concepts in thermodynamics many of the concepts and equations in chapter 18 are simply given to you and you are asked to accept them on fai

Concepts in Thermodynamics
Many of the concepts and equations in chapter 18 are simply given to
you and you are asked to accept them on faith. Here I try to fill in
many of the blanks so you can acquire a deeper understanding of
thermodynamics.
The second law of thermodynamics states that a process will occur if
it results in an increase in the entropy of the universe. Your
textbook defines entropy as a measure of energy dispersal and they
point out that dispersed energy represents a more probable state than
concentrated energy. (That’s why high energy molecules are very
valuable: they are rare!) Thus, the second law simply says that
processes occur spontaneously in the direction that corresponds to
going from a low probability state to a high probability state. Most
textbooks also equate entropy with “disorder” since a highly ordered
system is also highly improbable. Your text does not give the general
equation for S because it involves a difficult technical concept:
“reversible process”. The equation is:
where qrev represents the heat transfer due to a “reversible”
process
The origin of this equation is explained in sections I and II below.
In this context, “reversible” means a process that is carried out
infinitely slowly. Your textbook gives one special case in which this
equation can be used to calculate Ssurr:

A chemical reaction can transfer only a tiny amount of heat to or from
the surroundings (compared to the total heat content of the
surroundings). Thus, as far as the surroundings are concerned, a
reaction transfers a negligible amount of heat. This is equivalent to
transferring the heat reversibly. Thus the general equation can be
used to calculate Ssurr. However, Srxn can NOT be calculated using
this equation because Hrxn is NOT negligible with respect to the
system.
Section I tries to clarify the meaning of a reversible process and
demonstrates that doing a process “reversibly” (very slowly) results
in the maximum possible work. This idea is used in section II to show
how the equation for entropy comes from the analysis of the efficiency
of a steam engine that is run reversibly.
In section III, I expand on how one uses the third law of
thermodynamics to measure the absolute entropy of a pure substance in
a given state. Recall that the third law states that a perfect crystal
at absolute zero has zero entropy (perfect order).
In section IV, I show how the analysis of the efficiency of a steam
engine led to the second law of thermodynamics.
In section V, I use the expansion of an ideal gas to illustrate the
connection between entropy, energy dispersal, disorder, and
probability. It is not at all obvious how the equation for entropy is
related to the probability of a system being in a particular state.
This section tries to help you see that connection.
Without proof, your textbook makes the following statement: “G
represents the maximum useful work that can be done by a
product-favored system on its surroundings under conditions of
constant temperature and pressure.” In section VI, I show why this is
true. (As you will see, the statement in your textbook is not exactly
correct. G represents the maximum amount of “non-PV work” that a
reaction can do.)
Some of the explanations below require the use of simple calculus. I
don’t think this will cause you any problems.
I. Reversible processes and reversible chemical reactions
In thermodynamics, “reversible” has a technical meaning. A reversible
process is one that is infinitely slow! Obviously, no real process is
reversible but you can approximate a reversible process by doing it as
slowly as possible. For example, you can think of a reaction that is
at equilibrium as occurring infinitely slowly (nothing is happening!).
Imagine that the forward reaction occurs just barely faster than the
reverse reaction so, for all practical purposes, the reaction is at
equilibrium. You can make the reaction reverse by adding a tiny amount
of product or by removing a tiny amount of reactant. Thus, a reaction
at equilibrium is reversible! Conversely, a reaction that is not at
equilibrium is irreversible. All real processes are irreversible.
Reversible processes produce the maximum amount of work
If a process does work on the surroundings, you get more work out of
the process if it is done slowly. This is because less heat is lost to
the surroundings. So a reversible process (infinitely slow) does the
maximum work. There is one example of this that is easy to understand:
the work done by a gas expanding inside a piston. We will see that the
maximum amount of work is obtained when the gas is allowed to expand
very slowly (reversibly).

Let’s compare a reversible vs. an irreversible expansion of the gas.
Initially, let’s assume that the internal and external pressures are
the same (Pint = Pext). To expand the gas reversibly, we reduce the
external pressure very slowly (technically, infinitely slowly) so the
system never strays significantly from equilibrium. This is a
reversible process because a very small increase in the external
pressure would cause the gas to be compressed. To expand the gas
irreversibly, we reduce the pressure very quickly. As the gas expands
it does work on the surroundings (by moving the piston against the
external pressure). The plots of pressure vs. volume for these two
processes are shown below. Figure (a) represents the reversible
expansion. Figure (b) is the irreversible expansion.

The work done in these two processes is the area under the curves.
(Remember that PV has units of energy or work). Clearly, more work is
done by the reversible expansion. The reversible expansion does the
maximum amount of work because the gas is pushing against the maximum
possible external pressure. (Since the process remains at equilibrium
throughout the reversible expansion, Pext = Pint.) If the external
pressure was any higher, the process would reverse and the gas would
be compressed.
In general, energy released by a reversible process can do the maximum
amount of work because less of the energy is lost as heat. A process
that is done quickly (irreversibly) tends to generate turbulence and
friction resulting in heat loss to the surroundings.
II. Entropy and the efficiency of a steam engine: The origin of the
equation S = qrev/T
Thermodynamics was developed by people who wanted to maximize the
efficiency of a steam engine. A steam engine operates much like an
internal combustion engine of a car except high pressure steam is used
to move a piston rather than the combustion of gasoline. The movement
of the piston is then used to do useful work. After the piston has
moved, the remaining steam is ejected from the engine and the piston
returns to its original position and is ready for another cycle. This
cycle is represented schematically below.

In essence, a steam engine is a machine for converting heat into work.
Heat is extracted from a high temperature reservoir (hot steam at Th).
Some of the heat is converted to work (w) by allowing the steam to
expand. After expansion, the remaining heat (cooler steam at Tl) is
ejected into a low temperature reservoir. The goal is to convert as
much of the extracted heat into useful work as is possible. The
efficiency of a steam engine is:

Work (w) is a negative number because work is done by the system
(steam) on the surroundings. qh is positive because it is heat added
to the system from the surroundings. Thus, the negative sign makes the
efficiency a positive number.
Since the engine operates in a cycle (returns to its initial state): Esys
= 0 (E is a state function.)
From the first law of thermodynamics: Esys = qh + ql + w = 0 (note
that ql is negative with respect to the system)
Thus: qh + ql = -w

[Notice that efficiency = 1 (100%) only if ql = 0 which means that all
of the heat has been converted to work!]
The maximum efficiency results if the engine is operated reversibly
(see section I).
If the steam behaves as an ideal gas and the engine is operated
reversibly, then it can be shown (I won’t derive it here) that:
(If the engine is run irreversibly, then )
(Notice that efficiency = 1 only if Tl = 0 which means that all of the
heat has been converted to work so the “steam” is now at absolute
zero! To maximize the efficiency of an internal combustion engine, the
engine is designed to operate at very high temperature and to
discharge the products at the lowest possible temperature!)
Thus:

Rearranging gives:

This result shows that the function q/T is a state function (that we
call entropy!) since it sums to zero for a cyclic process. (The system
returns to its original state so all state functions describing the
system must not change). But understand that q/T is a state function
only if the engine is run reversibly so that q = qrev. If the engine
is run irreversibly, then the sum above would be less than zero!
Recall that q is not a state function so for q/T to be a state
function we must choose a particular path: a reversible path!
At constant temperature:
If T is allowed to vary, S = integrated between the initial
and final temperatures.
III. Measuring absolute entropy (where did the numbers in appendix 3
come from?)
From the third law of thermodynamics, we know that the entropy of a
perfect crystalline substance at
0 K is zero. The existence of a zero point on the entropy scale allows
us to assign an absolute entropy value (as opposed to a relative
entropy value) to a pure substance (see appendix 3). Notice that the
unit for entropy (J/K) is the same as that for heat capacity. Absolute
entropy is measured by determining how the heat capacity of a
substance changes with temperature. (The temperature is changed VERY
slowly to approximate a reversible process.) A substance with a higher
heat capacity also has more entropy (see section IV)!
Remember that heat capacity (C) is the amount of heat required to
raise the temperature of a substance 1 K (or 1 oC).
Thus the heat absorbed during a temperature change is:
q = CT
We have to use a little calculus here:
dq = CdT
The entropy change caused by a change in temperature is:
S = (note: C depends on temperature)
To determine the absolute entropy at temperature T, the integral is
evaluated from 0 K to T. This is equivalent to determining the area
under the curve produced by plotting C/T vs. T (see graph below). From
the third law, we know that the entropy at 0 K is zero so the area
under the curve is the absolute entropy at temperature T!! If phase
changes occur, then Hfus/T and Hvap/T must be added to obtain the
total entropy.

Assuming no phase changes, the area under the curve is the absolute
entropy of this substance at 300 K. The unit for heat capacity is J/K
so the area would also have this unit! If phase changes occur, then Hfus/T
and Hvap/T must be added to obtain the total entropy. The numbers in
appendix 3 were measured using one mole of each substance at a
pressure of one bar so they are standard molar entropies (So) whose
unit is J/mol-K.
IV. What does any of this have to do with the second law?
Based on the previous argument about the efficiency of a steam engine,
we will show that the entropy of the universe increases whenever an
irreversible process occurs (the second law!). It is easy to show that
no steam engine can have an efficiency greater than that calculated
above for a reversible engine. If there was such an engine, one could
transfer heat from a colder object to a hotter object without doing
any work. From our common every day experience, we know that this
never happens (heat only flows spontaneously from hot to cold)!
Pretend that you had two reversible steam engines and that engine 1 is
more efficient than engine 2. Hook them up as shown below. Engine 2
uses the work produced by engine 1 to pump heat from the low
temperature reservoir to the high temperature reservoir. For the
overall process w = 0. However, since engine 1 is more efficient than
engine 2, q1 the high temperature reservoir without any work being done!
IMPOSSIBLE!! Such an engine is called a “perpetual motion machine of
the second kind!” Think what you could do with it!! (A perpetual
motion machine of the first kind is one that violates the first law of
thermodynamics and produces work without extracting an equivalent
amount of heat. Notice that the system below does not violate the
first law!)
Ok, the next argument is a bit hairy but let’s give it a go! We want
to show that for an irreversible process the entropy of the universe
must increase.
From the argument above, an irreversible steam engine is less
efficient than a reversible engine. Thus:
=
Again by the first law –wirr = qh + ql.
Substituting and rearranging gives:

Now assume that extraction of heat from the high temperature reservoir
is done reversibly but the expulsion of heat into the low temperature
reservoir is done irreversibly.
Sh = qh/Th

Since S is a state function and the engine operates in a cycle:
Sh + Sl = 0
Sh = -Sl


We have shown that for an irreversible process the entropy change of
the system is greater than q/T. Now imagine that the system was the
entire universe! Since there is no way to transfer heat into or out of
the universe (where would it come from or go??), q = 0
Thus (FINALLY!), for an irreversible (spontaneous) process:
Suniv> 0
Here are 5 different ways to state the second law of thermodynamic:
1.
For any spontaneous process the entropy of the universe increases.
2.
It is impossible for heat to spontaneously flow from a cold object
to a hot object.
3.
No steam engine can be more efficient than a reversible steam
engine.
4.
It is impossible to build a perfectly efficient steam engine.
5.
It is impossible to build a perpetual motion machine of the second
kind.
The second law also explains why gases always expand to fill the
available space (see section V). You never see a gas spontaneously
contract!
(Note: It can be shown that the efficiency of a heat engine is
independent of the working material. So the second law applies to all
systems, not just those consisting of an ideal gas!)
V. Relationship between entropy, probability, and disorder
Another interpretation of the second law is that a spontaneous process
corresponds to a system changing from a less probable state to a more
probable state. It should be clear that a disordered state is more
probable than an ordered state since there are many more ways for a
system to be disordered than ordered.
Let’s use the expansion of an ideal gas to see the connection between
entropy and probability.
Consider two connected vessels filled with one mole of an ideal gas as
shown below:

It should be obvious that it is very unlikely that all of the gas will
be in just one of the vessels. Instead, it will spread out into both
vessels until the pressure is uniform throughout the system. If V = ½
V’, then the right-hand vessel will contain twice as much gas as the
left-hand vessel (2/3 of the gas will in the right-hand vessel and 1/3
in the left-hand vessel). In general, the most probable distribution
of the gas is for the left hand vessel to contain
molecules of gas (NA is Avogadro’s number)
Let’s start with all of the gas (one mole) in the left-hand vessel and
calculate the entropy change that occurs when this ideal gas expands
reversibly at constant temperature to fill the entire system. So the
gas is expanding from an initial volume of V to a final volume of V +
V’.
Since there are no attractive or repulsive forces between the
molecules of an ideal gas, its internal energy depends only on the
temperature (see note below for further discussion of this fact).
Since T is constant, E does not change upon expansion:
E = q + w = 0 so:
q = -w
Recall that for the expansion of a gas (see section I):
w = - (The area under the PV curve, the negative sign produces
the correct sign for w.)
For one mole of an ideal gas:
so:
w = -RT
If the gas expands from V1 to V2, then integrating gives.
w = -RTln
In our example, V1 = V and V2 = V + V’ so:
w = -RTln
Since q = -w and the gas was expanded reversibly, and
S = Rln
Now we can see the connection to probability:
The probability (P1) of one molecule of the gas being within a volume,
V, is:

The probability (P) of all NA molecules being in a volume, V, is:
(If V < V + V’, then this probability is VERY low as it should
be!)
Taking the natural log of both sides and multiplying by “Boltzmann’s
constant” (k) gives:
klnP = kNAln = -Rln = -S (k = R/NA)
We have shown that: S = -klnP (where P is the probability of the gas
occupying volume V)
As the gas expands form V to V + V’, P increases from a small fraction
to 1, and S decreases from a large positive number to zero. That is,
initially, all of the gas is in the left-hand vessel which represents
a very low probability, highly ordered state with very low entropy.
Thus, the gas expands spontaneously until the probability, disorder,
and entropy are as high as possible. When the gas has filled the
entire volume, the probability is one, the system is maximally
disordered which means that the system is at equilibrium!
Note: There is a subtle but important point that I did not emphasize
above. If we let the gas expand freely from the left hand vessel until
it fills the entire system, then no work will be done and no heat will
be absorbed. No work is done because the gas is expanding into a
vacuum so it is not pushing anything out of the way. Since no work is
done and the internal energy doesn’t change (constant T), then no heat
is absorbed and q/T = 0. Does this mean the S = 0? NO! The free
expansion of a gas is irreversible so q is NOT qrev and q/T is NOT S!
In order to calculate S, we have to imagine a reversible pathway and
determine how much heat is absorbed. So we imagined that the gas
expanded exactly as was done in section I! That is, we allowed it to
expand (at constant temperature) against an external pressure that was
reduced VERY slowly. Under these conditions, the gas does work as it
expands and therefore it must absorb an equivalent amount of heat from
the surroundings in order for the internal energy to remain constant.
Imagine the apparatus is submerged in a very large water bath at
temperature T. Initially, the pressure of the gas is equal to the
external pressure and the temperature of the gas is the same as the
temperature of the water bath. Now imagine the expansion occurring in
a series of tiny steps:
1.
The external pressure is reduced by a tiny bit.
2.
Since the pressure of the gas is now a little bit higher than the
external pressure, the gas expands a little bit and does a little
bit of work in the process.
3.
Recall that the internal energy (E) of a system is equal to the
sum of the potential and kinetic energy of all of the particles in
the system. Since an ideal gas is free of attractive or repulsive
forces, its potential energy is zero. Therefore, its internal
energy is equal to its kinetic energy. Because the gas does a
little bit of work, its internal energy and, therefore, its
kinetic energy drops a little bit. This was a long-winded way of
saying that the temperature of the gas drops a little bit.
4.
Now the temperature of the gas is a little bit lower than the
temperature of the water bath so heat flows into the gas from the
water bath until their temperatures are the same again. Since the
water bath is very large, its temperature does not change so the
final temperature of both the water bath and the gas are the same
as the original temperature. (If you prefer, the water bath can be
relatively small but maintained at a constant temperature by an
electrical heater equipped with a thermostat. So when heat flows
from the water bath into the gas, the temperature of the water
bath starts to drop but then the heater clicks on and heats the
bath back to its original temperature. In other words the heat
source is actually some huge power plant somewhere!) The point is
that the gas must absorb heat from the surroundings in order to
maintain a constant temperature as it expands.
5.
Repeat steps 1-4 until the gas has reached the desired final
volume.
We can then calculate S as we did above. Since S is a state function
this is also S for the free expansion of the gas! (S for an
isothermal expansion of a gas depends only on the initial and final
volumes!)
VI. Relationship between G and work
From the first law:
E = q + w
w = E – q
We showed in section IV that for any process at constant temperature:
q ≤ TS (The equality applies only to a reversible process.)
so:
w ≥ E – TS
at constant pressure:
w = -PV + w’ (w’ represents all non-PV work such as electrical work.
The negative sign is due to our convention that positive work means
work done on the system which corresponds to a decrease in volume)
-PV + w’ ≥ E – TS
w’ ≥ E – TS + PV
By definition, H = E + PV
w’ ≥ H – TS
w’ ≥ G
Remember that only reactions with negative G can occur and produce
work. For example, the last inequality says that if G = -10 kJ/mol,
then the reaction is spontaneous and w’ can be -10, -9, -8 etc. Thus,
-10 kJ/mol is the maximum amount of work that can be done by the
reaction. (Negative work is work done by the system on the
surroundings.) If G = +10 kJ/mol, the reaction is not spontaneous and
the number represents the minimum amount of work that must be done on
the system to force the reaction to occur.
We have shown that the change in Gibbs Free Energy represents the
maximum non-PV work obtainable from a spontaneous chemical reaction at
constant T and P. The maximum is obtained only if the reaction is done
reversibly (otherwise some of the released energy is lost as heat).
Remember that a large G means that the reaction is far from
equilibrium. Any system (e.g. a spring) that is far from equilibrium
is capable of doing a lot of work.
Here is another way to look at this:
G = H – TS
H = q (heat actually produced or absorbed by the reaction)
TS = qrev [heat that would be transferred if the reaction was done
reversibly (very slowly)]
G = q - qrev
This form of the equation say that for a spontaneous process (negative
G), the amount of heat transferred is always less than the amount of
heat that would be transferred by a reversible process. The difference
is the amount of energy available to do work. If a reaction is at
equilibrium, then it cannot do any work (G = 0) and all of the energy
produced by the reaction is lost as heat (q = qrev).
Let’s again put in some numbers to make it easier to understand this:
Let’s first consider a spontaneous reaction (G < 0).
If q = -10 kJ/mol (exothermic), then qrev could be -9, -8, -7, etc.
The differences ( q – qrev) would be negative (-1, -2, -3, etc.). This
represents the portion of q that can be converted into (non-PV) work.
The remainder of the heat is dispersed into the surrounding and is
wasted. Remember that the second law says that it is impossible to
convert heat into work with 100% efficiency.
If q = +10 kJ/mol (endothermic), then qrev could be 11, 12, 13, etc.
The differences ( q – qrev) would again be negative (-1, -2, -3, etc.)
and the same argument applies. Notice in this case, S would have to
be positive in order for the reaction to overcome the unfavorable H
and allow the reaction to occur. For an exothermic reaction, S can be
negative as long as it isn’t too negative!
Now let’s consider a non-spontaneous reaction (G > 0)
If q = -10 kJ/mol (exothermic), then qrev could be -11, -12, -13, etc.
The differences ( q – qrev) would be positive (1, 2, 3, etc.). This
represents the amount of work that must supplied by the surrounding to
force the non-spontaneous reaction to occur.
If q = +10 kJ/mol (endothermic), then qrev could be 9, 8, 7, etc. The
differences ( q – qrev) would again be positive (1, 2, 3, etc.) and
the same argument applies.
If a reaction is at equilibrium (G = 0) then it can do no work. In
this case q = qrev and all of the heat produced is dispersed into the
surroundings and none of it can be converted into work.
Note: You might have noticed a seeming contradiction. In part I, we
showed that a reversible mechanical process produces the most work.
Here we seem to be saying that a reversible reaction (one at
equilibrium) produces the least amount of work (zero!). What’s really
happening here is that the reaction is proceeding at equal rates in
both directions. The heat produced by the exothermic direction is
absorbed by the reverse reaction so no heat is available to do work. A
reaction at equilibrium does no work because there is no net reaction!
If a spontaneous reaction could be carried out reversibly, then more
of the heat produced could be converted into work. In some sense, you
could say that the heat produced by the exothermic reaction is driving
the endothermic reaction with 100% efficiency! Thus, the reaction IS
doing the greatest amount of work when it is at equilibrium and no
heat is lost to the surroundings! (I admit that this last argument is
a bit shaky!)
14

  • DODATEK Č 5 K NÁJEMNÍ SMLOUVĚ Č 115N1117 SMLUVNÍ
  • “EL FORJADOR DE HISTORIAS” LISTA DE DIÁLOGOS EN ESPAÑOL
  • BALKANLARDAN ULUĞ TÜRKİSTAN’A TÜRK HALK İNANÇLARI II KAŞKAYİLER –
  • 1 JUMADIL AWAL 1436 AH DENGAN MENYEBUT NAMA ALLAH
  • ÖĞRETIM TEKNOLOJILERI VE MATERYAL TASARIMI MATERYAL(PROJE) IŞILDAYAN ÖZDEŞLİK KAZANIM
  • VOORBEELD NIEUWE OUDERVERKLARING GEGEVENS LEERLING ACHTERNAAM VAN DE LEERLING
  • AL REFERENTE DELL’OCC DELL’ODCEC DI MILANO VIA PATTARI 6
  • DEAR   I’M CONFIDENT THAT ATTENDING THE NATIONAL
  • REPRESENTACIONS TEMPORADA 20102011 ANGLÈS – ELS PASTORETS ADAPTACIÓ
  • CONSTRAINTS CONSTRAINTS REFER TO ANY RESTRICTIONS OR LIMITATIONS PLACED
  • FIXTURE SERIE 43 HONOR 1º FECHA EQUIPOS RESULTADOS 1º
  • ESMATASANDI TERVISEKESKUSE ARENGUKAVA KOOSTAMISE JUHEND (ERF TOETUSE TAOTLUSE ESITAJA
  • “DESARROLLO DE NUEVOS MATERIALES EN TÉCNICAS DE SEPARACIÓN MEMBRANAS
  • INTERMEDIATE MACROECONOMICS ECON 207 FALL SEMESTER 2009 PROFESSOR JOHN
  • NA OSNOVU ČLANA 6 I ČLANA 84 STAV 1
  • UNIVERSIDADE DE BRASÍLIA DIRETORIA DE ADMINISTRAÇÃO ACADÊMICA INSTRUÇÕES GERAIS
  • HEADQUARTERS NETWORK ENTERPRISE TECHNOLOGY CENTER ARMY SIGNAL REGIMENT (P)
  • COLEGIO DE FARMACÉUTICOS PIDE A LA POBLACIÓN HACER UN
  • PLIEGO DE CONDICIONES PARA ADQUISICIÓN DE PRODUCTOS FARMACEUTICOS Y
  • KONSPEKT TEMAT 87 ORGANIZACJA I ZADANIA SYSTEMU WYKRYWANIA I
  • BUPATI KUNINGAN PERATURAN BUPATI KUNINGAN NOMOR 68 TAHUN 2012
  • LISTA DE ANEXOS ANEXO 1 ANEXO 1A ACUERDOS MULTILATERALES
  • ROK SZKOLNY 20102011 PRZEWODNICZĄCA KRYSTYNA SÓS ZCA Z
  • WWFCANADA IDENTIFIES AT LEAST FOUR MAJOR MEASURES NEEDED TO
  • WWWRECURSOSDIDACTICOSORG ¡HOLA AMIGOS! Y SEGUIMOS PUES CON EL ESTUDIO
  • CONVENTION CADRE D’ACCUEIL D’ENTREPRISES DANS DES LOCAUX UNIVERSITAIRES VU
  • A NORDLAND C KFT AZ ALÁBBI PROJEKTHEZ NYERT TÁMOGATÁST
  • POROZUMIENIE O PROGRAMIE ZAJĘĆ ROK AKADEMICKI 2020 KIERUNEK
  • “LA TELEVISIÓN HÍBRIDA OFRECE ENORMES POSIBILIDADES PARA LA INTEGRACIÓN
  • GSPSNUSA2713 0 NOTIFICATION 1 NOTIFYING MEMBER UNITED