Cap 7 relaciones termodinamica

7
Thermodynamic Relations
7.1. General aspects. 7.2. Fundamen tals of partial differen tiation. 7.3. Some general thermodynam ic
relations. 7.4. Entropy equation s (Tds equation s). 7.5. Equation s for internal energy and enthalp y .
7.6. Measurable quan tities : Equation of state, co-efficien t of expansion and compressibil i ty ,
speci fic heats, Joule-Thomso n co-efficien t 7.7. Clausiu s-Clap eryo n equatio n—Hig hligh t s —
Objectiv e Type Questio n s—Exercises.
7.1. GENERAL ASPECTS
In this chapter, some important thermodynamic relations are deduced ; principally those
which are useful when tables of properties are to be compiled from limited experimental data, those
which may be used when calculating the work and heat transfers associated with processes under-gone
by a liquid or solid. It should be noted that the relations only apply to a substance in the solid
phase when the stress, i.e. the pressure, is uniform in all directions ; if it is not, a single value for
the pressure cannot be alloted to the system as a whole.
Eight properties of a system, namely pressure (p), volume (v), temperature (T), internal
energy (u), enthalpy (h), entropy (s), Helmholtz function (f) and Gibbs function (g) have been
introduced in the previous chapters. h, f and g are sometimes referred to as thermodynami c
potentials. Both f and g are useful when considering chemical reactions, and the former is of
fundamental importance in statistical thermodynamics. The Gibbs function is also useful when
considering processes involving a change of phase.
Of the above eight properties only the first three, i.e., p, v and T are directly measurable.
We shall find it convenient to introduce other combination of properties which are relatively easily
measurable and which, together with measurements of p, v and T, enable the values of the
remaining properties to be determined. These combinations of properties might be called ‘thermo-dynamic
gradients’ ; they are all defined as the rate of change of one property with another while
a third is kept constant.
7.2. FUNDAMENTALS OF PARTIAL DIFFERENTIATION
Let three variables are represented by x, y and z. Their functional relationship may be
expressed in the following forms :
f(x, y, z) = 0
x = x(y, z)
y = y(x, z)
z = z(x, y)
Let x is a function of two independent variables y and z
x = x(y, z)
...(i)
...(ii)
...(iii)
...(iv)
...(7.1)
341

342 ENGINEERIN G THERMOD Y N AM ICS
Then the differential of the dependent variable x is given by Fx I F x I
y
z
z y
...(7.2)
where dx is called an exact differential.
F x I F I
H K
If G y J
z
= M and H z K y
= N
Then dx = Mdy + Ndz ...(7.3)
Partial differentiation of M and N with respect to z and y, respectively, gives
M 2 x
z yz
M N
or
z y
and
N 2 x
y zy
...(7.4)
dx is a perfect differential when eqn. (7.4) is satisfied for any function x.
Similarly if y = y(x, z) and z = z(x, y) ...(7.5)
then from these two relations, we have F y I F y dy = I
x z
dx +
z x
dz
z z
dz = x y
dx + y
x
dy
dy = G x
I
z
dx + G z
I
F
x
x N
I
y
F
y
dx 
I
x
dy
Q
F y I F y I F z I F y I F z I
L O M P M P H K G J H G K J
= N x
z
z
x
x
y Q dx + z x y x
dy
F y I F y I F z I
L O M P M P
= N x
z
z
x
x
y Q dx + dy
y y z
 F F
H H G G
I I
K K J J
F
H G
I
K J
or x z x = 0
F yI F z I F y I
or z x x y
= –
x
x F z I F y I
or y
z
x y z x
= – 1
In terms of p, v and T, the following relation holds good
FpI FT I F v I
v T p
v
T p
= – 1
...(7.6)
...(7.7)
...(7.8)
...(7.9)
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M-thermTh7-1.pm 5
dx = H K G J dy  H K G J dz

H G K J G H J
K H G F
J I
G F I
K H
J
K L O
 H H G K G K J J H G K J
 H H G K G K J J H G K J
z x y
z
F
H G
I
K J H K H K G J G J
H K G J H K G J H K G J
x G J


F y F y
H K H K J J
z z
H K G J H G K M J P M P
H K H K G J G J H G K J

THERMO D Y N A M IC RELATION S 343
7.3. SOME GENERAL THERMODYNAMIC RELATIONS
The first law applied to a closed system undergoing a reversible process states that
dQ = du + pdv
According to second law,
dQ
ds =
rev.
Combining these equations, we get
Tds = du + pdv
or du = Tds – pdv
The properties h, f and g may also be put in terms of T, s, p and v as follows :
dh = du + pdv + vdp = Tds + vdp
Helmholtz free energy function,
df = du – Tds – sdT
= – pdv – sdT
Gibb’s free energy function,
dg = dh – Tds – sdT = vdp – sdT
Each of these equations is a result of the two laws of thermodynamics.
Since du, dh, df and dg are the exact differentials, we can express them as F uI F uI du =
s v
ds +
v s
dv,
h h
dh = s p
ds + p dp,
F f I F f I
df =
v T
dv +
T v
dT,
g
dg = G pJ
T
g
dp + G T J
p
dT.
...(7.10)
...(7.11)
...(7.12)
...(7.13)
Comparing these equations with (7.10) to (7.13) we may equate the corresponding co-efficients.
For example, from the two equations for du, we have F uI F uI
s v
= T and
v s
= – p
The complete group of such relations may be summarised as follows : F uI F hI
s
= T = s
F uI F f I
= – p =
v s
v T
g
p
s
= v = p
T
f
H T Kv
= – s = G T J
p
...(7.14)
...(7.15)
...(7.16)
...(7.17)
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M-thermTh7-1.pm 5
T
F
H G
I
K J
HG J K HG J
K G
F
J I
H K H
F I
K G J
H H G K G K J J
F I F I
H H G K G K J J
H G K J
v H K G J
p
g
F I
F I F I
s
H K
H K
H G K J G H J
K H G
F
h
I
J
K G H K J
G J
H K

F T I F pI
Also,
HG K − HG K J J
H K s H K
v s
F T I F vI
p s p
F p I F sI
T v v T
v s
T p p
T
...(7.18)
...(7.19)
...(7.20)
...(7.21)
The equations (7.18) to (7.21) are known as Maxwell relations.
It must be emphasised that eqns. (7.14) to (7.21) do not refer to a process, but simply expres s
relations between properties which must be satisfied when any system is in a state of equilibrium.
Each partial differential co-efficient can itself be regarded as a property of state. The state may be
defined by a point on a three dimensional surface, the surface representing all possible states of
stable equilibrium.
7.4. ENTROPY EQUATIONS (Tds Equations)
Since entropy may be expressed as a function of any other two properties, e.g. temperature
T and specific volume v,
s = f(T, v) F s I F s I i.e., ds =
T v
dT +
v T
dv
s s
or Tds = T
T v
dT + T
v T
dv ...(7.22)
But for a reversible constant volume change
dq = cv (dT)v = T(ds)v
s
or cv = T
T
F s I F p I
But,
v T
=
T v
...(7.23)
[Maxwell’s eqn. (7.20)]
Hence, substituting in eqn. (7.22), we get
Tds = cv
dT + T H T Kv
dv ...(7.24)
This is known as the first form of entropy equation or the first Tds equation.
Similarly, writing
s = f(T, p)
Tds = T H T K p
dT + T H pKT
dp ...(7.25)
dharm
M-thermTh7-1.pm 5
s v
F
H G I
K J − F
H G
I
K J
v
F
H
I
K
G J  G J
H  H G K G K J J
H H G K G K J J
HF I HF I G K G K J J
G J
H H G K G K J J
F p G
I J
F G s J I G F s
J
I

where cp = T H T K p
s v G J
Also G pJ
T
= – H T K p
...(7.26)
[Maxwell’s eqn. (7.21)]
whence, substituting in eqn. (7.25)
Tds = cpdT – T H T K p
dp ...(7.27)
This is known as the second form of entropy equation or the second Tds equation.
7.5. EQUATIONS FOR INTERNAL ENERGY AND ENTHALPY
(i) Let
F uI
To evaluate
v T
Then
or
But
Hence
u = f(T, v) F uI F uI F u du = I
T v
dT +
v T
dv = cv dT +
v T
dv
let u = f (s, v)
F uI F uI
du =
s
ds +
v
dv
F uI FuI F s I F uI
=
v
s v v
F uI F s I F s I F uI
= T,
=
s v v
T
T v v s
= – p
F uI F p I
= T
v T
T v
– p
...(7.28)
...(7.29)
This is sometimes called the energy equation.
From equation (7.28), we get
| |
| |
du = cvdT + TT F p I
v
− pW dv
(ii) To evaluate dh we can follow similar steps as under
h = f(T, p)
dh = F h I dT + G hI
dp
= cpdT + H p
I
T
F
H K J
dp
T
...(7.30)
...(7.31)
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M-thermTh7-1.pm 5
F I F I
H H H G K G K G K J J J
v s H H G K G K J J
T v T s H K H K H K H K
H H H H G K G K G K G K J J J , J
H H G K G K J J
H  T K S G J V
T p H K G J
p
GF s IJ
H K
F v I G J
H G K J
G J G J G J  G J
R U
F h G K J

346
FhI
To find p
T
;
Then,
But
ENGINEERIN G THERMOD Y N AM ICS
let h = f(s, p)
F hI FhI
dh = s ds + p dp
FhI FhI F s I F hI
p
T
= s p p
T
+ p
s
F hI = T, G s J = – G vJ , G hJ = v
p T p s
F I
K
Hence G pJ
T
= v – T H T K p
...(7.32)
From eqn. (7.31), we get
dh = c
p
R U S V T| T W|
dT + |v −T H
v I
p
| dp ...(7.33)
7.6. MEASURABLE QUANTITIES
Out of eight thermodynamic properties, as earlier stated, only p, v and T are directly
measurable. Let us now examine the information that can be obtained from measurements of
these primary properties, and then see what other easily measurable quantities can be introduced.
The following will be discussed :
(i) Equation of state
(ii) Co-efficient of expansion and compressibility
(iii) Specific heats
(iv) Joule-Thomson co-efficient.
7.6.1. Equation of State
Let us imagine a series of experiments in which the volume of a substance is measured over
a range of temperatures while the pressure is maintained constant, this being repeated for various
pressures. The results might be represented graphically by a three-dimens ional surface, or by a
family of constant pressure lines on a v-T diagram. It is useful if an equation can be found to
express the relation between p, v and T, and this can always be done over a limited range of states.
No single equation will hold for all phases of a substance, and usually more than one equation is
required even in one phase if the accuracy of the equation is to match that of the experimental
results. Equations relating p, v and T are called equations of state or characteristic equations.
Accurate equations of state are usually complicated, a typical form being
pv = A +
B

C
+ ......
where A, B, C, ...... are functions of temperature which differ for different substances.
An equation of state of a particular substance is an empirical result, and it cannot be
deduced from the laws of thermodynamics. Nevertheless the general form of the equation may be
dharm
M-thermTh7-1.pm 5
p H K G J
s H K G J
F I F I F I
H G K J
H HG JK HG JK H G K G K J J
H s K G J
H p K H p K H p K
h
H
F G v
J
I F G K J
v v 2

predicted from hypotheses about the microscopic structure of matter. This type of prediction has
been developed to a high degree of precision for gases, and to a lesser extent for liquids and solids.
The simplest postulates about the molecular structure of gases lead to the concept of the perfect
gas which has the equation of state pv = RT. Experiments have shown that the behaviour of real
gases at low pressure with high temperature agrees well with this equation.
7.6.2. Co-efficient of Expansion and Compressibility
From p-v-T measurements, we find that an equation of state is not the only useful informa-tion
which can be obtained. When the experimental results are plotted as a series of constant
pressure lines on a v-T diagrams, as in Fig. 7.1 (a), the slope of a constant pressure line at any
v
given state is T . If the gradient is divided by the volume at that state, we have a value of a
property of the substance called its co-efficient of cubical expansion . That is,
Fig. 7.1. Determinatio n of co-efficien t of expansion from p-v-T data.
=
v H T K p
...(7.34)
Value of can be tabulated for a range of pressures and temperatures, or plotted graphically
as in Fig. 7.2 (b). For solids and liquids over the normal working range of pressure and tempera-ture,
the variation of  is small and can often be neglected. In tables of physical properties  is
usually quoted as an average value over a small range of temperature, the pressure being atmos-pheric.
This average co-efficient may be symbolised by  and it is defined by
v2 −v
v (T2 −T )
...(7.35)
Fig. 7.2 (a) can be replotted to show the variation of volume with pressure for various
v
constant values of temperature. In this case, the gradient of a curve at any state is p . When
this gradient is divided by the volume at that state, we have a property known as the compressibility
K of the substance. Since this gradient is always negative, i.e., the volume of a substance always
decreases with increase of pressure when the temperature is constant, the compress ibility is
usually made a positive quantity by defining it as
dharm
M-thermTh7-1.pm 5
= 1
F
H G I
K J
p
1 GF  v IJ
1 1
F
H G
I
K J
T

Fig. 7.2. Determinatio n of compressibi lity from p-T data.
K = –
F I
K
1 G vJ ...(7.36)
T
K can be regarded as a constant for many purposes for solids and liquids. In tables of
properties it is often quoted as an average a value over a small range of pressure at atmospheric
temperature, i.e.,
v2 −v
v ( p2 − p )
When  and K are known, we have F p I F TI v
T v v p p
T
v
K = v and G vJ = – Kv,
F p I 
T v K
Since H
...(7.37)
When the equation of state is known, the co-efficient of cubical expansion and compressibility
can be found by differentiation. For a perfect gas, for example, we have
F
T J =
R
v
K T
 G J 2
and H 

RT
v
Hence =
1 F v
K p
=
pv
=
T
,
and K = –
v HpKT
=
RT
=
p
.
7.6.3. Specific Heats
Following are the three differential co-efficients which can be relatively easily determined
experimentally.
dharm
M-thermTh7-1.pm 5
K = – 1
H K H K G J G J H G K J = – 1
F I F I
H G K J =
v H p
1 1
F I
G J
T p H K p
T
I F I
H G K
p p p p
v H T G
I J
R 1
1 F v I G J
p v 2
1

F uI
Consider the first quantity
T
. During a process at constant volume, the first law
informs us that an increase of internal energy is equal to heat supplied. If a calorimetric experi-ment
is conducted with a known mass of substance at constant volume, the quantity of heat Q
required to raise the temperature of unit mass by ΔT may be measured. We can then write :
H ΔT K= H ΔT Kv
v
. The quantity obtained this way is known as the mean specific heat at constant
volume over the temperature range ΔT. It is found to vary with the conditions of the experiment,
i.e., with the temperature range and the specific volume of the substance. As the temperature
range is reduced the value approaches that of H T K , and the true specific heat at constant
volume is defined by cv = G T K . This is a property of the substance and in general its value
varies with the state of the substance, e.g., with temperature and pressure.
According to first law of thermodynamics the heat supplied is equal to the increase of enthalpy
during a reversible constant pressure process. Therefore, a calorimetric experiment carried out
with a substance at constant pressure gives us, G ΔT J = G ΔT J
p
which is the mean specific heat
at constant pressure. As the range of temperature is made infinites imally small, this becomes the
rate of change of enthalpy with temperature at a particular state defined by T and p, and this is
h
true specific heat at constant pressure defined by c
p
=
T
. c
p
also varies with the state, e.g.,
with pressure and temperature.
The description of experimental methods of determining c and c can be found in texts on
physics. When solids and liquids are considered, it is not easy to measure c owing to the stresses
set up when such a substance is prevented from expanding. However, a relation between c , c ,
and K can be found as follows, from which c may be obtained if the remaining three properties
have been measured.
The First Law of Thermodynamics, for a reversible process states that
dQ = du + p dv
Since we may write u = (T, v), we have
du = H
du
K v
dT + H
u
K T
dv
 dQ = H
uJ
v
dT + Tp H v J
T W dv = cv dT + Tp H v J
T W dv
This is true for any reversible process, and so, for a reversible constant pressure process,
dQ = cp(dT)p = cv(dT)p + |p H
u
K | (dv)p
Hence cp – cv =
R |S F uJ | U G F V v I
T| p W| H J
K T p
Also H
p
Kv
= G s J
T
=
1 Sp H
u
KT
V , and therefore
cp – cv = T H T Kv H T K p
dharm
M-thermTh7-1.pm 5
H G K J
v
F Δu I F Q I G J G J
GF u
IJ
F
I v u
H
J
v
Δh
p
F
H
I
K
F Q
H
I
K
F
H G
I
K J
p
p v
v
p v
v
F I F I
 G J G J
T v
F I R| F I U| R| F I U|
T
u u G K G K G K |S |V |S |V
R| F I U|
T v S G J V T W
H v G
I
K
T
F I R| F I U| G J T
F
H
I
K T| G J v T v W|
G F p J I G F v
J
I

Now, from eqns. (7.34) and (7.37), we have
cp – cv =
2Tv
...(7.38)
Thus at any state defined by T and v, c can be found if c ,  and K are known for the
substance at that state. The values of T, v and K are always positive and, although may some-times
be negative (e.g., between 0° and 4°C water contracts on heating at constant pressure), 2 is
always positive. It follows that cp is always greater than cv.
The other expressions for cp and cv can be obtained by using the equation (7.14) as follows :
Since cv = H
u
Kv
= H
u
Kv H
s
Kv
We have cv = T H T J ...(7.39)
Similarly, cp = H T K p
= F
h s
s
I
p
F
T
I
p
Hence, cp = T H T K p
...(7.40)
Alternative Expressions for Internal Energy and Enthalpy
(i) Alternative expressions for equations (7.29) and (7.32) can be obtained as follows : G uI = T G p I – p ...(7.29)
v
F p I F T I F vI
T v v p
T
or H T J
v
= – H T K p H v KT
= +
Kv
=
K
Substituting in eqn. (7.29), we get
H v KT
= T
K
– p ...(7.41)
F I G J
Thus, du = cvdT + H
T
− pK dv ...[7.28 (a)]
Similarly, H pJ = v – T H T K ...(7.32)
F u I
But by definition, T p
= v
Hence H pKT
= v(1 – T) ...(7.42)
dharm
M-thermTh7-1.pm 5
K
v p
GF IJ GF IJ GF IJ
T s T
GF s
I K
v
F G h
J
I H K H K G J G J
F G s
J
I F F
H J J
v K H T
K T But H H G K G K J J H K G J = – 1
GF I F I F I K
p G J G J
v p v 
GF IJ
u 
K
F G
I
K
h
T
GF v
IJ
p
G H J
K F I G
h
J

Thus
(ii) Since
or
Hence
dh = c
p
dT + v(1 – T) dp
u = h – pv
G uJ = F hI – p G vJ – v
H p K HG p JK H p K
T T T
= v – vT + pKv – v
G uJ = pKv – vT
T
...[7.31 (a)]
...(7.43)
7.6.4. Joule-Thomson Co-efficient
Let us consider the partial differential co-efficient H p K . We know that if a fluid is flowing
through a pipe, and the pressure is reduced by a throttling process, the enthalpies on either side of
the restriction may be equal.
The throttling process is illustrated in Fig. 7.3 (a). The velocity increases at the restriction,
with a consequent decrease of enthalpy, but this increase of kinetic energy is dissipated by friction,
as the eddies die down after restriction. The steady-flow energy equation implies that the enthalpy
of the fluid is restored to its initial value if the flow is adiabatic and if the velocity before restriction
is equal to that downstream of it. These conditions are very nearly satisfied in the following experi-ment
which is usually referred to as the Joule-Thomson experiment.
T
Constant h
lines
p1,T 1 p2, T2
  

Fluid





p2, T2
p1, T1
Slope = 

p
(a) (b)
Fig. 7.3. Determination of Joule-Thomson co-efficien t.
Through a porous plug (inserted in a pipe) a fluid is allowed to flow steadily from a high
pressure to a low pressure. The pipe is well lagged so that any heat flow to or from the fluid is
negligible when steady conditions have been reached. Furthermore, the velocity of the flow is kept
low, and any difference between the kinetic energy upstream and downstream of the plug is negligible.
A porous plug is used because the local increase of directional kinetic energy, caused by the
restriction, is rapidly converted to random molecular energy by viscous friction in fine passages
of the plug. Irregularities in the flow die out in a very short distance downstream of the plug, and
dharm
M-thermTh7-1.pm 5
F I F I
F I
H p K
F G
I J
T
h

temperature and pressure measurements taken there will be values for the fluid in a state of
thermodynamic equilibrium.
By keeping the upstream pressure and temperature constant at p and T , the downstream
pressure p is reduced in steps and the corresponding temperature T is measured. The fluid in the
successive states defined by the values of p and T must always have the same value of the
enthalpy, namely the value of the enthalpy corresponding to the state defined by p and T . From
these results, points representing equilibrium states of the same enthalpy can be plotted on a T-s
diagram, and joined up to form a curve of constant enthalpy. The curve does not represent the
throttling process itself, which is irreversible. During the actual process, the fluid undergoes first
a decrease and then an increase of enthalpy, and no single value of the specific enthalpy can be
ascribed to all elements of the fluid. If the experiment is repeated with different values of p and T ,
a family of curves may be obtained (covering a range of values of enthalpy) as shown in Fig. 7.3 (b).
The slope of a curve [Fig. 7.3 (b)] at any point in the field is a function only of the state of the
fluid, it is the Joule-Thomson co-efficient , defined by = G p J . The change of temperature due
to a throttling process is small and, if the fluid is a gas, it may be an increase or decrease. At any
particular pressure there is a temperature, the temperature of inversion, above which a gas can
never be cooled by a throttling process.
Both c and , as it may be seen, are defined in terms of p, T and h. The third partial
differential co-efficient based on these three properties is given as follows : FhI F p I F T I
p T h h p
= – 1
F I
h
Hence p
T
= – cp ...(7.44)
may be expressed in terms of cp, p, v and T as follows :
The property relation for dh is dh = T ds + v dp
From second T ds equation, we have F v I
Tds = cp dT – T T dp
 dh = c dT –
L
T F v I −v
O
dp
p
For a constant enthalpy process dh = 0. Therefore,
| F v I |
0 = (cp dT)h + NT T
p W Qh
L M R U O M |S |V
P P
or (cp dT)h = MST G TJ −vV dpP
h
  = G T J
h
=
M M T G
H J P P
cp N
T K p
−v
Q
...(7.45)
...(7.46)
For an ideal gas, pv = RT ; v =
RT
dharm
M-thermTh7-1.pm 5
H G K J
p
NM QP
v −T dp H K G J
F I R|
T
U|
W
L
N
O
Q
F I L F I O
1 1
2 2
2 2
1 1
1 1
F T
I
H K
h
p
H K G J H H G K G K T J J
H K G J
p H  T G K M J P
p
v
M | H K | P
H p K
1  v
p

or

F v I R v
T
p p T
 =
cp HT 
T
−vK = 0.
Therefore, if an ideal gas is throttled, there will not be any change in temperature.
or
Let h = f(p, T)
h F h I Then dh = p dp + T dT
F h I
But T
p
F I
K
= cp
h
 dh = H pKT
dp + cp dT
For throttling process, dh = 0
h p
p
T
T h
p
1 h
p p
T
h
p
T
is known as the constant temperature co-efficient.
...(7.47)
...(7.48)
...(7.49)
7.7. CLAUSIUS-CLAPERYON EQUATION
Clausius-Claperyon equation is a relationship between the saturation pressure, tempera-ture,
the enthalpy of evaporation, and the specific volume of the two phases involved. This equa-tion
provides a basis for calculations of properties in a two-phase region. It gives the slope of a
curve separating the two phases in the p-T diagram.
p
Critical point
Liquid
Vapour
Solid Triple point
Sublimat ion
curve
T
Fig. 7.4. p-T diagram.
dharm
M-thermTh7-1.pm 5
H K G J = =
F I
H
T H K p
H K G J
F I
 0 = F I
H K G J
F
H G
I
K J + c
c = –
F
H G
I
K J
F
H G
I
K J
1 v G J
G J G J
G J

The Clausius-Claperyon equation can be derived in different ways. The method given below
involves the use of the Maxwell relation [eqn. (7.20)]
F I F I
H H G K G K J J
p s
=
T v
v T
Let us consider the change of state from saturated liquid to saturated vapour of a pure
substance which takes place at constant temperature. During the evaporation, the pressure and
temperature are independent of volume.
F I
dp sg −sf
dT vg −vf
 H G K J =
where, s
= Specific entropy of saturated vapour,
g
s= Specific entropy of saturated liquid,
f v= Specific volume of saturated vapour, and
g v= Specific volume of saturated liquid.
f Also, sg – sf = sfg =
hfg
and vg – vf = vfg
where sfg = Increase in specific entropy,
T
vfg = Increase in specific volume, and
hfg = Latent heat added during evaporation at saturation temperature T.
 dp sg −sf sfg hfg
dT vg −vf vfg T .vfg
...(7.50)
This is known as Clausius-Claperyon or Claperyon equation for evaporation of liquids.
The derivative
dT
is the slope of vapour pressure versus temperature curve. Knowing this slope
and the specific volume vg and vf from experimental data, we can determine the enthalpy of
evaporation, (hg – hf) which is relatively difficult to measure accurately.
Eqn. (7.50) is also valid for the change from a solid to liquid, and from solid to a vapour.
At very low pressures, if we assume v − v and the equation of the vapour is taken as
pv = RT, then eqn. (7.50) becomes
dp hfg hfg p
dT Tvg RT 2
...(7.51)
or hfg =
RT2 dp
p dT
...(7.52)
Eqn. (7.52) may be used to obtain the enthalpy of vapourisation. This equation can be
rearranged as follows :
dp
=
hfg .
dT
Integrating the above equation, we get
dp
=
hfg dT
h L
T
O
p R NT T2 Q
...(7.53)
dharm
  
= =
p R T 2
p z R 2 z
dp
~ g fg
ln
p 2 fg
1 1
 M − P
1 1

Knowing the vapour pressure p at temperature T we can find the vapour pressure p
1 1 2
corresponding to temperature T2 from eqn. (7.53).
From eqn. (7.50), we see that the slope of the vapour pressure curve is always +ve,
since v > v and h is always +ve. Consequently, the vapour pressure of any simple compressible
substance increases with temperature.
— It can be shown that the slope of the sublimation curve is also +ve for any pure substance.
— However, the slope of the melting curve could be +ve or –ve.
— For a substance that contracts on freezing, such as water, the slope of the melting
curve will be negative.
+Example 7.1. For a perfect gas, show that
c
p
– c
v
= Np H v K T Q H TK p
 pv v H v K T
where  is the co-efficient of cubical/volume expansion.
Solution. The first law of thermodynamics applied to a closed system undergoing a reversible
process states as follows :
dQ = du + pdv
As per second law of thermodynamics, FdQI
ds =
rev.
Combining these equations (i) and (ii), we have
Tds = du + pdv
...(i)
...(ii)
Also, since

Thus,
h = u + pv
dh = du + pdv + vdp = Tds + vdp
Tds = du + pdv = dh – vdp
Now, writing relation for u taking T and v as independent, we have
du = F u I
v
dT +
F
v
I
T
dv
u
= cv dT +
u
H G K J
v T
dv
Similarly, writing relation for h taking T and p as independent, we have F h I FhI dh = T p
dT + p dp
FhI
= cp dT + p
T
dp
In the equation for Tds, substituting the value of du and dh, we have
cv dT + F uJ
T
dv + pdv = cp dT + H pKT
dp – vdp
or c
v
dT + Mp FuI
N H v K Q N H p K
T
P dv = c
p
dT –
Lv −FhI
T
O
dp
dharm
M-thermTh7-1.pm 5
H T G K J
g f fg
L F I O F I F I M P
u  v u G J G J G J
H K G J
T H G K J
F I
H G K J H K G J T
H K G J
H G
I
v K
F h I
L O
M Q
P

Since the above equation is true for any process, therefore, it will also be true for the case
when dp = 0 and hence
(c
p
– c
v
) (dT)
p
= Np HuKT Q (dv)
p
or (cp – cp) = Mp H v K P HTK p
By definition, =
1 F v I
p
 The above equation becomes,
cp – cv = Mp H v K P v
or = pv+ v
F
v
I
T
Proved.
u
+Example 7.2. Find the value of co-efficient of volume expansion  and isothermal
compressibility K for a Van der Waals’ gas obeying
Hp 
v2 K (v −b) = RT.
Solution. Van der Waals equation is
Hp 
v2 K (v −b) = RT
Rearranging this equation, we can write
RT a
v −b v2
Now for we require HTK p
. This can be found by writing the cyclic relation,
H v K p HTKv HpKT
= – 1
Hence  p T
v
p
= –
H vKT
From the Van der Waals equation, F p I R
T
v v −b
Also H v KT
= –
(v −b)2
+
v3
Hence =
1 F v I =
1 M
–
G H
J
 GF  IJ
p
Kv P
p
N H v K T Q
dharm
M-thermTh7-1.pm 5
L F I O
M v P
L F I O F I
T
u
N Q
v
v H  T K
L O
T
F u I
N Q
H K
F a I
F a I
p = −
F u I
F I F I F I
T p v
F I F I
H K
H T
K
F  p
I
v
H K =
F I
p RT 2 a
L F I O
v H T K v
T
p
M M M
P
P P

or =
1 M
−
P
P P
v −b P .
Rv
2
(v −b)
. (Ans.)
RTv 2a(v b)
(v −b)2 v3
L O
Also, K = –
1 H
v K = –
v M 2a
1
RT P =
RTv
2
(v
a v
2
b
2 . (Ans.)
v3 (v −b)2
M M M
M P M P
Example 7.3. Prove that the internal energy of an ideal gas is a function of temperature alone.
Solution. The equation of state for an ideal gas is given by
p =
RT
But H v KT
= T HT Kv
− p [Eqn. (7.29)]
= T
v
− p = p – p = 0.
Thus, if the temperature remains constant, there is no change in internal energy with
volume (and therefore also with pressure). Hence internal energy (u) is a function of temperature
(T) alone. ...Proved.
Example 7.4. Prove that specific heat at constant volume (c ) of a Van der Waals’ gas is a
function of temperature alone.
Solution. The Van der Waals equation of state is given by,
RT a
v −b v2
or H p Kv
= v −b
or G 2p J = 0
v
Now
Fdc I
= T H
p
K

T
Hence
T H K G J  G J
F
v
I
H K v G J
T
= 0
v
Thus c of a Van der Waals gas is independent of volume (and therefore of pressure also).
Hence it is a function of temperature alone.
+Example 7.5. Determine the following when a gas obeys Van der Waals’ equation,
Hp 
v2 K (v – b) = RT
(i) Change in internal energy ; (ii) Change in enthalpy ;
(iii) Change in entropy.
Solution. (i) Change in internal energy :
The change in internal energy is given by
du = cvdT + MT HTKv
− pP dv
dharm
M-thermTh7-1.pm 5
v
R
− RT 2a

L
N
O
Q
3 2 − −
F  I 1 M P v − b
v p T − N Q
3
)
− 2 ( − )
v
F u I F  p I
R
v
p = −
F I
T
R
F I
H T 2 K
F I
dv
2
2
c v
v
F a I
L F  p I O
N Q

358
But,













F p I L  R RT a UO
T
v
= = R
T v −b v2
v −b
z 2
z 2
z 2 L
F G R I
J du c
v dT 
N T
H − K Q
M M P − p
P
NM T W QP
O
dv
= cv z2
dT z2 L
1 1 1
T
F
v −b
I
−
R
v −b
−
O
dv
a V
= cv
2
dT 
2 M U MP 1
v −b
2 R RT
H K v G J S
RT RT
−
v −b

a P dv
2 1 1
z 2 z
2
L O
N Q c dT .dv
1 1
u
2
– u
1
= c
v
(T
2
– T
1
) + a
F 1
−
1 I
. (Ans.)
1 2
(ii) Change in enthalpy :
The change in enthalpy is given by
dh = c dT + Mv −T F v I P dp
p
FhI
= 0 + v – T
F v I
...(1)
T p
Let us consider p = f(v, T)
 dp = H v KT
dv + HT Kv
dT
 (dp)T = H v KT
dv + 0 as dT = 0 ...(2)
From equation (1),
(dh)T = Nv −T HT K pQ (dp)T.
O
H
Substituting the value (dh) of (dp) = v from −T Heqn. K (2), we Kget T NT v p
T
dv
F I F I
M p v  p P M P
= Nv H vKT
−T HT K p H v KT Q dv ...(3)
Using the cyclic relation for p, v, T which is F v I FTI FpI
T
p
p
v
v
T

F v I G pJ −G p J
p T v
dharm
M-thermTh7-1.pm 5
H K
TS − WV N Q M P
v
v b
v
a
v
z  z
v v G J
= v 2
H K
L O
p NM H T K QP
H p K H  T K
F p I F p I
F p I
L F v I O M P M P
T
L M M v P p P Q

L F I F I F I O
H K H H G K G K J J − 1
F I F I H T K H v K H T K

Substituting this value in eqn. (3), we get
(dh) = Mv FpI T F p I P dv ...(4)
T v
For Van der Waals equation
p  LF RT I a O
T v v −b v2
T
= –
(v −b)2 
v3 ...(5)
p R
HT Kv
=
M P
T MH
RT
−
a
K Pv
=
v −b
...(6)
R
−
Substituting the values of eqns. (5) and (6) in equation (dh) (1), we get T = Mv
(v −b)2

v
aU
T H v
R
bK P dv

2 2 v
1
(dh)T = – RT
1 (v −b)2 dv + 2a
F I G J
2 2 dv
1
dv
+ RT
1 (v −b)
(h2 – h1)T = – RT Mloge
F
v −bJ −b R
v 2
G − I
H K TS 1 −
1
v −b
v −b
WV
UP
– 2a
F 1
−
1 I
RT log
F v2 −b I
2 1 1
= bRT M(v2 −b)
−
(v1 −b) Q – 2a M v2
−
v1
P . (Ans.)
(iii) Change in entropy :
The change in entropy is given by
ds = cp
dT

H
p Kv
. dv
For Van der Waals equation,
p R
T
v
v −b
...as per eqn. (6)
 ds = cv T

v −b
dv
2
 ds = cv
2
MdT P R
2
(v
dv
b)
 s2 – s1 = cv loge M
1
P + R loge N
v2 −b
Q . (Ans.)
Example 7.6. The equation of state in the given range of pressure and temperature is
given by
v =
RT
−
C
where C is constant.
Derive an expression for change of enthalpy and entropy for this substance during an
isothermal process.
dharm
M-thermTh7-1.pm 5
F I
F I
L O
T N H  v K H  T K Q
F I H v K H K
= − N Q M RT 2a P
F I  L F I O
N G J v − b 2
v Q
S RT T
V W −
L
NM
O
QP 3
2
z z z z
v2 b
1 2 1
L
N
O
Q
v v v b e H K H − K G J G J
L 1 1 O L O
N P
1 1
N Q
T T
H K 
dT R
z
z L O z
1
1 N T Q 1 −
L T O L O
T
2
N Q v 1 −b M P
p T 3

Solution. The general equation for finding dh is given by
dh = cp dT +
L
v −T F
M T
P M P
I
p
O
dp
2
dh = 0 + z2 MR
|
T|
v −T F
T
I
p
UP
T
as dT = 0 for isothermal change.
From the given equation of state, we have
HTK =
R

3C
...(i)
 h2 – h1 = z2
L M
O
N Q
RF RT
−
C I −
|
RT
−
3C U
dpPT
= M
2
1
|
T| V
G −
4C J dpP
T
 −
4C
[( p2 − p )]
The general equation for finding ds is given by
ds = cp
dT
−H v K p
dp
z2
ds =
L
−
2F
M M T
Q
I
p
dp
O
T
as dT = 0 for isothermal change.
(s2 – s1) = M 2
Substituting the value from eqn. (i), we get
−G R

3C I
dp
O
H p1 K G J
H T 4 K
4 1
= – R loge
Fp2 I
–
F3CI
(p2 – p1) (Ans.)
Example 7.7. For a perfect gas obeying pv = RT, show that c and c are independent of
pressure.
Solution. Let s = f(T, v)
Then ds = HT Kv
dT + Hv KT
dv
Also u = f(T, v)
Then du = HT Kv
dT + H v KT
dv = cv dT + H vKT
dv
Also, du = Tds – pdv
Tds – pdv = cv dT + HuKT
dv
ds = cv T

T MF
v
I
T
 pP dv
dharm
M-thermTh7-1.pm 5
v
N H K Q
z
L O
1
v
H G K S J
V
N W| Q 1 M P
F v I
p p T 4
H p T p T G K S J
|
W|
M P 3 3 1
N L z
H F K T 3 I Q
O 1
T
3 T
F I
T T
1 H G K J
N
z
v
P 1
P L F z H K J N Q
P
p T T
v p
F s I F s I
F u I F u I F u I
F I
v
dT L u O
N H  K Q
1

Equating the co-efficients of dT in the two equations of ds, we have
F I  H G K J
cv s
T T v
c
v
= T H s K
Fcv I 2 s
v T Tv
From eqn. (7.20), F s I F p I
v
T
T
v
2 s 2 p
vT T2
Fcv I F 2p I
v T T2
v
 H G K J
Also p =
RT
...(Given)
F p I R
T v
F
p I
T
v
F I
H K
0 or G  v J
T
0
2
c
v
This shows that cv is a function of T alone, or cv is independent of pressure.
Also, cp = T H s K p
H p K T
Tp
From eqn. (7.21),
F sI −F v I
T p
2s 2v
pT T2
p
G cp J −T G 2
2 J
T p
Again, v =
R
F v I R
T p F 2v I
and T2
p
= 0 ;
...(Given)
H p KT
= 0
This shows that cp is a function of T alone or cp is independent of pressure.
dharm
M-thermTh7-2.pm 5
p
H K G J 
H G K J
F I
T
v
H K G J  T
H K  H K
F I
v
H K G J H G K J T
v
H K

v

H G K J
2 
F I
T
F  I G J
c p  s
T
2
H p K H T K
F I − H K G J
F I F I
v
H p K H  T
K
p
F G c I p
J

Example 7.8. Using the first Maxwell equation, derive the remaining three.
Solution. The first Maxwell relation is as follows :
F T I F  p I
H v Ks
 −H s Kv
(1) Using the cyclic relation
T v s
H v Ks
. H s KT
. HT Kv
= – 1
s T s
 Hv KT
= – H v Ks
. HT Kv
Substituting the value from eqn. (i) in eqn. (ii), we get
s p s
Hv KT
= H s Kv
. HT Kv
Using the chain rule,
p s T
H s Kv
. HT Kv
. H p Kv
= 1
Substituting the value of eqn. (iv) in eqn. (iii), we get
F s I F p I
Hv KT
= HT Kv
This is Maxwell Third relation.
(2) Again using the cyclic relation
vI
F H s I F I F K H p
K H K
p
v
. . = – 1
v
s
s
p
v FpI F v I
s
 H K = – H K H K .
p s
v
p
s
Substituting the value from eqn. (i) into eqn. (v)
v T v
HsK p
= H v Ks
. HpKs
Again using the chain rule,
T v p
H v Ks
. HpKs
. HT Ks
= 1
Substituting the value of (vi) into (v), we get
F v I F  T I
HsK p
H p Ks
This is Maxwell second relation.
(3) HTK p
. H p Kv
. H vKT
= – 1
...(i) (Eqn. 7.18)
...(ii)
...(iii)
...(iv)
...(v)
...(vi )
dharm
F I F I F I
F I F I F I
F I F I F I
F I F I F I
F I
F I F I F I
F I F I F I
F I F I F I
v T p

F v I F p I F v I
T
H K = – H K H K .
p T
v
p
T
F p I F s I F s I F v I
= – H s Kv HT Kv HpKT H s KT
Substituting the value from eqn. (i), we get F v I F TI F s I F s I F vI
p v s T v p
H T K = G H K J G H K J G H J K T
H G K J
s T
RS F T I F v I F s I UV F s I F s
I
= TH v Ks
. H sKT
. HTKvW H pKT
= – HpKT
F I
 H v K p
= – HpKT
This is Maxwell fourth relation.
Example 7.9. Derive the following relations :
(i) u = a – T F
a
T
I (ii) h = g – T F g I
p
(iii) cv = – T G 2a J (iv) cp = – T
F 2 g J
G K 2 T
T H 
v p
H  K 2
where a = Helmholtz function (per unit mass), and
g = Gibbs function (per unit mass).
Solution. (i) Let a = f(v, T)
Then da = H v KT
dv + HT Kv
dT
Also da = – pdv – sdT
Comparing the co-efficients of dT, we get
HT Kv
= – s
Also
or
Hence
(ii) Let
Then
Also
a = u – Ts
u = a + Ts = a – T HT Kv
u = a – T HT Kv
. (Ans.)
g = f(p, T)
dg = HpKT
dp + HTK p
dT
dg = vdp – sdT
Comparing the co-efficients of dT, we get
HTK p
= – s
dharm
M-thermTh7-2.pm 5
F I
F I
F I F I

T
F s I
H K
v H  T K
F I I
F a I F a I
F a I
a
a
g g
F g I

Also h = g + Ts = g – T HTK p
Hence h = g – T HTK p
. (Ans.)
(iii) From eqn. (7.23), we have
c
v
= T H s K
Also
F a I
= – s
v
or HT Kv
2a
= – HT2 Kv
From eqns. (i) and (ii), we get
c
v
2a
= – T G T2 J
v
. (Ans.)
(iv) From eqn. (7.26), we have
cp = T H s K p
Also HTK = – s
or
F
T
I
p
s G J T 2
= – H
2g
K p
From eqns. (i) and (ii), we get
2 g
cp = – T HT2 K p
. (Ans.)
...(i)
...(ii)
...(i)
...(ii)
Example 7.10. Find the expression for ds in terms of dT and dp.
Solution. Let s = f(T, p)
Then ds = H s K p
. dT + HpKT
dp
As per Maxwell relation (7.21)
HpKT
= – HTK p
Substituting this in the above equation, we get
ds = H s K p
dT – H v K p
. dp ...(i)
The enthalpy is given by
dh = cpdT = Tds + vdp
Dividing by dT at constant pressure
HTK p
= cp = T HTK p
+ 0 (as dp = 0 when pressure is constant)
dharm
M-thermTh7-2.pm 5
F I
H T K
F I F I
F I
F I
F I
p
F I
F I
F g I
F g I
T
v
s G J
H K
T
g
H K
G J
F I F s
I
 T
F s I F v I
F I F I
T  T
F h I F s I

Now substituting this in eqn. (i), we get
But
ds = cp
dT
F I
−H
s
K . dp ...(ii)
1F v I
v T
p
Substituting this in eqn. (ii), we get
ds = cp
dT
– vdp (Ans.)
T
F T I F T
I
(i) H p K =
Tv
(ii) H v K = –
T
c K
.
where = Co-efficient of cubical expansion, and
K = Isothermal compressibility.
Solution. (i) Using the Maxwell relation F (7.19), we have T I FvI F v I FTI
s s
p T
p
s
p
Also cp = T HTK
From eqn. (7.34), =
v
F
T
I
p
H p Ks

vT
i.e., H p Ks

c
 . (Ans.)
(ii) Using the Maxwell relation (7.18) FT I FpI F p I FT I
s s
v
T
v
s
v
Also cv = T HT K
K = – 1 F v I
FTI T F p I
v c T
Also FpI F v I FTI = – 1
T p v
i.e., H p Kv
= – HvKT HTK p
= – H−vK K v =
K
(Eqn. 7.23)
(Eqn. 7.36)
dharm
M-thermTh7-2.pm 5
p
G J
T T
= H K
H p K = H  K = H K H K
F I
F I
F I
F I
v H  p K
s v v H K H K
F I F I F I F I
c p
s s v
s
p
1 v
H K
 T
c p
 T Tv
p
H v K = – H s K = – H K H K
v
T
Then = –
H v K H T K H p K
T
p v 1 

F I
v c K v
 HT Ks
=
−T
. (Ans.)
+Example 7.12. Derive the third Tds equation
F T I
Tds = cv H p Kv
dp + cp
F
v
I
p
dv
and also show that this may be written as :
Solution. Let
Then
or
But
Hence
Also
and
Tds =
cv
Kdp +
c
v
dv.
dss == f(p, v) F
p
T
H G K J
I dp + HsK dv
s
Tds = T F s I dp + T H
s
K dv
v
G J
p F I F I
T p T v
= T H s Kv HTKv
F I
+ T F s I
p
FTI
p
dv
H s Kv
= T and H s K p
= T
Tds = cvHpK dp cp HvK dv ...Proved.
F
I
F I F I
p
= FpI
−
F v I = – H v K
F
p
I
=
K
p H K H K
v
T
1 T
T
p
H v K p
= v
Substituting these values in the above Tds equation, we get
Tds =
c

K
dp 
v
dv ...Proved.
Example 7.13. Using Maxwell relation derive the following Tds equation
Tds = cp dT – T HTK p
dp. (U.P.S.C. 1988)
Solution.
where cp = T H TK p
Also,
s = f (T, p)
Tds = T HTK p
dT + T HpKT
dp
H s KT
= – F v I
p
...(i)
......Maxwell relation
Substituting these in eqn. (i), we get
Tds = cp dT – T HTK p
dp. (Ans.)
dharm

p
H K v
F I
p
H K v 
v p
H K H K
F I
v p
F I
F I
F I F I
F I
T
c v
 T
c p
T T
T
v H K H K
v
T 
T 1
c v p
F v I
GF s
IJ
s s
F I
p H T K
F v I

Example 7.14. Derive the following relations
F I
FT I TG p J − p
H K u
v
Solution.
FT I
H  K
T
c
v
= .
v
can be expressed as follows :
u
H v K = F vI
F I F I
Kv = – F uI
T u
u v T
− H H K
H K H K
u
T
T
v
H v K
F T I
u
Also Tds = du + pdv
or du = Tds – pdv
or HuK = T H
s
K – p H v K
or
v
F uI
T v T
= T
F s I
– p
T T
or H u Kv
= T H
s
Kv
F I
T T
Dividing eqn. (i) by eqn. (ii), we get

F I K
F I
FT I THvK − p
v
H K =
s
T
u s
T
v
s
Also cv = T HT K
and F
s
H G K J
v
I
T
= F
p
H K
F
H
T
I
v
u
T
... Maxwell relation
Substituting these value in eqn. (iii), we get
TG p J − p
v
u
=
cv
v ...Proved.
+Example 7.15. Prove that for any fluid
(i) HhKT
= v HpKT
+ T H p Kv
(ii) HhKT
= v – T HTK p
Show that for a fluid obeying van der Waal’s equation
p =
v −b
–
v2
...(i)
...(ii)
...(iii)
where R, a and b are constants
h (enthalpy) =
RTb
−
2a
+ f(T)
where f(T) is arbitrary.
dharm
F I F I F I
H v K H v K
I I
T
H
v
F I H K
T  T
I
K
F I F I F I F I F I
v v T p
v
RT a
v −b v

Solution. We know that
ds =
c v  p
T
dT  HT K dv
Also dh = Tds + vdp = T M v dT F p I dvP + vdp
v
i.e., dh = c
v
dT + T H p Kv
+ dv + vdp
Putting dT = 0, we get
G hJ T G p J v G pJ ...Proved.
T v T
(ii) HhKT
hKT H v KT
H 
= NT F
 

p I
v
vF pI
vJ
T Q G 
i.e.,
h GF IJ G J T p
F
p
I
T
= T H
p
K
F vI
+ v
Also F p I
v H
vJ
T
= – H v K p
 Eqn. (i) becomes
HpKT
= v – T H vKp
...Proved.
[Eqn. (7.24)]
...(i)
Now
and








i.e.,
RT a
v −b v2
p RT 2a
(v −b)2 v3
F p I R
T v −b
FhI L −RT 2aO F R I
T (v −b)2 v3 v −b
RTv 2a RT −RTv RT 2a
2
(v −b )
v
2 v −b (v −b)2 v −b v2
−RTv RT(v −b) 2a −RTv RTv −RTb 2a
(v −b)2 v2 (v −b)2 v2
FhI −RTb 2a
v
T (v −b)2 v2
or
RTb 2a
v −b v
This shows h depends on T and v.
...Proved.
(i) FhI = v – T
v
FTI FuI H p = – c(ii) = T K – p T p
h T
dharm
M-thermTh7-2.pm 5
v
F I
L
N
O
Q
F I
F I F I F I
F I F I F I L O F I
H K v T
F I F I
F I F I
p = −
F I H v K = – 
H K v =
N Q H K
=  = 
H K = 
h = −  f(T)
c
T H  T K
T
H  v K H T K H  v K

p v p
M G H J K G H J K P
T v H p K
T H K
H K G T J G p K  T
 h
 T
T
H v K = v M  P T
= –   =  
F I F I
H T K
H K p H K H K p p v T
v

With the aid of eqn. (ii) show that
F u I F v I F I
H pKT
= – T HTK – p HpKT
The quantity cp
p
FTI is known as Joule-Thomson cooling effect. Show that this cooling
h
effect for a gas obeying the equation of state (v – b) =
RT
–
C
is equal to H
3C
K −b.
Solution. We know that F
p
I
= – cp
Also =
cp
L
HT K p
−v
O

FhI
= – T F v I −v
O
= v – T F v I
T p p
Also = H p Kh
 H pKT
= – cp H p Kh
.
...[Eqn. (7.44)]
...[Eqn. (7.46)]
... Proved.
(ii) Let
Also
u = f(T, v)
du = HT Kv
dT + H v KT
dv
= c
v
dT + H v KT
dv ...(i)
du = Tds – pdv
Substituting the value of Tds [from eqn. 7.24], we get
du = c
v
dT + T
F p I
G J v
N  T Q
dv – pdv
= cv dT + MT H p Kv
− pP dv ...(ii)
From (i) and (ii), we get
HuK = T HT K – p ...Proved.
FuI FuI F v I
p
T
v
T
p
T
or
F
p
I
u v = HpKp
T
T
GF IJ
T HTKv
− pQ
or H pKT
= T F
H G J K G H J F I T
K H K
I
v
F
p
I
T
– p G pJ
T
...Proved.
dharm
M-thermTh7-2.pm 5
H K T
NM QP
L
NM QP
F I
F I F I
F I F I
F I
v
H p K
F I
p T 2
T G 2 J
h
T 1 M F  v I P
H p K H  T K M P H T K
T
h T
u u
u
H K G J
T
L F I O
F I F I
v
T
G p
J
v
Also H K = H K H K G J G J
F I L F I O
H K G J
G J G  J N
M P
u  p  v
v

370
We know that
or
Also
and
FpI F v I FTI
v
H K H K H K = – 1
T
T
p
p
v
I
K J − F F
p v v
T p T
F
p
I
H K H K G J G J u v
T
= – T F
v
T
I
p
– p F
p
I
T
= 1
L O M P N Q
c T p H  K
M
F uI
p
−vP
...Already proved.
...[Eqn. (7.46)]
Now v – b = RT
−
C ...[Given]
HTK p
=
R

2C
Substituting this value in the expression of above, we get
=
1 L
T F H
G I J N
R

2C
K −v
O
M Q
or cp = T G R
H 
p 3 K T p T 2
2CJ –
RT

C
– b =
3
2
−b
or cp H p Kh
=
T2
– b ...Proved.
Example 7.17. The pressure on the block of copper of 1 kg is increased from 20 bar to 800
bar in a reversible process maintaining the temperature constant at 15°C. Determine the following :
(i) Work done on the copper during the process,
(ii) Change in entropy, (iii) The heat transfer,
(iv) Change in internal energy, and (v) (cp – cv) for this change of state.
Given :  (Volume expansitivity = 5 × 10–5/K, K (thermal compressibility) = 8.6 × 10 12 m2/N
and v (specific volume) = 0.114 × 10–3 m3/kg.
Solution. (i) Work done on the copper, W :
Work done during isothermal compression is given by
2
W = pdv
1
The isothermal compressibility is given by
K = –
1 F v I
T
 dv = – K(v.dp)T
2 2
 W = – pKv.dp = – vK pdp
1 1
Since v and K remain essentially constant
 W = – vK (p2
2 – p1
2)
= –
0.114 10−3 8.6 10−12
[(800 × 105)2 – (20 × 105)2]
dharm
M-thermTh7-2.pm 5
F
H G
H H G G I I
K K J J
v T p
H K
T
p T 2
F v I
p T 3
c p 3 p T
P
F I
C
T
GF IJ T 3 C
–
z
v H  p K
z z
2
2

= –
0.114 8.610−15
× 1010 [(800)2 – (20)2]
= –
0.114 8.6 10−5
(640000 – 400) = – 3.135 J/kg. (Ans.)
The negative sign indicates that the work is done on the copper block.
(ii) Change in entropy :
The change in entropy can be found by using the following Maxwell relation :
G pJ
T
= – H v K p
F I = –
F I
v
F
T
I
p
= – vas
1 H v K p
=  
 (ds)
T
= – v (dp)
T
Integrating the above equation, assuming v and  remaining constant, we get
s2 – s1 = – v (p2 – p1)T
= – 0.114 × 10–3 × 5 × 10–5 [800 × 105 – 20 × 105]
= – 0.114 × 10–3 × 5 (800 – 20) = – 0.446 J/kg K. (Ans.)
(iii) The heat transfer, Q :
For a reversible isothermal process, the heat transfer is given by :
Q = T(s2 – s1) = (15 + 273)(– 0.4446) = – 128 J/kg. (Ans.)
(iv) Change in internal energy, du :
The change in internal energy is given by :
du = Q – W
= – 128 – (– 3.135) = – 124.8 J/kg. (Ans.)
(v) cp – cv :
The difference between the specific heat is given by :
c
p
– c
v
=
2Tv
... [Eqn. (7.38)]
(5 10−5)2 (15 273)  0.114 10− 3
=
8.6 10−12 = 9.54 J/kg K. (Ans.)
Example 7.18. Using Clausius-Claperyon’s equation, estimate the enthalpy of vapourisation.
The following data is given :
At 200°C : vg = 0.1274 m3/kg ; vf = 0.001157 m3/kg ; H dT K = 32 kPa/K.
Solution. Using the equation
F dp I fg
dT T (vg −v )
where, hfg = Enthalpy of vapourisation.
Substituting the various values, we get
32 × 103 =
(200 273)(0.1274 −0.001157)
 hfg = 32 × 103 (200 + 273)(0.1274 – 0.001157) J
= 1910.8 × 103 J/kg = 1910.8 kJ/kg. (Ans.)
dharm
M-thermTh7-2.pm 5
2
2
F I
H K
s v v
T H K v T
K
F dp I G J
H G K J =
h
s f
h fg

Example 7.19. An ice skate is able to glide over the ice because the skate blade exerts
sufficient pressure on the ice that a thin layer of ice is melted. The skate blade then glides over
this thin melted water layer. Determine the pressure an ice skate blade must exert to allow
smooth ice skate at – 10°C.
The following data is given for the range of temperatures and pressures involved :
hfg( ice) = 334 kJ/kg ; vliq. = 1 × 10 m3/kg ; vice = 1.01 × 103 m3/kg.
Solution. Since it is a problem of phase change from solid to liquid, therefore, we can use
Clausius-Claperyon equation given below :
dp fg 1
dT fg T
Multiplying both the sides by dT and integrating, we get
p2
dp fg
T2 dT
1 v g 1
T
or (p2 – p1) =
fg
loge GT2 J ...(i)
But at p1 = 1 atm., t1 = 0°C
Thus, p1 = 1.013 bar, T1 = 0 + 273 = 273 K
p2 = ?, T2 = – 10 + 273 = 263 K
Substituting these values in eqn. (i), we get
334 103
(p2 – 1.013 × 10 ) =
1− 1.01
× loge 273
= 334

0
H 263 K
. 103
× loge
F273I
= 12.46 × 105 N/m2
or p2 = 12.46 × 105 + 1.013 × 105
= 13.47 × 105 N/m2 or 13.47 bar. (Ans.)
This pressure is considerably high. It can be achieved with ice skate blade by having only a
small portion of the blade surface in contact with the ice at any given time. If the temperature
drops lower than – 10°C, say – 15°C, then it is not possible to generate sufficient pressure to melt
the ice and conventional ice skating will not be possible.
Example 7.20. For mercury, the following relation exists between saturation pressure
(bar) and saturation temperature (K) :
log10 p = 7.0323 – 3276.6/T– 0.652 log10 T
Calculate the specific volume v
g
of saturation mercury vapour at 0.1 bar.
Given that the latent heat of vapourisation at 0.1 bar is 294.54 kJ/kg.
Neglect the specific volume of saturated mercury liquid.
Solution. Latent heat of vapourisation, hfg = 294.54 kJ/kg (at 0.1 bar)
Using Clausius-Claperyon equation
dp h fg h
fg
dT v gT (vg −v )T
Since vf is neglected, therefore eqn. (i) becomes
dp fg
dT vgT
...(given)
...(i)
Now, log
10
p = 7.0323 –
3276.6
– 0.652 log
10
T
dharm
M-thermTh7-2.pm 5
=
h
v .
z h
= z
p f T
h F I
fg
v
H T1 K
F I 5 b g
263 H K
0 1
= =
f f
=
h
T

Differentiating both sides, we get
1 dp 3276.6 0.652
2.302p dT T 2.302 T
or
dT
= 2.302 × 3276.6 ×
T2 – 0.652
T
...(ii)
From (i) and (ii), we have
or
fg
vgT
= 2.302 × 3276.6 ×
T2 – 0.652
T
...(iii)
We know that
At p = 0.1 bar,
log
10
p = 7.0323 –
3276.6
– 0.652 log
10
T ... (given)
log10 (0.1) = 7.0332 –
3276.6
– 0.652 log10 T
– 1 = 7.0323 –
3276.6
– 0.652 log10 T
or 0.652 log10 T = 8.0323 –
3276.6
or log10 T = 12.319 –
5025.4
Solving by hit and trial method, we get
T = 523 K
Substituting this value in eqn. (iii), we get
294.54 103
vg 523
= 2.302 × 3276.6 ×
0.1105
– 0.652 ×
0.1105
(523)
563.17
= 275.75 – 12.46
g
i.e., vg = 2.139 m3/kg. (Ans.)
HIGHLIGHTS
1. Maxwell relations are given by
FTI
= – FpJ ; FTJ =
F
s
I
v s H  H  G I G I
F p I FsI F v I F sI
T
v
v
T
T
p
p
T
2. The specific heat relation s are
cp – cv =
vT2
; cv = T
F s I
v
; cp = T G T J
p
.
3. Joule-T homson co-efficien t is expressed as
= H p Kh
.
dharm
M-thermTh7-2.pm 5
T
T
. = –
dp p p
h p p
T
T
T
2 523
v
H  G K J
s K p K v s H G K J v
p
HG KJ  HG KJ H  − H G K G K J J ; .
F I
K H K G J
s
T H K
GF  T JI

4. Entropy equation s (Td s equatio n s) :
Tds = c dT + T
F p I
dv
v
Tds = cpdT – T H T K p
dp
5. Equatio n s for interna l energy and enthalpy :
F F
H H
I I
v K T K
G u J = T G p J – p
T v
F p I du = c dT + Tv
−pV dv
SR T
G p K = v – T G v J
R U
dh = cpdT + |v T
T
p| dp
...(1)
...(2)
...(1)
...[1 (a)]
...(2)
...[2 (a)]
| |
OBJECTIVE TYPE QUESTIONS
Choose the Correct Answer :
1. The specific heat at constan t pressure (c ) is given by
(a) cp = T
F s I
p
(b) cp = T
F T I
p
(c) cp = T
F
T
I
p
(d) cp = T
F
T
I
p
.
2. The specific heat relation is
(a) (cp – cv) =
vT2
(c) (cp – cv) =
pTK
3. The relation of intern al energy is
(a) du =
F K
cv
I
F I G J
dp + H
cp −pK dv
(c) du = H 
cpJ dp + G v
−vJ dv
4. Tds equatio n is
(a) Tds = cvdT +
T
dv
(b) (cp – cv) =
vTK
(d) (cp – cv) =
v T
.
F
H K J v 
(b) du = G K
c I dp + H
cp pK dv
GF K
JI F HG cv I (d) du = H cpK dp + −pKJ
v dv.
 (b) Tds = cpdT –
T
dv
(c) Tds = cvdT +
TK
dv (d) Tds = cvdT +
T
dp.
Answers
1. (a) 2. (a) 3. (a) 4. (a).
dharm
M-thermTh7-2.pm 5
F I
 
v H  T K
U
W
F
H
I
T
F I
H K p
v − F 
H G I
K S J V
K  2
 2 K
H  K
v HG  KJ
T
G v J
G J
 h J
T
T W
p
H H G K G K J J
T s
H H G K G K v J v J
G J v 
GF K I
K
c F p I
H K
2
v 
F I G J

K K
 K

1. Define the co-efficien t of :
(i) Volume expansion
EXERCISES
(ii ) Isothermal compressibil ity
(iii ) Adiabatic compressibili ty.
2. Derive the Maxwell relation s and explain their importance in thermo dy namics.
3. Show that the equatio n of state of a substance may be written in the form
dv
= – Kdp + dT.
v
4. A substance has the volume expansivity and isothermal compressibi li ty :
=
T
; K = p
Find the equation of state. NAns.
pv
constantQ
5. For a perfect gas, show that the difference in specific heats is
cp – cv =
T
.
6. For the followin g given differen tial equatio n s,
du = Tds – pdv
and dh = Tds + vdp
prove that for perfect gas equation ,
G u J = 0 and
T
F
H
G h J = 0.
T
7. Using the cyclic equation, prove that
HTKv
= KT .
8. Prove that the change in entropy is given by
ds =
cv L KT
. dp 
M 
Q
p O
dv.
9. Deduce the followin g thermo dy namic relation s :
(i)
Fh I
= v – T G v J = – c
FT I
(ii )
Fu I
= T
F p I
– p.
T p h T v
10. Show that for a Van der Waals gas
cp – cv =
1 −2a (v −b)2 / RTv3 .
11. A gas obeys p(v – b) = RT, where b is positive constan t. Find the expression for the Joule-Thomson co-efficien
t of this gas. Could this gas be cooled effectiv ely by throttlin g ?
12. The pressure on the block of copper of 1 kg is increased from 10 bar to 1000 bar in a reversible process
maintainin g the temperatu re constant at 15°C. Determin e :
(i) Work done on the copper durin g the process (ii ) Change in entropy
(iii ) The heat transfer (iv) Change in internal energy
(v) (cp – cv) for this change of state.
The followin g data may be assumed :
Volume expansivity () = 5 × 10–5/K
Isothermal compressibi lity (K) = 8.6 × 10–12 m2/N
Specific volume (v) = 0.114 × 10–3 m3/kg
[Ans. (i) – 4.9 J/kg ; (ii ) – 0.57 J/kg K ; (iii ) – 164 J/kg ; (iv) – 159.1 J/kg ; 9.5 J/kg K]
dharm
M-thermTh7-2.pm 5
F I
1 1
L O
M T P
R
H  p K 
I
p K
GF  p JI 
T
c
N  v
P
F
H
I
H G J p  K K G H K J G J G J
T p p  H  v K H  T
K R

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