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Advanced Physical Chemistry
Systems and Surroundings
C HAPTER 2
H EAT, W ORK , I NTERNAL E NERGY, E NTHALPY,
AND THE F IRST L AW OF T HERMODYNAMICS
In general, chemists can the divide the universe, in a virtual manner, into two parts:
the system and the surroundings.
A system is that part of the universe about which the chemist performs
measurements.
AND
C HAPTER 3
T HE I MPORTANCE OF S TATE F UNCTIONS :
I NTERNAL E NERGY AND E NTHALPY
•
•
Professor Angelo R. Rossi
http://homepages.uconn.edu/rossi
Department of Chemistry, Room CHMT214
The University of Connecticut
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Isolated system: nothing can cross the boundaries of the system, nor can the
system change its size.
An isolated system is completely cut off from its surroundings.
In a closed system none of the contents of a closed system can enter or leave,
but the walls do allow both the conduction of heat and the volume of such a system
to change.
An open system the volume changes, heat can enter or leave, and the contents of
the system can enter and leave as well.
The surroundings are chosen sufficiently large so that nothing that goes on inside
the system can change the surroundings very much.
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[email protected]
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Systems and Surroundings
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Boundaries Between System and Surroundings
A boundary separates the system from the from the rest of the universe by a real or
idealized construct.
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The boundary between the system and surroundings may conduct heat, or
insulate them completely.
The boundary may be rigid or movable.
The boundary may be impermeable to the transfer of matter, or it may be
permeable to certain species.
The boundary walls determine the type of system and can be
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A system is separated from its surroundings by a boundary, either real
or idealized as shown in (a).
In (b) is shown a simplification where the system is surroundings by a
single wall that may be either an insulator or heat conductor.
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rigid or movable.
diathermal and allow heat to pass through them.
adiabatic and not allow heat to pass.
impermeable to matter or permeable to all or only some forms of matter.
A closed system, for instance, would have walls that are rigid, adiabatic, and
impermeable, while walls that were movable, diathermal, and permeable would be
imaginary boundaries prohibiting nothing.
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Thermodynamic State of a System
Examples of Thermodynamic Systems
The term state of a system implies enough knowledge of the general condition of a
system to allow that system to be exactly duplicated.
The simplest systems to study in thermodynamics are closed systems containing
gases.
A consequence of this definition is that the system must be at equilibrium so that the
properties of the system do not change with time.
For example, it is useful to think of a gas contained in a cylinder fitted with a piston as
the system, but the cylinder and piston are not taken as part of the system but simply
serve as a container.
How much information is needed in order for a system to be duplicated exactly?
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Useful properties of a system such as the temperature, pressure, volume, number
of moles of each constituent, the density, the total mass, ... are important to report.
For two systems to become identical, only a few properties have to be
communicated.
It turns out experimentally that for a single component system no more than three
properties are ever needed, as long as one is extensive.
For more complex systems more properties are needed. The actual number
depends on the number of components and the number of phases present, e.g.
solid, liquid, etc..
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Although the volume of such a system can change by moving the piston, it is
possible to determine the volume exactly at any point in a process.
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The type of a system (adiabatic, isothermal, etc.) can be fixed by choosing proper
materials for the walls of the cylinder and the piston.
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If it is assumed that such a system is undergoing a non-reversible process, then
there is always a definite pressure inside the cylinder but, the pressure outside the
cylinder may be different than the inside pressure.
The mechanical work done by a system depends on the outside pressure, not that
inside the system.
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Extensive and Intensive Thermodynamic Properties
Thermodynamic Processes
Thermodynamic properties are classed into one of two groups: extensive or
intensive.
Our interest is primarily in systems which undergo changes from an initial state to a final state which is
called a thermodynamic process.
It is straightforward to determine an intensive or extensive property:
Consider two separate solutions consisting of sodium chloride and silver nitrate which are in an initial state.
1.
2.
3.
2.
A precipitate forms and settles on the bottom leading to the final state.
3.
By definition, the initial and final states are equilibrium states.
4.
As the system moves from the initial state to the final state, it passes through a number of intermediate
points which are NOT equilibrium states.
5.
Intensive variables of a system are not really defined when the system is not at equilibrium.
The pressure, temperature, and density may vary from point to point inside the system.
6.
Extensive variables such as total energy, a total volume, and total mass still have values no matter
how disturbed a system becomes.
These values may constantly change but do have specific values at any instant of time.
Make an imaginary copy of the system under consideration.
Put the two systems in equilibrium and remove any barriers between them.
Either the property will remain the same and is an intensive property.
Or it will double and is then an extensive property.
1.
For example, the pressure in each of two systems at equilibrium must be the same
since they are identical. Putting them together will result in no change in pressure.
Thus, pressure is an intensive property.
On the other hand, the volume will clearly double and is an extensive property.
Intensive properties include properties such as pressure, temperature, and density,
while extensive properties include volume, number of moles, and energy.
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The two solutions are allowed to mix: Ag+ (aq) + Cl− (aq) −→ AgCl (s)
The sequence of situations the system goes through in passing from the initial state to the final state is
called the path taken by the system.
Because the intensive variables often have no values during a process, it is usually not possible to exactly
specify the path a process takes in terms of them.
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Mechanical Work
Work
Mechanical work involves raising or lowering a weight to a specific height.
Work is a scalar quantity. The differential pV work done on a system is given by:
Assume that there is a gas inside a cylinder fitted with a weightless, frictionless piston.
1.
2.
3.
4.
dw = −Pext dV
Place a weight of mass m on the top of the piston.
Because of various processes taking place inside the cylinder, the piston and the
mass are pushed up or down until a final resting place (i.e. the final state), is
reached.
To measure the amount of work performed, determine the net height through
which the weight has been raised or lowered. The weight can be raised a distance
h1 , lowered a distance h2 , and raised a third distance h3 .
The work done then depends on h the net distance traveled:
where Pext is the external or applied pressure.
Work (w) done by the surroundings on a system is positive (+), and work (w) done by
the system on the surroundings is negative (-).
(a) Work is done on a system by the surroundings.
The stops are pulled out, and the system is compressed to a new equilibrium state.
h = h1 − h2 + h3
(b) Work is done on the surroundings by the system.
and is given by:
When the stops are pulled out, the system expands
w = −mgh
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to a new equilibrium state.
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Work
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Work Depends on Path
The total work w on a system when there is a finite change in volume is obtained by
summing the infinitesimal amounts of work given by Equation 1
w=−
Z
2
Pext dV
For the change of state of a mole of gas
from (2P0 , V0 ) to (P0 , 2V0 ), the work done
on the gas depends on path
(2)
1
By the upper path, w = −2P0 V0 .
If the expansion or compression is carried out very slowly (i.e. reversibly), the
pressure throughout the gas will be uniform, and equal to Pext , and the maximum work
of expansion (-) or compression (+) will be obtained.
In this case, Pext = P , and the pressure is a function of temperature and pressure:
w=−
Z
2
P (T, V ) dV
1
By the lower path, w = +P0 V0 .
For a clockwise cyclic process
w = −2P0 V0 + P0 V0 = −P0 V0
(3)
1
The integral in Equation 2 is called a line integral because its value depends on the
path.
When the gas is allowed to expand or compress rapidly (i.e. irreversibly), the
substitution Pext = P cannot be made, and
Z 2
w = −Pext
dV = Pext (V2 − V1 )
(4)
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(1)
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Adiabatic Processes and Internal Energy
Heat
When work is done on a system that is thermally insulated so that there is no
exchange of heat with the surroundings, the thermodynamic state of the system is
changed and called an adiabatic process.
A given change in the state of a system can be accomplished in ways other than performing work under adiabatic conditions.
A hot object can be immersed in water, and after the
experiment, the temperature of the water is higher
and has a greater internal energy, U .
Work (w) and heat (q) are both forms of energy
crossing a boundary.
The change in internal energy U produced by the
transfer of heat q to a system when no work is done
is given by
Joule performed experiments during the period 1840 - 1849 which demonstrated that
the change in state of water in an adiabatic process is independent of path and
depends only on the amount of work performed.
The property of the system whose change is calculated in this way is called internal
energy, U .
The increase in internal energy from the work w done on a system to change it from
one state to another in an adiabatic process is given by
∆U = w
∆U = q
The heat absorbed by a closed system in a process
in which no work is done is equal to an increase in
the internal energy of the system.
(For an adiabatic process)
∆U indicates the internal energy in the final state minus the internal energy in the
initial state, ∆U = U2 − U1 where U2 and U1 are the final and initial states, respectively.
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If the system does work on the surroundings, w is negative AND ∆U is negative
indicating the internal energy of the system decreases when the process is adiabatic.
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(no work done)
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A postive value of q (+) indicates that heat is absorbed by the system from its surroundings.
A negative value of q (-) indicates that the systems releases heat to its surroundings.
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(a) A system in state 1 is insulated from the
heat resevoir. (b) The system is brought
into contact with the heat resevoir through
a heat-conducting wall. (c) The system is
then insulated from the heat resevoir and is
found to be in state 2.
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Heat
Obtaining Heat Flow, q
As opposed to work, heat is a very subtle concept.
Consider a process that goes between two states, 1 and 2.
On the molecular level heat is manifested as increased random motions
of atoms and molecules.
1.
Heat is measured indirectly.
2.
3.
The total internal energy change during a process is the sum of all of
the changes in the various forms of energy during that process.
4.
Thus, if w is the work done during a process and q is the heat absorbed
or released during that process as well as the only forms of energy
involved
∆U = q + w
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The state goes from 1 to 2 adiabatically so that no heat flow is involved. In such a
process q is zero, and the change in internal energy is
∆U = w
On a macroscopic level heat is seemingly mysterious because it can’t
be seen or touched.
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By measuring the work, it is possible to know the change in internal energy for this
process.
The system returns from state 2 to state 1 by any path that does involve heat.
The heat involved can be measured indirectly because the change in internal
energy in going from 1 to 2 is already known from the adiabatic measurements,
and ∆U will not change if we take a different path.
Thus, it is necessary to measure only w along this new path to obtain q.
qpath = ∆Uadiabatic − wpath
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First Law of Thermodynamics and Internal Energy
First Law of Thermodynamics and Internal Energy
Since the internal energy can be changed by a given amount by either heat or work,
these quantities are equivalent.
If both heat and work are added to a system,
Since the internal energy is a function of the state of the system, there is no change in
internal energy when a system is taken through a series of changes that return it to its
initial state.
∆U = q + w
(5)
dU = dq + dw
(6)
This is expressed by setting the cyclic integral to zero:
I
dU = 0
For an infinitesimal change in state
The internal energy U of a system is an extensive property. It the system is doubled,
the internal energy is doubled.
The above equations are statements of the first law of thermodynamics.
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Since the internal energy is a state function, it is necessary to know how many
variables have to be specified to describe the state of the system.
This law is the postulate that states there exists a property U , the internal energy,
that is a function of the state variables of a system and ∆U for a process in a
closed system may be calculated by Equation 6
If ∆U is negative, the system loses energy in heat that is evolved and work that is
done by the system.
The first law has nothing to say about how much heat is evolved and how much
work is done except that Equation 6 is obeyed.
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If the system involves only P V work, the internal energy of a mass of pure substance
can be described by a mathematical function of T, V, n or T, P, n represented by
U (T, V, n) and U (T, P, n)
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Thermodynamic Pathways
Behavior of Pressure During an Irreversible Process
The line connecting the initial and final states is called the path the
process takes.
A process goes from state 1 to state 2 by means of a series of non-equilibrium steps.
There is no way to show the path on a P − V diagram because the system does not
have a single uniform pressure.
In general, any line connecting the initial and final states on a
pressure-volume diagram is a path.
If the process is reversible, its progress can be plotted on a suitable
diagram such as a pressure-volume diagram.
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P
It may not be possible to showthe progress of an irreversible process.
1
2
b
b
V
The initial and final states are labelled 1 and 2.
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Reversible Processes
Thermodynamic Pathway for Internal Energy (U )
If a system traverses through a series of reversible states, the path can be shown on a
diagram because the pressure is defined.
A process goes from State 1 to State 2 along the red path and go back from State 2 to State 1 along a
different path i esignated the blue path.
P
1
State 1 to State 2 and Back Again
2
b
b
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The internal energy change for the red path is ∆Ured while that for for the blue path is ∆Ublue .
∆Ured + ∆Ublue = ∆U = 0
V
The initial and final states are labelled 1 and 2.
•
The total internal energy change, ∆U for this entire process must be zero, since U is a state function.
No matter what path we chose from State 2 back to State 1, the internal energy change for the return path
will have the same value but opposite sign.
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Several special processes whose resulting paths may be reversible or irreversible.
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Work of Compression for a Gas at Constant Temperature
Irreversible Process
More Thermodynamic Pathways
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Consider the compression of a gas at constant temperature and contained in a rigid
cylinder by a frictionless and weightless piston.
An isothermal process takes place at constant temperature.
Isothermal systems have walls that conduct heat and their surroundings have to be
at a constant temperature.
A constant volume (isochoric) process is obtained by having rigid walls around
the system, and these walls may or may not conduct heat.
A constant pressure (isobaric) process takes place in systems having flexible
walls whose surroundings are at a constant pressure.
A typical example is the path taken by a process that goes on in a flexibly-walled
system surrounded by the atmosphere.
An adiabatic process takes place in a system whose walls are impermeable to
heat. No heat passes into or out of the system.
The cylinder is immersed in a thermostat at temperature T , and the space above the
cylinder is evacuated so that the final pressure P2 is due only to the mass m.
The gas is initially confined to volume V1 because the piston is held by the stops.
Compression of a gas from P1 , V1 , T to P2 , V2 , T in a single step.
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Work of Compression for a Gas at Constant Temperature
Irreversible Process
Work of Compression for a Gas at Constant Temperature
Reversible Process
When the stops are pulled out, the piston falls to an equilibrium position, and the gas is compressed to a
volume V2 .
The compression can be carried out with less work if it is performed in two or more steps.
Compression of a gas from P1 , V1 , T to P2 , V2 , T in many steps.
Compression of a gas from P1 , V1 , T to P2 , V2 , T in a single step.
The gas can be compressed in two steps by using a mass m′ to compress the gas to a volume
the first step, and then using a mass m′′ for the second step.
The pressure at the end of the process is given by
P2 =
f
mg
force
=
=
area
A
A
V1 +V2
2
in
By using more and more steps, it is possible to arrive at the point (c) in the diagram above which shows that
the minimum amount of work is required in the limit of an infinite number of steps.
For each step, the pressure is changed by an infinitesimal amount, and the work is given by Equation 1 at
constant temperature:
Z
Z V2
w=
dw = −
P dV
(7)
The amount of work lost to the surroundings is
w = mgh = mg × (h2 − h1 ) = P2 × A × (h2 − h1 ) = −P2 (V2 − V1 )
V1
Since V2 < V1 , the work done on the system is postive and is the smallest amount of work that can be used
to compress the gas from V1 to V2 in a single step.
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Work of Expansion of a Gas at Constant Temperature
Reversible Processes
The work obtained from the surroundings from a single-step expansion is clearly not great enough to
compress the gas back to its initial state in a single-step compression.
The infinite-step compression at constant temperature described above is referred to as a reversible
process because the energy accumulated in the system in the compression is exactly the same amount
required to expand the gas back to the initial state.
wcycle = −
Z
V2
V1
P dV −
Z
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V1
P dV = 0
V2
A reversible process is one which a system in an initial state arrives at a final state through a series of
equilibrium states.
Expansion of a gas from P1 , V1 , T to P2 , V2 , T in a single step.
•
However, the work obtained from the surroundings in the infinite-step expansion is exactly the amount
required to compress the gas back to its initial state.
It is sufficiently slow such that a very slight change in the external conditions can change the direction of
the process.
If a process takes place such that it is reversible along a path, then the entire process is said to be
reversible.
•
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The reversible process requires that the internal and external conditions be balanced.
Reversible processes are not real, since infinitely slow process cannot be carried out.
However, they are useful idealizations of real processes.
Expansion of a gas from P1 , V1 , T to P2 , V2 , T in many steps.
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Reversible Isothermal Compression or Expansion
Compression of an Ideal Gas at Constant Pressure
In a reversible process, the change is accomplished in infinitesimal steps.
Two moles of a gas at 1 bar and 298 K are compressed at constant
temperature by use of a constant pressure at 5 bar.
Since the gas is at its equilibrium pressure at each step of an expansion or
compression, the pressure given by an equation of state can be substituted, into
Equation 7.
Since an ideal gas has been assumed, p can be replaced by
Z
wrev = −nRT
For a compression
For an expansion,
V2
V1
V2
V1
V2
V1
nRT
V
How much work is done on the gas?
which yields:
w = −P2 (V2 − V1 ) = −P2
dV
V2
= −nRT ln
V
V1
nRT
nRT
−
P2
P1
= nRT
P2
−1
P1
5 bar
− 1) = 19820 J
1 bar
The work is performed on the system and is therefore positive.
w = (2 mol)(8.3145 J mol−1 K−1 )(298 K)(
< 1, the logarithm is negative, and the work is positive.
> 1, the logarithm is positive, and the work and is negative.
For an ideal gas at constant pressure, P1 V1 = P2 V2 , and the reversible work is also
given by:
P2
wrev = nRT ln
P1
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Reversible Expansion of An Ideal Gas
Reversible Compression or Expansion for a van der Waals Gas
One mole of an ideal gas expands from 5 bar to 1 bar at 298 K.
If a gas obeys van der Waals equation, the internal pressure is given by:
(a) Calculate w for a reversible expansion.
wrev = nRT ln
1 bar
P2
= (1 mol)(8.3145 J mol−1 K−1 )(298 K) ln
= −3988 J
P1
5 bar
wirrev = −P2 (V2 − V1 ) = −P2
−1
wirrev = (1 mol)(8.3145 J mol
nRT
nRT
−
P2
P1
K
−1
= nRT
P2
−1
P1
nRT
n2 a
− 2
V − nb
V
Z
The integral then becomes:
(b) Calculate w for an expansion against a constant external pressure of 1 bar.
P =
w=−
V2
V1
n2 a
nRT
− 2
V − nb
V
dV
which integrates to:
1 bar
)(298 K)(
− 1) = −1982 J
5 bar
w = −nRT ln
V2 − nb
V1 − nb
+ n2 a
1
1
−
V2
V1
The work is negative because this is and expansion, and the system performs work.
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Heat Capacity
Constant Volume (Cv ) Heat Capacity
The quantity of heat q can be measured by determining the change in
temperature of a mass of material that absorbs heat.
The first law is given by
The heat capacity C is defined as
If the process
R is carried out at constant volume without expansion work,
then w = − Pexternal dV = 0, and
C = lim
∆T →0
∆U = q + w
d✁ q
q
=
Tf − Ti
dT
∆U = qV =
but d✁ q is an inexact differential because heat is not a state function
and depends on path.
Z
Tsys,f
Tsys,i
CVsys (T )dT = −
Z
Tsurr,f
Tsurr,i
CVsurr (T )dT
Because of this, a constant-volume path or constant-pressure path
must be considered.
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Change in State at Constant Volume
Heat Capacity Temperature Dependence
Since the heat capacity at constant volume can be measured, it can be integrated to yield:
CV can be a function of temperature:
∆UV =
Z
T2
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CV = a + bT + cT 2
CV dT = qV
T1
where a, b, and c are constants.
The integral then becomes
Z T2
Z
Z U2
CV dT =
dU =
for a finite change in temperature at constant volume. This is illustrated in the graph below
U1
T1
T2
(a + bT + cT 2 )dT
T1
which yields
c
b
∆U = a(T2 − T1 ) + (T22 − T12 ) + (T23 − T13 )
2
3
Over a small temperature range, CV may be nearly constant
∆UV = CV (T2 − T1 ) = CV ∆T
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Enthalpy and Constant Pressure (CP ) Heat Capacity
State Functions, Paths, Exact Differentials
Constant-pressure process are more common in Chemistry than constant-volume process, since many
reactions are carried out in open vessels.
If only pressure-volume work is done, and the pressure is constant and equal to the applied pressure, then
∆U = ✁
dqP − Pexternal ∆V = ✁
dqP − P dV
where ✁
dqP is the heat for the constant pressure process.
If the initial and final state are designated by 1 and 2, respectively then
U2 − U1 = qP − P (V2 − V1 )
. The heat absorbed is given by
qP = (U2 + P V2 ) − (U1 + P V1 )
It is possible now to introduce a new state function called the enthalpy H defined by
Adiabatic Path (1) and Nonadiabatic Path (2)
H = U + PV
and
∆H = qP = H2 − H1 =
Z
Tsys,f
Tsys,i
sys
CP
(T )dT = −
Z
Tsurr,f
Tsurr,i
The internal energy of both the initial and final states of paths 1 and 2 are the same because U is a state
function.
Z
f inal
surr
CP
(T )dT
∆U =
dU is an exact differential because it is independent of the path between the initial and final states.
The heat absorbed at constant pressure is equal to the change in enthalpy.
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dU
initial
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Path Functions and Inexact Differentials
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Derivatives in Thermodynamics
The total energy transferred as heat q is the sum of the individual contributions of each
point along the path.
Therefore,
✘
∆q = ✘
qf ✘
−✘
qi
cannot be written because q is not a state function.
The path of integration must be specified for q because it depends on path:
q=
Z
f
dq
i,path
For example, q = 0 for the adiabatic path, and q ′ 6= 0 for the nonadiabatic path.
An inexact differential is an infinitesimal quantity which gives a result that depends
on path when it is integrated between the initial and final states.
The partial derivative
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∂U
∂V
T
is the slope of U with respect to V
and the temperature T held constant.
The heat flow (q) and work (w) are both inexact differentials because their values are
different if the path is different.
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Derivatives in Thermodynamics
U is State Function: Thermodynamic Consequences
Assume U is a function of volume and temperature,
U = U (V, T )
Suppose V and T change infinitesimally:
•
U′ = U +
dU =
The partial derivative
∂U
∂T
is the slope of U with respect to T
V
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U = U (V, T ) - Infinitesimal Changes
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Euler Chain Relation
Find
∂Vm
∂T
P
RT
P =
∂x
∂y
z
∂y
∂z
x
∂z
∂x
dT
V
∂U
∂V
dV +
T
∂U
∂T
dT
V
= -1
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for a gas which obeys the following equation of state:
For a differentiable function: z = z(x, y)
∂U
∂T
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Derivatives in Thermodynamics
T
in Volume and Temperature
Derivatives in Thermodynamics
dV +
Generally state functions depend on two independent variables, and the dependence on the
third variable is subsumed into these two by the
equation of state.
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and the volume V held constant.
∂U
∂V
The internal energy U ′ differs from U by an infinitesimal amount: dU = U ′ − U
•
Vm
+
a + bT
2
Vm
For the gas write the molar volume as a function of temperature and pressure:
y
Vm = f (T, P )
The variables permute in cycles.
then
Reciprocal Identity
dVm =
∂y
∂x
z
= 1
∂x
∂y
0=
z
−1 =
The order of differentiation does not matter:
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∂z
∂y
x y
∂
=
∂y
∂Vm
∂T
dT +
P
∂Vm
∂P
dP
T
Now variations of T and p are restricted to values which leave Vm constant (dVm = 0):
Relationship Between Partial Derivatives
∂
∂x
"
∂z
∂x
∂Vm
∂P
T
∂Vm
∂T
×
#
y x
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Fall 2013
∂P
∂T
+
P
∂Vm
∂T
Vm
P
∂P
×
×
∂Vm
;
∂Vm
T
∂T
= −
∂P
∂T
∂P
∂Vm
∂P
∂T
Vm
Euler Chain Relation
P
Vm
T
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More on Exact and Inexact Differentials
More on Exact and Inexact Differentials
Consider the path from a to b in the figure below:
Consider the path from a to b in the figure below:
Path of a System in Going from State a to State b
Path of a System in Going from State a to State b
The differential dz = ydx + xdy = d(xy) is an exact differential.
The differential dz = ydx is not an exact differential:
Z
b
dz = z =
a
Z
b
Z b
dz = ∆z =
a
ydx = area I
a
Z b
a
d(xy) = xb yb − xa ya
The reason dz = ydx + xdy is an exact differential may be seen from the above figure. The integral from state a to state b is
because this area depends on the path between a and b.
Z b
a
dz = ∆z =
Z b
a
ydx +
Z b
xdy = area I + area II
a
The sum of these areas is independent of the shape of the curve (path) between a and b.
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Euler’s Criterion for Exactness
Change in State at Constant Volume
If z has a definite value at each point in the xy plane, then it must be a function of x and y. If z = f (x, y) then in general,
When a system changes from one state to another at constant volume,
the change in internal energy U can be calculated from the heat q
evolved and the work w done on the system by the surroundings.
It is convenient to take U as a function of T and V . Since U is a state
function,
U = U (V, T )
dz = M (x, y)dx + N (x, y)dy
where M (x, y) and N (x, y) are functions of the variables x and y.
dz =
Comparing the above two equations,
M (x, y) =
Since the mixed partial derivatives are equal,
∂
∂y
then
"
∂x
∂x
∂z
∂z
∂z
∂x
dy
x
N (x, y) =
=
y x
∂y
∂z
∂y
y
#
∂M
dx +
y
=
x
∂
∂x
"
∂N
∂x
∂z
∂y
∂z
∂y
the differential dU is given by
∂U
∂U
dU =
dT +
dV
∂T V
∂V T
x
#
x y
•
y
If dz is an exact differential, then the above equation must be satisfied, and this is Euler’s criterion for exactness.
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The first term is the change in internal energy because of
temperature changes.
The second term is the change in internal energy because of the
change in volume.
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Change in State at Constant Volume
Change in State at Constant Pressure
If only pressure-volume work is involved, then
∂U
∂U
dT + Pext +
dV
d✁ q =
∂T V
∂V T
For an infinitesmal change at constant pressure
dqP = dH
(8)
where dH is an exact differential, since the enthalpy is a function of the state of the
system.
Since H = H(T, P ), the enthalpy can be expressed as a function of temperature and
pressure
∂H
∂H
dT +
dP
(11)
dH =
∂T P
∂P T
If a change in state of a system X takes place at constant volume
X(V1 , T1 ) −→ X(V1 , T2 )
then Equation 8 reduces to
d✁ qV =
∂U
∂T
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If a change in state of a system X takes place at constant pressurer:
X(P1 , T1 ) −→ X(P1 , T2 ). then Equation 10 and Equation 11 can be combined to
obtain
∂H
dT
dqP =
∂T P
dT
V
It is convenient to define the heat capacity CV at constant volume as
d✁ qV
∂U
CV =
=
(9)
dT
∂T V
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(10)
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Change in State at Constant Pressure
The Relation Between CV and CP for a Perfect Gas
The heat capacity at constant pressure CP is
∂H
dqP
=
CP =
dT
∂T P
The constant-pressure heat capacity differs from the constant-volume heat capacity by the work needed to
change the volume of the system.
The relation between heat capacities at constant pressure and constant volume can be derived using
Equation 8
∂U
dqP = CV dT + P +
dV
∂V T
(12)
The heat capacity at constant pressure can be measured so that Equation 12 can be
integrated for a finite change in temperature at constant pressure:
∆HP =
Z
Dividing by dT and setting
dqP
dT
= CP
∂V
∂U
CP − CV = P +
∂V T
∂T P
T2
CP dT
T1
(13)
The quantity on the right-hand side of Equation 13 is positive so that CP > CV .
This work is the result of two components:
1.
2.
driving back the atmosphere
stretching bonds in a material including weak intermolecular forces
∂U
For an ideal gas, ∂V
= 0 and ∂V
= nR
Thus,
∂T
P
T
P
CP − CV = nR
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Irreversible Adiabatic Processes with Gases
Reversible Adiabatic Processes with Gases
When an adiabatic expansion is carried out reversibly,
For the compression and expansion of gases in isolated systems, no heat is gained or lost by the gas
(q = 0), so the process is adiabatic, and the first law becomes dU = dw.
CVm dT = −P dVm = −
If the expansion is opposed by an external pressure, work is done on the surroundings, and the temperature
drops as internal energy is converted to work.
Since q = 0, then ∆U = w,
Z
U2
U1
dU = −
Z
Pext dV
V1
CV m
∆U = U2 − U1 = w
T2
T1
Z
T2
T1
CVm ln
T2
=
T1
where γ =
CP,m
.
CV,m
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P2
P1
γ−1
γ
Changes in Internal Energy
The internal energy U can be expressed as a function of U (V, T ), and for infinitesimal changes in U , the
exact differential is given by
∂U
∂U
dU =
dV +
dT
∂V T
∂T V
∂U
dU =
dV + CV dT
∂V T
∂U
CV =
(the constant volume heat capacity)
∂T V
∂U
πT =
(internal pressure)
∂V T
The internal energy increases (dU > 0) as the volume of the sample expands isothermally (dV > 0),
if attractions are dominant because the molecules
become further apart on average.
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γ−1
and
γ
γ
P1 Vm,1
= P2 Vm,2
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Changes in Internal Energy
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Vm,1
Vm,2
Using the ideal gas law, the following alternative forms can be obtained:
T2
=
T1
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Vm,1
T2
= R ln
T1
Vm,2
Since CP,m − CV,m = R, the above equation can be written as
CV dT = CV (T2 − T1 ) = w
This relation applies to the adiabatic expansion of an ideal gas with CV independent of temperature
whether the process is reversible or irreversible.
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dVm
dT
= −R
T
V
Z Vm,2
dT
dVm
= −R
T
V
Vm,1
CV m
V2
For an ideal gas, the internal energy is a function of only T and dU = CV dT .
Z
RT
dVm
V
πT =
∂U
∂V
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>0
T
If repulsions are dominant, the internal energy decreases (dU < 0) as the volume of the sample expands isothermally (dV > 0).
πT is a measure of the variation of internal energy of a substance as its volume is changed at constant
temperature.
πT is called the internal pressure and is a measure of the cohesive forces in the sample.
πT =
∂U
∂V
<0
T
For a perfect gas, the internal energy is independent of the volume at constant temperature:
The equation can then be written as
dU = πT dV + CV dT
∂U
∂V
=0
T
The internal energy of a perfect gas is independent
of the volume occupied by the sample.
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Changes in U at Constant Pressure
Changes in U at Constant Pressure
We want to determine how the internal energy varies with temperature
as the pressure is kept constant:
The expansion coefficient is defined as
1
V
α=
dU = πT dV + CV dT
which yields
Divide by both sides of the above equation by dT
dU
dV
= πT
+ CV
dT
dT
∂U
∂T
P
= απT V + CV
P
which is completely general for a closed system of constant composition.
For a perfect gas πT = 0 and
and impose the condition of constant pressure on the resulting
differentials:
∂U
∂V
= πT
+ CV
∂T P
∂T P
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∂V
∂T
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∂U
∂T
= CV
P
which means that the constant-volume heat capacity of a perfect gas is the slope of internal energy
versus temperature at constant pressure AS WELL AS the slope of internal energy versus
temperature at constant volume.
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Temperature Dependence of the Enthalpy
Temperature Dependence of the Enthalpy
Assume that the enthalpy (H) is a function of the temperature and pressure:
For the equation,
H = H(P, T )
∂H
∂T
=
V
αµ
1−
CP
κT
κT is defined as the isothermal compressibility:
Then, we can write
dH =
and it is recognized that CP =
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∂H
∂T
P
dP +
∂H
∂P
=
∂H
∂P
T
∂H
∂T
dT
κT = −
P
to yield
dH =
∂V
∂P
∂T
∂P
1
V
T
and µ is defined as the Joule-Thomson coefficient:
dP + CP dT
µ=
T
H
The above equation can be recast to give
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∂H
∂T
V
αµ
CP
1−
κT
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Isothermal Compressibility
The Joule-Thomson Expansion
The negative sign below in the definition of κT ensures that compressibility is a
positive quantity.
1 ∂V
κT = −
V ∂P T
Joule and William Thomson (Lord Kelvin) devised
an experiment which had the constraint of constant
enthalpy:
A tube with a porous plate separating it into two
parts is constructed.
An increase in pressure (a positive dP ) produces a decrease in volume (negative dV )
The isothermal compressibility is obtained from the slope of the plot of volume versus
pressure at constant temperature.
κT (perfect gas) =
1
P
The porous plate acts as a throttle allowing the gas
to go through it slowly.
On each side of the plate there is a piston that fits
the tube snuggly, and each piston can be pushed
against the porous plug.
(the higher the pressure, the lower the compressibility)
A quantity of gas is placed in the left side of the tube
between the piston and the porous plate, but the
piston on the right side of the tube is flush against
the porous plate.
The whole apparatus was insolated and adiabatic
so that no heat can enter or leave.
Joule and Thomson measured the following property
∂T
µJ−T =
∂P H
Experimental Apparatus
for Joule-Thomson Experiment
which is isenthalpic.
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The Joule-Thomson Expansion
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Analysis of the Joule-Thomson Expansion
The process starts with a volume (Vi ) on the left side and zero volume on the right side (Vi = 0).
During the experiment gas is pushed through the
porous plate by pushing the piston on the left side.
The expansion then ends with zero volume on the left side (Vf = 0), and a finite volume on the right side
(Vf ).
At the same time, the piston on the right side is
pulled in such a way that the pressure on the right
side is Pf which is always less than Pi .
Work done on left side = wlef t = −Pi (0 − Vi ); work done on right side = wright = −Pf (Vf − 0).
All changes to the gas occur adiabatically (q = 0) and
On the left-hand side of the tube, the initial volume,
pressure, and temperature of the gas are Vi , Pi , Ti .
∆U = w
At the end of the experiment all the gas has been
pushed through the porous plate.
The total work performed is
The final volume on the left side is 0. On the right
side the final volume is Vf , the pressure is Pf , and
the temperature is Tf .
which yields
w = wlef t + wright = Pi Vi − Pf Vf
∆U = Uf − Ui = Pi Vi − Pf Vf
Careful measurement shows that Tf is not equal
to Ti .
The above equation can be rearranged to yield
Uf + Pf V f = U i + Pi V i
Experimental Apparatus
and finally
for Joule-Thomson Experiment
Hf = Hi
which is an isenthalpic process.
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The Joule-Thomson Expansion
The Joule-Thomson Coefficient for Gases
What is measured experimentally is
Ideal Gases
µJ−T = lim
∆P →0
Tf − Ti
Pf − Pi
For a perfect gas H = U + nRT , and there is no Joule-Thomson effect as shown below:
µJ−T = −
which becomes the experimental quantity that is measured
µJ−T =
∂T
∂P
CP
∂H
∂P
T
=−
∂U
∂P
=
T
1
CP


∂U
∂P
+
T
✟
✯0∂V ∂U ✟
✟
✟∂V T ∂P T

✿
∂(nRT ✘
✘) ✘
=0
✘✘∂P
T
0
=0
The molecular interpretation of this relation is that there is no interaction between the molecules of a perfect gas, and so the internal energy
does not change with the distance between molecules.
Joule incorrectly concluded that the internal energy of the gas is independent of the volume.
H
Real Gases
At low temperatures, the attractive forces dominate while at high temperatures the repulsive forces dominate. The change in µ from a positive
value at low temperatures and negative value at high temperatures can be explained as follows
This means that a cooling effect is obtained below the inversion temperature, and a
heating effect is obtained above the inversion temperature.
•
For µ > 0 implies dT < 0 when dP < 0 and the gas cools on expansion.
•
The inversion temperature is found experimentally to also depend on pressure.
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because
It is found experimentally that µJT is positive at low temperatures and negative at
high temperatures giving rise to to an inversion temperature (TI ).
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1
At low temperatures, intermolecular attractions are the most important interactions.
When the cold gas is expanded, the average distance between the molecules is increased and the molecules are pulled apart which
requires energy.
Since the process is adiabatic, some of the internal energy is consumed, and the gas cools.
At sufficiently high temperatures, the predominant interactions are repulsions, and gas molecules want to separate.
When the gas expands, energy is obtained as the molecules separate increasing the internal energy of the gas as well as heating it.
The internal energy of a real gas depends on volume at constant temperature because of attractive and repulsive forces.
The conclusion is that the Joule-Thomson effect depends on the interactions between gas molecules.
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