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p. 1 of 5
At 27C the reading on a manometer filled with mercury is 60.5
cm. The local acceleration of gravity is 9.784 m s-2. To what
pressure does this height of mercury correspond? Solution: The
pressure exerted by the column of mercury is equal to its height
times its density times the local acceleration of gravity, or in
symbols P = hg. We have h = 60.5 cm = 0.605 m. The density of
mercury at 27C is 13.53 g cm-3 = 13530 kg m-3 (where would I find
this?). So, we have P = 0.605 m 13530 kg m-3 9.784 m s-2 = 80088 kg
m-1 s-2 = 8.01104 Pa = 0.801 bar Work: Usually, mechanical work can
be defined as the product of a force times a distance or the
integral of a force over some distance through which it acts. If F
is the component of force along the direction of motion and dl is a
differential displacement in that direction, then this can be
expressed as dW = F dl
hclTypewritten textLECTURE NOTES
hclTypewritten TextCHEMICAL ENGINEERING THERMODYNAMICS
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hclTypewritten TextWhat is thermodynamics? One answer: a field
of science and engineering that describes, macroscopically, the
driving forces for the flow of energy (as heat and/or work) and the
conditions for equilibrium within a system (where equilibrium is
the state where these driving forces are absent and the system has
no tendency to change). Thermodynamics does not deal with rates of
change, but only with the driving forces that cause change. To do
thermodynamics, we must understand basic physics (like that learned
in the mechanics portion of a first-year college physics course)
and be adept at using and converting units of measurement. In
engineering, most numbers are meaningless without units. We will
not review units and unit conversions here, but you should read the
review of dimensions and units in chapter 1 of Smith, Van Ness, and
Abbott (SVA). Pay particular attention to the section on pressure.
In general, while I will post detailed lecture notes like the ones
you are reading, and these will follow the textbook fairly closely,
these are not an adequate substitute for also reading the textbook
chapters as we cover them. Example 1.3 from SVA
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where dW is the differential amount of work performed when the
force F acts through the displacement dl.
In thermodynamics, we often deal with the work done by an
expanding fluid (or work done on a fluid to compress it). In this
case, the force F is the pressure multiplied by the area over which
the pressure is applied (F = PA). If the compression is done by a
cylinder of constant area A then the change in total volume of the
fluid is dVt = A dl. Using these expressions for dl and F gives
dW = -P dV t
The cross-sectional area cancels out (so the work is independent
of the shape of the piston and cylinder). The negative sign arises
if we use the sign convention that work done on the gas by the
piston is positive. When the piston does work on the gas, the
change in volume is negative and the pressure is positive, so the
negative sign is required to make the work come out to be positive.
In integral form, then, the work on a gas in compressing it from
pressure P1 to pressure P2 is:
2
1
P
PW PdV=
How would we compute this, though? The limits of the integral
are in terms of pressure, while the variable of integration is dV.
We need an equation of state that relates the pressure and volume
(and temperature, if temperature changes) in order to evaluate
this. We will spend substantial time this semester considering
equations of state.
Forms of Mechanical Energy (Kinetic Energy and Gravitational
Potential Energy)
Kinetic energy (EK = mu2 where m is the mass of an object and u
is its speed) and gravitational potential energy (EP = mgz where m
is the mass of an object, z is its elevation relative to a
reference elevation, and g is the local acceleration of gravity)
are forms of mechanical energy that can be converted (fully) to
work. Note that only changes in or relative amounts of kinetic and
potential energy are meaningful, since the objects speed and
elevation must be defined relative to some stationary reference
frame and reference position, respectively.
Mechanical Energy Conservation: Example 1.4 in SVA.
An elevator with a mass of 2,500 kg rests at a level 10 m above
the base of an elevator shaft. It is raised 100m above the base of
the shaft, where the cable holding it breaks. The elevator falls
freely to the base of the shaft and strikes a strong spring. The
spring is designed to bring the elevator to rest and, by means of a
catch arrangement, to hold the elevator at the position of maximum
spring compression. Assuming the entire process to be frictionless,
and taking g = 9.8 m s-2, calculate:
(a) The potential energy of the elevator in its initial
position, relative to the base of the shaft.
(b) The work done in raising the elevator
Lecture 1
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(c) The potential energy of the elevator in its highest position
relative to the base of the shaft.
(d) The velocity and kinetic energy of the elevator just before
it strikes the shaft.
(e) The potential energy of the compressed spring.
(f) The energy of the system consisting of the elevator and
spring (1) at the start of the process, (2) when the elevator
reaches its maximum height, (3) just before the elevator strikes
the spring, and (4) after the elevator has come to rest.
Solution:
The whole situation will probably be clearer if we draw a sketch
of it:
10 m
100 m
(a) If our reference elevation is the bottom of the shaft, and
the initial elevation of the elevator is 10 m above the bottom of
the shaft, then the potential energy of the elevator (relative to
the bottom of the shaft) is
Ep = mgz = 2500 kg 9.8 m s-2 10 m = 245,000 kg m2 s-2 = 245
kJ
(b) The work done to raise the elevator is equal to the integral
of the force applied to the elevator over the distance it is
raised. In this case, the force is constant F = mg (or more
formally, F = mg/gc to make it work in English units). So, the work
done is
( )2 2
1 1
2 1
z z
z z
W Fdz mgdz mg z z= = = W = 2500 kg 9.8 m s-2 90 m = 2,205,000 kg
m2 s-2 = 2,205 kJ
Lecture 1
-
(c) Just as in part (a), we have
Ep = mgz = 2500 kg 9.8 m s-2 100 m = 2,450,000 kg m2 s-2 = 2,450
kJ
(d) We use the principle of conservation of mechanical energy
(in this frictionless system) to equate the change in gravitational
potential energy to the change in kinetic energy.
( ) ( ),2 -2
,
0
0 2,450 kJ 0 0
2,450 kJ = 2,450,000 kg m s
P K
K after
K after
E E
E
E
+ =
+ =
=
That is, the 2,450 kJ of potential energy that the elevator has
when it is at a height of 100 m and its kinetic energy is zero
(because its velocity is zero) is converted completely to kinetic
energy when it falls to a height of zero (where its potential
energy is zero). The velocity corresponding to this kinetic energy
is obtained from
( )
2 -2 21, 2
2 -2
12
2,450,000 kg m s
2,450,000 kg m s 44.3 m/s2500
K afterE mu
ukg
= =
= =
Note that we could get this same result by solving Newton's laws
of motion for the elevator. At t = 0, we have z = 100 m and u = 0.
The acceleration of the elevator (by gravity) is a = g = 9.8 m s-2.
Integrating this to get the velocity gives:
-29.8 m s
, where C is a constant of integration
dv adtv at C
= =
= +
At t = 0, v = 0, so C = 0, and v = -9.8 t m/s
Integrating again,
-2
2 212
9.8 m s
4.9 , where C is a constant of integration
dz v tdtz vt C C t
= =
= + =
At t = 0, z = 100 m, so C = 100 m, and z = (100 - 4.9t2) m
To find the velocity when the elevator reaches z = 0, we first
must find the time when z = 0 by solving 100 - 4.9t2 = 0, which
gives t = 4.518 s. At this time, the velocity is then v = -9.8 m
s-2 4.517 s = 44.3 m/s. We could show (but won't) that the law of
conservation of mechanical and potential energy can be derived from
Newton's laws of motion.
This is one of many examples in the course in which there is
more than one logical and appropriate way to reach the correct
answer. On occasion, I will show methods that differ from
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those in the textbook. In such cases, you are encouraged to try
applying both methods, which will hopefully deepen your
understanding of the problem and its solution.
(e) The change in potential energy of the spring plus the change
of kinetic energy of the elevator must sum to zero (again, for a
frictionless system), so EP,spring = 2,450 kJ after the elevator
comes to rest (converting all of its kinetic energy to potential
energy of the spring). Note that since the spring is described as
'very stiff', we are neglecting any change in the elevation of the
elevator during the compression of the spring. If the spring
compresses a significant distance, then the gravitational potential
energy would become negative (since the elevation of the elevator
would be less than the reference elevation) and the potential
energy of the spring would be slightly more than 2,450 kJ.
(f) If the elevator and spring are taken together as the system,
then the energy of the system only changes when something outside
the system does work on it, which only happens when the elevator is
initially raised from 10 m to 100 m. So, initially, the total
energy is 245 kJ (as in part (a)). After the elevator is raised to
100 m, the total energy of the spring plus the elevator is 2,450
kJ. When the elevator falls and compresses the spring, energy is
converted from one form to another within the system, but no work
is done on or by objects outside the system, so the total energy
remains 2,450 kJ.
Heat: See discussion in SVA section 1.9.
Heat is what flows when energy is transferred from a warmer
object to a cooler one, raising the internal energy of the cooler
object and lowering the internal energy of the warmer one, until
their temperatures are equal. A more precise definition will
require improved definitions of temperature and internal energy.
The first law of thermodynamics (up next) will show us that heat
and work are equivalent in some senses but not in others. As a
result, heat is measured in the same units (joules or btu) as
work.
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When we add heat to or do work on a substance, we may increase
its internal energy. From a macroscopic point of view, this is
energy stored within the material that can later be recovered as
heat transferred to a cooler object. Microscopically, this energy
is stored in the motions of the molecules that make up the material
and in the potential energy of the interactions between the
molecules. In classical thermodynamics, absolute values of internal
energy aren't known, just relative values, which are all that are
required for thermodynamic analysis anyway.
The First Law: Energy is Conserved
Although energy assumes many forms, the total quantity of energy
is constant, and when energy disappears in one form it appears
simultaneously in other forms.
We often analyze problems by dividing the universe into the
system (the part we are explicitly studying) and the surroundings
(everything else).
Any changes in the energy of the system must be accompanied by
changes in the energy of the surroundings. That is
(Energy of the System) + (Energy of the Surroundings) = 0
Energy Balances for Closed Systems:
If our system is closed (no mass flows across its boundaries)
then energy can only enter or leave the system as heat or as work,
and we have
(Energy of the System) = Q + W
The System (the part of the universe
were looking at)
The Surroundings (Everything Else)
Q
W
FIRST LAW OF THERMODYNAMICS (chapter 2 in SVA)
Internal energy:
-
Where Q is energy transferred from the surroundings to the
system by heat, and W is energy transferred from the surroundings
to the system as work (work done on the system by the
surroundings). This implies that
(Energy of the Surroundings) = -Q - W
The sign convention for Q and W used here is the one used in
SVA, but is not universal. You should always be sure you understand
which way energy is flowing and not just rely on the sign in the
equations to get it right.
If all energy transferred into the system goes to increase the
internal energy of the system (as opposed to increasing its kinetic
or potential energy by changing its overall position or velocity),
then we have
Ut = Q + W
where Ut is the total internal energy of the system.
For infinitesimally small (differential) changes, this is
dUt = dQ + dW
The quantity Ut is an extensive quantity - it is the total
internal energy of the system, which if the system is homogeneous,
is proportional to the total amount of matter in the system.
Homogeneous means "the same throughout" - at every point within
the system the properties (temperature, pressure, density,
composition, etc.) are the same as at every other point.
If the system is homogeneous, then we can define intensive
properties like the internal energy per mole or per mass
U = Ut/n = internal energy per mole = total internal energy
divided by the number of moles
U = Ut/m = internal energy per mass = total internal energy
divided by the total mass
Then, for a homogeneous closed system containing n moles of
material, we can write
(nU) = nU = Q + W
d(nU) = ndU = dQ + dW
Often, we simply write thermodynamic equations for a
representative amount of a homogeneous material, in which case, we
would just write (for one mole of material, for example)
U = Q + W
dU = dQ + dW
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This last expression, in particular, allows us to keep track of
the molar internal energy of a substance and compute changes in it
by computing the amount of work done on it by its surroundings and
the amount of heat that flows to it from its surroundings. By
observing how its temperature and other properties change as its
internal energy changes we can also relate changes in internal
energy (a property of the substance) to changes in its other
properties (like density, etc.) that are observable.
Thermodynamic State and State Properties
At a given thermodynamic state, a homogenous system always has
the same thermodynamic properties. For a homogeneous pure
substance, fixing two properties (i.e. temperature and pressure or
temperature and density or density and pressure) fixes all of its
other thermodynamic properties (temperature, pressure, density,
internal energy, etc.). For more complex systems, more than 2
thermodynamic properties may have to be specified to specify the
state of the system, but once the required number is specified, the
thermodynamic state of the system is fixed and all other
thermodynamic properties are also fixed.
State functions depend only on the thermodynamic state of the
system and do not depend on the path by which the system arrived at
that state. Internal energy is a state function.
Q and W are not state functions. They do not represent
properties of the system, but represent interactions of the system
with its surroundings. As such, they depend on the path. The system
can get from one state to another through different paths, where by
different paths we mean through different interactions with its
surroundings. If two paths move the system between the same two
states (they cause the same change in the state of the system) then
the sum Q + W will be the same for both paths, since the change in
internal energy must be the same. However, Q and W, individually,
will be different for the two paths.
For example, suppose I have some water at a temperature of 280 K
and a pressure of 1 atmosphere. Fixing the temperature and pressure
specifies the state of the system. I could stir the water violently
until its temperature increased to 300 K (still at 1 atmosphere)
while keeping it thermally insulated. In this case, Q would be
zero, and we would have U = W. Alternatively, I could heat the
water from 280 K to 300K by sitting the (non-insulated) beaker on a
warm surface. In this case, no work would be done on the water (W =
0), and we would have U = Q. Both methods of heating the water
result in the same change in the state of the water (they go from
the same initial thermodynamic state to the same final
thermodynamic state), so U is the same in both cases. However, Q
and W are different for the two cases.
The Phase Rule:
For any non-reacting system at thermodynamic equilibrium, the
number of independent variables that must be fixed to establish its
intensive state is given by
F = 2 - + N
where is the number of phases present, N is the number of
chemical species present, and F is the number of degrees of freedom
that the system has. The number of degrees of freedom of the
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system is the number of phase rule variables (temperature,
pressure, composition of each phase, etc.) that must be fixed to
completely specify the state of the system, including all other
phase-rule variables.
See example 2.5 in SVA.
Reversible Processes
In thermodynamic analyses, we often make use of the idealized
concept of a reversible process, much like in first-year physics we
make use of a lot of frictionless processes, even though such a
process can only be approximately realized in a real system.
A reversible process is one whose direction can be changed at
any point by infinitesimal changes in external conditions
(which means there must be no friction, no inertia (so it must
happen infinitely slowly), no heat transfer through finite
temperature differences, and so on).
A reversible process
Is frictionless
Is never more than infinitesimally away from equilibrium
Goes through a set of equilibrium states
Is driven by infinitesimally small force or temperature
imbalances
Can be reversed at any point by an infinitesimal change in
external conditions
When reversed, retraces its original path back to the original
state of the system
Constant Volume processes, Constant Pressure processes, and
Enthalpy
For a homogeneous, closed system containing n moles of a
substance, the energy balance was (in differential form)
d(nU) = dQ + dW
For a mechanically-reversible closed-system process, the work
term is
dW = -PdVt = -Pd(nV) = -nPdV
So, for a constant volume process, the work is zero and we
have
-
dQ = n dU (at constant V)
or, in integral form
Q = nU (at constant V)
At constant pressure, the energy balance becomes
d(nU) = dQ - nPdV (at constant pressure)
or
dQ = n(dU + PdV) = nd(U + PV) (at constant pressure)
This motivates the definition of enthalpy as H = U + PV, so that
at constant pressure dQ = n(dH), just as at constant volume dQ =
n(dU).
Note that even though Q is not a state function, H is defined in
terms of U, P, and V, which are state functions, so H is a state
function as well.
See example 2.8 in SVA.
Heat Capacity
The constant volume heat capacity is defined as
vV
dUCdT
For a constant volume process, we can then write
dU = CvdT (at constant V)
Or, integrating over a finite change in temperature
2
1
T
v
T
U C dT = (at constant V) For a mechanically reversible
constant-volume process, we then have
2
1
T
v
T
Q n U n C dT= = (for a mechanically reversible, constant volume
process) Likewise, we define the heat capacity at constant pressure
as
-
pP
dHCdT
Then, for a constant pressure process on a fixed amount of a
homogeneous substance, we have
dH = CpdT (at constant P)
and for finite temperature changes
2
1
T
p
T
H C dT = (at constant P) And finally, for a mechanically
reversible, constant pressure process, this tells us that
2
1
T
p
T
Q n H n C dT= = (for a mechanically reversible, constant
pressure process) See example 2.9 in SVA
Mass and Energy Balances for Open Systems
An open system is one in which mass, as well as energy, crosses
the system boundaries. For an open system, energy is carried in and
out of the system with the mass that enters and leaves the system,
as well. In general, we can write a mass balance on a system as
accumulation = in - out
Short of nuclear reactions, we cannot have a 'generation' term
in a mass balance (mass is conserved). The same is true for a mole
balance in a non-reacting system, but of course in a reacting
system we can have a change in the total number of moles due to
reaction (leading to a 'generation' term in the balance
equation).
The total mass balance on a fixed control volume with streams
entering and leaving it can be written in SVA's notation as
( ) 0cv fsdm mdt + =
Where the first term is the time derivative of the total mass in
the control volume (accumulation) and the second term is the
difference between the total inflow and outflow (out - in) in the
various streams. This should be very simple compared to things done
last semester in CE 212, so don't let it seem harder than it really
is!
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Similarly, the energy balance can be written as
( ) ( )( )212cv fsd mU H u zg m Q Wdt + + + = +
where U is the internal energy per unit mass (specific internal
energy) within the (homogeneous) control volume. H is the specific
enthalpy of each inflow and outflow stream, u is the average flow
velocity of each inlet and outlet stream, and z is the elevation of
each inlet and outlet stream. In many, many cases, kinetic energy
and gravitational potential energy can be neglected, and this
becomes simply
( ) ( )cv fs
d mUmH Q W
dt+ = + (when kinetic and potential energy changes are
negligible)
Furthermore, we are often interested in systems at steady state,
in which case the time derivative is zero, and we have simply
( ) fsmH Q W = + (at steady-state AND when kinetic and potential
energy changes are negligible)
There are many possible variations on this energy balance, and
one must carefully assess, in a given problem which form should be
used (which terms should be included).
Example 2.12 in SVA
Example 2.15 in SVA
hclTypewritten textUnsolved Problems (SVA)prob 2.6 to 2.15 and
2.25 to 2.27
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You should be able to define, and locate on such a diagram, the
solid region, liquid region, gas region, supercritical fluid
region, fusion curve (melting curve), sublimation curve,
vaporization curve, critical point, and triple point.
Here is the PT diagram for water, which has the unusual feature
of a fusion curve with a negative slope:
PVT behavior of pure substances
Let's start by looking at PT and PV diagrams for a 'typical'
substance. Here is a PT diagram:
Volumetric Properties of Pure Fluids (chapter 3 in SVA)
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The PT diagrams don't contain any information about the volume
of the fluid, which is an important piece of information for many
thermodynamic analyses. So, we should look at some PV diagrams too.
Here is a generic one that I drew myself. Again, you should be able
to identify all of the regions
An isotherm (path of constant temperature) on such a diagram
might look like this:
The temperature for that isotherm is far below the critical
temperature, so as the pressure is decreased, there is first a
small increase in the specific volume (as the liquid phase expands
just a
Solid + Vapor
Sol
id
Sol
id +
Liq
uid
Liquid + Vapor
Liquid
Vapor
Specific Volume
Pre
ssur
e
critical point
Solid + Vapor
Sol
id
Sol
id +
Liq
uid
Liquid + Vapor
Liquid
Vapor
Specific Volume
Pre
ssur
e
critical point
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little), then there is a big change in specific volume at
constant pressure, corresponding to the phase change from liquid to
gas (note that this will require substantial heat input to keep the
temperature constant). After that, the specific volume continues to
change with pressure as the gas expands.
For a temperature above the critical point, the isotherm would
look more like
At temperatures above the critical point, there is no
vapor-liquid phase transition. An isotherm just at the critical
temperature would have a point of inflection at the critical point,
as shown here:
Solid + Vapor
Sol
id
Sol
id +
Liq
uid
Liquid + Vapor
Liquid
Vapor
Specific Volume
Pre
ssur
e
critical point
Solid + Vapor
Sol
id
Sol
id +
Liq
uid
Liquid + Vapor
Liquid
Vapor
Specific Volume
Pre
ssur
e
critical point
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Equations of State:
For a single-component, single-phase substance, we know from the
phase rule (and from the diagrams at which we were just looking)
that two intensive thermodynamic quantities can be specified, and
then the state of the system (and therefore all other intensive
thermodynamic quantities) is specified. This implies that for a
single-phase, single-component system there exists a unique
relationship between the temperature, pressure, and specific volume
(T, P, and V). We call this relationship an equation of state for
the substance. Once we specify two of the three values of P, V, and
T, the third can be found from the equation of state. We can
consider any one of the three quantities to be a function of the
other two. For example, we can consider specific volume to be a
function of temperature and pressure
V = V(T,P), which implies that (by definition of the partial
derivatives)
P T
V VdV dT dPT P
= +
The partial derivatives are related to the isothermal
compressibility and the volume expansivity, which are physical
properties that are sometimes available from tables in handbooks
(for liquids and solids). Volume expansivity () is defined as
1P
VV T
=
Isothermal compressibility () is defined as
1T
VV P
=
So, we can write
dV dT dPV
=
If the isothermal compressibility and the volume expansivity
were both zero, then the fluid would be incompressible, which is a
common approximation for liquids. Using constant values for and is
the next-better approximation, which accounts, on average, for the
pressure and temperature dependence of the specific volume of
liquids. An exact treatment would require pressure and temperature
dependent values of and (but this approach is rarely required). See
example 3.1 in SVA
For gases, we are familiar with the ideal gas equation of state,
which can be written as
RTVP
=
-
or
RTPV
=
or
PVTR
=
Next, we'll consider more realistic (and therefore
mathematically more complex) equations of state for gases.
-
At sufficiently low pressure, all gases approach ideal gas
behavior, which means that they obey the ideal gas equation of
state:
PV = RT
where V is the molar volume (volume per mole). The ideal gas
equation of state provides the basis for virtually all equations of
state for gases. Equations of state that are more accurate at
higher pressures usually add corrections to the ideal gas equation
of state. Since it is the baseline against which other equations
are derived, we will start by going through important properties of
ideal gases and and how to do process calculations on them.
An important property of an ideal gas is that its internal
energy depends only on temperature (and not on pressure). This, in
turn, implies that the constant volume heat capacity is a function
only of temperature (and not of pressure or specific volume). That
is
( )
( )v vV
U U TU dUC C TT dT
=
= =
From the definition of enthalpy and the constant pressure heat
capacity, we can show that H is also a function only of
temperature, and we can derive the relationship Cp = Cv + R as
follows:
( ) ( )
( )p vP
H U PV U T RT H TH dH dUC R C T RT dT dT
+ = + =
= = + = +
Because enthalpy and internal energy are functions only of
temperature, we have
dU = CvdT
and dH = CpdT
for all processes (not just the usual constant volume and
constant pressure processes, respectively).
So, for an ideal gas undergoing any process we have
2
1
2
1
T
v
T
T
p
T
U C dT
H C dT
=
=
Ideal Gas Equation,
-
regardless of the details of the process.
Process calculations with ideal gases
Calculating the heat and work required to affect changes in an
ideal gas is particularly simple (at least compared to similar
calculations with more complex equations of state). For an ideal
gas in a closed system undergoing a mechanically reversible
process, we have (on a per mole or per mass basis)
dU = CvdT = dQ + dW
We also have (again for a mechanically reversible, closed-system
process)
dW = -PdV
So we can write
dQ = CvdT + PdV
We can rearrange these relationships and use the ideal gas law
to eliminate one of the 3 variables (P, V, or T) to solve
particular problems. If we substitute P = RT/V, we get
dW = -(RT/V) dV
dQ = CvdT + (RT/V) dV
If we substitute V = RT/P, from which 2R RTdV dT dPP P
= , we get
RTdW RdT dPP
= +
v pRT RTdQ C dT RdT dP C dT dPP P
= + =
Finally, if we substitute T = PV/R, from which V PdT dP dVR
R
= + , we get
dW = -PdV
pvvC PC VV PdQ C dP dV PdV dP dV
R R R R
= + + = +
We can use these relationships to solve for the heat and work
requirements for all sorts of processes.
Isothermal Processes:
-
Since the internal energy and enthalpy of an ideal gas are
functions only of temperature, they do not change in an isothermal
process
U = H = 0 for an isothermal process
Using the equations from above with temperature and volume as
the independent variables:
dW = -(RT/V) dV
dQ = CvdT + (RT/V) dV = (RT/V) dV
We can integrate to get the following expressions for the heat
and work
W = -RT ln(V2/V1) = RT ln(P2/P1)
Q = RT ln(V2/V1) = -RT ln(P2/P1)
Isobaric Processes:
For a constant pressure (isobaric) process, we can start from
the equations written above in terms of pressure and
temperature
RTdW RdT dP RdTP
= + =
p pRTdQ C dT dP C dTP
= =
Integrating these, we get
( ) ( )2 1 2 1W R T T P V V= =
2
1
T
p
T
Q C dT= We also have, as always,
2
1
2
1
T
v
T
T
p
T
U C dT
H C dT
=
=
-
Isochoric Processes:
For constant volume (isochoric) processes, we can use the
equations in terms of T and V to get
dW = -(RT/V) dV = 0
dQ = CvdT + (RT/V) dV = CvdT
From which
W = 0
2
1
T
v
T
Q C dT= We also have, as always,
2
1
2
1
T
v
T
T
p
T
U C dT
H C dT
=
=
Adiabatic Processes: Ideal gas with constant heat capacity
For an adiabatic process, by definition no heat enters or leaves
the system, so dQ = 0.
We can therefore take any of the expressions we had above for dQ
and set it equal to zero. Doing so gives
dQ = CvdT + (RT/V) dV = 0, or CvdT = - (RT/V) dV
and 0, or p pRT RTdQ C dT dP C dT dPP P
= = =
and 0, or p pv vC P C PC V C VdQ dP dV dP dV
R R R R= + = =
If the heat capacities are constant, we can integrate each of
these to get
-
2 21 1
2 1
1 2
ln ln
v
v
vR
C
dT R dVT C V
T VRT C V
T VT V
=
=
=
and
2 21 1
2 2
1 1
ln ln
p
p
p
RC
dT R dPT C P
T PRT C P
T PT P
=
=
=
and
2 21 1
2 1
1 2
ln ln
p
v
p
v
p
vC
C
CdP dVP C V
CP VP C V
P VP V
=
=
=
Often, we use the relationship Cp = Cv + R and we define the
heat capacity ratio as
p pvv v p
C CC RC C C R
+= = =
To get
1
1 11
v
p v
p p
RC
C CRC C
=
= = =
Which then gives
-
( )
( )
12 1
1 2
1
2 2
1 1
2 1
1 2
T VT V
T PT P
P VP V
=
=
=
for an ideal gas with constant heat capacity in an adiabatic
process
which we can also write as
( )
( )1
1
constant
constant
constant
TV
TP
PV
=
=
=
for an ideal gas with constant heat capacity in an adiabatic
process
Polytropic Processes:
We define a polytropic process as one for which
PV = constant
For some value of . For this model, SVA give equations for
relating T and V, for relating T and P, and for computing heat and
work requirements. Rather than derive them, we will now work
through examples 3.2 through 3.6 (or as many as possible before the
end of class).
-
For gases, we know that at constant temperature, the product of
pressure and specific volume is approximately constant. Therefore,
we might try to write an equation of state for gases that looks
like:
2 3 ...PV a bP cP dP= + + + +
where the coefficients a, b, c, etc. are functions only of
temperature (and not of pressure). This is an infinite series, but
if the PV product is a weak function of pressure, then it can be
truncated after a few terms. We could also write it as
( )2 31 ' ' ' ...PV a B P C P D P= + + + + where we've simply
re-defined the constants as b = aB', etc.
It turns out that a in the above equation is the same
temperature-dependent function for all gases, though the other
constants are, in general, different for each particular gas.
Experimentally, as the pressure becomes very small (mathematically,
in the limit that P goes to zero) the pressure-volume product (PV)
goes to the same value for all gases. This is used to establish the
ideal gas temperature scale and, in turn, the value of the
Universal Gas Constant. If we denote the gas-independent value of
PV as the pressure goes to zero as (PV)*, then we can define the
ideal gas temperature scale by assigning the quantity (PV)* to be
proportional to temperature
( )*PV a RT= = where the proportionality constant is the
Universal Gas Constant. R is then defined by setting the
triple-point of water to be at T = 273.16 K and using the
experimental value of (PV)* at the triple-point of water, which is
(PV)* = 22711.8 cm3 bar mol-1. This gives
3 -1 -1 3 -1 -1 -1 -122711.8 83.1447 cm bar mol K 8.31447 m Pa
mol K 8.31447 J mol K273.16
R = = = =
So, this gives us the first term in our virial equation of state
- the pressure (and specific volume) independent term that is the
same for all gases. Since it will always be the same, we could move
it to the other side of the equation and write
2 31 ' ' ' ...PV PVZ B P C P D PRT a
= = + + + +
We define Z as the compressibility, which is equal to 1 for an
ideal gas. Variations in Z away from 1 then indicate non-ideal
behavior. Sometimes the virial equation of state is written as an
infinite series expansion in inverse volume rather than in
pressure, in which case the virial coefficients are different.
VIRIAL EQUATIONS OF STATE:
-
2 31 ...PV B C DZRT V V V
= + + + +
where, with some mathematical manipulation, we could relate the
two forms of the virial coefficients as
( )2
2
'
'
BBRTC BC
RT
=
=
and so forth.
In either form (volume-expansion or pressure-expansion), the
virial equations will only be useful for practical applications
when they converge within a few terms. This is true both because
the computations would become inconvenient and because only the
first few coefficients are likely to be available from experiments.
So, for practical applications, we will truncate these to a one-,
two-, or three-term form. The one-term form, in which we keep only
the first term is just the ideal gas law:
1PVZRT
=
Truncating the pressure-expansion version after two terms
gives
1 ' 1PV BPZ B PRT RT
= + = +
where, in the second version, we have substituted ' BBRT
= . Alternatively, we could truncate
volume expansion as
1PV BZRT V
= +
but the other expression in terms of P is more convenient and
more commonly used.
We could also truncate the volume expansion after 3 terms to get
a 3-term virial equation
21PV B CZRT V V
= + +
This equation gives P in terms of V directly, but is cubic in V
and therefore if one wants to solve for V, one generally does so
iteratively. Remember that the coefficients B and C are temperature
dependent, and therefore the solution for T in terms of P and V
(which we usually don't want to do anyway) may be simple or complex
depending on the temperature dependence.
-
Virial coefficients beyond the 3rd virial coefficient are rarely
known. If we look at a plot of the compressibility of a typical
substance like this one for CO2, which I've prepared from data
downloaded from the NIST WebBook, we can get an idea of the range
over which a 1, 2, or 3-term virial expansion may be appropriate. A
similar plot for methane is given on p. 87 of SVA.
Here is the same plot with the different truncations of the
virial equation superimposed on it.
Compressibility of CO2
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
0 10 20 30 40 50 60 70 80 90 100Pressure (MPa)
Z =
(PV)
/(RT) 600 K
400 K
Compressibility of CO2
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
0 10 20 30 40 50 60 70 80 90 100Pressure (MPa)
Z =
(PV)
/(RT)
400 K, actual
600 K, actual
600 K, 3-term
600 K, 2-term
1-term
400 K, 2-term and 3-term
-
The values of the virial coefficients are obtained from the
slope and curvature of the curves as P goes to zero:
0
2
20
'
1'2
P
P
dZBdP
d ZCdP
=
=
For this particular case, we see that at 400 K, the 3-term
expansion doesn't work any better than the 2-term expansion, and
both are good to about 10 or 15 MPa (100 to 150 atm). At 600K, the
2-term expansion is only good to about 10 MPa, but the 3-term
expansion is good out to more than 50 Mpa. Clearly both the values
of the virial coefficients and the range over which a given
expansion is useful depend on temperature.
For precise fitting of measured PVT data, one can use an
extended virial equation in which additional terms are added to the
3-term virial expansion to allow more flexibility in data fitting.
An example is the Benedict/Webb/Rubin (BWR) equation of state:
2
0 0 02 3 6 3 2 2 21 exp
B RT A C TRT bRT a a cPV V V V V T V V
= + + + + +
This has 8 parameters, and sometimes some of these are taken to
have additional temperature dependence. We will not make much use
of the BWR equation of state in this course, but for substances for
which the parameters are available, it can provide high
accuracy.
Example 3.7 from SVA
Cubic Equations of State:
So far, we have talked about equations of state that are
basically good for gases - ideal or non-ideal, but still gas-like
(that is, with compressibility reasonably near 1). There are many
situations where we would like to have an equation of state that
describes both the liquid and the vapor phase of a substance. The
simplest and most widely used of such equations are cubic equations
of state. By cubic, we just mean that they are cubic polynomials in
the specific volume. The first (historically) and simplest of these
is the van der Waals equation of state.
The van der Waals Equation of State:
This was the first practical cubic equation of state, and it
is:
2RT aP
V b V=
-
where a and b are positive. If a and b were both zero, this
would become the ideal gas equation of state.
Some general features of cubic equations of state are
illustrated for the van der Waals EOS as plotted for CO2 below.
For this, and all cubic equations of state, there is always a
single pressure corresponding to a given volume (at any
temperature), but there may be either one or three volumes
corresponding to a given pressure. For temperatures above the
critical temperature, there is only a single volume for a given
pressure (or at least only a single physically meaningful volume,
which must be greater than b). For temperatures below the critical
temperature, there are 3. The smallest corresponds to the liquid
phase, while the largest corresponds to the vapor phase. The middle
one doesn't have a direct physical meaning. We will consider this
all in more detail next time.
van der Waals EOS for CO2
0
50
100
150
200
250
0 200 400 600 800 1000 1200 1400
Volume (cm3/mol)
Pres
sure
(bar
)
600 K
400 K
300 K
250 K
200 K
-
r
c
rc
TTTPPP
That is, if we scale the temperature and pressure by the
critical temperature and pressure, all fluids are about the same.
This is nearly true for simple fluids (monatomic gases except
helium), but less nearly true for other substances. The description
of other fluids can be improved by introducing a third parameter.
The most popular 3rd parameter is the acentric factor, . The
acentric factor of a pure substance is defined in terms of its
vapor pressure. If we plot the log of the vapor pressure vs.
inverse temperature, we get approximately a straight line. If we
plot this as reduced pressure vs. inverse reduced temperature, we
get a plot like the one shown below.
of the word theorem. It might better be termed a 'coarse
generalization of experimental observations' or something like
that. It is not a statement that is mathematically provable or that
is always true. The reduced temperature and reduced pressure are
defined as
The Theorem of Corresponding States The two-parameter theorem of
corresponding states, which is based on experimental observations,
says that all fluids, when compared at the same reduced pressure
and temperature, have about the same compressibility. Note that
this is an unusual use
-
Vapor Pressure Curves
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
01 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
1/Tr (1/K)
log 1
0(Pr,s
at)
Argon (blue)
Methane (magenta)
Hexane
Ethanol
Tr=0.7
The acentric factor is defined as the difference in the value of
this curve from the value of -1.0 that is observed at T
r = 0.7 for simple fluids (argon, neon, krypton). Thus, to find
it one needs
one vapor pressure measurement (at Tr = 0.7) as well as the
critical properties. This is the basis
of the three-parameter theorem of corresponding states: all
fluids that have the same acentric factor, when compared at the
same reduced pressure and temperature, have about the same
compressibility. This is more generally true than is the
2-parameter theorem of corresponding states. That is, it agrees
with reality for more cases, because it contains one more parameter
and is therefore more flexible. The acentric factor is used in some
equations of state, like the Soave-Redlich-Kwong equation of state
and the Peng-Robinson equation of state, where it shows up in the
function a(T). It is also used in the generalized correlations for
the compressibility of gases that we will discuss shortly.
-
Z = Z0 + Z1
where Z is the compressibility of the gas (or supercritical
fluid), is the acentric factor that we just discussed, and Z0 and
Z1 are both functions of the reduced temperature and reduced
pressure. The most popular of these correlations is the Lee/Kesler
correlation. For it, values of Z0 and Z1 are found in appendix E of
the textbook. To use these, you first look up the critical
temperature, critical pressure, and acentric factor for the
substance of interest, then compute the reduced pressure and
temperature, then look up Z0 and Z1 in the tables (interpolating
between entries) and then finally put them in the above equation
for Z. For nonpolar or slightly polar gases, this correlation
should give the compressibility to with 2 or 3 percent.
Generalized correlations for gases: There are a variety of
generalized correlations that can be used to estimate the PVT
behavior of real (non-ideal) gases. These are often of the form
hclTypewritten textUnsolved Problems
prob. 3.6, 3.8, 3.14, 3.27, 3.46
-
Over a limited range of reduced temperatures and pressures
(shown on p. 103 of SVA), this correlation can be represented by a
two-term truncated virial equation of state as follows. The
compressibility is written as
1 1 c rc r
BP PBPZRT RT T
= + = +
with the correlation
0 1cc
BP B BRT
= +
which gives
0 11 r rr r
P PZ B BT T
= + +
or, in the form of the original Lee/Kesler correlation:
0 1
0 0
1 1
1 rr
r
r
Z Z ZPZ BT
PZ BT
= +
= +
=
The coefficients B0 and B1 are functions only of temperature
(remember that the virial coefficients themselves are functions
only of temperature). They are well represented by:
01.6
14.2
0.4220.083
0.1720.139
r
r
BT
BT
=
=
Generalized correlations for liquids The Lee/Kesler correlation
discussed above for gases also includes data for subcooled liquids.
Figure 3.14 on p. 101 of SVA shows its predictions for both liquids
and gases. For saturated liquids (those in equilibrium with the
vapor) some simpler correlations work. The Rackett equation gives
good results and only requires the critical parameters as
inputs:
( )0.28571 rTsat
c cV V Z
= Another graphical correlation for liquid densities is shown in
figure 3.17 on p. 109 of SVA. So, now we have a variety of
equations of state and correlations to compute the PVT behavior of
fluids. In the homework and in-class examples, we will practice
applying these for such direct calculations (i.e. compute volume
given temperature and pressure), and then we will go on to use them
throughout the rest of the semester to solve more complicated and
more interesting problems.
-
and vV T T
U U UdU dT dV C dT dVT V V
= + = +
where we have used the definition of Cv that we have already
discussed in previous lectures. The last term (the dV term) is zero
for Any constant-volume process, with any fluid Any fluid where
internal energy does not depend on molar volume, including
Ideal gases Incompressible liquids
In those cases, we have simply
2
1
vT
v
T
dU C dT
U C dT
=
= Similarly for enthalpy, we write H = H (T,P)
and pP T T
H H HdH dT dP C dT dPT P P
= + = +
where we have used the definition of Cp that we have already
discussed. The last term (the dP term) is zero for Any
constant-pressure process, with any fluid
When we put heat into (or remove heat from) a fluid, this heat
might be used (in whole or in part) to do work on the surroundings,
to change the temperature of the fluid, to change the phase of the
fluid, or to provide the energy required to carry out a chemical
reaction. In the next couple of lectures, we will talk about these
heat effects. The first ones that we will consider are sensible
heat effects, which involve heating that changes the temperature of
the system. Sensible Heat Effects If we consider a closed system
where there are no phase transitions, no chemical reactions, and no
changes in the composition of the system, then adding heat to or
removing heat from the system will change its temperature and/or
cause it to do work on the surroundings. Our goal here is to relate
the temperature change and work done to the amount of heat added.
For a homogeneous substance of constant composition, the phase rule
tells us that two intensive properties must be fixed to specify the
state of the fluid completely. Thus the internal energy or enthalpy
can be written as a function of two intensive variables, like
temperature, pressure, and molar volume. Thus, using T and V as the
independent variables for U, we write U = U (T,V)
hclTypewritten textSENSIBLE HEAT AND LATENT HEAT
-
Any fluid where the enthalpy does not depend on pressure,
particularly ideal gases In those cases, we have simply
2
1
p
T
p
T
dH C dT
H C dT
=
= Furthermore, we know that for constant-volume processes, Q =
U, and for constant-pressure processes, Q = H. This lets us
calculate things like the amount of heat required to change the
temperature of a substance by a certain amount. Heat capacities are
often expressed in the form of polynomials in temperature, or as
other simple functions. The one used in SVA most commonly is
2 2pC
A BT CT DTR
= + + +
Note that for ideal gases, the constant volume and constant
pressure heat capacities are related by Cp = Cv + R So for an ideal
gas, if Cp is given by the form shown above, then Cv is given
by
2 21vC A BT CT DTR
= + + +
For incompressible liquids, Cv and Cp are the same (a constant
pressure process on an incompressible substance is also a
constant-volume process, since the volume is always constant for an
incompressible substance). The above form of the heat capacity is
easily integrated - for example
( )
( ) ( ) ( )
2 2
1 1
2 2
2 2 3 32 1 2 1 2 1
2 1
1 12 3
T T
p
T T
H C dT R A BT CT DT dT
B CH R A T T T T T T DT T
= = + + +
= + +
,all
p i p ii
C y C=
SVA gives a longer description of evaluation of this, but I
assume you can all do integrals, so let's not spend time on it.
They also give a representative computer program for doing this in
appendix D. We might similarly have a handy function in an Excel
spreadsheet For mixtures of gases at low pressure, one can simply
take the heat capacity of the mixture as the average of the heat
capacity of the components, weighted by their mole fractions in the
mixture. That is, if y
i is the mole fraction of species i in the mixture, then the
molar heat capacity of the
mixture is simply
-
Latent Heats of Pure Substances When a pure substance is
converted from one phase to another (solid to liquid, liquid to
vapor, etc.) heat flows to or from the substance, but there is no
change of temperature. The energy per mole required to vaporize a
liquid is called the latent heat of vaporization, and the energy
per mole required to melt a solid is called the latent heat of
fusion. We know from the phase rule that for a 1-component, 2-phase
system, specifying one intensive variable specifies the state of
the system. Thus, the latent heat of a phase change is a function
only of one intensive variable, and can therefore be written as a
function of temperature alone. This function can be written as
satdPH T V
dT =
where H is the latent heat for the phase change (the difference
between the molar enthalpies of the two phases), V is the volume
change for the phase change (big for vaporization, small for
melting), and Psat is the pressure of the two-phase system (which
is a function only of temperature, since specifying the temperature
specifies all other intensive properties of the 2-phase system).
For a vaporization phase change, Psat is the vapor pressure of the
liquid. This equation, which we will derive in a few weeks, is
called the Clapeyron equation. So, to get the heat of vaporization,
we could either make a direct measurement (using a calorimeter), or
we could compute it from the vapor pressure curve and the
volumetric properties of the liquid and gas. Some correlations for
estimating heats of vaporization are given on p. 131 of SVA. I
won't bother to repeat them here, but clearly you should be able to
apply them.
-
The process is carried out at constant pressure, and no non-flow
work is done, so Q = H That is, the heat added or removed is equal
to the change in enthalpy of the stream flowing through the
calorimeter. This, in turn, is equal to the difference in enthalpy
between the products and the reactants, at constant temperature.
The heat of reaction for the reaction is defined as this change in
enthalpy at constant temperature. Now, consider the generic
balanced chemical reaction
aA + bB lL + mM The standard heat of reaction for this reaction
at temperature T is defined as the enthalpy change when a moles of
A plus b moles of B in their standard states at temperature T react
to form l moles of L and m moles of M in their standard states at
the same temperature T. A standard state is a particular state of a
species at temperature T and at specified conditions of pressure,
composition, and physical condition (e.g. gas, liquid, or solid).
In SVA, the standard state conditions for computing heats of
reaction are: Gases: the pure substance in the ideal-gas state at 1
bar. Liquids and solids: the pure substance at 1 bar. Older data
tabulations often use 1 atm rather than 1 bar as the standard
pressure, but this makes a negligible difference for most
applications. It is particularly important to note and remember
Reactants in at To Products out at To
Q (in or out)
Isothermal Flow
Calorimeter
Heats of Reaction So far, the heat effects we have discussed
(sensible heat and latent heat of pure substances) have been for
physical processes (processes where no chemical reactions occur).
In fact, pretty much everything that we have done so far this
semester has been for physical processes. However, as you might
guess, chemically reacting systems are an important part of
chemical engineering. Heat effects due to chemical reaction can be
quite large compared to those associated with physical processes.
During a chemical process, energy is stored in and released from
the chemical bonds that form between atoms in molecules. Since
these bonds are, usually, much stronger than the physical
interactions between molecules, they can store and release much
greater amounts of energy. Consider an isothermal flow calorimeter
for measuring the heat released or consumed by reaction. In such a
device, a stoichiometric mixture of reactants flows in and reacts,
and we measure the heat (addition or removal) required to return
the reaction products to the inlet temperature of the
reactants.
-
that the heat of reaction is defined for a particular set of
stoichiometric coefficients of the reaction. For example, we could
write the reaction for the partial oxidation of methane to CO and
H2 (synthesis gas) as CH4 + O2 CO + 2 H2 or as 2 CH4 + O2 2 CO + 4
H2 Either of these is a valid way to write the reaction, but the
heat of reaction for the second version is twice that of the first
version, so it is important that one know which version is being
used. A heat of reaction can only be interpreted with respect to a
particular writing of the reaction. Standard Heats of Formation: It
would be impossible (or at least impractical) to tabulate the heats
of reaction for all possible reactions of interest. Therefore, they
are calculated from a standard set of heats of reaction. This set
is the heats of formation of the chemical species. The heat of
formation of a substance is defined as the heat of reaction when
the substance is formed, in its standard state at temperature T,
from the elements in their standard states at temperature T. For
example, the heat of formation of ethanol (CH3CH2OH) is equal to
the heat of reaction for the reaction 2 C(s) + 3 H2(g) + O2(g)
CH3CH2OH(l) This is the formation reaction for ethanol. In this
hypothetical reaction, two moles of graphite (the standard state of
carbon) plus 3 moles of hydrogen gas plus one-half of a mole of
oxygen gas produce one mole of liquid ethanol. Heats of formation
are defined as the heat of reaction per mole of the species formed,
and therefore these reactions are written so that the
stoichiometric coefficient of the species formed is one. The value
of heats of formation is that we can write any reaction as the sum
of formation reactions. Consider again the partial oxidation of
methane to give synthesis gas CH4 + O2 CO + 2 H2 This can be
written as the sum of the following formation reactions: C + O2 CO
CH4 C + 2 H2 CH4 + O2 CO + 2 H2 Note that we don't have to include
formation reactions for H2 or O2, since they are elements. The heat
of reaction of the first reaction (C + O2 CO) is the heat of
formation of CO (Hf(CO)). The heat of reaction of the second
reaction is minus the heat of formation of CH4 (-Hf(CH4))
(reversing the direction of the reaction reverses the sign of the
heat of reaction). So, the heat of reaction for the partial
oxidation is Hrxn = Hf(CO) - Hf(CH4). In general, any heat of
reaction can be computed as the heat of formation of the products
minus the heat of formation of the reactants. That is, for the
generic reaction
aA + bB lL + mM the heat of reaction is
Hrxn = lHf(L) + mHf(M) - aHf(A) - bHf(B)
-
Thus, if we have a table of the enthalpies of formation of
species (at some standard temperature), then we can compute the
heat of reaction for any reaction involving those species (at the
same standard temperature). The heats of formation of elements are
zero by definition. They form the reference point for defining the
heats of formation of everything else. Usually, heats of formation
are tabulated at a standard temperature of 298 or 298.15 K. To get
heats of formation or heats of reaction at other temperatures, we
must integrate the heat capacities of the reactants and products to
see how the enthalpies of the reactions and products change with
temperature. Heats of Combustion: Most formation reactions cannot
actually be carried out in practice. For example, we probably could
not devise a calorimeter where we form ethanol (in 100% yield) from
a stoichiometric mixture of graphite, hydrogen, and oxygen. So,
heats of formation usually have to be determined from calorimetric
measurements that can easily be carried out. The most common class
of reactions that goes nicely to completion is combustion
reactions. Imagine that we have a flow calorimeter in which we can
burn things. Then we could first burn some hydrogen H2(g) + O2(g)
H2O(l) From which we could get the heat of formation of water (this
is a formation reaction that we can measure directly). Next, we
might be able to burn some graphite directly, as C(s) + O2(g)
CO2(g) From which we would get the heat of formation of CO2 (this
is another formation reaction that we can measure directly).
Knowing these, we could boldly start burning all sorts of compounds
made up of only C, H, and O, and could determine their heats of
formation from the heats of reaction for the combustion reaction
CxHyOz + (x + y - z) O2(g) xCO2(g) + y H2O(l) and the known heats
of formation for carbon dioxide and water. This particular heat of
reaction (to convert 1 mole of the substance burned plus a
stoichiometric amount of oxygen to carbon dioxide and water) is
known as the heat of combustion of the material. For example, we
could burn ethanol as CH3CH2OH(l) + 3 O2(g) 2 CO2(g) + 2 H2O(l) The
standard heat of reaction for this reaction at 298 K is ( ) ( ) (
),298 ,298 2 ,298 2 ,298 3 22 ( ) 2 ( ) ( )o o o orxn f f fH H CO g
H H O l H CH CH OH l = + This is the heat of combustion for
ethanol. So, we might also write it as
( ) ( ) ( ) ( ),298 3 2 ,298 2 ,298 2 ,298 3 2( ) 2 ( ) 2 ( ) (
)o o o ocomb f f fH CH CH OH l H CO g H H O l H CH CH OH l = + Note
that in these reactions we have explicitly indicated the phase (s
for solid, l for liquid, or g for gas) that is being taken as the
standard state for each species. From the above From the above
relationship between the heat of formation and the heat of
combustion, we have
( ) ( ) ( ) ( ),298 3 2 ,298 2 ,298 2 ,298 3 2( ) 2 ( ) 2 ( ) (
)o o o of f f combH CH CH OH l H CO g H H O l H CH CH OH l = + So,
let's imagine for a moment that we don't know the heats of
formation of anything, but we have a flow calorimeter that we
operate at a constant pressure of 1 bar and an inlet and outlet
temperature of 25 C. We first burn H2 in it, and find a heat of
combustion of -68.3174 kcal/mol. Next, we burn some graphite and
find a heat of combustion of -94.0518 kcal/mol. Finally, we
-
burn some ethanol and find a heat of combustion of -258.3884
kcal/mol. The formation reactions for H2O and CO2 are the same as
the combustion reactions of H2 and C, respectively, so the heats of
formation for H2O and CO2 are equal to the heats of reaction that
we measure for the H2 and C combustion reactions (both are at the
same standard state conditions of 25 C and 1 bar). So, we get
( )( )
,298 2
,298 2
( ) 95.0518 kcal/mol
( ) 68.3174 kcal/mol
of
of
H CO g
H H O l
=
=
Then, we can compute the heat of formation of ethanol from
these, and from the heat of combustion that we measured for it:
( ) ( ) ( ) ( )( )
,298 3 2
,298 3 2
( ) 2 94.0518 2 68.3174 267.6988 kcal/mol
( ) 66.35 kcal/mol = -277.61 kJ/mol
of
of
H CH CH OH l
H CH CH OH l
= +
=
Reassuringly, this agrees with the value in the table in the
back of SVA to within 0.1 kJ/mol. I took the data for the example
from Perry's Handbook. The Temperature Dependence of Heats of
Reaction: So, now we know how to use heats of formation to compute
heats of reaction at a given temperature, but what if we want to
know a heat of reaction at some temperature other than the standard
temperature of 25 C at which standard heats of formation are
generally tabulated? One way that we could compute this would be to
compute the enthalpy change for a hypothetical process in which we
heat or cool the reactants to the standard state temperature, then
carry out the reaction isothermally, then cool or heat the products
back to the temperature of interest. The total enthalpy change for
this 3-step process would be equal to the entropy change for the
1-step process of carrying out the reaction at the initial
temperature. For the first step of heating or cooling the reactants
to 298.15 K, we have
298.15 K
total1 ,reactantso
pT
H C dT = where total,reactantspC is the total heat capacity of
the reactants for the reaction as written. For example, if we had
the generic reaction
aA + bB lL + mM it would be total,reactants ( ) ( )p p pC aC A
bC B= + For the second step, carrying out the reaction at 298.15 K,
we have 2 ,298.15 Ko orxnH H = Finally, for the 3rd step, bringing
the products back to the initial temperature, we have
total3 ,products298.15 K
To
pH C dT = So, for the overall process, we have
298.15 K
total total, ,reactants ,298.15 K ,products
298.15 K
To o ooverall rxn T p rxn p
TH H C dT H C dT = = + +
Remembering that switching the limits of an integral (changing
the direction of integration) just changes its sign, we can combine
the two integrals to get
-
( )total total, ,298.15 K ,products ,reactants298.15 K
To orxn T rxn p pH H C C dT = +
That is, the heat of reaction at some other temperature is equal
to the heat of reaction at the standard temperature of 298.15 K
plus the integral from the standard temperature to the temperature
of interest of the difference in heat capacity between the
reactants and the products. Just like the enthalpy of reaction is
the difference between the enthalpy of the products and reactants,
we can define the change in heat capacity of reaction as the
difference between the total heat capacity of the products and the
reactants. So, for the generic reaction
aA + bB lL + mM For which the heat of reaction is
Hrxn = lHf(L) + mHf(M) - aHf(A) - bHf(B) The difference in heat
capacity on reaction would be
Cp = lCp(L) + mCp(M) - aCp(A) - bCp(B) where all of the heat
capacities can be temperature dependent, so their difference is
also temperature dependent. The heat of reaction at some arbitrary
temperature T is then
, ,298.15 K298.15 K
To orxn T rxn pH H C dT = +
We could further generalize this by writing the general chemical
reaction with an arbitrary number of reactants and products as done
by SVA on p. 137 1 1 2 2 3 3 4 4... ...A A A A + + + + where i is
the stoichiometric coefficient of species Ai. The Ai (A1, A2, etc.)
are just the names of the species participating in the reaction.
The convention we will use here (and that most widely used) is that
the stoichiometric coefficients are positive for products and
negative for reactants. Then, the above equation can be formally
written as: 1 1 2 2 3 3 4 4 ... 0A A A A + + + + = or simply
0i ii
v A = With this notation, the heat of reaction can be written
as
( )o orxn i f ii
H v H A = The change in heat capacity can be written as
,p i p ii
C v C = and so on. This is very handy for doing things in an
automated way on the computer. If the individual heat capacities of
the species are written in the usual polynomial form that we used
last time
, 2 2p i i i i iC
A B T C T D TR
= + + +
Then we can write
-
( )2 2,
2 2
p i p i i i i i ii i
p
C v C R v A B T C T D T
CA BT CT DT
R
= = + + +
= + + +
where we define
, , , i i i i i i i ii i i i
A v A B v B C v C D v D = = = = Using this notation with the
heat capacity integral to find the heat of reaction at some
arbitrary temperature T, we have
( )( ) ( )( ) ( )( )
, ,298.15 K298.15 K
2 2, ,298.15 K
298.15 K
2 32 3, ,298.15 K
1 1298.15 K 298.15 K 298.15 K2 3 298.15 K
To orxn T rxn p
To orxn T rxn
o orxn T rxn
H H C dT
H H R A BT CT DT dT
B CH H R A T T T DT
= +
= + + + +
= + + +
As was the case for the usual heat capacity integral, SVA have
given a defined function and an example computer program for this
in the appendices of the text. Similarly, we might use a
spreadsheet . This example spreadsheet is set up to evaluate the
heat of reaction for methanol synthesis from synthesis gas, as
shown in example 4.6 in SVA. The final section in chapter 4 of SVA
is on 'Heat Effects of Industrial Reactions', which we will study
by working a couple of the examples from that section, which also
illustrate the other things we've been looking at today.
hclTypewritten textUnsolved Problems
Prob No. 4.8 and 4.19
-
The first law of thermodynamics says that energy is conserved,
and does not differentiate between energy added to or removed from
a system as heat and energy added to or removed from a system as
work. In the first law, heat and work are equivalent. However, our
everyday experience, as well as centuries of scientific studies and
engineering experience shows that heat and work are not equivalent
in all ways. Work is spontaneously converted to heat all around us,
and we can find many examples where work is fully converted to
heat. The classical example of this is the historic experiments of
Joule where he stirred fluids and measured the amount of work done
and the temperature rise of the fluids. However, there are no
examples of energy in the form of heat being spontaneously and
completely converted to work, despite centuries of effort directed
at constructing a device that would do so. This suggests that there
is some sense in which work is a higher or more valuable form of
energy than heat, and in fact if that were not the case, there
would not be much need to differentiate between the two. This idea
is formalized and quantified in the second law of
thermodynamics.
There are many ways to state the second law of thermodynamics.
Given one of them, it is possible (sometimes by simple means,
sometimes by more elaborate means) to derive the others. Two
possible statements of it (from SVA, p. 156) are:
No device can operate such that its only effect is to convert
heat absorbed by a system completely into work done by the
system.
Or
No process is possible that consists solely of the transfer of
heat from one temperature level to a higher one. We will provide
more mathematical statements of the second law shortly. This will
all be done quickly, on the assumption that it is review of
material recently learned in EAS 204. Heat Engines, Carnot Engines,
and so forth: A heat engine is a device that converts heat to work.
Much of classical thermodynamics was developed in the context of
trying to understand and improve steam engines, which powered
locomotives, steamboats, etc. in times past. In these and in other
heat engines, heat is absorbed by the system at some high
temperature and partially converted to work and partially
transferred out of the system at some lower temperature. For
example, in a steam engine, water flows into a boiler where it is
vaporized and superheated (absorbing heat at a high temperature),
then the steam expands through a mechanical device like a turbine,
converting part of this heat to work. The steam then condenses
(transferring heat out of the system at a low temperature) and is
returned to the boiler, so that the overall process is cyclic. The
working fluid of the heat engine (i.e. the steam) absorbs some heat
|QH| at a high temperature, does some net amount of work |W| on the
surroundings, then rejects some heat |QC| at low temperature. Since
the overall process is
THE SECOND LAW OF THERMODYNAMICS
-
cyclic, the overall change in the state of the system (and
therefore in the internal energy of the working fluid) is zero. So,
the first law tells us that |W| = |QH| - |QC| That is, the net work
done is equal to the difference between the heat flow in at high
temperature and the heat flow out at low temperature. Note that for
the example of the steam engine, the net work is some relatively
large amount of work done by the expanding steam minus some
relatively small amount of work that must be done to pump the water
up to the high pressure of the boiler. The thermal efficiency of a
heat engine is simply defined as the fraction of the heat input
that is converted to work. That is
net work output 1heat absorbed
H C C
H H H
W Q Q QQ Q Q
= = =
If we could just eliminate the rejection of heat at from the
system (make |QC| zero), we could achieve unity (100%) efficiency.
However, the second law of thermodynamics forbids this. In fact,
the most efficient possible engine is one that operates in an
entirely reversible fashion (with no irreversibilities). Such an
idealized engine is called a Carnot engine, after N.L.S. Carnot,
who first described it. In such a (hypothetical) device, the
following four steps make up the engine cycle: 1. Reversible,
adiabatic compression from temperature TC to temperature TH. 2.
Reversible, isothermal heat transfer during which |QH| is absorbed
from the surroundings. 3. Reversible, adiabatic expansion from
temperature TH to TC. 4. Reversible, isothermal heat transfer
during which |QC| is rejected to the surroundings,
bringing the working fluid back to its original state. Any
engine that operates in a fully reversible way is considered a
Carnot engine. In SVA, they demonstrate that (a) no engine can have
a higher thermal efficiency than a Carnot engine (for the same high
and low temperature levels) and (b) the thermal efficiency of a
Carnot engine depends only on the high and low temperature levels.
Thermodynamic Temperature Scales: The fact that the efficiency of a
Carnot engine depends only on temperatures and not on any
properties of the working fluid allows us to establish a
thermodynamic temperature scale that is independent of any material
properties. Conveniently, this will work out to be the same as the
ideal gas temperature scale (and the Kelvin scale) that we have
already been using. Assuming that this was well covered in EAS204,
we will skip the derivations. In SVA, they demonstrate that if
temperature is defined by the ideal gas temperature scale, then the
thermal efficiency of a Carnot engine is
1 CH H
W TQ T
= =
and the heat absorbed and heat rejected are related by
H HC C
Q TQ T
=
To get 100% efficiency, we would have to either have a cold
reservoir at 0K, or a hot reservoir at infinite temperature.
Obviously, neither of these is achievable.
-
Entropy: From the above discussion of Carnot engines, we get
(among other things) a relationship between the heat absorbed at TH
and the heat rejected at TC in a Carnot cycle:
H HC C
Q TQ T
=
or
H CH C
Q QT T
=
If we take the working fluid as our system and use our
convention that heat flow into the system is positive, then QH is
positive and QC is negative, and we have
CHH C
QQT T
=
or
0CHH C
QQT T
+ =
Thus, for a complete cycle, the two quantities Q/T associated
with heat transfer into and out of the system sum to zero. A key
characteristic of a thermodynamic property (like temperature,
pressure, internal energy, etc.) is that for any cyclic process,
its final value is the same as its initial value. This suggests
that there may be another thermodynamic property of the system
whose changes are given by Q/T. In SVA they approach this by
pointing out that an arbitrary reversible, cyclic process can be
constructed from many Carnot cycles in which an infinitesimal
amount of heat dQH is absorbed at temperature TH and another
infinitesimal (but different) amount of heat dQC is rejected at
temperature TC. From such an analysis, one can conclude that
0revdQT
=v That is, the integral over any cyclic, reversible process of
dQ/T is zero. This is then used to define a new property entropy,
denoted by the symbol St, as
t revdQdST
=
or trevdQ TdS= The superscript t is used here to indicate that
this is the total entropy of the system (an extensive quantity) not
the molar or specific entropy (an intensive quantity). We can show
that entropy is a state function, since we could evaluate the
change in entropy from state A to state B by doing the integral
1A to B via path 1
t revdQST
= along one reversible path, and then do the integral from state
B back to A along another reversible path
2B to A via path 2 A to B via path 2
t rev revdQ dQST T
= = Since the total integral over the cyclic path from A to B
and back to A must be zero, we have
-
1 2 0t tS S + = Since we are free to pick an arbritrary
reversible path for either part 1 or part 2, the entropy change
must be independent of the path selected, and
1 2A to B by any reversible path
t t t trevB A
dQS S S ST
= = = If the system goes from state A to state B by an
irreversible process, then the entropy change will still be the
same, but it will not be given by the integral, of dQ/T along the
irreversible path. So, even for irreversible processes, we can
always compute the entropy change along some hypothetical
reversible path that connects the same initial and final states as
the actual irreversible path. If a process is both reversible and
adiabatic, then dQrev will be zero throughout the process, and dSt
will be zero throughout, and finally, St will be zero. So, a
reversible, adiabatic process is also called an isentropic process.
The discussion of entropy so far can be summarized as: There exists
a property that we will call entropy and denote by S. It is related
to observable properties of the system. For a reversible process,
changes in this property are given by
t revdQdST
=
The change in entropy for any process that takes a system from
state A to B (whether the actual process is reversible or not) is
given by
A to B by any reversible path
t revdQST
= Mechanically reversible processes are processes where all
driving forces for change in the state of the system are
infinitesimal except for temperature differences for heat transfer
between the system and the surroundings. For mechanically
reversible processes, the entropy change can still
be calculated as t dQST
= where T is the temperature of the system (which must be
uniform). In this case, the irreversibilities occur outside the
system, so from the point of view of the system, it is immaterial
what temperature the heat finally flows to. The process is
internally reversible - there are no irreversibilities (no finite
driving forces) within the system. Entropy Changes for an Ideal Gas
To get started with computing entropy changes, we will consider
entropy changes for processes involving ideal gasses. For one mole
of fluid undergoing a mechanically reversible process in a closed
system, the first law tells us that: dU = dQrev - PdV Since (by
differentiation of H = U + PV) enthalpy is related to internal
energy by dH = dU + PdV + VdP
-
we see that for a mechanically reversible process dH = dQrev -
PdV + PdV + VdP = dQrev + VdP or dQrev = dH - VdP The above is true
in general (not just for an ideal gas), but if we consider the
particular case of one mole of an ideal gas, we have dH = CpdT and
V = RT/P, so dQrev = CpdT - RT/PdP so
rev pdQ dT dPC R
T T P=
and therefore
pdT dPdS C RT P
=
Integrating this from some initial state T0, P0 to some final
state T and P gives
0
0ln
T
p
T
dT PS C RT P
=
Although we used a mechanically reversible process to derive
this, the final two equations relate only properties of the system
(state functions) and are therefore independent of path and apply
to all processes, reversible or irreversible. For an ideal gas with
constant heat capacity, the above equation is even simpler
0 0
ln lnpT PS C RT P
=
For a gas in which the heat capacity is defined by the sort of
polynomial that we have been using:
2 2pC
A BT CT DTR
= + + +
then
( ) ( ) ( )0
30
2 2 2 20 0 0
0 0
ln
ln ln2 2
T
T
S A D PB CT dTR T PT
S T C D PA B T T T T T TR T P
= + + +
= + +
( ) ( ) ( )0
2 2 2 20 0 0
0ln
2 2
Tp
T
C T C DICPS dT A B T T T T T TRT T
= + +
As they did for enthalpy and internal energy changes, SVA also
express this in terms of a mean heat capacity and provide a defined
function and example computer program for it in the appendices. We
can also do this calculation in a spreadsheet that evaluates the
first part of this:
-
Stotal 0 for all processes That is, the total entropy of the
universe (the system plus the surroundings) cannot decrease. For
reversible processes, it does not change. For irreversible
processes, it increases. This is, in essence, a statement that
things always happen in a certain direction. Heat always flows from
hot to cold because that increases the total entropy. Or, from a
different perspective, we define entropy such that its total value
increases when heat flows from hot to cold. The fact that
spontaneous heat transfer always occurs from hot to cold is then
consistent with the version of the second law as written above. If
heat leaves a hot object (at TH) and flows into a cooler object (at
TC) then the entropy of the hot object decreases by Q/TH while that
of the cooler object increases by Q/TC. So, the total change of
entropy of the universe for this process is Q(1/TC 1/TH), which is
positive. If the temperatures were only infinitesimally different,
the process would be reversible, and this difference would approach
zero. Notice that while the first law of thermodynamics is an
equality stating that the total amount of energy in the system plus
the surroundings remains constant the second law of thermodynamics
is an inequality stating that the total amount of entropy in the
system plus the surroundings can only remain constant or increase.
Entropy balances for open systems: We can write an entropy balance
for an open system in the same way that we wrote an energy balance
and a mass balance except that entropy is not conserved so we must
include generation of entropy as well as its flow in and out of the
system. Mathematically, we can write the entropy balance as:
( ) ( ) 0tsurrcv
gfs
d mS dSSm Sdt dt
+ + = The first term is the net rate of entropy transport out of
the system via flow. It is the difference between the entropy
carried out by streams flowing out of the control volume and that
carried in by streams flowing into the control volume. The second
term is the rate of accumulation of entropy within the control
volume. The third term is the rate of change of the entropy of the
surroundings due to the process of interest. gS is the total rate
of entropy generation. It is zero for reversible processes and
positive for irreversible processes. If the surface of the control
volume is made up of a number of areas through each of which an
amount of heat Qj flows from the surroundings at temperature Tj,
then the rate of entropy change of the surroundings due to this
heat flow will be:
The Second Law of Thermodynamics A mathematical statement of the
2
nd
law of thermodynamics: Given the definition of entropy that we
developed/adopted last time, either of the two statements of the
second law of thermodynamics given there can be shown to be
equivalent to:
-
t
jsurr
j j
QdSdt T
= Substituting this, the entropy balance becomes
( ) ( ) 0jcv gfsj j
d mS QSm S
dt T + =
If the system is at steady state, this simplifies to
( ) 0j gfsj j
QSm S
T =
If there is a single inlet and a single outlet from the system,
with equal mass flowrates, we can write
0jout in gj j
QS S S
T =
In this final equation, the heat flow and entropy generation
terms are per unit mass of material flowing through the system,
rather than the total heat flows and total entropy generation used
in the previous equations. Examples 5.5 and 5.6 from SVA
hclTypewritten textUnsolved Problems
Prob 5.8, 5.10, 5.12, 5.19, 5.29, 5.35 and others, u can solve
for practice
-
The reversible work done on the closed system is dWrev = -Pd(nV)
which is our usual expression developed in the first chapter of
SVA. The reversible heat flow is (as developed in the previous
lecture and chapter 5 of SVA) dQrev = Td(nS) Combining these last 3
equations gives d(nU) = Td(nS) - Pd(nV) or dUt = TdSt - PdVt These
last two equations give a general relationship between the primary
thermodynamic variables for the fluid - U, S, T, P and V. Although
we derived it for a reversible process, the final expressions
relate only state variables that are path-independent. Therefore,
this relationship holds for any process occurring in a closed
system between equilibrium states. The system can be made up of
multiple phases, and it can undergo reactions, as long as no
material enters or leaves it and the initial and final states that
we wish to relate are equilibrium states. For convenience, we
define three more thermodynamic properties related to the internal
energy. These are the enthalpy (which we have been using) H U PV +
the Helmholtz energy (or Helmholtz free energy) A U TS and the
Gibbs energy (or Gibbs free energy) G H TS
In lectures 3 through 6 of this course, we looked fairly
intensively at how to compute the volumetric properties (i.e.
volume given T and P) for fluids from equations of state and from
correlations based on the theorem of corresponding states. After
that, we looked at heat effects, and computed enthalpy and internal
energy changes when the temperature of a substance is changed using
the heat capacity. Then we defined entropy, and computed entropy
changes when the temperature of a substance was changed, again
using heat capacity data. Now, we want to bring this all together
to see how, in general, not just for ideal gases, we can compute
changes in the thermodynamic properties of a fluid (internal
energy, enthalpy, and entropy) for a change in state of the fluid,
using PVT and heat capacity data. We will also define and use two
new thermodynamic quantities, the Helmholtz free energy and the
Gibbs free energy. Property Relations for Homogeneous Phases First,
we will combine the first and second laws to derive a fundamental
property relationship between U, S, T, P and V - all of the primary
thermodynamic variables. We can write the first law for a closed
system containing n moles of fluid as: d(nU) = dQ + dW For the
special case of a reversible process, this is d(nU) = dQ
rev + dW
rev
hclTypewritten textTHERMODYNAMIC PROPERTIES OF FLUIDS,
-
For each of these, we can transform the thermodynamic property
relation for internal energy using the definitions above to get a
new thermodynamic property relation where internal energy is
replaced by enthalpy, Helmholtz energy, or Gibbs energy. For
enthalpy, we have nH = nU + nPV so d(nH) = d(nU) + Pd(nV) + nVdP or
d(nU) = d(nH) - Pd(nV) - nVdP equating this to the fundamental
thermodynamic property relationship for d(nU) d(nU) = d(nH) -
Pd(nV) - nVdP = Td(nS) - Pd(nV) so d(nH) = Td(nS) + nVdP or dHt =
TdSt + VtdP Likewise, differentiating the definition of A gives
d(nA) = d(nU) - Td(nS) - nSdT so d(nU) = d(nA) + Td(nS) + nSdT =
Td(nS) - Pd(nV) so d(nA) = -nSdT - Pd(nV) or dAt = -StdT - PdVt
Finally, for the Gibbs energy, d(nG) = d(nH) - Td(nS) - nSdT so
d(nH) = d(nG) +Td(nS) + nSdT = Td(nS) +nVdP so d(nG) = -nSdT + nVdP
or dGt = -StdT + VtdP The above are all written for the total mass
of a closed system. What we are often interested in is the
corresponding relationsh
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