution of speeds:
RTMvev
RT
Mvf 2223 2
2
π4)(
1) Molecular speeds
1.3 The kinetic model of gases
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218
MRTc?
The mean speed,c
The most probable speed,?c
212
M
RTc
21
r e l
8
πμ
kTc
The relative mean speed,
relc
BA
BA
mm
mmμ
213
MRTc
The root mean square speed
1.3 The kinetic model of gases
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2),The collision frequency,z
When there are N molecules in a volume V,The
collision frequency,z,the number of collisions made
by one molecule divided by the time interval during
which the collisions are counted,is expressed as
2πdσ?Nσ
r e lcz?
where,is collision cross-section and d is
the collision diameter.
σVN?N
) / 2( BA ddd
1.3 The kinetic model of gases
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The equations tell us that the collision frequency
increases with increasing temperature in a sample
held at constant volume,The reason is that the mean
relative speed increases with temperature,And at
constant temperature,the collision frequency is
proportional to the pressure,
kT
pcz r e lσ?
In terms of pressure
1.3 The kinetic model of gases
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zcλ?
3),The mea free path,λ
Since and,
kT
pcz r e lσ? cc 212?r e l
The mean free path is the average distance a
molecule travels between collisions,If a molecule
collides with a frequency z,it spends a time 1/z
in free flight between collisions,and therefore
travels a distance (l/z),The mean free path is:
1.3 The kinetic model of gases
c
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3),The mea free path,λ
then:
σp
kTλ
212
In a sample of constant volume,the pressure is
proportional to T,so T/p remains constant when the
temperature is increased,So the mean free path is
independent of the temperature.The distance
between collisions is determined by the number of
molecules present in the given volume,not by the
speed at which they travel.
1.3 The kinetic model of gases
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The perfect gas
1.1 The states of gases
1.2 The gas laws
1.3 The kinetic model of gases
Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
1,The properties of gases
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Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
Real gases do not obey the perfect gas law
exactly because molecules interact with each
other,Two important inter-molecular forces
exist,repulsive forces attractive forces.
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Repulsive forces between molecules assist expansion
and attractive forces assist compression,They are
significant only when molecules are almost in contact:
short-range interactions,
Attractive forces have a relatively long range and
are effective over several molecular diameters,They
are important when the molecules are fairly close
together but not necessarily touching,Attractive
forces are ineffective when the molecules are far
apart.
1.4 Molecular interactions
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1),The compression factor
The compression factor,Z,is defined as
RT
pVZ m?
For a perfect gas,Z = 1 under all conditions; the
deviation of Z from 1 is a measure of departure
from perfect behaviour.
Z R TpV m?
1.4 Molecular interactions
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Results by plotting the
compression factor,Z,
against pressure for
several gases at 0 ℃,
At very low pressures,Z? 1.
At high pressures,Z> 1,
Repulsive forces are now dominant.
At intermediate pressures,Z<1
For most gases,the attractive forces
are dominant.,
A perfect gas has Z = 1 at all
pressures.
1.4 Molecular interactions
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2),Virial coefficients
At large molar volumes and high
temperatures the real isotherms
do not differ greatly from perfect
isotherms,The small differences
suggest that the perfect gas law is
in fact the first term in the
expression of the form:
1.4 Molecular interactions
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These are two versions of the
virial equation of state,The
coefficients B,C,...,which
depend on the T,are the second,
third,..,virial coefficients; the
first virial coefficient is 1,
)''1( 2 pCpBRTpV m
)1( 2
mm
m V
C
V
BRTpV
1.4 Molecular interactions
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0)( 0, PTpZ
> At higher T
0)( 0, PTpZ
< At lower T
0)( 0,PTpZ
At TB
TB is so called the Boyle
Temperature
0)( 0,PTppV m
At TB
Boyle Temperature,TB
1.4 Molecular interactions
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Boyle Temperature,TB
At Boyle temperature,TB
the properties of the real
gas coincide with those of a
perfect gas as p? 0; and
pVm? R TB
1.4 Molecular interactions
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3),Condensation
At C,all similarity to perfect behaviour
is lost,This stage is represented by the
horizontal line CDE,At the point just
to the left of C,a liquid appears,and
there are two phases in the system in
this stage,
Experimental isotherms of carbon
dioxide at several temperatures
1.4 Molecular interactions
The pressure corresponding to the line
CDE,when both liquid and vapour are
present in equilibrium,is called the
vapour pressure of the liquid at
the temperature of the experiment.
At E,the sample is entirely liquid,
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4),Critical constants
At and above Tc,the sample
has a single phase,The liquid
phase of a substance does not
form ab ove the cri tical
temperature.
Critical point
1.4 Molecular interactions
Tc,the critical point of the
gas,Critical constants are
different from gas to gas,
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Critical point
0
TV
p
m
m
02
2
TV
p
1.4 Molecular interactions
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4),Critical constants
Critical point The temperature,pressure,and molar volume at the critical point
are called,
Critical temperature,Tc
Critical pressure,pc
Critical molar volume,Vc
Tc,pc,Vc -critical constants
1.4 Molecular interactions
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1.5 The van der Waals equation
For the perfect gas:
1),The size of molecules
is negligible;
2),Molecules do not
interact;
…
pV=nRT
For real gases:
1),The molecule itself
occupies a volume;
2),There are interactions
among molecules;
…
(V － △ V ) =nRT(p+△ p)
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n R TnbVV anp
)(
2
2
2
mm V
a
bV
RTp?
1),Expressions the van der Waals equation
RTVp
)(
m2
mV
a? b?
2
mV
ap
bV
(p+△ p) (V -△ V ) =nRT
1.5 The van der Waals equation
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1),Expressions the van der Waals equation
The constants a and b are the van der Waals
coefficients,They are characteristic of each gas
but independent of the temperature.
,the internal pressure due to intermolecular
forces,
2
mV
a
1.5 The van der Waals equation
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2),The features of the equation
Perfect gas isotherms are obtained at high
temperatures and large molar volumes.
(1)
2
mm V
a
bV
RTp?
T RT?
RTpV?mV Vm b? Vm -b? Vm
2mVa
1.5 The van der Waals equation
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2),The features of the equation
The critical constants are related to the van
der Waals coefficients.
(2)
At the critical point:
1.5 The van der Waals equation
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Critical point
1.5 The van der Waals equation
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2),The features of the equation
02)( 3
m
2
mm
V abV RTV p 062
432
2
mmm )( V
a
bV
RT
V
p
2
mm V
a
bV
RTp?
0
TV
p
02
2
TV
p
1.5 The van der Waals equation
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The critical compression factor
3 750
8
3,
c
cc
c RT
VpZ
bV 3?c 227 bap?c
Rb
aT
27
8?
c
c
cc
T
VpR
3
8?
cTcp
RcV
For all gases,the critical compression factor,Zc,is
constant.
1.5 The van der Waals equation
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Example
Using the van der Waals equation to estimate a molar
volume,the molar volume of CO2 at 500K and 100 atm
by treating it as van der Waals gas.
Method,To rearrange the van der Waals equation as a
form for the molar volume,multiplying the both sides of
the van der Waals equation by,)( bVV?
m3m
02
mm
V
a
bV
RTp)( bVV?
m3m
0m2m3m
p
abV
p
aV
p
RTbV
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The van der Waals’s coefficients,a and b are as following
a = 3.640 atmL2 mol-2,b = 4.267× 10-2 L mol-1,and
RT/p= 0.410 L mol-1,
0105511064345230 3m22m3m,.,VVV
The acceptable root is L mol-1.
3700m,?V
Answer:
Example
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1.6 The principle of corresponding states
An important general technique in science for
comparing the properties of objects is to choose
a related fundamental property of the same kind
and to set up a relative scale on that basis,The
critical constants are characteristic properties of
gases,so a scale can be set up by using them as
yardsticks.
1),Relative variables
2),The corresponding states
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The reduced variables of a gas may be obtained by
dividing the actual variable by the corresponding
critical constant:
1),Reduced variables
cr ppp?
cmr VVV?
cr TTT?
The reduced pressure:
The reduced temperature,
The reduced volume:
1.6 The principle of corresponding states
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The corresponding form in reduced variables:
22
crcr
cr
cr VV
a
bVV
TRTpp?
2
mm V
a
bV
RTp?
The van der Waals Equation:
cr ppp? crm VVV? cr TTT?
2),The generalized van der Waals equation
1.6 The principle of corresponding states
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Substitute the following critical constants for the above
equation
bV 3?c 227 bap?c
Rb
aT
27
8?
c
229327
8
27 2 rr
rr
)( Vb
a
bbVb
aT
b
ap?
22
crcr
cr
cr VV
a
bVV
TRTpp?
then
1.6 The principle of corresponding states
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2
3
13
8
rr
r
r VV
Tp?
The generalized van der Waals equation:
2
mm V
a
bV
RTp?
mV
RTp?
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
Physical Chemistry
Peter Atkins
(Sixth edition)
Bilingual Program
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Part 1,Equilibrium
Part 3,Change
Part 2,Statistics
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
An Outline
Part 1,Equilibrium
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Part 1,Equilibrium
Part 3,Change
Part 2,Statistics
An Outline
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Part 2,Statistics
11,Quantum theory,introduction and principles
12,Quantum theory,techniques and applications
13,Atomic structure and atomic spectra
14,Molecular structure
15,Molecular symmetry
16,Spectroscopy 1
17,Spectroscopy 2
18,Spectroscopy 3
19,Statistical thermodynamics,the concepts
20,Statistical thermodynamics,the machinery
21,Diffraction techniques
22,The electric and magnetic properties of molecules
23,Macromolecules and colloids
An Outline
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Part 1,Equilibrium
Part 3,Change
Part 2,Statistics
An Outline
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24,Molecules in motion
25,The rates of chemical reactions
26,The kinetics of complex reactions
27,Molecular reaction dynamics
28,Processes at solid surfaces
29,Dynamic electrochemistry
Part 3,Change
An Outline
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Following each chapter,there are Exercise and
Problem sections,An Exercise is a straightforward,
direct application of an item in the text,A problem
is more complex and may draw on the literature.
The further problems are summarized in Micro-
Projects in the end of each Part,These Micro-
Projccts are designed to draw on knowledge from
all the chapters in each part,The MicroProjccts are
intended to be helpful when reviewing the material
of each part of the text,and also provide some
interesting applications,
An Outline
9版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
0,Introduction
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Introduction
Physical chemistry is the branch of chemistry that
develops principles of the subject,Its concepts are used
to explain observations on the physical and chemical
properties of matter,Physical chemistry is also essen-
tial for developing and explaining the modern techni-
ques used to determine the structure and properties of
matter,such as new synthetic materials and biological
macromolecules.
In this section,we will concentrate on concepts of:
Matter and Energy
System and its surroundings
State and Equations of states
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0,Basic concepts
0-1 Fundamentals
1),Law,a summary of experience.
2),Hypothesis,a guess at an explanation in
terms of more fundamental concepts.
3),Model,a simplified version of the system
that focuses on the essentials of the
problem.
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0-2 Matter
A substance is a distinct,pure form of matter,The amount
of substance,n,in a sample is reported in terms of a unit
called a mole (mol),The formal definition of 1 mol is that it
is the amount of substance that contains as many objects as
there are atoms in exactly 12 g of carbon-12,This number
is found experimentally to be around 6.02× 1023,If a
sample contains N entities,the amount of substance it
contains is n = N/NA,where NA is the Avogadro constant:
NA = 6.02× 1023 mol-1,Note that NA is not a pure number.
1),Substance:
0,Basic concepts
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0-2 Matter
2),Intensive property:
a property that is independent of the amount of
the substance,
3),Extensive property:
a property that depends on the amount of the
substance.
1),Substance:
0,Basic concepts
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0-3 Energy
The central concept of all explanations in physical
chemistry is that of energy,Briefly,energy is the
capacity to do work.
There are two contributions to the total energy of
a system from the matter it contains,The kinetic
energy and the potential energy.
0,Basic concepts
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0-3 Energy
1),Kinetic energy:
the energy it possesses as a result of its motion.
2),Potential energy:
the energy it possesses as a result of its position.
3),The unit of energy,
Joule (J),SI unit; 1 J = 1 kg m2 s-2,
0,Basic concepts
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0- 4 Systems and surroundings
For the purposes of physical chemistry,the universe is
divided into two parts,the system and its surroundings.
The system is the part of the world in which we have a
special interest,The surroundings are where we make
our measurements,The type of system depends on the
characteristics of the boundary that divides it from the
surroundings.
0,Basic concepts
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(a),An open system:
can exchange matter and
energy with its surroundings.
(b),A closed system:
can exchange energy with its
surroundings,but it cannot
exchange matter.
(c),An isolated system:
can exchange neither energy nor
matter with its surroundings.
Systems
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0-5 The equation of States
1),State:
the entire property of a system,
2),Function of states:
the property determined by sole value of states.
3),Equation of states:
a mathematical relation that interrelates
variables of function of states,
0,Basic concepts
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure
substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
20版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
1,The properties of gases
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This chapter establishes the properties of gases that
will be used throughout the text,It begins with an
account of an idealized version of a gas,a perfect gas,
and shows how its equation of state may be assembled
experimentally,We then see how this relation between
the properties of the gas can be explained in terms of
the kinetic model,in which the gas is represented by a
collection of point masses in continuous random
motion,Finally,we see how the properties of real
gases differ from those of a perfect gas,and construct
an equation of state that describes their properties,
1,The properties of gases
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The perfect gas
1.1 The states of gases
1.2 The gas laws
1.3 The kinetic model of gases
Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
1,The properties of gases
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The perfect gas
The simplest state of matter is a gas,a form of matter
that fills any container it occupies,A gas may be
pictured as a collection of molecules in continuous
random motion,with speeds that increase as the
temperature is raised,A gas differs from a liquid in
that the molecules of a gas are widely separated from
one another and move in paths that are largely
unaffected by intermolecular forces.
1,The properties of gases
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1.1 The states of gases
The physical state of a sample of a substance is defined by
its physical properties,and two samples of a substance that
have the same physical properties are in the same state.The
state of a pure gas is specified by giving the values of its
volume,V,amount of substance,n,pressure,p,and temp-
erature,T,However,it has been established experimentally
that it is sufficient to specify only three of these variables,
for then the fourth variable is fixed,That is,it is an experi-
mental fact that each substance is described by an equation
of state,an equation that interrelates these four variables.
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1),Pressure,p
Definition,Pressure is defined as force divided
by the area to which the force is applied.
Unit,Pascal (Pa),1 Pa= 1 Nm-2; the SI unit,
It is defined as 1 newton per meter squared;
Standard pressure:,?p Pa10
5p
1.1 The states of gases
The general form of an equation of state is,p= f (T,V,n)
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Mechanical equilibrium
Two gases are in separate
containers that share a
common movable wall
The stage when the two pressures
are equal and the wall has no
further tendency to move,This
condition of equality of pressure
on either side of a movable wall is
a state of mechanical equilibrium
between the two gases,
p1= p2
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2),Temperature,T
Temperature scales:
The Celsius scale of temperature,?,℃
The thermodynamic temperature scale,T,K
T / K=? / ℃ +273.15
The temperature is the property that tells us the
direction of the flow of energy,
1.1 The states of gases
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Thermal equilibrium
Energy flows as heat from a region
at a higher T to one at a lower T if
the two are in contact through a
diathermic wall,as in (a) and (c).
However,if the two regions have
identical temperatures,there is no
net transfer of energy as heat even
t h ou gh t h e t wo regi on s are
separated by a diathermic wall (b).
The latter condition corresponds to
t h e t w o r e g i o n s b e i n g at
thermal equilibrium,T1= T2
1.1 The states of gases
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The Zeroth Law of
thermodynamics
Zeroth Law of
thermodynamics
If an object A is in
thermal equilibrium
with B and if B is in
thermal equilibrium
with C,then C is in
thermal equilibrium
with A.
1.1 The states of gases
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3),Volume,V,m3
4),The amount of substance
n,mole (mol)
1.1 The states of gases
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1),Three important gas Laws
3),Mixtures of gases
2),The combined gas Law
1.2 The gas laws
4),Mole fractions and partial pressures
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a),Boyle’s Law- at constant T
At constant T,p and V of a fixed
amount of gas are related by
pV = constant
Each of the curves in the graph
corresponds to a single T and is
called an isotherm,
According to Boyle's law,each of
the isotherms is hyperbola.
1.2 The gas laws — 1).Three important gas Laws
The p-V dependence of a fixed
amount of perfect gas at different
T.
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Vp
1?
pV
1?
At constant T,the pressure
of a sample of gas is inversely
proportional to its volume,
and the volume it occupies is
inversely proportional to its
pressure.p against 1/V
1.2 The gas laws — 1).Three important gas Laws
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The limiting law
Boyle's law is valid only at low pressures;and that
real gases obey it only in the limit of the pressure
approaching zero ( p?0).
Equations that are valid in this limiting sense will
be signaled by a oon the equation number in our
text book.
1.2 The gas laws — 1).Three important gas Laws
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b),Charls’s Law- at constant p
The variation of V with T
c o n s t a n t?TV
The volume of any gas
should extrapolate to zero
at? = -273 ℃ or T= 0 K.
1.2 The gas laws — 1).Three important gas Laws
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At a given pressure and temperature,the molar volume
of a gas is approximately the same; the volume of a
sample of gas is proportional to the amount of molecules
present,
Avogadro's principle,The equal volumes of gases at the
same pressure and temperature contain the same
numbers of molecules.
c),Avogadro’s Principle
nV c o n s t a n t
at constant T and p:
nVV m?
1.2 The gas laws — 1).Three important gas Laws
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b),Charls’s Law
c o n s t a n t?Tp
The pressure varies
linearly with the
temperature,and
extrapolates to zero
at T= 0K
The variation of p with T
1.2 The gas laws — 1).Three important gas Laws
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When combine the Boyle’s Law,Charle’s Law and
Avogadro’s Principle,then:
2),The combined gas Law
the perfect gas equation
nTpV c o n s t a n t
n R TpV?
1.2 The gas laws
RT
M
mpV? RTpV?m
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2),The combined gas Law
the perfect gas equation
A gas that obeys this Equation exactly under
all conditions is called a perfect gas,A real gas
behaves more like a perfect gas the lower the
pressure,and is described exactly by this
equation in the limit p? 0,
n R TpV?
1.2 The gas laws
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When combine the Boyle’s Law,Charle’s Law and
Avogadro’s Principle,then:
2),The combined gas Law
the perfect gas equation
The constant,R,is called gas constant,and is the
same for all gases.
1-1-
1-1-2
K m o l J314.8
K m o l a t m L1020578.8
K15.273m o l1
L414.22a t m1
R
1.2 The gas laws
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2),The combined gas Law
R,The gas constant
8.31451 JK-1mol-1
8.20578 10-2LatmK-1mol-1
8.31451 10-2LbarK-1mol-1
1.98722 cal K-1mol-1
The perfect gas equation
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2),The combined gas Law
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The pressure exerted by a mixture of perfect gases
is the sum of the partial pressures of the gases.
Dalton’s Law
BA ppp
V
RTnp J
J?
:Js u b s t a n c ee a c hf o r
1.2 The gas laws — 3),Mixtures of gases
Generally?
ipp
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The partial pressure of a perfect gas is the pressure that
it would exert if it occupied the container alone..
a),Dalton’s Law
gas mixture
n=nA+ nB
P V
gas A
nA
P V
gas B
nA
P V
p= pA+ pB
V
RTnnp )(
BA BABA ppVRTnVRTn
1.2 The gas laws — 3),Mixtures of gases
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A container of volume 10.0 L holds 1.00 mol N2 and
3.00 mol H2 at 298 K,What is the total pressure in
atmospheres if each component behaves as a perfect
gas?
Example;BA ppp
V
RTn
p JJ?
Since
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a t m879
10.0L
298K)m olL a t m K10( 8,20 6
3,00 m ol )( 1,00 m ol
)(
1-1-2-
BABA
.?
V
RT
nnpppWe have
Example
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The mole fraction,xJ,is the amount of
component J expressed as a fraction of the
total amount of molecules,n,in the sample:
4),Mole fractions and partial pressures
BAJJ nnnnnx
1.2 The gas laws
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1AA xx
xJ = 0,no J molecules are present;
xJ = 1,only J molecules are present
For example,a mixture of 1.0 mol N2 and 3.0
mol H2,and therefore of 4.0 mol molecules in
all,consists of mole fractions 0.25 of N2 and
0.75 of H2,It follows from the definition xJ,
that,whatever the composition of the mixture
1.2 The gas laws
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It follows that:
4),Mole fractions and partial pressures
p
pxx
pp
)( BA
BA
pxp JJ?
The partial pressure of a gas J in a
mixture is defined as:
The partial pressures pA and
pB of a binary mixture of
gases of total pressure p as
the composition changes
from pure A to pure B,
1.2 The gas laws
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The perfect gas
1.1 The states of gases
1.2 The gas laws
1.3 The kinetic model of gases
Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
1,The properties of gases
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1),Molecular speeds
3),The mean free path
2),The collision frequency
1.3 The kinetic model of gases
Three assumptions for the kinetic model,
1),The size of the molecules is negligible,in the sense
that their diameters are much smaller than the
average distance travelled between collisions.
2),The molecules do not interact,except that they
make perfectly elastic collisions,
3),The gas consists of molecules of mass m in ceaseless
random motion.
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Maxwell distribution of speeds:
RTMvev
RT
Mvf 2223 2
2
π4)(
1) Molecular speeds
1.3 The kinetic model of gases
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218
MRTc?
The mean speed,c
The most probable speed,?c
212
M
RTc
21
r e l
8
πμ
kTc
The relative mean speed,
relc
BA
BA
mm
mmμ
213
MRTc
The root mean square speed
1.3 The kinetic model of gases
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2),The collision frequency,z
When there are N molecules in a volume V,The
collision frequency,z,the number of collisions made
by one molecule divided by the time interval during
which the collisions are counted,is expressed as
2πdσ?Nσ
r e lcz?
where,is collision cross-section and d is
the collision diameter.
σVN?N
) / 2( BA ddd
1.3 The kinetic model of gases
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The equations tell us that the collision frequency
increases with increasing temperature in a sample
held at constant volume,The reason is that the mean
relative speed increases with temperature,And at
constant temperature,the collision frequency is
proportional to the pressure,
kT
pcz r e lσ?
In terms of pressure
1.3 The kinetic model of gases
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zcλ?
3),The mea free path,λ
Since and,
kT
pcz r e lσ? cc 212?r e l
The mean free path is the average distance a
molecule travels between collisions,If a molecule
collides with a frequency z,it spends a time 1/z
in free flight between collisions,and therefore
travels a distance (l/z),The mean free path is:
1.3 The kinetic model of gases
c
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3),The mea free path,λ
then:
σp
kTλ
212
In a sample of constant volume,the pressure is
proportional to T,so T/p remains constant when the
temperature is increased,So the mean free path is
independent of the temperature.The distance
between collisions is determined by the number of
molecules present in the given volume,not by the
speed at which they travel.
1.3 The kinetic model of gases
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The perfect gas
1.1 The states of gases
1.2 The gas laws
1.3 The kinetic model of gases
Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
1,The properties of gases
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Real gases
1.4 Molecular interactions
1.5 The van de Waals equation
1.6 The principle of corresponding states
Real gases do not obey the perfect gas law
exactly because molecules interact with each
other,Two important inter-molecular forces
exist,repulsive forces attractive forces.
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Repulsive forces between molecules assist expansion
and attractive forces assist compression,They are
significant only when molecules are almost in contact:
short-range interactions,
Attractive forces have a relatively long range and
are effective over several molecular diameters,They
are important when the molecules are fairly close
together but not necessarily touching,Attractive
forces are ineffective when the molecules are far
apart.
1.4 Molecular interactions
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1),The compression factor
The compression factor,Z,is defined as
RT
pVZ m?
For a perfect gas,Z = 1 under all conditions; the
deviation of Z from 1 is a measure of departure
from perfect behaviour.
Z R TpV m?
1.4 Molecular interactions
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Results by plotting the
compression factor,Z,
against pressure for
several gases at 0 ℃,
At very low pressures,Z? 1.
At high pressures,Z> 1,
Repulsive forces are now dominant.
At intermediate pressures,Z<1
For most gases,the attractive forces
are dominant.,
A perfect gas has Z = 1 at all
pressures.
1.4 Molecular interactions
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2),Virial coefficients
At large molar volumes and high
temperatures the real isotherms
do not differ greatly from perfect
isotherms,The small differences
suggest that the perfect gas law is
in fact the first term in the
expression of the form:
1.4 Molecular interactions
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These are two versions of the
virial equation of state,The
coefficients B,C,...,which
depend on the T,are the second,
third,..,virial coefficients; the
first virial coefficient is 1,
)''1( 2 pCpBRTpV m
)1( 2
mm
m V
C
V
BRTpV
1.4 Molecular interactions
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0)( 0, PTpZ
> At higher T
0)( 0, PTpZ
< At lower T
0)( 0,PTpZ
At TB
TB is so called the Boyle
Temperature
0)( 0,PTppV m
At TB
Boyle Temperature,TB
1.4 Molecular interactions
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Boyle Temperature,TB
At Boyle temperature,TB
the properties of the real
gas coincide with those of a
perfect gas as p? 0; and
pVm? R TB
1.4 Molecular interactions
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3),Condensation
At C,all similarity to perfect behaviour
is lost,This stage is represented by the
horizontal line CDE,At the point just
to the left of C,a liquid appears,and
there are two phases in the system in
this stage,
Experimental isotherms of carbon
dioxide at several temperatures
1.4 Molecular interactions
The pressure corresponding to the line
CDE,when both liquid and vapour are
present in equilibrium,is called the
vapour pressure of the liquid at
the temperature of the experiment.
At E,the sample is entirely liquid,
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4),Critical constants
At and above Tc,the sample
has a single phase,The liquid
phase of a substance does not
form ab ove the cri tical
temperature.
Critical point
1.4 Molecular interactions
Tc,the critical point of the
gas,Critical constants are
different from gas to gas,
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Critical point
0
TV
p
m
m
02
2
TV
p
1.4 Molecular interactions
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4),Critical constants
Critical point The temperature,pressure,and molar volume at the critical point
are called,
Critical temperature,Tc
Critical pressure,pc
Critical molar volume,Vc
Tc,pc,Vc -critical constants
1.4 Molecular interactions
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1.5 The van der Waals equation
For the perfect gas:
1),The size of molecules
is negligible;
2),Molecules do not
interact;
…
pV=nRT
For real gases:
1),The molecule itself
occupies a volume;
2),There are interactions
among molecules;
…
(V － △ V ) =nRT(p+△ p)
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n R TnbVV anp
)(
2
2
2
mm V
a
bV
RTp?
1),Expressions the van der Waals equation
RTVp
)(
m2
mV
a? b?
2
mV
ap
bV
(p+△ p) (V -△ V ) =nRT
1.5 The van der Waals equation
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1),Expressions the van der Waals equation
The constants a and b are the van der Waals
coefficients,They are characteristic of each gas
but independent of the temperature.
,the internal pressure due to intermolecular
forces,
2
mV
a
1.5 The van der Waals equation
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2),The features of the equation
Perfect gas isotherms are obtained at high
temperatures and large molar volumes.
(1)
2
mm V
a
bV
RTp?
T RT?
RTpV?mV Vm b? Vm -b? Vm
2mVa
1.5 The van der Waals equation
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2),The features of the equation
The critical constants are related to the van
der Waals coefficients.
(2)
At the critical point:
1.5 The van der Waals equation
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Critical point
1.5 The van der Waals equation
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2),The features of the equation
02)( 3
m
2
mm
V abV RTV p 062
432
2
mmm )( V
a
bV
RT
V
p
2
mm V
a
bV
RTp?
0
TV
p
02
2
TV
p
1.5 The van der Waals equation
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The critical compression factor
3 750
8
3,
c
cc
c RT
VpZ
bV 3?c 227 bap?c
Rb
aT
27
8?
c
c
cc
T
VpR
3
8?
cTcp
RcV
For all gases,the critical compression factor,Zc,is
constant.
1.5 The van der Waals equation
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Example
Using the van der Waals equation to estimate a molar
volume,the molar volume of CO2 at 500K and 100 atm
by treating it as van der Waals gas.
Method,To rearrange the van der Waals equation as a
form for the molar volume,multiplying the both sides of
the van der Waals equation by,)( bVV?
m3m
02
mm
V
a
bV
RTp)( bVV?
m3m
0m2m3m
p
abV
p
aV
p
RTbV
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The van der Waals’s coefficients,a and b are as following
a = 3.640 atmL2 mol-2,b = 4.267× 10-2 L mol-1,and
RT/p= 0.410 L mol-1,
0105511064345230 3m22m3m,.,VVV
The acceptable root is L mol-1.
3700m,?V
Answer:
Example
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1.6 The principle of corresponding states
An important general technique in science for
comparing the properties of objects is to choose
a related fundamental property of the same kind
and to set up a relative scale on that basis,The
critical constants are characteristic properties of
gases,so a scale can be set up by using them as
yardsticks.
1),Relative variables
2),The corresponding states
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The reduced variables of a gas may be obtained by
dividing the actual variable by the corresponding
critical constant:
1),Reduced variables
cr ppp?
cmr VVV?
cr TTT?
The reduced pressure:
The reduced temperature,
The reduced volume:
1.6 The principle of corresponding states
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The corresponding form in reduced variables:
22
crcr
cr
cr VV
a
bVV
TRTpp?
2
mm V
a
bV
RTp?
The van der Waals Equation:
cr ppp? crm VVV? cr TTT?
2),The generalized van der Waals equation
1.6 The principle of corresponding states
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Substitute the following critical constants for the above
equation
bV 3?c 227 bap?c
Rb
aT
27
8?
c
229327
8
27 2 rr
rr
)( Vb
a
bbVb
aT
b
ap?
22
crcr
cr
cr VV
a
bVV
TRTpp?
then
1.6 The principle of corresponding states
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2
3
13
8
rr
r
r VV
Tp?
The generalized van der Waals equation:
2
mm V
a
bV
RTp?
mV
RTp?
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
