Download The Second Law: Definition of Entropy

9/23/2008
Spontaneity of Chemical Reactions
Chemistry 433
One might be tempted based on the results of thermochemistry to predict that all exothermic reactions would
be spontaneous. The corollary this would be the statement
that no endothermic reactions are spontaneous. However,
this is not the case. There are numerous examples of
p
Of course,,
endothermic reactions that are spontaneous.
heat must be taken up from the surroundings in order for
such processes to occur. Nonetheless, the enthalpy of the
reaction does not determine whether or not the reaction
will occur, only how much heat will be required or
generated by the reaction. The observation that gases
expand to fill a vacuum and that different substances
spontaneously mix when introduced into the same vessel
are further examples that require quantitative explanation.
Lecture 9
Entropy and the Second Law
NC State University
Spontaneity of Chemical Reactions
As you might guess by now, we are going to define a
new state function that will explain all of these observations
and define the direction of spontaneous processes.
This state function is the entropy.
py is related to heat and heat flow and yyet heat is not
Entropy
a state function. Recall that q is a path function. It turns
out that the state function needed to describe spontaneous
change is the heat divided by the temperature.
Here we simply state this result.
qrev
T
ΔS =
We will prove that entropy is a state function in this lecture.
A cyclic heat engine
2
1
I
IV
Phase
I.
II.
III.
IV.
II
3
4
III
Transition
1→2
2→3
3→4
4→1
Path
Isothermal
Adiabatic
Isothermal
Adiabatic
Engines
Historically, people were interested in understanding the
efficiency with which heat is converted into work. This was
a very important question at the dawn of the industrial
revolution since it was easy to conceive of an engine
powered by steam, but it turned out to be quite difficult to
build one that was efficient enough to get anything done!
In an engine
engine, there is a cycle in which fuel is burned to heat
gas inside the piston. The expansion of the piston leads to
cooling and work. Compression readies the piston for the
next cycle. A state function should have zero net change
for the cycle. It is only the state that matters to such a
function, not the path required to get there. Heat is a path
function. As we all know in an internal combustion engine
(or a steam engine), there is a net release of heat.
Therefore, we all understand that δq ≠ 0 for the cycle.
Work and Heat for the Cycle
Condition
w = -q
ΔU = w
w = -q
ΔU = w
The work is
w = wI + wII + wIII + wIV
q = qI + qIII
= - wI - wIII
For the adiabatic steps
qII = qIV = 0
For the isothermal steps
ΔU = 0
Neither the work nor the heat is a state function. Neither one
is zero for the cycle as should be the case for a state function.
The work is:
w = wI + wII + wIII + wIV
=-nRThotln(V2/V1)–Cv(Tcold–Thot)–nRTcoldln(V4/V3)–Cv(Thot – Tcold)
w = -nRThotln(V2/V1) – nRTcoldln(V4/V3) [since wII = - wIV]
w = -nRT
nRThotln(V2/V1) – nRTcoldln(V1/V2) [since V4/V3 = V1/V2]
w = -nRThotln(V2/V1) + nRTcoldln(V2/V1) [property of logarithms]
The heat is:
q = qI + qIII [since qII = qIV = 0 for adiabatic processes]
= - wI - wIII [since dU = 0 for isothermal steps]
q = nRThotln(V2/V1) + nRTcoldln(V4/V3)
q = nRThotln(V2/V1) + nRTcoldln(V1/V2) [since V4/V3 = V1/V2]
q = nRThotln(V2/V1) - nRTcoldln(V2/V1) [property of logarithms]
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A new state function: Entropy
Thermodynamics of an Engine
The heat is not a state function. The sum qI + qIII is not zero.
From this point on we will make the following definitions:
qI = qhot
qIII = qcold
The cycle just described could be the cycle for a piston
in a steam engine or in an internal combustion engine.
The hot gas that expands following combustion of a small
quantity of fossil fuel drives the cycle. If you think about
the fact that the piston is connected to the crankshaft you
pressure on the p
piston is
will realize that the external p
changing as a function of time and is helping to realize
an expansion that as close to an ideal reversible expansion
as the designers can get. If we ignore friction and assume
that the expansion is perfectly reversible we can apply the
above reasoning to your car. The formalism above for the
entropy can be used to tell us the thermodynamic efficiency
of the engine.
V2
V
– nRTcold ln 2 ≠ 0
V1
V1
V2
V2
= nRln
– nRln
=0
V1
V1
q = qhot + qcold = nRThotln
qrev qhot qcold
=
+
T
Thot Tcold
However, the heat divided by temperature is a state function.
This reasoning leads to the idea of a state function called the
entropy. We can write:
q
ΔS = rev
T
Question
Thermodynamic Efficiency
We define the efficiency as the work extracted divided by
the total heat input.
efficiency = work done
heat used
|w | |nR(Tcold – Thot)ln (V2 / V1)| Thot – Tcold
=
η = qtotal =
nRThotln (V2 / V1)
Thot
hot
Your car has an operating temperature of 400 K. If the
ambient temperature is 300 K, what is the thermodynamic
efficiency of the the engine?
A.
B.
C.
D.
75%
50%
25%
5%
The efficiency defined here is the ideal best case. It assumes
a reversible process with no losses due to friction. The
temperature Thot is the temperature of the expansion in the
engine. The temperature Tcold is the temperature of the
exhaust. Tcold cannot be less than the temperature of the
surroundings.
Question
Question
Your car has an operating temperature of 400 K. If the
ambient temperature is 300 K, what is the thermodynamic
efficiency of the the engine?
The thermodynamic cycle was derived for reversible
expansions. What are the consequences if the cycle is
not perfectly reversible?
A.
B.
C.
D.
A. The work of expansion will decrease
B. The work of compression will decrease
C. There will be no adiabatic expansion
D. There can be no cycle
75%
50%
25%
5%
|w | T – Tcold
T
η = qtotal = hot
= 1 – cold = 1 – 300K = 0.25
400K
Thot
Thot
hot
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The Thermodynamic
Temperature Scale
Question
The thermodynamic cycle was derived for reversible
expansions. What are the consequences if the cycle is
not perfectly reversible?
A. The work of expansion will decrease
B. The work of compression will decrease
C. There will be no adiabatic expansion
D. There can be no cycle
The definition of entropy is qhot/Thot + qcold/Tcold = 0.
We can write this as
qhot/Thot = - qcold/Tcold
Since qcold is negative we can combine the minus sign
with qcold and write the expression as
| hot|/Thot = |q
|q
| cold|/Tcold
and finally
|qhot|/|qcold| = Thot/Tcold
The ratio of the heats is equal to the ratio of temperatures
for two steps in a thermodynamic cycle. This defines a
temperature scale and allows one to measure temperature
as well (i.e. this scheme represents a thermometer). Both
this expression and the thermodynamic efficiency further
imply that there is an absolute zero of temperature.
Heat Transfer
To examine the function that we have just defined, let us
imagine that we place to identical metal bricks in contact
with one another. If one of the bricks is at equilibrium at
300 K and the other at 500 K, what will the new equilibrium
temperature be? Intuitively, you would say 400 K and you
would imagine
g
that heat flows spontaneously
p
y from the
warmer brick to the colder brick. The entropy function
makes these ideas quantitative.
q
q
q1 q2
ΔS = 2 + 1
T2 T1
T1= 500 K
T2= 300 K
Using this definition
of entropy change as
the heat flow divided
by the temperature.
System and surroundings
Up to this point we have considered the system, but we have
not concerned ourselves with the relationship between the
system and the surroundings. When we consider heat flow
In the state function, entropy, we need to carefully account
for how the heat flow takes place in order to determine the
effect We will show that for reversible processes the
effect.
entropy change is zero. However, for irreversible processes
the entropy change can be positive (spontaneous) or negative
(reverse process is spontaneous). For isolated systems the
entropy change is also zero. However, when there is heat
flow between the system and surroundings the entropy can
be non-zero. In the case that the heat flow is irreversible the
entropy change will be non-zero.
Heat Transfer
Let’s assume that heat flows from the hot body to the cold
body. Then q1 is negative (the flow from the hot body) and
q2 is positive (the flow into the cold body). Moreover,
q2 = -q1 = q
This means that we can substitute in q to obtain:
ΔS =
q
q
–
= q 1 – 1 = q 1 – 1 = 0.0013 q
T2 T1
T2 T1
300 500
For this calculation it does not matter how big q is, but only
that it is a postive number so that ΔS is positive.
T2= 300 K
T1= 500 K
q
Using this definition
ΔS > 0, which says
that the process is
spontaneous.
Calculating reversible and
irreversible paths
It is important to reiterate that the calculation of the entropy
of the system always follows a reversible path.
You might ask, well what happens if the process is not
reversible?
To consider this let us the example of expansion of gas
in a cylinder. The process can occur along different paths.
a. constant pressure expansion, w = -PextΔV = - Pext(Vf – Vi)
b. reversible isothermal expansion the work w = -nRTln(Vf/Vi).
For both a. (irreversible) and b. (reversible) we will calculate
the entropy of the system along a reversible path.
ΔSsys = q/T = -w/T = nRln(Vf/Vi)
For the reversible path, we can use the fact that ΔSsurr = - ΔSsys
to obtain ΔSsurr = - nRln(Vf/Vi).
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System and surroundings
The heat transfer example shows us that we always must
consider the surroundings. Any time the system releases
heat it goes into the surroundings and this contributes to the
overall entropy change. Thus, the total entropy is:
ΔStotal = ΔSsurr + ΔSsys
For a reversible process the total entropy is zero.
zero If a
process is irreversible and spontaneous the entropy change
is positive. This implies that we must treat the system and
the surroundings differently when we calculate entropy.
The rule is: always calculate the entropy of the system along
a reversible path.
If the process is truly reversible then:
ΔSsurr = - ΔSsys
ΔStotal = 0 and
Understanding the irreversible path
For the irreversible path we use the actual work of the constant
pressure expansion, w = -PextΔV = - Pext(Vf – Vi) to calculate
ΔSsurr = -q/T = w/T = -Pext(Vf – Vi)/T where T is the same
temperature we used for the isothermal expansion. Note that
the sign is opposite since the heat is flowing into the surroundings (and out of the system).
We see that in this case the entropy change ΔSsurr is smaller in
magnitude than ΔSsys. We know this from the first law where
we saw that the irreversible work of expansion is always less
than the reversible work of expansion. Thus,
ΔStotal > 0
for the irreversible process.
The dependence of the
entropy on volume
Statements of the second law
ΔSrev = qrev/T = 0
for ΔSirr = qirr/T > 0
The entropy of an irreversible process is greater than zero
If the process is spontaneous.
Clausius: the entropy of the universe tends towards a maximum.
For a constant temperature (isothermal)expansion we have:
dS = δqrev/T = - δwrev/T.
The logic behind this statement is that the internal energy
change is zero for a constant temperature process and
so δqrev = - δwrev. To calculate the reversible work we
simply
i l plug
l iin δwrev = -PdV.
PdV A
According
di to the
h id
ideall gas llaw
P = nRT/V so dS = nRdV/V.
S2
Famous words:
“Die Entropie des Universums strebt ein Maximum zu.”
The dependence of the
entropy on temperature
The entropy change as a function of the temperature is
derived at constant volume using the fact that
dU = δqV = nCVdT.
The reversible heat in this case, qrev, is a constant volume
heat and so it can be replaced by
dS = δqrev/T = nC
CvdT/T at constant volume
l
To obtain ΔS we need to integrate both sides
S2
S1
dS = nCv
T2
T1
dT
T
We obtain:
ΔS = S2 – S1 = nCvln(T2/T1).
Exactly the same reasoning applies at constant pressure,
so that ΔS = nCpln(T2/T1).
S1
dS = nR
V2
V1
dV
V
The result of this equation is that ΔS = nRln(V2/V1)
at constant temperature.
History of Refrigeration
In prehistoric times game was stored in a cave or packed
in snow to preserve the game for times when food was
not available. Later, ice was harvested in the winter to be
used in the summer. Ice is still used for cooling and
storing food. The intermediate stage in the history of
cooling
li foods
f d was to
t add
dd chemicals
h i l like
lik sodium
di
nitrate
it t or
potassium nitrate to water causing the temperature to fall.
Cooling wine via this method was recorded in 1550, as
were the words "to refrigerate." The evolution to
mechanical refrigeration, a compressor with refrigerant,
was a long, slow process and was introduced in the last
quarter of the 19th century.
Source : U.S. Department of Agriculture
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Importance of Refrigeration
Refrigeration slows bacterial growth. Bacteria exist
everywhere in nature. They are in the soil, air, water, and
the foods we eat. When they have nutrients (food),
moisture, and favorable temperatures, they grow rapidly,
increasing in numbers to the point where some types of
bacteria can cause illness. Bacteria grow most rapidly in
the range of temperatures between 40 and 140 °F, the
"Danger Zone," some doubling in number in as little as 20
minutes. A refrigerator set at 40 °F or below will protect
most foods.
Coefficient of Performance
The coefficient of performance, or COP (sometimes CP),
of a heat pump is the ratio of the output heat to the supplied
work or
q
COP = hot
|wtotal|
pp
by
y the condenser and w
where q is the useful heat supplied
is the work consumed by the compressor. (Note: COP has
no units, therefore heat and work must be expressed in the
same units.)
According to the first law of thermodynamics, in a reversible
system we can show that qhot = qcold + w and w = qhot − qcold,
where qhot is the heat taken in by the cold heat reservoir
and qcold is the heat given off by the hot heat reservoir.
Cyclic Refrigeration
The refrigeration cycle uses a fluid, called a refrigerant, to
move heat from one place to another. The key to
understanding how it works is recognizing that at the same
pressure, the refrigerant boils at a much lower temperature
p , the refrigerant
g
commonlyy used in
than water. For example,
home refrigerators boils between 40 and 50°F as compared to
water's boiling point of 212°F. Let's look at the process to see
how boiling and condensing a refrigerant can move heat. The
process is the same whether it is operating a refrigerator, an
air conditioner or a heat pump. This example illustrates a
closed-loop system.
Cyclic Refrigeration
This consists of a refrigeration cycle, where heat is
removed from a low-temperature space or source and
rejected to a high-temperature sink with the help of
external work, and its inverse, the thermodynamic power
cycle.
l In
I the
th power cycle,
l heat
h t is
i supplied
li d from
f
a highhi h
temperature source to the engine, part of the heat being
used to produce work and the rest being rejected to a
low-temperature sink. This satisfies the second law of
thermodynamics.
Coefficient of Performance
Therefore, by substituting for w,
q
COPheating = q –hotq
hot
cold
Therefore,
COPheating =
Thot
Thot – Tcold
temperature Thot is the temperature of the expansion in the
engine. The temperature Tcold is the temperature of the
exhaust. Tcold cannot be less than the temperature of the
surroundings. For refrigeration, the COP is:
q
Tcold
COPcooting = q –coldq =
Thot – Tcold
hot
cold
Cyclic Refrigeration
We'll begin with the cool, liquid refrigerant entering the indoor
coil, operating as the evaporator during cooling. As its name
implies, refrigerant in the evaporator "evaporates." Upon
entering the evaporator, the liquid refrigerant's temperature is
between 40 and 50°F and without changing its temperature, it
absorbs heat as it changes state from a liquid to a vapor. The
heat comes from the warm, moist room air blown across the
evaporator coil. As it passes over the cool coil, it gives up some
of its heat and moisture may condense from it. The cooler,
drier room air is re-circulated by a blower into the space to be
cooled.
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Cyclic Refrigeration
The vapor refrigerant now moves into the compressor, which is
basically a pump that raises the pressure so it will move
through the system. Once it passes through the compressor,
the refrigerant is said to be on the "high" side of the system.
Like anything that is put under pressure, the increased
pressure from the compressor causes the temperature of the
refrigerant to rise. As it leaves the compressor, the refrigerant
is a hot vapor, roughly 120 to 140°F.
It now flows into the refrigerant-to-water heat exchanger,
operating as the condenser during cooling. Again, as the name
suggests, the refrigerant condenses here. As it condenses, it
gives up heat to the loop, which is circulated by a pump.
Cyclic Refrigeration
In summary, the indoor coil and refrigerant-to-water heat
exchanger is where the refrigerant changes phase, absorbing
or releasing heat through boiling and condensing. The
compressor and expansion valve facilitate the pressure
changes, increased by the compressor and reduced by the
expansion valve.
Cyclic Refrigeration
The loop water is able to pick up heat from the coils because it
is still cooler than the 120 degree coils.
As the refrigerant leaves the condenser, it is cooler, but still
under pressure provided by the compressor. It then reaches
the expansion valve. The expansion valve allows the high
pressure refrigerant to "flash" through becoming a lower
pressure, cooled liquid. When pressure is reduced, as with
spraying an aerosol can or a fire extinguisher, it cools. The
cycle is complete as the cool, liquid refrigerant re-enters the
evaporator to pick up room heat. In winter, the reversing valve
switches the indoor coil to operate as the condenser and the
heat exchanger as the evaporator.
Vapor-compression cycle
A simple stylized diagram of the refrigeration cycle:
1) condensing coil
2) expansion valve
3) evaporator coil
4) compressor.
Cyclic Refrigeration
A refrigeration cycle describes
the changes that take place in
the refrigerant as it alternately
absorbs and rejects heat
as it circulates through a
refrigerator. Heat naturally
flows from hot to cold. Work
is applied to cool a space by
pumping heat from a lower temperature heat source into
a higher temperature heat sink. Insulation is used to reduce
reduce the work and energy required to achieve and
maintain a lower temperature in the cooled space.
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