1
1
2
1
1
1
dV
dV
dS
dV
dV
dS
dS +=
(13) 0
1
2
2
1
1
≥








−= dV
T
P

T
P
dS
(14)
T
H
S
∆
=∆
KJ/mol 96.108
K 15.373
J/mol 657,40
⋅==∆
vap
S
It is assumed that T = T and thus the total entropy is unchanged by a transfer of thermal
5
1 2
energy between the two parts. The constant-energy restriction of equation 12 is satisfied because
any work done by one part on the other can be compensated by such a transfer of thermal energy.
7/10/07 2- 40
Again consider a system constrained to constant total volume, and to have a uniform temperature.
Then, as before, we can write the total change in entropy that might result from a change in
volume of either part in the form
(T, V)

and, because total volume is constant, dV = - dV . Substitution of equation 12 gives
1 2
5
This proves that two bodies in equilibrium must have not only the same temperature but also the

same pressure. It also proves that, when two bodies in contact, at the same temperature, have
different pressures, the body at the higher pressure will tend to expand and compress the body at
the lower pressure.
ISOTHERMAL ENTROPY CHANGES. Entropy changes at a single temperature follow directly from
equation 5. (Calculations involving a temperature difference or temperature change will be
treated later.)
A change of phase, under equilibrium conditions, will be at constant temperature and
constant pressure. The thermal energy transfer, Q = Q , will be ∆H. Therefore,
rev
For example, the entropy change for the melting of ice was found to be 1.22 J/g·K or 22.0
J/mol·K. The entropy of vaporization of water at 100 C is
o
More typical liquids, not involving such strong intermolecular forces, have ∆S values of about
vap
92 J/mol·K at their normal boiling points. This is known as Trouton’s rule.
The work done in a reversible, isothermal expansion of an ideal gas was found to be
(equation 10, Chapter 1) W = - nRT ln V /V . For an ideal gas at constant temperature ∆E = 0
rev 2 1
and therefore Q = -W . The entropy change is thus
rev rev
( )
(15) /ln rev. I.G., ,
12
VVnR
T
Q
ST
rev
==∆
Diabatic, like its better-known twin diabetic, implies that something “passes through”

6
(thermal energy or sugar). Thus adiabatic tells us there is no thermal energy transfer; Q = 0. In
practice we achieve adiabatic conditions by insulating the system, by maintaining system and
surroundings at the same temperature, or by carrying out the process very rapidly. Most gas
expansions are fast enough to be nearly adiabatic.
One sometimes sees the statement that ∆S = Q/T for a reversible process. This is not
7
really wrong, but it is doubly misleading. It de-emphasizes the need to find Q (which is often
rev
different from Q) and it provides no clue to finding ∆S for an irreversible process. Equation 5, ∆S
= Q /T, covers both irreversible and reversible processes.
rev
7/10/07 2- 41
If the gas expands through a pinhole or stopcock (an effusion-controlled leak) into an
evacuated container (Figure 2), no work is done because the gas exerts forces only on the
immovable walls of the container. By insulating the containers we can ensure that Q = 0. The
energy is unchanged and therefore, if the gas is ideal, the temperature remains constant. This
serves as an experimental demonstration (performed by Joule, in 1844) of the dependence of
energy only on the temperature, although it is not as sensitive as a later method by Joule and
Thomson. The entropy change of an ideal gas in such an irreversible, adiabatic expansion can be
6
calculated as follows.
Entropy is a state function, so the entropy change depends only on the initial state and the
final state. Entropy changes are independent of the path taken between the initial and final states.
When an ideal gas expands at constant temperature from V to V , the final state is the same
1 2
whether the process is reversible or irreversible. Therefore, the entropy change for the adiabatic
expansion of the ideal gas into a vacuum is
(I.G., T) ∆S = nR ln V /V (16)
2 1

even though Q = -W = 0 in this process.
The example just cited illustrates a general method for evaluating entropy changes in
irreversible processes. Having determined what are the initial and final states, a path is found
between those states that will be entirely reversible. Then equation 5 tells us ∆S = Q /T.
rev
7
It is of interest to calculate the entropy change for the surroundings in the reversible and
irreversible expansions above. In the reversible expansion, Q = Q = - Q = -(Q ) .
rev surr rev surr
7/10/07 2- 42
Therefore
(rev) ∆S = - nR ln V /V
surr 2 1
The adiabatic expansion into a vacuum produces no change at all in the surroundings, and
therefore
∆S = 0
surr
Adding together the entropy changes for system and surroundings, we obtain, for the reversible
expansion,
(∆S) = 0
system + surroundings
and for the irreversible expansion, into a vacuum,
(∆S) = nR ln V /V > 0
system + surroundings 2 1
This result agrees with the requirements of the second law.
The particular irreversible path chosen is an extreme case. If the gas had been allowed to
expand against a constant external pressure (as in Figure 1, Chapter 1) the entropy change for
system plus surroundings would have been positive, but less than that for the expansion into a
vacuum. The magnitude of the total entropy change (system plus surroundings) can be taken as a
measure of the degree of thermodynamic irreversibility of any process.

INTERPRETATION OF ENTROPY. We have seen that we can measure the increase in entropy by
measuring the amount of energy transferred to the system, as thermal energy transfer, Q, divided
by the (absolute) temperature, provided we measure Q along a path between initial and final
states that is completely reversible. To know whether a process will actually occur, we must
make a similar measurement for changes in the surroundings, as we will emphasize below.
Without getting involved in the detailed calculations of statistical mechanics, it is important
to look more carefully at the meaning of entropy as a spread function. In particular, there are two
ways of measuring the spreading, or the randomness, of a state. We have seen that entropy is
appropriately called the spread function because it is a measure of the amount of spreading of
energy among the molecules of a substance. However, in addition to considering where the
packets of energy are located we should look also at where the molecules are (like the yellow and
white rice molecules we mixed earlier). Remarkably, the same measurement of Q /T gives
rev
information on this second type of spreading.
Consider a sample of an ideal gas, occupying an initial volume V (e.g., 1 m ). Assume, quite
i
3
arbitrarily for the moment, that each molecule requires some very small volume, V (e.g., 2 Å ,
o
3
most of which volume elements are unoccupied at any given instant). Then we can say there are
spaces for V /V molecules (5 x 10 in our example). If we let the gas expand to a larger volume,
i o
29
V , then by the same reasoning there are now spaces for V /V molecules. The important thing is
f f o
7/10/07 2- 43
that we have increased the number of available spaces by the ratio V /V . But we already know
f i
that if we expand an ideal gas, the entropy increases by ∆S = nR ln V /V . In other words, when

f i
we increased the number of available, or accessible, spaces by the ratio V /V , the entropy
f i
increased by the logarithm of this same ratio.
We could equally well make the same kind of argument for molecules in a crystal, where the
number of available spaces is more readily counted. If some number of “stray” molecules are
introduced and given an opportunity to diffuse throughout the crystal, we would obtain a
substantial increase in entropy per molecule. Or we may simply drop a crystal of salt into water.
When we return, very likely the salt molecules will have diffused throughout the water. The salt
molecules, or their constituent ions, are attracted to each other and roughly equally attracted to
water molecules, so there is negligible energy difference, but there is a very high probability of
the salt diffusing into the water, never to return to the crystal unless external conditions are
changed. It is like a drop of water on the pavement. Even though energetically the water drop
would prefer to remain with its neighbors, the lure of wide open spaces is enough to make the
drop evaporate. The process is driven by an increase in entropy, or in probability.
The crystal is highly ordered; the solution is highly disordered, or random. There are more
states accessible to the salt when it is dissolved in water, or accessible to gas molecules when a
larger volume is available to the molecules. More available, or accessible, states can be equated
to a greater probability. Thus entropy is variously described as measuring disorder, or
randomness, or probability. We can always organize molecules, forming crystals or condensing a
gas to a liquid or arranging conditions such that molecules will form living structures. To do so
always has a cost involved, specifically the cost of increased randomness in the surroundings.
The second law of thermodynamics does not tell us we cannot decrease entropy in any given
sample. It tells us only that, for the universe as a whole (system plus surroundings), entropy will
always increase.
A study of how signals may be communicated from one place to another gave rise to a new
understanding of probability under the heading of information theory. Information theory turned
out to be of major importance in understanding the meaning of entropy as well as in discovering
better ways to organize the storage and distribution of information over telephone lines, as radio
or television signals, or in computers.

Gibbs Free Energy
When investigating equilibrium or reversibility, it was necessary to find the change for the
system but also for the surroundings. Usually we prefer to ignore the surroundings most of the
time. The free energy functions provide a means of doing just that.
We began with the variables P, V, and T, to which we added energy, E, and subsequently
have added enthalpy, H, entropy, S, and now the two free energy functions, F and G, as shown in
equations 17-19 and in Figure 3.
H = E + PV (17)
F = E - TS (18)
G = H - TS (19)
(22) V
P
G
T
=






∂
∂
(23) S
T
G
P
−=







∂
∂
For many years the Gibbs free energy and the Helmholtz free energy were not clearly
8
distinguished, and thus both were designated by the symbol F. In recent decades the symbol F
has been selected for Helmholtz free energy and the symbol G for Gibbs free energy. Unless
specifically indicated otherwise, the term “free energy” will always mean the Gibbs free energy in
this book. [Fig. 3: A becomes F.]
7/10/07 2- 44
G = F + PV (19a)
G = E + PV - TS (19b)
We introduced the Helmholtz free energy from the relationship (eqn. 4) ∆A = ∆F = W and
8
rev
the observation that a change in F represents the minimum amount of work that must be done on
the system, at constant temperature, to get from the initial to the final state. For example, you
cannot put a box on a shelf without doing some work on the box (and gravitational field), nor can
you compress a gas without doing work on the gas. There is good reason for picking the correct
function. Recall that we found enthalpy, H, to be much more convenient than energy, E, when
measuring thermal energy transfers under conditions of constant pressure. For much the same
reasons, we will find G to be generally more convenient than F. The Helmholtz free energy is
more appropriate at constant volume; Gibbs free energy is better for constant pressure.
FREE ENERGY AND EQUILIBRIUM. Assume for the moment that the only work done is work of
expansion or compression, and that the composition of the system does not change. Then, from
equation 19b,
dG = dE + d(PV) - d(TS)

As we have seen, because E is a state function, dE is independent of path, so
dE = q + w = q + w
rev rev
regardless of path. We may therefore substitute q + w for dE, to obtain
rev rev
dG = q + w + P dV + V dP - T dS - S dT (20)
rev rev
But q = T dS and w = - P dV. Therefore, quite generally,
rev rev
dG = V dP - S dT (21)
We effectively split this equation into two parts, by holding constant first the temperature,
then the pressure:
We introduced the assumption that w # w by considering an expansion or compression
9
rev
of an ideal gas, and from that and the independence of dE on path, we concluded that q = dE -
rev
w $ q. Now we may work backward from the second law, which is one of our basic postulates,
rev
(dS) = q /T + (q ) /T $ 0. This inequality must be valid for any process occurring in
syst + surr rev rev surr surr
the system, including if q = (q ) , which may therefore be assumed without loss of generality.
surr rev surr
But we know that q = - q (by combining the first law with the first-law equation), and because
surr
we are assuming temperature is constant, we may set T = T. Therefore (dS) = q /T -
surr syst + surr rev
q/T $ 0 and thus q $ q. It follows from this that w # w .
rev rev
7/10/07 2- 45

Note, of course, that P and V naturally go together, as do
T and S.
An equally valid alternative to equation 20 is obtained
by substituting dE = q + w, giving
dG = q + w + P dV + V dP - T dS - S dT
Then, for constant temperature and pressure,
(T,P) dG = q + w + P dV - T dS
(24)
But at constant pressure we may safely assume the pressure of the system is the same as the
pressure of the surroundings (meaning the part of the surroundings that is in actual contact with,
and therefore may affect, the system), so w = - P dV, where, as usual, P is the pressure of the
system. Also, we know that q $ q. Therefore,
9
rev
(T,P) dG = q - T dS = q - q # 0
rev
Thus
(T,P) dG # 0 (25)
in general, and for the limiting case of the reversible, or equilibrium, process,
(T,P) dG = 0 (25a)
This shows that ∆G must be negative, or, in the limit of a reversible process, zero. That is, G is a
minimum for an equilibrium state. This is the more convenient criterion for equilibrium that was
sought. Looking for the change in free energy, G, under our normal operating conditions of
∫ ∫ ∫∫
===







∂
∂
==∆
12
/ln / PPnRTPdPnRTdPVdP
P
G
dGG
T
7/10/07 2- 46
constant temperature and constant pressure, we need only look at changes of the system. It is not
necessary to make separate calculations for the surroundings.
We began this section with the assumption that the only work done was P dV work, or work
of expansion or compression. Now we revisit that question. Retain the conditions of constant
temperature and constant pressure, but from equation 20 we obtain
(T, P) dG = P dV + w
rev
The total work done on the system may be broken into expansion/compression work and other
forms (e.g., electrical work) which we label w’. For reversible processes,

w = - P dV + w’
rev rev
It follows that

(T,P) dG = w’ (26)
rev
Thus, apart from providing a convenient measure of whether the system is or is not at equilibrium
(with only P dV work), the free energy tells us how much other work can be expected from a
process. In an electrochemical cell, w’ is the electrical energy generated by the chemical reaction.

Work against a gravitational field or against a magnetic field would also be included in w’.
IDEAL-GAS EXPANSIONS. The change in free energy of an ideal gas in an isothermal expansion or
compression is particularly important because it has been found possible to express free energy
changes of all systems, whether gases or not, in this same form. The broader theory will be
developed in the discussion of physical equilibria.
Start with the definition of G, equation 19. At constant temperature,
(T) ∆G = ∆H - T ∆S (27)
In an isothermal, reversible expansion of an ideal gas, ∆E = 0 and ∆H = 0.
q = - w = InRT dV/V = nRT ln V /V
rev rev 2 1
∆S = q /T = nR ln V /V
rev 2 1
∆G = - T ∆S = - nRT ln V /V
2 1
or
(T, I.G.) ∆G = nRT ln P /P (28)
2 1
The same result could be obtained from equation 22.
FREE-ENERGY CHANGES IN CHEMICAL REACTIONS. A very important application of the free-
energy function is to the description of chemical reactions. Because G is a state function, free
energies can be added and subtracted in the same manner as enthalpies. It is not possible to
7/10/07 2- 47
assign meaningful absolute values to free energies, but handbook tables give values of G relative
to the elements from which compounds are formed. Thus the free energy of formation of any
element is, by definition, zero. From Table 1 we can see that the standard free energy of
formation of SO in its standard state (gas, at 1 atm pressure, at 25 C) is - 370.4 kJ/mol, whereas
3
o
∆G (SO ) = - 300.2 kJ/mol. Subtracting (and setting ∆G (O ) = 0) gives ∆G = - 70.2
o o

f 2 f 2 reaction
kJ/mole. The negative value indicates that the reaction is spontaneous, showing that oxygen gas
will combine with sulfur dioxide to form sulfur trioxide at this temperature and pressure, at least
in the presence of a suitable catalyst.
Table 1 STANDARD ENTHALPIES AND FREE ENERGIES OF FORMATION*
Compound State ∆H ∆G Compound State ∆H ∆G
f f f f
o o o o
AgBr c -100.4 -96.9 H O g -241.8 -228.6
2
AgCl c -127.0 -109.8 lq -285.8 -237.1
AgI c -61.8 -66.2 H O g -187.8 -120.4
2 2
Ag O c -31.1 -11.2 lq -136.3 -105.6
2
Al O c -1675.7 -1582.3 H S g -20.6 -33.4
2 3 2
Br g 30.9 3.1 I c(rh) 0 0
2 2
C diamond 1.9 2.9 NH g -45.9 -16.4
3
C graphite 0 0 aq -362.5 -236.5
CF g -933.6(639.9) 66.2 NH Cl c -314.4 -202.9
4 4
CCl lq -139.3 -68.6 N O g 81.6 103.2
4 2
CO g -110.5 -137.2 NO g 91.3 87.6
CO g -393.5 -394.4 NO g 33.2 51.3
2 2
CaCO calcite -1207.6 -1129.1 N O g 11.1 99.8

3 2 4
aragonite -1207.8 -1128.2 NaCl c -411.2 -384.1
CaCl c -795.4 -748.8 aq -407.3 -393.1
2
aq -877.2 -816. NaOH c -425.8 -379.5
CaF c -1228.0 -1175.6 aq -470.1 -419.1
2
aq -1208 -1031.2 O g 142.7 163.2
3
CaO c -634.9 -603.3 SO g -296.8 -300.1
2
Ca(OH) c -985.2 -897.5 SO g -395.7 -371.1
2 3
aq -1002.8 -858. CH g -74.6 -50.5
4
FeO c -272.0 -244.3 C H g 227.4 209.9
2 2
Fe O c -824. -742.2 C H g 52.4 68.4
2 3 2 4
Fe O c -1118.4 -1015.4 C H g -84.0 -32.0
3 4 2 6
HBr g -36.3 -53.4 CH OH g -201.0 -162.3
3
aq -121.6 -104.0 lq -239.2 -166.6
HF g -273.3 -275.4 C H OH g -234.8 -167.9
2 5
HI g 26.5 1.7 lq -277.6 -174.8
C H lq 49.1 124.5
6 6
g 82.9 129.7

*Values are in kJ/mol, for 25 C, 1 atm. The standard states for liquids (lq) and crystalline solids (c) are the pure
o
materials at 1 atm pressure and the standard states for gases (g) are the pure gases at 1 atm pressure in the “ideal
gas state” (that is, extrapolated from low pressures assuming ideal gas behavior). “aq” refers to the (hypothetical
ideal) 1 molal solution in water.
(
)
T
reactantsproducts
T
P
GG
P
G








∂
−∂
=







∂
∆∂
T
reactants
T
products
T
P
G
P
G
P
G






∂
∂
−









∂
∂
=






∂
∆∂
reactantsproducts
T
VV
P
G
−=






∂
∆∂
(29) V
P
G
T

∆=






∂
∆∂
7/10/07 2- 48
Free energy values do not tell us anything about rates of reaction. There may be a significant
activation energy required to get a reaction to proceed, even when it is thermodynamically
allowed. (That is what catalysts are good for. By offering a different path, they may involve a
substantially lower activation energy and therefore a faster rate.) On the other hand, if ∆G > 0,
the reaction cannot proceed, under the specified conditions, regardless of catalysts or time.
LECHATELIER’S PRINCIPLE AND EQUILIBRIUM. For every chemical reaction there exists an
equilibrium point at which there is no further tendency for the reaction to proceed forward or
backward. Although the equilibrium point of most reactions is far to one side or the other, a true
equilibrium point can be calculated, and for nearly all systems the equilibrium point can be
demonstrated if the experimental techniques are sufficiently sensitive.
LeChatelier’s principle states that when any stress is applied to a system at equilibrium, the
system will respond in a manner that will reduce the stress. For example, the reaction
2 NO + O !6 2 NO
2 2
releases thermal energy (∆H = - 56.53 kJ/mol(NO )) but causes a decrease of volume, or
2
pressure. Therefore an increase of temperature will tend to drive the reaction to the left, so that
the reaction will absorb thermal energy, but an increase of pressure, by compression, will tend to
drive the reaction to the right to decrease the number of moles of gas and hence the pressure.
These relationships can be derived, quantitatively, from the second law.

The change in ∆G = G - G with change of pressure is
reaction products reactants
The derivative of the sum is the sum of the derivatives:
From equation 22, these derivatives are the respective volumes:
or
(30) S
T
G
P
∆−=






∂
∆∂
(30a)
T
HG
T
G
P
∆−∆
=







∂
∆∂
(31)
T
H
T
G
P
∆
−=






∂
∆∂
It is assumed that there are no other changes that might cause a reverse effect. For
10
example, changes in the relative concentrations of the reactants, or addition of an inert gas, are
not within the scope of equations 29 and 30 or 31.
7/10/07 2- 49
Similarly, from equation 23,
This equation tells how ∆G (measured at some constant temperature and pressure) depends upon
the temperature at which the process occurs. The ∆S on the right-hand side is similarly to be
measured at a constant pressure and at a constant temperature (the temperature at which the
derivative, or the slope of ∆G vs. T, is determined). We may therefore substitute for - ∆S the

quantity (∆G - ∆H)/T.
To find the shift of equilibrium point we set ∆G = 0 and obtain
The first result, equation 29, says that if there is a volume increase during the process, a pressure
increase will cause an increase in ∆G, making the process less spontaneous. The second equation
(31) says that if the process is endothermic (∆H positive), a temperature increase will make ∆G
more negative, thus making the process more spontaneous. The equations apply to phase
changes or other equilibrium processes as well as to chemical reactions. The importance of the
10
equations is that they give a quantitative description of how ∆G changes, and therefore of how
the equilibrium point changes.
Free Energy and the Energy-Entropy Battle
For the formation of ammonia, we may write the equation
½ N +3/2 H &6 NH
2 2 3

∆G = - 16.4 kJ/mol, ∆H = - 45.9 kJ/mol, and ∆S = -99.2 J/mol·K. The reaction clearly is
spontaneous at room temperature and pressure, but it doesn’t go. The activation barrier is too
great. From LeChatelier’s rule, we can see that an increase in temperature, to overcome the
7/10/07 2- 50
activation barrier and reaction kinetics, will make the reaction less spontaneous, but an increase in
pressure will drive the reaction back to the right. The synthesis of ammonia is therefore carried
out at high temperature (to get the reaction to go) and at high pressure (to recover yield).
We can look at the reaction in a somewhat different way. Write

∆G = ∆H - T∆S = -45.9 kJ/mol - T x (-99.05 J/mol·K)

Examining the numbers suggests that the reaction as written (25 C and 1 atm of each gas) is
o
spontaneous because the reactants give off thermal energy (∆H = Q < 0 at const. P) without a
major change in the thermal energy of the system (T is constant). The system, as a whole, is

falling into a potential well, giving off energy that is conducted to the surroundings as thermal
energy. On the other hand, this drop in internal energy drives the system to a more ordered
internal arrangement — two gas molecules collapse into one gas molecule, something like a
crystallization (loss of internal degrees of freedom, or a more orderly arrangement). Thus if we
raise the temperature (neglecting for the moment the change in values of ∆H and ∆S), the extra
degrees of freedom of the molecules of the elements become more important (for example, at
constant pressure they will occupy a larger volume and as speeds increase, the spread in speeds
will increase as well), so the condensation of elements to compound(s) becomes less probable.
They must be driven together by increasing the pressure.
Of course, common sense tells us much the same thing, qualitatively. At high temperatures,
compounds tend to decompose. At low temperatures, substances tend to condense, to liquids or
to crystals. Energy differences are important at low temperatures. The spread function, entropy,
is increasingly important as the temperature is raised.
A rough rule of thumb is that formation of 1 mole of gas molecules raises the entropy by
about 125 J/mol·K (compare Trouton’s rule, p. 33, which is for vaporization only at the normal
boiling point). At room temperature, the resultant increase in T∆S would be on the order of 40
kJ/mol of gas formed. This is a small term compared to the enthalpy change of very exothermic
reactions, such as most combustions, but it is not small for many other reactions.
Entropies of reaction for a few representative chemical reactions are given in Table 2.

Table 2. STANDARD ENTROPIES OF REACTION, 25 C
o
Reaction ∆S , J/mol·K
o
2 Al (c) + ½ O Al O (c) -313.3
2 2 3
C (gr) + O CO +3.0
2 2
CO + ½ O CO -86.36
2 2

H + Cl 2 HCl +19.6
2 2
H + F 2 HF +14.1
2 2
H + I (c) 2 HI +166.2
2 2
H + ½ O H O (lq) -163.1
2 2 2
H + S (rh) H S +43.14
2 2
N + 3H 2 NH -198.2
2 2 3
S (rh) + O SO +11.2
2 2
3 C H C H (lq) -333.3
2 2 6 6
Entropy Changes with Change of Temperature
∫ ∫
==∆ (32)
T
q
dSS
rev
∫ ∫ ∫ ∫
====∆ (33)
T
dT
C
T
dT

dT
q
dT
dT
T
q
T
q
S
revrevrev
(34) ln
1
2
T
T
CS =∆
7/10/07 2- 51
The fundamental equation for entropy change, equation 5, gives the infinitesimal entropy
increase for the absorption of an infinitesimal amount of thermal energy at some temperature T.
If the temperature changes continuously during the process, the sum of the entropy changes can
be written as the integral
It is not at all obvious from equation 32 how this integral is to be evaluated, but it can be handled
quite easily as follows:

The heat capacity, C, is often nearly independent of temperature (over temperature intervals of
tens of degrees) so the equation can be simplified to the form


Note that the heat capacity has not been given a subscript in equations 33 or 34; this is because
no conditions such as constant pressure or constant volume have thus far been specified. If

pressure is constant, C is inserted in these equations; if volume is constant, C is required. If
P V
neither pressure nor volume is constant, it may be necessary to find an alternative path consisting
of constant pressure and constant temperature segments, or constant volume and constant
temperature segments. It may be, too, that it will be simpler to return to equation 32. For
example, in a reversible, adiabatic expansion or compression, q = q = 0, and therefore it is clear
rev
that the integral, and the entropy change, must be identically zero. To apply equation 34 one
would have to define an “adiabatic heat capacity” that would always be zero.
1
1
T
Q
S
−
=∆
2
2
T
Q
S =∆
(35) 0
11
2221
21
>









−=+
−
=∆+∆=∆
TT
Q
T
Q
T
Q
SSS
Actually, the entropy change of the insulator is zero, after the steady state has been
11
reached, because the state of the insulator is not changing.
7/10/07 2- 52
The entropy change for the transfer of thermal energy between two bodies at different
temperatures requires a different approach. In Figure 4, the body at the left is at a temperature
T , the body at the right is at a lower temperature T , and each body is assumed large enough that
1 2
its temperature will not be appreciably affected by the flow of thermal energy.
The primary method of energy transfer in such a situation (e.g., between the Sun and the
Earth) is by radiation. But the radiation process is not at equilibrium. (Radiation at equilibrium is
called black body radiation and requires the radiation field to be in equilibrium with the radiating
body, which cannot be the case if radiation is being “leaked” to the low-temperature body.) We
must therefore look for a reasonable substitute that will give us an answer suitably close to reality.
We achieve this approximation to reality by a thought experiment. Add a thermal insulator
connecting the two bodies. When a steady state has been achieved, the temperature will vary

continuously with the distance, as shown in the lower part of Figure 4, but the temperature will be
constant at each point in the system even though a small amount of thermal energy is flowing past
each point. The entropy change is the sum of three terms: the entropy change of the left-hand
body at T ; the entropy change for the insulator; and the entropy change for the right-hand body
1
at temperature T
2.

∆S = ∆S + ∆S + ∆S
1 insulator 2
For an amount of thermal energy transferred, Q, the first term is


where ∆S is the entropy change of the body at T . Similarly, the entropy change for the body at
1 1
temperature T is
2

The insulator may be considered part of the surroundings and need not be further included in the
calculation for the system. Then for the system, which consists of the two bodies at the two
11
temperatures,
Superficially the thermal insulator added to the problem of Figure 4 plays no significant part
in the calculation. It can make little difference to the state, or the entropy, of a body that loses a
given amount of thermal energy at the temperature T what happens to this energy after it is gone.
1