15

The Third Law of

Thermodynamics
The First and Second laws of
entropy. The third law, however, does thermodynamics have led to new concepts of energy conter
Va
not lead
of the entropy of a crystalline solid. Some to any new concept. It only places a limitation on and
the
its applications do lead to conclusions which are scientists hesitate to call it a 'law' at all. All the same,
borne out by experience and hence, in accorVance with
scientific terminology, it is a law, like the other two laws of
(The Nernst Heat Theorem. ) Before passing on to thermodynamics.
consider briefly the Nernst heat theorem. This is an old the Third law of thermodynamics, we may
forerunner of the third law of thermodynamics. generalisation but still has relevance as the
From the Gibbs-Helmholtz equation, viz..
AG- AH= T(84GVÝ)P ) ...(1)
where AG is the change in free energy and AH is the change in
a chemical reaction, it is seen that at the enthalpy accompanying any process including
absolute zero (i.e., T=0), AG=AH.
Richards, by measuring EMFs of cells at different
found that the value of 8AG/OT decreases with
temperatures,
and, therefore, concluded that AG and AH decrease in temperature
more and more closely as the temperature is tend
to approach each other
lowered. Nernst, relying on
this data, made an important suggestion that the value
zero gradually as the temperature is lowered towards of (AG)/oT approaches AG
This is known as the Nernst heat the absolute zero. or

and AH not only become equal to theoremAccording this theorem, AG AH

to
each other at absolute zerO but also
approach each other asymptotically, that is, gradually
close to absotute zero)This is illustrated in Fig. 1. Thus, atthetemperatures
AG and AH towards each other is as represented by the full lineapproach of
and not as
represented by the dotted line. TEMPERATURE (K)
KFig. 1. Variation of AG and AH
Mathematically, the theorem may be expressed as with temperature.
u [aAGyTIp = T’0u [8A)iÝT ]p = 0
T’0 ...(2)
where L stands for the limiting value.
We also know fromn the Second law of thermodynamics that
(8AGJ8)p = - AS ...(3)
and (aAH)VOTDp = ACp (Kirchhoff equation) ...(4)
where AS is the entropy change of the reaction and ACp is the difference in the heat capacities of the
products and the reactants.
It follows from Eqs. 2, 3 and 4 that
Lt AS =0 ...(5)
T’0

553
 554 THE THIRD LAW OF THERMODYNAMICS

and Lt ACp = 0 ...(6)

T’0

The significance of these equations is that the entropy change of'a reaction tends to approach zero
and that the difference between the heat capacities of products and rèactants also tends toapproach zero
as the temperature is lowered towards the absolute zero.
The Nerst theoremn holds good only in the case of pure solids...
As
(The Third Law of Thermodynamics. According to Eq. 6, ACp-tends to approach zero at 0 K.. This
means that at absolute zero, the heat capacities of products and feáctants in solid state are identical. This
leads to the suggestion that at absolute zero, all substances have the same heat capacity. The quantum
theory, as applied to heat capacities of solids, has shown that heat capacities of solids tend to become
zero at 0 K. The Nernst heat theorem may, therefore, be written as
Lt Cp =)
T’0
(7)

According to Eq. 5, AS becomes zero at absolute zero, i.e., the entropy change of a process
involving solids becomes zerO at 0 K. In other words, the absolute entropies of products and reactants in
the solid state are identical. Planck, therefore, suggested that entropies of all pure solids approach zero
at 0 K, i. e.,
stalemeut Lt S =0 ...(8)
T’0

This statement has led to the followingenunciation of the Third law of thermodynamics :
(At the absolute zero of temperature, the entropy of every substance may become zero and it does
become zero in the case of aperfectly crystalline solid.
In a perfect crystal, at absolute zero temperature, there is astate of perfect order, i.e., zero disorder
and hence of zero entropyy Walther Nernst (1864-1941), the German chemist, was awarded the 1920
Chemistry Nobel Prize for his work in thermochenmistry.
Nernst's work on heat theorem and electrochemistry has had a very great impact on the physical
sciences. He was one of the great pioneers of physical chemistry. In a lecture he delivered at Oxford in
1937, he said that it had taken three people to formulate the First lawof thermodynamics, two for the
Second law, but that he had been obliged to do the Third law allby himself. He added that it followed by
extrapolation that there could never be a Fourth law.
Determination of Absolute Entropies of Solids, Liquids and Gases. We know that for an infinitesimally
small change of state of a substance or a system, the entropy change is given by
dS = dg/T ...(9)
If the change takes place at constant pressure, then
(OS)p = (09)p/T ...(10)
(OS/YT)p = (0ql0T)pX 1/T ...(11)
By definition, (oql0T)p = Cp ...(12)
(OS/8T)p =CpX1/T ...(13)
or at constant pressure, ds = (Cp/T)dT ...(14)
For a perfectly crystalline substance, the absolute entropy S=0at T=0. Therefore, we may write :
S=S T=T
dS =
Js-0 JT=0 ...(15)

...(16)
where S is the absolute entropy of the crystalline solid under examination at the temperature T
 THE THIRD LAW OF THERMODYNAMICS 555

The integral u e4- 16 1s

desired temperature Tand thenevaluated
plotting by measuring
Cp against Cp at determining
In Tand various temperatures between
the area under T=0between
the curve and the
T=0andthe required temperature T. This area givess directly
S.
the valueof
Cince it is not possible to obtain the value of Cp at the
absolute zero, heat capacities are measured up to as low a
temperature as possible, usually up to 15K and the
value at
the absolute zero is obtained by extrapolation. This method.,
therefore, consists in determining heat capacities of the Cp
substance under examination at
temperatures
approximately 15 Kto the required temperaturevarying from Area
T. Agraph
fC VS In T is then plotted and extrapolated to the absolute
ero of temperature, as shown in Fig. 2. The area under
the graph gives the required value of ST, the absolute ln T
entropy of the substance at temperature T.
Fig. 2. Determination of absolute entropy.
Eq. 16. is thus written as follows :

S, = ...(17)
where 0< T <15 K.
The first integral is evaluated with the help of the Debye theory of heat capacities of crystalline
substances according to which, at very low temperatures (0 <T< 15K),
Cp z Cy aT3 ...(18)
where a is an empirical constant. Eq. 18 is known as the Debye T³ law.
Accordingly, Eq. 17 may be writen as
S = JO ar + ..(19)

The second integral in Eq. 17 is evaluated from experimental measurements of heat capacities.
ombining the heat capacity data with the enthalpy data on phase transformations, the absolute entropy of a
uDSance, whether solid, liquid or gas, at temperature T, can be determined, as illustrated below. In every
ese, the start is made with the substance in the crystalline solid state at the absolute zero when its absolute
opy is taken as zero, Then the total absolute entropy of the substance in the given state and at a given
erature is taken as the sum of all the entropy changes that the substance has to undergo in order to
qure the given state at the given temperature starting from the crystalline solid at absolute zero.
OuPpose, it is required to determine absolute entropy of a gas at 25°C under atmospheric pressure.
This would be equal to the sum of the entropy changes involved in the following processes each of which
0is brought about reversibly. It is assumed for general discussion that the substance in solid state exists in
WO allotropic forms a. and B.
ntaing the crystalline solid absolute zero to temperature T" where 0<T<15K, and evaluating
fromDebye's
eentropy Change with the aid of the theory. Let the entropy change be AS;.Then

AS, =
Jo
[ars -a(r"s ...(20)
2. Heating the crystalline solid form T to T,r where T, is the transition temperature at which the
Ctystaline solid changes from allotropic form ato allotropic form ß. The entropy change in this process
556 THE THIRD LAW OF THERMODYNAMICS

is given by

AS, -+Cps(a)dT ...(21)

where Cps (a) is the heat capacity of the solid in allotropic form a. AS, is evaluated by the integration of
above.
heat capacity data graphically, as described
3. Transition of the solid from allotropic forma to allotropic form Bat the transition temperature
T,. The entropy change in this process is given by
AS = AH, /Tr ..(22)
where AH, is the molar enthalpy of transition.
4. Heating the solid in allotropic form B up to its fusion point, Tius- The entropy change in this
process is given by

AS =
eus Cps(B) dT ...23)

where Cps (B) is the heat capacity of the solid in allotropic form p.
5. Changing the solid in allotropic form B into the liquid state at the fusion temperature TAus. The
entropy change of this process (entropy of fusion) is given by
ASs = AHfus Ifus ...(24)
where AHns is the molar enthalpy of fusion of the substance.
6. Heating the liquid from its freezing point (Tus) to its boiling point (,). The entropy change
involved in this case is given by

AS6 = Cp,l din T ..(25)

where Cp is the heat capacity of the substance in the liquid state. This can be evaluated by plotting Cpi
vS In Tbetween temperatures Ts and T, and noting the area below the graph, as described before.
7. Changing the liquid into the gaseous state at the temperature T,. The entropy change involved
here, AS, is the molar entropy of vaporisation and is given by
AS, = AHyapTb ...26)
where AH, is the enthalpy of vaporisation per mole of the substance.
8. Heatingthe gas from T, to the required temperature, i.e., 25°C (298-15 K). The entropy change
involved in this process is given by
298-15
CPpg d In T ..27)
JT,
where Cp, is the heat capacity of the substance in thè gaseous state at constant pressure. ASg is evaluated
by ploting Cpo vs In Tbetween temperatures T, and 298-15 Kand noting the area below the curve.
The absolute entropy of the gas at 298·15 K (25°C), ST, is equalto the sum of all the entropy changes
listed above. Thus,
ST = AS; t AS t AS; + AS t ASs + AS% + AS, + AS
 THE THIRD LAW OF THERMODYNAMICS
Table 1 gives the absolute entropy of HCl at 25°C determined by the procedure outlined a!u.
TABLE 1
Determination of Absolute Entropy of HCI at 25°C
Contribution JK-! moH
1. Extrapolation from 0to 15 K[using the Debye 7° law] 1-3
Cp d hT for solid a from 15 to 98-36 K
29:5
Transition, solid a ’ solid B, 1190/98-36
12-1
A [Cp dn Tfor solid ß from 98-36 to 158-91 K 21-1
Fusion, 1992/158-91
5. 12-6
6 (Cp din Tfor liquid from 158-91 to 188-07 K 9-9
7. Vaporization, 16150/188-07 85-9
8 ( Cp din Tfrom 188-07 to 298-15 K 13-5
Sh98-15 = 185-9
Absolute entropies of some of the elements and compounds in their standard states at 25°C,
calculated from the Third law of thermodynamics, are given in Table 2.
TABLE 2
Standard Absolute Entropies (S) of Elenents and Compounds at 25° C
Substance Absolute entropy Substance Absolute entrupy
(JK-' mol-1 J K-' mo-)
Hydrogen (g) 130-60 Mercury () 77-40
Nitrogen (g) 191-62 Mercury (g) 174-83
Oxygen (g) 205-01 Mercuric chloride (s) 144-76
Hydrogen chloride (e) 186-22 Mercurous chloride (s) 98-32
Hydrogen bromide (e) 199-15 Cuprous iodide (s) 6-65
Hydrogen iodide (g) 206-27 Lead bromide (s) 161-50
Carbon (diarnond) 2-43 Silver bromide (s) 96-11
Carbon (graphite) 5-69 Silver chloride (s) 107-15
Water () 70-29 Silver iodide (s) 115-57
Water (g) 188-74 Silver oxide (s) 121-75
Ammonia (g) 192-46 Ferric oxide (s) 89-95
Carbon monoxide (e) 197-90 Cupric oxide (s) 43-55
Carbon dioxide (e) 213-80 Magnesium oxide (S) 27-00
Nitric oxide (g) 210-45 Mercuric oxide (s) 71·46
Sulphur dioxide (g) 247-86 Sodium chloride (s) 72:38
Sodium (s) 51-04 Potassium chloride (s) 82-62
Magnesium () 32-51 Potassium bromide (s) 93·72
Sulphur (rhombic) 31-88 Methane (g) 18614
Sulphur (monoclinic) 32-55 Ethane (e) 229-49
Chlorine (g) 222-96 Ethylene (g) 185-35

Bromine (g)
Bromine ()
245-34
153·97
Acetylene (g)
Methanol ()
201·10
126-69
160-66
Aluminium (s) 28-32 Ethanol ()
Iron ($) 27-15 Benzene () 172-79
Copper (s) 83-34 Phenol (0) 142-24
Silver (s) 42-67 Acetone () 200-03
lodine (s) 116-73 Acetic acid () 159:82

Lodine (g)
Zinc (s)
260-62
41-00
Methyl chloride (g)
Methyl chloride ()
234-05
245-22
Lead (s) 64-85 Ether (/) 85-27
558
THE THIRD LAW OF THERMODYNAMICS

Example 1. Show that the entropy of any substance at very low temperatures (0 < T< 20K), where Debye's
relation for heat capacities of crystals is valid, is one-third of the molar heat capacity.
Solution: At low temperatures (0 < T< 20 K),
(Debye T law]
According to Eq. 14,

Sp- S = =a[ raT - r

Sp = aT/3 [: S6 = 0j
Evidently, Sp= Cp3. Hence, proved.
Example 2. C for uranium metal is 3-04 J K-l mol-! at 20 K. Calculate the absolute entropy of the metal in
JK mo- at 20K.
Solution : At lowtemperatures (0 < T< 20 K), Cp= Cy= aT [Debye's T law]
a= C/T³ = 3-04 J K- mol-'/(20 K)³ = 38-03 x10-5JK4 mol-!
Hence, Cp = aT = (38-03 X10-5 J mo-! K) T³
From Eq. 14, ` = (CJTaT = 38-03 x 10-5 J mol-! K4 T2 aT
O So-S = aT/3 =38-03 ×10-5 J K mo-! (20 K)³/3
O So = 1-01 J K- mot- (: S= 0]
Example 3. The heat capacity, Cp (in J K' mot) of asubstance is given by the following equations :
CAs) = 16-74x10-5 T°(0 < T< 50 K)
Cls) = 20-92 (50 <T< 150 K)
Cp) = 25-10 (150 <T<400 K)
At the melting point (150K), AH, = 1255-2 J mo'. Caleulate the absolute entropy of the substance in the liquid
state at 300 K.

Solution :
= 16-74 Tx10-TAT (The Debye T³ law)
JoT
16:-74 x10-5T3 0 = 558x10x(50 = 6-97 J K- mol-!
150 C,
AS =

J50
aT - 20-92) K-Imol-! In3O
50

- 20-92JK- mol- x1-09861 = 22-98 J K- mot-!

AS, = AS, =1255-2 Jmol-/150 K= 8:37 JK-' mol-!
=25-10JK-! mol-! /n/300
150
= 25-10 JK- mol-l x0-6931 = 17-40 JK-l mol-!
Hence, AS = Sf-S= AS + AS + AS; + AS
S, = 697 + 22-98 + 8-37 + 17-40 = 55-73 J K- mo (:: So = 0)
Experimental Verification of the Third Law of Thermodynamics
The heat capacity and enthalpy data on substances that exist in two different crystalline forms may
serve to verify the Thid law of thermodynamics. For reversible isothermal transition, a -’ B. we can write
AS = Sp-S=AH,/T,, ..(28)
562 THE THIRD LAW OF THERMODYNAMICS

AS,
(373-1 75-2JK-'mol-! -dT
J273-1 T
= 752J K molx2-303 log (373-1 K/273-1 K) = 23-5 J K mol
AS = AH/T, = 40600 J mol-/373-1 K = 108-8 J K- mol-!
Adding, we get AS = 22-0 + 23· + 108-8 = 154-3 JK- mol-1

Derivation of the Boltzmann Entropy Equation. The Boltzmann entropy equation, viz., S = kin W.
isprobably the most famous equation in statistical thermodynamics. Its derivation is intuitively simple
and appealing. Before we derive it, we must distinguish between two kinds of probabilities, viz.
mathematical probability and thermodynamic probability.
Mathematical probability is the ratio of the number of cases favourable to the occurrence of an
event to the total number of equally probable cases. This probability always lies between 0 and 1. On the
other hand, thermodynamic probability is the number of microstates corresponding to a given macrostate
when we are dealing with the distribution of molecules amongst an extremely large number of energy
levels. This probability, W, is a very large number, tending to be infinite.
Boltzmann suggested that entropy can be related to the thermodynamic probability Was
S =fW) ...51)
Consider two systems A and B whose entropies and microstates (also called complexions) are SA, SR.
W and WB, respectively. When the two systems are combined,
S= SA t S ...52)
By definition, the probability of a thermodynamic state is proportional to the number of complexions
(microstates) required to achieve it. Mathemåically, we know that the total probability is the product of
the probabilities of the independent events. This is also true of the complexions. Hence,
W = W. WB ...53)
and S4 + Sg =SW,WB) .54)
This relationship suggests that entropies are additive and probabilities are multiplicative. This can be
true only if S is a logarithmic function of W, i.e.,
S =k In W .(55)
which is the Boltzmann entropy equation. Eq. 55 gives the quantitative definition of entropy as disorder.
WResidual Entropy. Entropies calculated using the Third law are called thermal entropies. However,
the statistical entropies, calculated by the method of statistical mechanics (for instance, by using the
Boltzmann entropy equation at 0 K, viz., S= kIn W) are more rigorous. It is found that the thermal
entropies are somewhat smaller than the statistical entropies, the deviation ranging from 3-1 to
4-8 JK-l mol-!, We thus conclude that entropies of substance (such as H2, D), CO, N0, N,O, H,0, etc.)
are not zero at 0 K, as the Third law formulates, but are finite. These entropies are called residual
entropies. The existence of residual entropy in a crystal at 0 K is presumably due to the alternative
arrangements of molecules in the solid. Such arrangements are shown in Fig. 3 for CO and N,O :
CO
CO
CO
CO
CO
CO
CO
CO
88
CO CO
0C |88 OC
CO

NNO NNO NNO NNO NNÌ NNO ONN ONN

NNO NNO NNO NNO NNO ONN ONN NNO
(a) (b)
Eig. 3. Alternative molecular arrangements (a) Perfect crystal (b) Actual crystal
 THE THIRD LAW OF THERMODYNAMICS 563

Since both the arrangements are equally likely, from the Boltzmann entropy equation (55).
S = k ln W, with W= 2^A (where N, is Avogadro's
number), we find that
s = k In 24 = N, k In 2 = R In 2
=(8:314 J K- mo-l)0-693 = 5-76 J K- mol
Since the residual entropies are found experimentally to be less than this value, it is evident that the
oun alternative orientations of the CO and N,O molecules in the solid state at 0 K are
not
andom. For Hz and D2, to0, the thermal entropies at 0 K are less than the correspondingcompletely
entropies. The calculation of the statistical entropy assumes that there exists an equilibriumstatistical
between
ortho H,and para H) at all temperatures. The ASmix of ortho H, and para H, is found to be 18-37
mol-l in the vicinity of 0K. When this value is added to the thermal entropy (calculated fromJ K-l
canacity measurements), the agreement with the statistical entropy is very good. heat
Ludwig Boltzmann (1844-1906), the great German mathematical physicist, made outstanding contributions to the kinetic
theoryof gases, statistical mechanics, and the atomic nature of matter. However, his work on the atomic theory was criticized
hy several eminent scientists (among them W. Ostwald), and he suffered from severe depression. He committed suicide in 1906
by drowning. It was a tragic loss to physics. After his death his work on atomic theory was corroborated. On his tombstone is
inscribed his famous equation :S= k In W.

QUESTIONS AND PROBLEMS

LReview Questions
1. Explain the Nernst heat theorem. How does it lead to the enunciation of the third law of thermody namics?
2., State and explain the third law of thermodynamics. How can it be verified experimentally ?
3. Explain how the absolute entropy of a substance is determined with the help of the third law of thermodynamics.
4. Show that the entropy of any substance at low temperatures when the Debye T law is valid for the molar heat capacity, is
one-third of the molar heat capacity.

II. Problems
Calculate the entropy of SO, (2) at 400 K from the following data on contribution to entropy :
0) 0to 15 K(using Debye's law) = 1255 J K- mol
b) 15 Kto 197-64 K, solid = 84-182 J K- mol-!
(c) 197-64 K to 263-08 K, liguid = 24-937 J K-l mol
) AH, andAH, are 7-402 kJ mol- and 24-937 kJ mol-', respectively.
(e) The mclting point is 197-64K and the boiling point is 263-08 K.
) The temperature-dependence of C, of SO2 (8) is given by
G=49-769 + 4-56 X10- T (Ans. 264-09 J K- mot'
2.
Calculate with the help of entropy values given in Table 2,
Whe sandard entropy change for the reaction Pb(s) + Br2() ’ PbBrz(s)
ü) the standard entropy change for the formation of methyl alcohol. (Ans. () - S7-32 J K-1 ) - 242-7J K-' mot)
3, Write down the relation between AG° and AS for a chemical reaction. Calculate the free energy change of the reaction
Cotg) + ;0; (8) ’ CO(g): AH° = - 282-84 kJ;with the help of entropy values given in Table 2
[Ans. - 257-03 kJ]