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
2版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
3,The First Law,the machinery
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In this chapter we begin to unfold some of the
power of thermodynamics by showing how to
establish relations between different
properties of a system,The procedure we use
is based on the experimental fact that the
internal energy and the enthalpy are state
functions,and we derive a number of
relations between observables by exploring
the mathematical consequences of these facts.
3,The First Law,the machinery
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3.1 State functions
1),Exact and inexact differentials
2),Changes in internal energy
3),Expansion coefficient
3.2 The temperature dependence of the enthalpy
1),Changes in enthalpy at constant volume
2),The isothermal compressibility
3),The Joule -Thomson effect
3.3 The reaction between Cv and Cp
3,The First Law,the machinery
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The initial state of the system is i and in
this state the internal energy is Ui,Work
is done by the system as it expands
adiabatically to a state f,In this state the
system has an internal energy Uf and the
work done on the system as it changes
along Path 1 from i to f is w,U is a
property of the state; w is a property of
the path,
3.1 State functions
1),Exact and inexact differential
fi UU d
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In Path 2,the initial and final states are
the same but in which the expansion is
not adiabatic,In this path an energy q'
enters the system as heat and the work
w' is not the same as w,The work and
the heat are path functions,
1),Exact and inexact differential
fi qq p a t h,d
3.1 State functions
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△ U,independent of the path,an exact differential.An exact
differential is an infinitesimal quantity which,when integra-
ted,gives a result that is independent of the path between
the initial and final states.
q or w,dependent on the path,an inexact differential.When a
system is heated,the total energy transferred as heat is the
sum of all individual contributions at each point of the path.
1),Exact and inexact differential
△ q qf - qi dq dq q
3.1 State functions
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Consider a perfect gas inside a cylinder fitted with a
piston,Let the initial state be T,Vi and the final state
be T,Vf,The change of state can be brought about in
many ways,of which the two simplest are following:
Path 1,in which there is free expansion against zero
external pressure; Path 2,in which there is reversible,
isothermal expansion,Calculate w,q,and?U for each
process.
Example -Calculating work,heat,and internal energy
版权所有：华东理工大学物理化学教研室 9
Method,To find a starting point for a calculation in
thermodynamics,it is often a good idea to go back to
first principles,and to look for a way of expressing
the quantity we are asked to calculate in terms of
other quantities that are easier to calculate,Because
the internal energy of a perfect gas arises only from
the kinetic energy of its molecules,it is independent of
volume; therefore,for any isothermal change,in
general?U = q+w,
Example -Calculating work,heat,and internal energy
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Answer,Since?U = 0 for both paths and?U = q+w.
The work of free expansion is zero,
in Path 1,w = 0 and q = 0;
for Path 2,the work is given by
so w = - nRT ln (Vf /Vi),and
q = nRT ln (Vf /Vi),
i
flndf
i V
VnR T
V
VnR Tw V
V
Example -Calculating work,heat,and internal energy
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For a closed system of constant composition,U is a func-
tion of V and T,When V changes to V + dV at constant T,
U changes to
VVUUU T d'
2),Changes in internal energy
TTUUU V d'
TTUVVUUU VT dd'
TTUVVUU VT ddd
When T changes to T+dT at constant V,U changes to
3.1 State functions
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,the internal pressure,the same dimension as p
T
T V
Uπ?
TCVVUU V
T
ddd
VT
U?
the slope of a plot of U against T at constant V.
TCVπU VT ddd
TTUVVUU VT ddd
CV
In a closed system of constant composition,dU is propor-
tional to dV and dT,the coefficients of proportionality
being the partial derivatives.
3.1 State functions
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If dU > 0 as the volume of the sample
expands isothermal,dV > 0,which is
the case when there are attractive
forces between the particles?T >0
TCVπU VT ddd
When there are no interactions be-
tween the molecules,U is indepen-
dent of their separation and hence
independent of the volume the sample
occupies; hence?T = 0 for a perfect
gas.
3.1 State functions
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The expansion coefficient,?,is expressed as:
3),Changes in enthalpy at constant pressure
TCVπU VT ddd
pT
U
p
T T
Vπ?
VC?
pT
V
V
α?
1
Dividing both sides of the equation by dT,it follows that:
A large value of? means that the volume of the sample
responds strongly to changes in temperature
3.1 State functions
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Introducing the definition of? into the equation:
VT
p
CVTU
3),expansion coefficient
pT
U
p
T T
Vπ?
VC?
pT
V
V
α?
1
3.1 State functions
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The constant-volume heat capacity of a perfect gas is
equal to the slope of a plot of internal energy against
temperature at constant pressure as well as to the slope
at constant volume.
3),expansion coefficient
VT
p
CVαπTU
For a perfect gas,?T=0:
V
p
CTU
3.1 State functions
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Example
Method,To use the expression of the expansion coefficient
of a gas,we simply substitute the expression for V in terms
of T obtained from the equation of state for the gas,The
pressure is treated as a constant.
Derive an expression for the expansion coefficient for a
perfect gas by using the expansion coefficient of a gas,
Answer,Because pV = nRT
pT
pnR T
V
α?
)(1
TpV
nR 1
The higher the temperature,the less responsive is its volume
to a change in temperature.
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3.1 State functions
1),Exact and inexact differentials
2),Changes in internal energy
3),Expansion coefficient
3.2 The temperature dependence of the enthalpy
1),Changes in enthalpy at constant volume
2),The isothermal compressibility
3),The Joule -Thomson effect
3.3 The reaction between CV and Cp
3,The First Law,the machinery
版权所有：华东理工大学物理化学教研室 19
1),Changes in enthalpy at constant volume
For a closed system of constant composition,H is a function
of p and T,and it follows that
T
T
Hp
p
HH
pT
ddd?
3.2 The temperature dependence of the enthalpy
and
TCp
p
HH
p
T
ddd
This expression implies that:
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pCk
αμ
T
H
TV
1
The isothermal compressibility,κT,is defined as:
T
T p
V
V?
1?
The Joule-Thomson coefficient,?,is defined as:
Hp
Tμ
3.2 The temperature dependence of the enthalpy
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2),The isothermal compressibility
The negative sign in the definition of κT ensures that it is
positive,because an increase of pressure,implying a posi-
tive dp,brings about a reduction of volume,a negative dV,
For a perfect gas
It shows that the higher the pressure of the gas,the lower its
compressibility.
ppV
nR T
p
pnR T
V TT
111
2
T
T p
V
V?
1?
3.2 The temperature dependence of the enthalpy
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Example- Using the isothermal compressibility
The isothermal compressibility of water at 20℃ and 1
atm is 4.94 × 10-6 atm-1,What change of volume
occurs when a sample of volume 50 cm3 is subjected to
an additional 1000 atm at constant temperature?
Method,for an infinitesimal change of pressure at
constant temperature,from the definition of kT,the
volume changes:
T
T p
V
V?
1?
pVV T dd
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Answer,Integrate both sides:
pVV p
p T
V
V
dd f
i
f
i
pVV p
p T
V
V
dd f
i
f
i
Suppose that kT,and V are approximately constant over
the range of pressures of interest:
pVpVV Tp
pT
ΔdΔ f
i
V =-(4.94× 10-6 atm-1)× (50cm3)× (1000atm)
=-0.25cm3
Example - Using the isothermal compressibility
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3),The Joule -Thomson effect
Adiabatic expansion,Constant H,
let a gas expand through a porous barrier
from one constant pressure to another,
and monitored the difference of T that
arose from the expansion,The process
was adiabatic,It was observed a lower T
on the low-pressure side,the difference in
T being proportional to the p difference
maintained,This cooling by adiabatic
e x p a n s i o n is n o w c a l l e d t h e
Joule-Thomson effect.The the r modynamicbasis of Joule-Thomson
expansion,
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect
Because all changes to the gas
occur adiabatically,q = 0,
The gas on the left is compressed
isothermally and the work done:
wL = - pi (0- Vi) = piVi
The work done in right,
wR = - pf (Vf - 0) = -pfVf
The total work done:
w = wL + wR = piVi -pfVf
3.2 The temperature dependence of the enthalpy
The the r modynamic
basis of Joule-Thomson
expansion,
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3),The Joule -Thomson effect
Since q = 0,then U= w + q = w,
the change of U of the gas as it moves
from one side of the throttle to the
other is:
Uf- Ui = w = piVi - pfVf
Uf + pfVf = Ui + piVi
Hf = Hi
The expansion occurs without change
of enthalpy,an isenthalpic process.
3.2 The temperature dependence of the enthalpy
The the r modynamic
basis of Joule-Thomson
expansion,
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3),The Joule -Thomson effect
The property is the ratio of the T change to the change of p,
ΔT/Δp,Adding the constraint of constant enthalpy and taking
the limit of small Δp implies the quantity measured is (dT/dp)H,
which is the Joule-Thomson coefficient,μ,And μ is the ratio of
the change in temperature to the change in pressure when a
gas expands under adiabatic conditions.
HpTμ
For measurement,the isothermal Joule-Thomson coefficient,
μT,is used:
TT pHμ μC p
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect,for real gases
For a perfect gas,? = 0,But Real gases
have nonzero? and,depending on the
identity of the gas,p and T,the relative
magnitudes of the attractive and
repulsive intermolecular forces,the sign
of the coefficient may be either positive
or negative.
The sign of the Joule-
Thomson coefficient,?
A positive sign implies that dT is negative
when dp is negative,in which case the gas
cools on expansion,A negative sign implies
heating process.
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect,for real gases
Inside the boundary,?> 0 and outside it
is negative,The T corresponding to the
boundary at a given p is the inversion T
of the gas at that p,Reduction of p under
adiabatic conditions moves the system
along one of the isenthalps,The inversion
T curve runs through the points of the
isenthalps where their slopes change
from negative to positive.
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect,for real gases
The sign of the Joule-Thomson
coefficient,?
Gases that show a heating effect
(?<0) at one T show a cooling
effect (?> 0) when the T is below
their upper inversion temperature,
TI,A gas typically has two
inversion temperatures,one at
high temperature and the other at
low.
3.2 The temperature dependence of the enthalpy
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We know that
To introduce H=U+pV
3.3 The reaction between CV and Cp
Vp
Vp T
U
T
HCC?
Vpp
Vp T
U
T
pV
T
UCC?
)(
From the equation
VT
p
CVTU
VT
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VπpαVπαpVCC TTVp )(
Since
pTpTπ
V
T
pVTVpTpV
pp
)()(
pT
V
V
α?
1
V
Vp T
pVTαCC?
3.3 The reaction between CV and Cp
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T
Vp κ
TVαCC 2
T
T p
V
V
κ
1
V
Vp T
pTVαCC
pT
V
V
α?
1
3.3 The reaction between CV and Cp
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For a perfect gas
pκTα T
1 a n d 1
nR
κ
TVαCC
T
Vp
2
RCC mVmp,,
3.3 The reaction between CV and Cp
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A summary
pT
V
Vα
1
1),Internal pressure,πT
T
T V
Uπ?
For a perfect gas,?T = 0.
TCVπU VT ddd
Tα
1?For a perfect gas:
2),Expansion coefficient,?
3) Isothermal compressibility,κT
For a perfect gas:
T
T p
V
V
1
pκ T
1? For a perfect gas,? = 0
4),Joule-Thomson coefficient,?
Hp
Tμ
版权所有：华东理工大学物理化学教研室 36
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
2版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
3,The First Law,the machinery
版权所有：华东理工大学物理化学教研室 3
In this chapter we begin to unfold some of the
power of thermodynamics by showing how to
establish relations between different
properties of a system,The procedure we use
is based on the experimental fact that the
internal energy and the enthalpy are state
functions,and we derive a number of
relations between observables by exploring
the mathematical consequences of these facts.
3,The First Law,the machinery
版权所有：华东理工大学物理化学教研室 4
3.1 State functions
1),Exact and inexact differentials
2),Changes in internal energy
3),Expansion coefficient
3.2 The temperature dependence of the enthalpy
1),Changes in enthalpy at constant volume
2),The isothermal compressibility
3),The Joule -Thomson effect
3.3 The reaction between Cv and Cp
3,The First Law,the machinery
版权所有：华东理工大学物理化学教研室 5
The initial state of the system is i and in
this state the internal energy is Ui,Work
is done by the system as it expands
adiabatically to a state f,In this state the
system has an internal energy Uf and the
work done on the system as it changes
along Path 1 from i to f is w,U is a
property of the state; w is a property of
the path,
3.1 State functions
1),Exact and inexact differential
fi UU d
版权所有：华东理工大学物理化学教研室 6
In Path 2,the initial and final states are
the same but in which the expansion is
not adiabatic,In this path an energy q'
enters the system as heat and the work
w' is not the same as w,The work and
the heat are path functions,
1),Exact and inexact differential
fi qq p a t h,d
3.1 State functions
版权所有：华东理工大学物理化学教研室 7
△ U,independent of the path,an exact differential.An exact
differential is an infinitesimal quantity which,when integra-
ted,gives a result that is independent of the path between
the initial and final states.
q or w,dependent on the path,an inexact differential.When a
system is heated,the total energy transferred as heat is the
sum of all individual contributions at each point of the path.
1),Exact and inexact differential
△ q qf - qi dq dq q
3.1 State functions
版权所有：华东理工大学物理化学教研室 8
Consider a perfect gas inside a cylinder fitted with a
piston,Let the initial state be T,Vi and the final state
be T,Vf,The change of state can be brought about in
many ways,of which the two simplest are following:
Path 1,in which there is free expansion against zero
external pressure; Path 2,in which there is reversible,
isothermal expansion,Calculate w,q,and?U for each
process.
Example -Calculating work,heat,and internal energy
版权所有：华东理工大学物理化学教研室 9
Method,To find a starting point for a calculation in
thermodynamics,it is often a good idea to go back to
first principles,and to look for a way of expressing
the quantity we are asked to calculate in terms of
other quantities that are easier to calculate,Because
the internal energy of a perfect gas arises only from
the kinetic energy of its molecules,it is independent of
volume; therefore,for any isothermal change,in
general?U = q+w,
Example -Calculating work,heat,and internal energy
版权所有：华东理工大学物理化学教研室 10
Answer,Since?U = 0 for both paths and?U = q+w.
The work of free expansion is zero,
in Path 1,w = 0 and q = 0;
for Path 2,the work is given by
so w = - nRT ln (Vf /Vi),and
q = nRT ln (Vf /Vi),
i
flndf
i V
VnR T
V
VnR Tw V
V
Example -Calculating work,heat,and internal energy
版权所有：华东理工大学物理化学教研室 11
For a closed system of constant composition,U is a func-
tion of V and T,When V changes to V + dV at constant T,
U changes to
VVUUU T d'
2),Changes in internal energy
TTUUU V d'
TTUVVUUU VT dd'
TTUVVUU VT ddd
When T changes to T+dT at constant V,U changes to
3.1 State functions
版权所有：华东理工大学物理化学教研室 12
,the internal pressure,the same dimension as p
T
T V
Uπ?
TCVVUU V
T
ddd
VT
U?
the slope of a plot of U against T at constant V.
TCVπU VT ddd
TTUVVUU VT ddd
CV
In a closed system of constant composition,dU is propor-
tional to dV and dT,the coefficients of proportionality
being the partial derivatives.
3.1 State functions
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If dU > 0 as the volume of the sample
expands isothermal,dV > 0,which is
the case when there are attractive
forces between the particles?T >0
TCVπU VT ddd
When there are no interactions be-
tween the molecules,U is indepen-
dent of their separation and hence
independent of the volume the sample
occupies; hence?T = 0 for a perfect
gas.
3.1 State functions
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The expansion coefficient,?,is expressed as:
3),Changes in enthalpy at constant pressure
TCVπU VT ddd
pT
U
p
T T
Vπ?
VC?
pT
V
V
α?
1
Dividing both sides of the equation by dT,it follows that:
A large value of? means that the volume of the sample
responds strongly to changes in temperature
3.1 State functions
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Introducing the definition of? into the equation:
VT
p
CVTU
3),expansion coefficient
pT
U
p
T T
Vπ?
VC?
pT
V
V
α?
1
3.1 State functions
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The constant-volume heat capacity of a perfect gas is
equal to the slope of a plot of internal energy against
temperature at constant pressure as well as to the slope
at constant volume.
3),expansion coefficient
VT
p
CVαπTU
For a perfect gas,?T=0:
V
p
CTU
3.1 State functions
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Example
Method,To use the expression of the expansion coefficient
of a gas,we simply substitute the expression for V in terms
of T obtained from the equation of state for the gas,The
pressure is treated as a constant.
Derive an expression for the expansion coefficient for a
perfect gas by using the expansion coefficient of a gas,
Answer,Because pV = nRT
pT
pnR T
V
α?
)(1
TpV
nR 1
The higher the temperature,the less responsive is its volume
to a change in temperature.
版权所有：华东理工大学物理化学教研室 18
3.1 State functions
1),Exact and inexact differentials
2),Changes in internal energy
3),Expansion coefficient
3.2 The temperature dependence of the enthalpy
1),Changes in enthalpy at constant volume
2),The isothermal compressibility
3),The Joule -Thomson effect
3.3 The reaction between CV and Cp
3,The First Law,the machinery
版权所有：华东理工大学物理化学教研室 19
1),Changes in enthalpy at constant volume
For a closed system of constant composition,H is a function
of p and T,and it follows that
T
T
Hp
p
HH
pT
ddd?
3.2 The temperature dependence of the enthalpy
and
TCp
p
HH
p
T
ddd
This expression implies that:
版权所有：华东理工大学物理化学教研室 20
pCk
αμ
T
H
TV
1
The isothermal compressibility,κT,is defined as:
T
T p
V
V?
1?
The Joule-Thomson coefficient,?,is defined as:
Hp
Tμ
3.2 The temperature dependence of the enthalpy
版权所有：华东理工大学物理化学教研室 21
2),The isothermal compressibility
The negative sign in the definition of κT ensures that it is
positive,because an increase of pressure,implying a posi-
tive dp,brings about a reduction of volume,a negative dV,
For a perfect gas
It shows that the higher the pressure of the gas,the lower its
compressibility.
ppV
nR T
p
pnR T
V TT
111
2
T
T p
V
V?
1?
3.2 The temperature dependence of the enthalpy
版权所有：华东理工大学物理化学教研室 22
Example- Using the isothermal compressibility
The isothermal compressibility of water at 20℃ and 1
atm is 4.94 × 10-6 atm-1,What change of volume
occurs when a sample of volume 50 cm3 is subjected to
an additional 1000 atm at constant temperature?
Method,for an infinitesimal change of pressure at
constant temperature,from the definition of kT,the
volume changes:
T
T p
V
V?
1?
pVV T dd
版权所有：华东理工大学物理化学教研室 23
Answer,Integrate both sides:
pVV p
p T
V
V
dd f
i
f
i
pVV p
p T
V
V
dd f
i
f
i
Suppose that kT,and V are approximately constant over
the range of pressures of interest:
pVpVV Tp
pT
ΔdΔ f
i
V =-(4.94× 10-6 atm-1)× (50cm3)× (1000atm)
=-0.25cm3
Example - Using the isothermal compressibility
版权所有：华东理工大学物理化学教研室 24
3),The Joule -Thomson effect
Adiabatic expansion,Constant H,
let a gas expand through a porous barrier
from one constant pressure to another,
and monitored the difference of T that
arose from the expansion,The process
was adiabatic,It was observed a lower T
on the low-pressure side,the difference in
T being proportional to the p difference
maintained,This cooling by adiabatic
e x p a n s i o n is n o w c a l l e d t h e
Joule-Thomson effect.The the r modynamicbasis of Joule-Thomson
expansion,
3.2 The temperature dependence of the enthalpy
版权所有：华东理工大学物理化学教研室 25
3),The Joule -Thomson effect
Because all changes to the gas
occur adiabatically,q = 0,
The gas on the left is compressed
isothermally and the work done:
wL = - pi (0- Vi) = piVi
The work done in right,
wR = - pf (Vf - 0) = -pfVf
The total work done:
w = wL + wR = piVi -pfVf
3.2 The temperature dependence of the enthalpy
The the r modynamic
basis of Joule-Thomson
expansion,
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3),The Joule -Thomson effect
Since q = 0,then U= w + q = w,
the change of U of the gas as it moves
from one side of the throttle to the
other is:
Uf- Ui = w = piVi - pfVf
Uf + pfVf = Ui + piVi
Hf = Hi
The expansion occurs without change
of enthalpy,an isenthalpic process.
3.2 The temperature dependence of the enthalpy
The the r modynamic
basis of Joule-Thomson
expansion,
版权所有：华东理工大学物理化学教研室 27
3),The Joule -Thomson effect
The property is the ratio of the T change to the change of p,
ΔT/Δp,Adding the constraint of constant enthalpy and taking
the limit of small Δp implies the quantity measured is (dT/dp)H,
which is the Joule-Thomson coefficient,μ,And μ is the ratio of
the change in temperature to the change in pressure when a
gas expands under adiabatic conditions.
HpTμ
For measurement,the isothermal Joule-Thomson coefficient,
μT,is used:
TT pHμ μC p
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect,for real gases
For a perfect gas,? = 0,But Real gases
have nonzero? and,depending on the
identity of the gas,p and T,the relative
magnitudes of the attractive and
repulsive intermolecular forces,the sign
of the coefficient may be either positive
or negative.
The sign of the Joule-
Thomson coefficient,?
A positive sign implies that dT is negative
when dp is negative,in which case the gas
cools on expansion,A negative sign implies
heating process.
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect,for real gases
Inside the boundary,?> 0 and outside it
is negative,The T corresponding to the
boundary at a given p is the inversion T
of the gas at that p,Reduction of p under
adiabatic conditions moves the system
along one of the isenthalps,The inversion
T curve runs through the points of the
isenthalps where their slopes change
from negative to positive.
3.2 The temperature dependence of the enthalpy
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3),The Joule -Thomson effect,for real gases
The sign of the Joule-Thomson
coefficient,?
Gases that show a heating effect
(?<0) at one T show a cooling
effect (?> 0) when the T is below
their upper inversion temperature,
TI,A gas typically has two
inversion temperatures,one at
high temperature and the other at
low.
3.2 The temperature dependence of the enthalpy
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We know that
To introduce H=U+pV
3.3 The reaction between CV and Cp
Vp
Vp T
U
T
HCC?
Vpp
Vp T
U
T
pV
T
UCC?
)(
From the equation
VT
p
CVTU
VT
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VπpαVπαpVCC TTVp )(
Since
pTpTπ
V
T
pVTVpTpV
pp
)()(
pT
V
V
α?
1
V
Vp T
pVTαCC?
3.3 The reaction between CV and Cp
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T
Vp κ
TVαCC 2
T
T p
V
V
κ
1
V
Vp T
pTVαCC
pT
V
V
α?
1
3.3 The reaction between CV and Cp
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For a perfect gas
pκTα T
1 a n d 1
nR
κ
TVαCC
T
Vp
2
RCC mVmp,,
3.3 The reaction between CV and Cp
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A summary
pT
V
Vα
1
1),Internal pressure,πT
T
T V
Uπ?
For a perfect gas,?T = 0.
TCVπU VT ddd
Tα
1?For a perfect gas:
2),Expansion coefficient,?
3) Isothermal compressibility,κT
For a perfect gas:
T
T p
V
V
1
pκ T
1? For a perfect gas,? = 0
4),Joule-Thomson coefficient,?
Hp
Tμ
版权所有：华东理工大学物理化学教研室 36
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
