An Introduction to Thermodynamics phần 3 docx

without changing their effect, and therefore decreases the pressure on the walls, requiring the
“supplement” to P to fit the ideal gas equation form; see Phys. Teach. 34, 248-249 (April, 1996).
The reader with some knowledge of special relativity may recognize that the total
16
energy of any system is measured by its mass, multiplied by the square of the speed of light.
However, it would be necessary to measure masses about a million times more accurately than is
now possible to be able to determine energies to the accuracy required in thermochemistry.
7/10/07 1- 27
It should also be possible to relate the volume of any liquid or solid to its temperature and
pressure, or to express such other properties as refractive index, heat capacity at constant volume
or pressure, thermal conductivity, heats of vaporization or fusion, or vapor pressures of solids or
liquids, in terms of the temperature and pressure. Some of these equations will be encountered in
later chapters.
Thermochemistry

The application to chemical reactions of the principles developed thus far is called
thermochemistry. In particular, the heats of reaction are measured and tabulated and from these
and from measured heat capacities the enthalpy changes are calculated for other reactions or for
other experimental conditions.
HESS’S LAW. The enthalpy change for a chemical reaction, such as the oxidation of sulfur dioxide
to sulfur trioxide —
2 SO (g) + O &6 2 SO (liq)
2 2 3
— can be expressed as the difference between the enthalpies of the initial and final states.
∆H = H - H
reaction final initial
= H(2 SO ) - H(2 SO ) - H(O )
3 2 2
There is no way within thermodynamics of measuring an absolute energy, or an absolute
16
enthalpy. Only energy, and enthalpy, changes can be determined. However, knowing that these

energy and enthalpy changes depend only on the initial and final states, it is possible to add and
subtract chemical reactions and add and subtract the corresponding enthalpy changes. That is, we
may quite arbitrarily select a reference energy and/or enthalpy level and measure all values from
that arbitrary level. In particular, it is possible to tabulate “heats of formation”, the enthalpy
changes in the reaction of the elements to form each compound, and from these to calculate
enthalpies of other reactions. This principle is known as Hess’s law.
The reactions for the formation of the gases SO and SO from the elements are
2 3
S + O SO
2 2
S + 3/2 O SO (liq)
2 3
7/10/07 1- 28
The measured enthalpy changes for these reactions at 25 C and 1 atm pressure are -296.90 kJ/mol
o
and -437.94 kJ/mol. Subtraction of the first reaction from the second gives
SO + ½ O &6 SO (liq)
2 2 3
and subtraction of the enthalpy changes gives -141.04 kJ/mol, which is the heat of reaction for the
oxidation of SO to SO (liq).
2 3
Exactly the same elements, in the same quantities, always appear on both sides of a chemical
equation (which is why reactions as written are called “equations”). Subtraction of the elements
from both sides of an equation will yield, on each side, product minus reactants for the reactions
of formation of each of the substances appearing in the original equation. In the example above,
the original equation was SO + ½ O &6 SO . Subtract 1 mol of S and 3/2 mol of O from each
2 2 3 2
side. The equation can then be written as the formation of each compound (i.e., of SO , O , and
2 2
SO ) from the elements.

3

(SO - S - O ) + (½ O - ½ O ) &6 (SO - S - 3/2 O )
2 2 2 2 3 2
and therefore
∆H = ∆H (SO ) - ∆H (SO ) - ∆H (½ O )
reaction form form form
3 2 2
= -437.94 kJ/mol -296.90 kJ/mol - 0
= -141.04 kJ/mol(SO liq)
3
(Notice that the heat of formation of any element, in its standard state, is necessarily zero.)
An entirely equivalent way of obtaining the same numbers is to consider the enthalpy of each
compound on a scale taken with reference to the elements. Such enthalpy values are called
standard enthalpies of the compounds; they are identical with the standard enthalpies of
formation.
Hess’s law can often be applied to find heats of reaction that could not be directly measured
experimentally. For example, the reaction of two molecules of ethylene, C H , to form
2 4
cyclobutane, C H , would not readily occur quantitatively under conditions conducive to
4 8
measurement of the heat of reaction. But both ethylene and cyclobutane can be burned in
oxygen, and subtraction of these reactions gives the reaction equation desired.
2 C H + 8 O &6 4 CO + 4 H O
2 4 2 2 2
C H + 8 O &6 4 CO + 4 H O
4 8 2 2 2
Subtraction of the second from the first gives

2 C H &6 C H

2 4 4 8
and, therefore, subtraction of the ∆H for the second combustion from the ∆H for the first
combustion gives ∆H for the condensation reaction. Heats of combustion (equal to - ∆H )
reaction
are comparatively easy to measure and are often tabulated.
KIRCHHOFF’S LAW. The heat of reaction at a temperature other than that given in a table can be
( ) ( )
∫ ∫
+∆+=∆
1
2
2
1




products1reactants2
(17)
T
T
T
T
PP
dTCHdTCH
( )
[
]
∫
−+∆=∆

2
1

) reactants(products12
T
T
PP
dTCCHH
7/10/07 1- 29
found by calculating enthalpy changes along an arbitrary path. The total enthalpy change is
independent of this choice of path. The method is known as Kirchhoff’s law.
Assume that ∆H is known for a reaction at a temperature T and the ∆H at another temperature,
1
T , is to be found. Starting with the hot reactants at T (Figure 4), the reaction could be carried
2 2
out isothermally to obtain products at the same temperature. An alternative path would be to
cool the reactants to the temperature T , carry out the reaction isothermally at T , and warm the
1 1
products to T . The heat of reaction at T is already known and if the heat capacities at constant
2 1
pressure are known, the enthalpy changes can be calculated for the processes of cooling reactants
and warming products. This path must give the same ∆H as the isothermal reaction at T .
2

or,
because interchanging limits of an integral will change the sign,

If the difference in heat capacities is independent of temperature, this may be rewritten in the form
∆H = ∆H + [C - C ](T - T ) (18)
2 1 P 2 1

p (products) (reactants)
For example, given that the heat of reaction for rhombic sulfur burning in oxygen to yield
sulfur dioxide gas is - 296.9 kJ/mol at 25 C (298 K), find ∆H at 95 C (368 K). The heat
o o
capacities are given in Table 2. Insertion of the numerical values into equation 18 gives

Table 2 HEAT CAPACITIES '
Average values (in J/mol-K) for temperature ranges indicated
Compound C Temperature, C
P
o
He 20.8 -200 up
H 28.8 25 to 200
2
O 29.4 25 to 200
2
H O(g)* 36.4 25 to 200
2
SO (g) 41.9 25 to 200
2
S (r) 23.7 25 to 200
S (m) 25.9 95 to 120
*For rough calculations it is sufficient to set
C (steam) = C (ice) = ½ C (liq H O).
P P P 2
This property associated with state functions has sometimes been confused with a
17
conservation principle. Enthalpy is not conserved.
7/10/07 1- 30
∆H = - 296,900 J/mol + (41.9- 29.4 - 23.7) x 70 J/mol

368
= - 296.1 J/mol
Sometimes there will be a phase transition during the warming or cooling process. Sulfur has
a phase change at 95 C, at which point rhombic sulfur goes to monoclinic sulfur; the monoclinic
o
sulfur melts at 119 C. The enthalpy changes are 11.78 and 39.24 kJ/mol. The heat of reaction
o
for liquid sulfur burning in oxygen to form SO at 119 C (392 K) can be calculated as follows (see
2
o
Figure 5).

∆H = - ∆H - C (m) (119 - 95) - ∆H - C (r) (95 - 25) - C (O ) (119 - 25)
392 fusion P tr P P 2
+ ∆H + C (SO ) (119 - 25)
298 P 2
∆H = - 39,240 - 25.9 x 24 - 11,780 - 23.7 x 70 - 29.4 x 94 - 296,900 + 41.9 x 94 J/mol
392
= - 349.0 kJ/mol

Note that temperature differences can be found without conversion to the Kelvin scale.
Both Hess’s law and Kirchhoff’s law are simply applications of the principle that changes in
a state function, such as the enthalpy, are completely determined by the initial and final states.
17
This principle is combined with the equation arising from the first law that shows that i f the
pressure is constant, the enthalpy change will be equal to the heat absorbed by the system. Thus
the “heat of reaction,” by which we mean ∆H (at a particular temperature, pressure, and
reaction
concentrations of reactants and products), is only equal to the heat absorbed i f the reaction
proceeds at constant pressure (and at the specified temperature and concentrations). It is

sometimes more convenient to carry out a reaction at constant volume. Then the heat absorbed is
not equal to the “heat of reaction” (that is, to ∆H ), but it is still determinate because heat
reaction
absorbed equals the change in energy when the system follows a constant-volume path and
because ∆E is fixed by the initial and final states.
reaction
7/10/07 1- 31
The experimental determination of a heat of reaction is called calorimetry. A typical
calorimeter (Figure 6) consists of a reaction chamber, surrounded by a layer of water, enclosed
by sufficient insulation to prevent heat loss to the surrounding laboratory. The reactants, at room
temperature, are placed in the reaction chamber, the calorimeter is closed, and the reaction is
initiated by an electrically heated wire or other controlled energy source. The reaction will
normally be exothermic and the reaction chamber will therefore become quite hot, but the heat is
conducted into the surrounding water layer so that the products, and the water, reach a final
temperature only slightly above the initial temperature. The ∆E is the same as if the entire
reaction
process had occurred at the initial temperature even though the materials may have become quite
hot during the course of the reaction. The heat given off by the reaction is calculated by
observing the temperature rise of the water, using the condition that all heat given off by the
reaction must have been absorbed by the water. Small corrections are required for the change of
temperature, from the initial room temperature, of the products, and for the small amount of
energy added by the hot wire or other initiation method. Some systems may also require a
correction for changes of concentrations during the reaction.
7/10/07 1- 32
Problems
1. At room temperature the heat capacity (C ) of most solid elements (except the very light ones)
V
is about 3R. Specific heat is the heat capacity per unit mass, or the value in J/g·K. (For solids the
difference between constant volume and constant pressure conditions is small and may be
neglected here.)

a. Find the specific heat of Pb (molar mass = 207.19).
b. Find the specific heat of Cu (molar mass = 63.546).
c. Find ∆E for 25 g of Cu when it is warmed from 20 C to 35 C.
o o
2. Estimate the final temperature when 10 cm of iron (density 7.86 g/cm ) at 80 C is added to 30
3 3 o
ml of water at 20 C.
o
3. Find the work done when 3 mol of N gas at 4 atm pressure expands slowly at 25 C against a
2
o
constant pressure of 1 atm.
4. Calculate the work done when 0.25 mol of SO at 27 C and 1 atm expands reversibly and
2
o
isothermally to a final pressure of 0.20 atm. Find Q for the gas during this process.
5. Calculate the final temperature if an ice cube (25 g) at - 5 C is added to a cup of coffee (150
o
ml, or “2/3 cup”) at 90 C, neglecting heat loss to the surroundings.
o
6. The heat of vaporization of benzene, C H , is 30.72 kJ/mol and the normal boiling point is
6 6
80.1 C.
o
a. What is ∆E if the vapor is an ideal gas?
vap
b. Part of the thermal energy absorbed by a liquid as it vaporizes to a gas is required to
perform work on the atmosphere as the substance expands to a vapor. What fraction of the total
Q goes into work against the atmosphere when benzene (density about 0.88 g/cm ) vaporizes at
3

its normal boiling point?
7. Find ∆E at 25 C for the oxidation of S (rhombic) to give SO (gas) and for the oxidation of S
o
2
to give SO (liq). The heats of reaction are - 296.81 and - 441.0 kJ/mol. Assume gases are ideal.
3
8. Calculate ∆H and ∆E for the reaction, at 25 C,
o
SO (g) + ½ O 6 SO (g)
2 2 3
The heat of vaporization of SO at 25 C is 43.14 kJ/mol.
3
o
9. Manganese can be prepared by a thermite process,
3 Mn O + 8 Al 6 9 Mn + 4 Al O
3 4 2 3
The standard enthalpy of formation of Mn O is - 1387.8 kJ/mol and for Al O , - 1657.7 kJ/mol.
3 4 2 3
Find the amount of thermal energy given off by the reaction as written above, starting with the
reactants at room temperature and ending with the products at room temperature.
10. The interconversion of graphite and diamond does not occur at room temperature. Explain
how you could determine ∆H for this transition by measurements in the laboratory.
11. Calculate the heat of vaporization of water at 50 C.
o
12. Find ∆H for the hydrogenation of ethylene to produce ethane,
C H + H 6 C H
2 4 2 2 6
The torr is a unit of pressure equal to 1/760 atm. It is numerically equivalent to the unit
18
“mm of Hg” but is less awkward, especially when mercury is one of several vapors present. The

torr is named after Evangelista Torricelli, who invented the barometer in the 17 century.
th
7/10/07 2- 33
at 150 C. The standard heats of formation, at 25 C, of ethylene and ethane are 52.4 and - 84.0
o o
kJ/mol, respectively. Heat capacities (C ) of ethylene, hydrogen, and ethane within this
P
temperature interval are approximately 42.9, 28.8, and 52.5 J/mol·K. All three compounds are
gases at room temperature and above.
13. Benzene, C H , is reduced commercially with hydrogen to give cyclohexane, C H .
6 6 6 12
C H + 3 H 6 C H
6 6 2 6 12
At room temperature (25 C) the benzene and cyclohexane are liquids and the heat of reaction is
o
- 205.5 kJ/mol. Benzene boils at 80.1 C with a heat of vaporization of 47.8 kJ/mol; cyclohexane
o
boils at 80.7 C with a heat of vaporization of 33.0 kJ/mol. The average heat capacity of liquid
o
benzene is 136.0 J/mol·K, of liquid cyclohexane 154.9 J/mol·K, of benzene vapor 82.4 J/mol·K,
of cyclohexane vapor 150 J/mol·K, and of hydrogen 28.8 J/mol·K. Find ∆H for the reduction of
benzene with hydrogen at 150 C.
o
[0.44 cal/g·K; 0.41; 26.7 35.8 cal/mol·K; 6.94 cal/mol·K]Check numbers
14. Air is approximately 20% O and 80% N , by volume. Find the effective molar mass of air.
2 2
Calculate the density of air at 25 C and 1 atm in kg/m , assuming it acts as an ideal gas.
o 3
15. A gaseous compound, containing only sulfur and fluorine, is 62.7% (by weight) sulfur. At
27 C and 750 torr the density of the gas is 4.09 g/L. What is the molecular formula of the

o 18
compound? Assume the gas is ideal.
16. Find the pressure exerted by 32 g of methane in a 2 L steel bomb at 127 C, assuming the gas
o
is ideal.
17. Find the pressure exerted by 32 g of methane in a 2 L steel bomb at 127 C assuming the gas
o
obeys van der Waals’ equation. The constants may be taken as a = 2.25 L atm/mol and b =
2 2
0.0428 L/mol.
18. Find the volume occupied by 32 g of methane at a pressure of 5 atm at 27 C if the gas is
o
ideal. Calculate, with this value, the n a/V correction term in the van der Waals equation and
2 2
find, from this, an approximate value for the volume occupied by the methane if it obeys van der
Waals’ equation. Recalculate the correction term, with this better volume, and recalculate the
volume if necessary.
19. A good vacuum obtained with a mechanical pump and a mercury diffusion pump will
measure 10 torr, or better, of “noncondensable” gases (such as air), but there is 10 torr of
-5 -3
mercury vapor present unless this has been trapped out. Very good vacuum systems can give 10
-
torr.
9
a. How many molecules/cm are there at a pressure of 10 torr?
3 -3
b. How many molecules/cm are there at a pressure of 10 torr?
3 -9
c. What pressure would be required to achieve 1 molecule/cm ?
3

The first law prohibits certain perpetual-motion machines, called “perpetual-motion
1
machines of the first kind.” The more interesting attempts, called “perpetual-motion machines of
the second kind,” do satisfy the first law.
7/10/07 2- 34
2 The Second Law of Thermodynamics
During the 19 century several men, of quite different backgrounds and interests, struggled
th
with the basic problems of thermodynamics. Brilliant flashes of understanding were followed by
years of doubting, testing, and interpreting. Among the fundamental difficulties was a confusion
between two concepts. One of these was the concept of energy. The other, which is related to
the general concept of equilibrium and the direction of changes with respect to equilibrium, was
stated first, but the language was such that it was long misunderstood and therefore rejected. It is
now accepted as the second law of thermodynamics.
Very few people today who have any acquaintance with modern science would doubt the
following generalizations:
1. Perpetual-motion machines don’t work.
2. Bodies in equilibrium have the same temperature. When two bodies in contact have
different temperatures, energy flows, as heat (Q), from the warmer to the cooler body.
3. Bodies in equilibrium have the same pressure. When two bodies in contact, at the same
temperature, have different pressures, the body at the higher pressure tends to expand and
compress the body at the lower pressure.
None of these statements is required by the first law, which requires energy balance in any
1
process but does not say whether a process will actually occur. For example, both exothermic
and endothermic reactions are known that do proceed without external forcing. There is, despite
the variety in these generalizations, a similarity of pattern and intent — saying, for example,
whether a specific process will or will not occur — that suggests a common basis. The basis is
found in the postulate known as the second law of thermodynamics.
The Spread Function

We have seen that thermodynamics largely ignores the atomic structure of matter. It is quite
sufficient, for almost all purposes, to measure the macroscopic (large scale) variables such as
temperature, pressure, volume, and energy. The difficulty with this approach is that it effectively
hides from us one key property that may be called the “spread function,” S.
Consider the “thought experiment” of adding 10 grains of yellow rice to a large bag of white
rice, which will subsequently be mixed and poured and just thoroughly scrambled. Initially the
yellow grains were carefully arranged, on top in one corner. Where will they be a little later after
mixing?
There is no energy advantage for the yellow grains to move to the bottom, or back to the
top, or to any other specific location. We know that, once mixed, there is close to zero
(1)
T
Q
S
rev
=∆
If the ten yellow grains were regarded as indistinguishable, we should remove 10! = 3.6 x
2
10 possibilities that only interchange yellow grains among themselves. That would still leave
6
more than 10 possibilities, scarcely different qualitatively from 10 . The problem of overlapping
53 60
positions (two yellow grains being assigned to the same location) may safely be ignored at this
level of probability.
7/10/07 2- 35
probability that they will spontaneously reassemble. Even if the bag of rice is only 100 grains
deep, 100 grains wide, and 100 grains in the third direction, each yellow grain has 100 x 100 x
100 = 10 possible locations, which means that two grains have 10 x 10 = 10 possible locations
6 6 6 12
and the 10 grains have (10 ) = 10 possible locations. The probability they will all collect in

6 10 60
one location is not much more than 1 in 10 , which is awfully close to zero for most purposes.
60 2
Now imagine a very small number of molecules (say 10 ) to which we add 10 molecules that
6
are “excited” — perhaps they are each vibrating, or rotating, or otherwise have some extra energy
that can be transferred to a neighboring molecule. After a very short time, the extra energy will
be spread throughout the 10 molecules, so there is no significant chance the energy will ever
6
again (in the lifetime of the universe) be collected in the original 10 molecules. With a more
typical number of molecules (such as 6 x 10 ) and many ways for each to store energy, it is not at
23
all unreasonable to say the probability of “unspreading” the energy goes quickly to zero.
If we have a small steel ball (which will typically have 10 or more atoms), one way of
23
storing energy in the ball is to start it rolling. Another very large group of ways of storing energy
in the ball is to let some of the atoms begin to vibrate. The mechanical energy of rolling may be
converted by rolling friction to thermal energy of internal motions. There are so many possible
internal motion states that we don’t expect the energy to ever spontaneously reappear as
mechanical energy of rolling.
Entropy
Fortunately, there is an easy way to keep track of S, the spread function, without even having
to count molecules. Any energy passing into an object by thermal energy transfer, Q, is “gone” in
terms of having been spread around. It shows up afterward only as an increase (perhaps slight) in
the temperature of the system. So the first step is to keep track of Q, and in particular, we want
Q for a reversible path, or one that is as close as possible to equilibrium.
Then we want to know whether this amount of energy is important to the system, or whether
there is already so much energy spread around that this addition is relatively unimportant. That
average energy we will find is nicely measured by the (absolute) temperature of the system. So
we measure the increase in average energy as Q and divide by the temperature, T.

rev
Definition of Entropy Change
The name was chosen by Clausius from the Greek verb entrepo (εντρεπω), meaning “to
3
turn”, implying “change”, and because it somewhat resembles the word energy, one of the
important quantities involved in finding the number of states Entropy, S, is proportional to (or for
suitable units, equal to) the logarithm of the relative probability, i.e., to the number of states
accessible to the system. Unfortunately, it is also vaguely suggestive of enthalpy, H, which has
different dimensions, units, and meaning.
We also assume here that the process satisfies the first-law equation, ∆E = Q + W.
4
However, equations 1 and 5, and equation 6 (obtained below), are not subject to this restriction
against other forms of work.
7/10/07 2- 36
The “official” name of the spread function, S, is entropy . By the methods of statistical
3
mechanics we can show that S is a measure of the actual number of individual states over which
the energy of the system is spread. The greater the number of states (hence S), the greater the
probability associated with the overall state.
It is, of course, possible to transfer energy to our system without increasing the spread of
energy. For example, we can transfer just enough energy, in the proper form, to make the steel
ball start rolling. In a reversible process, this energy transfer is measured as work. As we saw for
ideal gas expansions and compressions, the work done on the system is a minimum (and Q is a
maximum) when the process is reversible. When the process is purely one of transferring energy
to a mechanical mode, such as rolling or moving in a straight line, the amount of thermal energy
transfer, Q = Q , is equal to zero. There is no increase in entropy.
max
An alternative way of approaching the second law is to recognize that ∆E is independent of
path, so we can write, for any process at constant temperature, between fixed end points
4

(T) ∆E = Q + W = Q + W = Q + W (2)
rev rev max min
There can be only one value of Q and one value of W , so we can represent these, also, as
max min
values “independent of path”, which for the moment we will label as ∆A (= W ) and as ∆B (=
rev
Q ) and regard A and B as state functions.
rev
(T) ∆E = ∆A + ∆B
or
E = A + B (3)
In any isothermal process in which the energy of the system changes by ∆E, some part of this
energy change, equal to ∆A, must appear as work done on the system. The second energy term,
called ∆B for the moment, cannot be lost, because of the first law (conservation of energy);
however, it may be regarded as “spilled energy”, which is lost for useful purposes during the
transfer operation because it has become spread through the system.
From equation 3, two important functions are obtained. One, temporarily given the symbol
A, is called the Helmholtz free energy, or sometimes the work function, and is characterized, as
above, by the constant temperature equation
(5)
T
q
dS
rev
=
KJ/g 22.1
K 15.273
J/g 6.333
⋅===∆
T

Q
S
rev
KJ/g 22.1
K 15.273
(ice) J/g 6.333
⋅−=
−
=∆
thermostat
S
7/10/07 2- 37
(T) ∆A = W (4)
rev
This function (which we will subsequently designate by the more current notation, F), and an
even more useful quantity derived from it, will be investigated later.
The spread function, S, is obtained from the second function, B, when we divide by the
temperature, T. That is, S = B/T or B = TS, where S is the spread function, entropy, previously
introduced. Then a small change in B breaks into two parts.
d(TS) = T dS + S dT
and at constant temperature
(T) d(TS) = T dS = q
rev
or, quite generally,
Statistical mechanics tells us something about the value of S, representing the number of
molecular states, but thermodynamics (which never quite admits the importance of molecules)
tells us only how to find changes in entropy, S.
ENTROPY, EQUILIBRIUM, AND SPONTANEOUS CHANGE. Consider the process of melting 1 g of
ice, at 0 C and 1 atm pressure, by bringing it into contact with a thermostat, or heat reservoir.
o

The thermostat might be a very large block of metal, well insulated from its surroundings and
large enough so that the reasonable amount of thermal energy added or withdrawn will not
change its temperature. For the ice,
If the thermostat is at exactly the same temperature,
Taking the ice as the system and the thermostat as the surroundings, (∆S) = 0. Thus
system + surroundings
far it might appear as if entropy also is conserved (compare equation 1, Chapter 1). But we know
that unless there is at least a small temperature differential between the thermostat and the ice, the
ice will not melt. We cannot melt an ice cube by bringing it into contact with a large block of ice;
it must be touched to something warmer. If the temperature of the thermostat is higher than
273.15 K, ∆S will be smaller in magnitude than ∆S , and ∆S for the ice plus the
thermostat ice
thermostat, or system plus surroundings, will be positive. This is the essence of the second law of
thermodynamics, which states
(8) T
S
E
V
=






∂
∂
(8a)
1
TE

S
V
=






∂
∂
7/10/07 2- 38
(∆S) > 0 (6)
system + surroundings
for any process whatever that actually occurs. The similarity, and contrast, to the statement of
the first law (equation 2, Chapter 1) should be noted. In the limit of a process that is truly
reversible (for example, thermostat and ice at the same temperature, so that the ice can melt or
refreeze by exchange of thermal energy with the thermostat), the total entropy change will
approach the value zero. Recognizing the importance of this limiting condition we may write
Second Law (∆S) $ 0 (6a)
system + surroundings
with the understanding that the equality can be approached as closely as we wish, if we have
sufficient patience.
From equation 6 or 6a, taken as the second basic postulate, it is possible to derive results that
are well established from our experience. There are, in general, many different paths that a
system may follow between any two specified states, of which at least one (actually, an infinite
number) will be thermodynamically reversible. Let q and w be the amount of thermal energy
i i
transferred and the amount of work done along the i path, or more generally, q , q , ··· q are
th

1 2 rev
the various amounts of thermal energy transferred and w , w , ··· w the various amounts of
1 2 rev
work done along these paths. Then, in general, q …q … ··· …q and w …w … ··· …w . But dE
1 2 rev 1 2 rev
is the same for each of these paths. That is, regardless of which path the system actually follows,

dE = q + w = q + w = ··· = q + w
1 1 2 2 rev rev
and, because q = T dS and w = - P dV (if the only work is work of expansion or compression),
rev rev
it follows that, for any path,
dE = T dS - P dV (7)
This equation is valid for any process occurring in a system of constant composition. (If there is a
change of chemical composition, as would be required if electrical work were produced, or a
change of concentrations, additional terms should be added to the first-law equation and to
equation 7.) At constant volume the second term of equation 7 becomes zero and we obtain
or
(10)
1
2
2
1
1
1
1
1
2
1
1

1
1
1
dE
dE
dS
dE
dE
dS
dE
dE
dS
dE
dE
dS
dE
dE
dS
dS
−=
+==
(11) 0
11
1
21
≥









−= dE
TT
dS
dV
T
P
dE
T
dS +=
1
(12)
T
P
V
S
E
=






∂
∂
7/10/07 2- 39

Take an arbitrary system, at constant volume, that
is thermally isolated from its surroundings (so that Q =
0 and ∆E = 0). The second law then requires that dS $
0 for any changes within this system. Divide the system
into two parts in any fashion (Figure 1). For example,
the total system might be a cup of water, which is
considered to be divided into two equal parts, or it
might be a bar of lead, with one cm in the center
3
considered as one part and the remainder as the second
part. Let E be the energy of the first part and E the
1 2
energy of the second part, and consider a flow of
energy (as thermal energy transfer, Q) from one part to
the other, holding the two sub-system volumes
constant. Then
dE = dE + dE = 0 (9)
1 2
dE = - dE
1 2
The total entropy change, dS, arising from a flow of energy between the two parts, is
(E, V)
Substitution of equation 8a and application of the second law gives
The necessary condition for equilibrium is that dS = 0 for the process in which q = dE flows
1 1
between the two parts. This can be satisfied only if T = T . This proves that bodies in
1 2
equilibrium have the same temperature. (This is sometimes called the zeroth law of
thermodynamics, for historical reasons, although it is a consequence of the second law.)
Suppose that T > T . Then (1/T - 1/T ) is negative and dE must also be negative, showing

1 2 1 2 1
that part one loses energy to part two. This proves that when two bodies in contact have
different temperatures, energy flows, as a thermal energy flow (Q), from the warmer to the cooler
body.
From equation 7 we can derive a relationship for the dependence of entropy on volume.
and therefore