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-effic ien t of expansion and compressibili ty ,

specific heats, Joule -Tho m so 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 thermodynamic

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 ENGI NEERIN G T HERMOD Y N AM ICS

Then the differential of the dependent variable x is given by Fx I F x I y

z z y

...(7.2)

where dx is called an exact differential.

If Gy

Jz

= M and H z Ky

= 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 yz M N

or z y

and N 2 x

y zy

...(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 I dy = x z

dx + z x

dz

z z dz = x y

dx + y x

dy

dy = Gx

Iz

dx + Gz

Ix N

Fx

Iy

dx Fy

Ix

dy

Q F y I F y I F z I F y I F z I

= N x

z z

x x

y Q dx + z x y x

dy

F y I F y I F z I =

N x z

z x

x y Q

dx + dy

y y z 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

Fp I FT I F v I v T p

v T p

= – 1

...(7.6)

...(7.7)

...(7.8) ...(7.9)

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dx = H K G J dy H K G J dz

H H G G K K J J

F H G I

K J H F I

K G J

L O

H H G G K K J J H G K J L O M P M P H K G J H G K J

H H G G K K J J H G K J L O M P M P

z x y F F

H H G G I I K K J J F

H G I K J

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

F I F I x

H K x G J

y y F F H H K K J J

z z

H K G J H G K J M P M P

H K H K G J G J H G K J

T HERMO D Y N A M IC RELAT ION 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,

dg = Gp

JT

dp + GT

Jp

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 h I s

= T = s

F uI F f I v s

= – p = v T

g

p s

= v = p T

H T Kv

= – s = GT

Jp

...(7.14) ...(7.15) ...(7.16)

...(7.17)

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T

F H G I

K J

H H G G K K J J

F H G I

K J H F I

K G J

H H G G K K J J

F I F I

H H G G K K J J

H G K J v H K G J

p

F I

F I F I

s

H K g

H K g

H H G G K K J J

F H G I

K J h H G K J

G J f

H K g


F T I F p I Also,

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 = cvdT + T H T K

v

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 Kp

dT + T H p KT

dp ...(7.25)

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s v − H H G G K K J J

s H K H K

F H G I

K J − F H G

I K J

v

F H

I K

G J G J

H H G G K K J J

H H G G K K J J F F I I H H G G K K J J

G J

H H G G K K J J

p F G I J

F I F I s G J G J s

T HERMO D Y N A M IC RELAT ION S 345

where cp = T H T Kp

Also Gp

JT

= – H T Kp

...(7.26) [Maxwell’s eqn. (7.21)]

whence, substituting in eqn. (7.25)

Tds = cpdT – T H T Kp

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 uI du = 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 FuI F s I F uI v

= s v v

F uI F s I F s I F uI s v

= T, v T

= T v v s

= – p

F uI F p I v T

= 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 Iv

−p

W 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 pIT

dp

T

...(7.30)

...(7.31)

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F I F I

H H H G G G K K K J J J

v s H H G G K K J J

T v T s H K H H H K K K

H H H H G G G G K K K K J J J J ,

H H G G K K J J

T H K G J S V | | | |

T p H K G J

p F H K J

s F I G J

H K s v G J

v F I G J

H G K J

G J G G G J J J

R U

h F G K J

346

FhI To find p

T ;

Then,

But

ENGI NEERIN G T HERMOD Y N AM ICS

let h = f(s, p)

F h I FhI dh = s ds + p dp

FhI FhI 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

Hence Gp

JT

= v – T H T Kp

...(7.32)

From eqn. (7.31), we get

dh = cp

dT + |v −T H

v Ip

|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

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p H K G J s H K G J

F I F I F I

H G K J

H K G J H G K J H H G G K K J J

s H K G J p H K H K H K p p

h

H F I

K F I G J v

F G K J R U S V T W | | T

v v 2

T HERMO D Y N A M IC RELAT ION S 347 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 ib ility is

usually made a positive quantity by defining it as dharm

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= 1

F H G I

K J p

1 v F G I J

1 1

F H G I

K J T


Fig. 7.2. Determinatio n of compressibility from p-T data.

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

Since Hv

K = v and G vJ = – Kv,

F p I T v K

...(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

FT

J = R

and H v

KT

RT

Hence = 1 F v

K p

= pv

= T

,

and K = – v Hp K

T

= RT

= p

.

7.6.3. Specific Heats

Following are the three differential co-efficients which can be relatively easily determined

experimentally .

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K = – 1

H K H K G J G J H G K J = – 1

F I F I

H G K J =

v p H F I

K

1 1

F I

G J T

p H K p T

I F I H G K v

p p p p

G J 2

v T H G I J R 1

F I 1 v 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 ∆T may be measured. We can then write :

H ∆T Kv

= H ∆T Kv

. The quantity obtained this way is known as the mean specific heat at constant

volume over the temperature range ∆T. 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 = GT 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∆T

J = G∆T

Jp

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 cp

= T

. cp

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 = Hdu

Kv

dT + Hu

KT

dv

dQ = HuJ

v dT + T

p H v JT W dv = cv dT + T

p H v JT 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 Hu

K | (dv)p

Hence cp – cv = |

p F uJ | F v I

T p

Also Hp

Kv

= G s JT

= 1 Sp H

u

KT

V , and therefore

cp – cv = T H T Kv

H T Kp

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H G K J v

F F I I ∆u Q G G J J

F I G J u

v F H

I J u

v

∆h

p

F H

I K

Q F H

I K

F H G I

K J p

p v v

p v v

I F F I

G J G J T v

F F F I I I R U R U | | | | T

u u G G G K K K S V S V | | | |

F I R U | | T v G J S V

T W

v H G I K

R U S V T W | | H K G J

T

F I F I R U | | G J T

F H

I K v T v G J

T W | | F I F I G J G J p v


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 = Hu

Kv

= Hu

Kv

Hs

Kv

We have cv = T H T J ...(7.39)

Similarly , cp = H T Kp

= Fs

Ip

FT

Ip

Hence, cp = T H T Kp

...(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)

F p I F T I F vIv

T v v p T

or H T Jv

= – H T Kp

H v KT

= + Kv

= K

Substituting in eqn. (7.29), we get

H v KT

= T K

– p ...(7.41)

Thus, du = cvdT + HT

−pK dv ...[7.28 (a)]

Similarly , H p J = v – T H T K ...(7.32)

F u I But by definition, T p

= v

Hence H p KT

= v(1 – T) ...(7.42)

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K

v p

F F F I I I G G G J J J T s T

F I G K s

v

F I G J h

H K H K G J G J h s

F I G J s

F F T H H K K J J

v T But H H G G K K J J H K G J = – 1

F I F F I I G K p

G G J J v p v

F G I J u

K F I G J

F G I K

h

T

F I G J v

p

H K G J

F G

I J

h


Thus

(ii) Since

or

Hence

dh = cp

dT + v(1 – T) dp

u = h – pv

G uJ = F hI

– p G vJ – v T T T

= v – vT + pKv – v

G uJ = pKv – vT 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. Determ ination of Joule -Tho m son 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 negligib le.

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-therm\Th7-1.pm 5

F I F I

F I

H p H G K J p K H K p

H K p

F G

I J

T

h

352 ENGI NEERIN G T HERMOD Y N AM ICS 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 = Gp

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 : FhI F p I F T I p T h h p

= – 1

F I 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 –

LT

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

or (cp dT)h = MST GT

J −vV dpP h

= G T Jh

= cp N

T GT K

p

−v

Q

...(7.45)

...(7.46)

For an ideal gas, pv = RT ; v = RT

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H G K J p

N Q M P

dp v −T H K G J R U S V | |

L O M P M P

F I R | T

U | W

L

N

O

Q F I F I L O

1 1 2 2

2 2 1 1

1 1

F I H K T

h

p

H K G J H H G G K K J J T

H K G J h

p T H G K J M P

p

v H K | | M P

H K p 1 v

H J M P M P

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 = cp

dh = H p KT

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

Sublimation

curve

T

Fig. 7.4. p-T diagram. dharm

\M-therm\Th7-1.pm 5

H K G J = =

F I

H F I

K 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 h


The Clausius-Claperyon equation can be derived in different ways. The method given below

involves the use of the Maxwell relation [eqn. (7.20)]

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.

dp sg −sf

dT vg −vf

where, sg

= Specific entropy of saturated vapour,

sf = Specific entropy of saturated liquid, vg = Specific volume of saturated vapour, and

vf = Specific volume of saturated liquid.

Also, sg – sf = sfg = hfg

and vg – vf = vfg

where sfg = Increase in specific entropy, 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 RT2

...(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

F F I I H H G G K K J J

F I H G K J =

T

dp

~ g fg

ln p fg 2

1 1

1 1 − M P

\M-therm\Th7-1.pm 5


Knowing the vapour pressure p at temperature T we can find the vapour pressure p

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

cp – c

v =

Np H v K

T Q H T Kp

pvv H v KT

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, F dQI 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 + Fv

IT

dv

u = cv dT +

v T dv

Similarly, writing relation for h taking T and p as independent, we have F h I FhI

dh = T p dT + p dp

FhI = 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 cv dT + M p

F uIT

P dv = cp

dT – Lv −

F hIT

O dp

dharm

\M-therm\Th7-1.pm 5

T H G K J

1 1 2

g f fg

L O F I F I F I M P u v u G J G J G J

H K G J T H G K J

u

F I H G K J

H G K J H K G J T

H K G J

H G I K v

F I h

L O v H K N Q p H K N 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

(cp – c

v) (dT)

p = N

p Hu

KT Q (dv)

p

or (cp – cp) = M p H v K P HTKp

By definition, = 1 F v I

p

The above equation becomes,

cp – cv = M p H v K P v

or = pv+ vFv

IT

Proved.

+Example 7.2. Find the value of co-efficient of volume expansion and isothermal

compressibility K for a Van der Waals’ gas obeying

H p v2 K (v −b) = RT.

Solution. Van der Waals equation is

H p v2 K (v −b) = RT

Rearranging this equation, we can write

RT a

v −b v2

Now for we require HTKp

. This can be found by writing the cyclic relation,

Hv

Kp

HT

Kv

Hp

KT

= – 1

v p Hence T

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–

Hp

Kv P

p

N H v KT Q

dharm

\M-therm\Th7-1.pm 5

F I L O v M P

F I L O F I T

u

N Q v

v T H K

L O T

u F I N Q

H K u

F I a

F I a

p = −

F I u

F I F I F I T p v

F I F I H K

H K F I T

p v

H K = F I p RT 2 a

F I L O

v T H K v T p

G J F I G J

M M M

P P P


or = 1 M

−v −b P .

Rv2

(v −b) . (Ans.)

RTv 2a(v b)

(v −b)2

v3

L O Also, K = –

1

Hv

K = – v

M2a

1

RT P =

RTv

2 (v

a v

2

b 2

. (Ans.) v3 (v −b)2

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 HT 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 v

2

or Hp

Kv

= v −b

or G 2 p J = 0

v

Now F dc I

= T H

T

p

K

Hence Fv

IT

= 0

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,

H p 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 HT Kv

− pP dv

dharm

\M-therm\Th7-1.pm 5

v

R

RT 2a −

L

N

M M M

O

Q

P P P

3 2 − −

F I 1 M P v b − v p T −

N Q M P M P

3 )

2 ( ) − −

v u F I F I p

R

v

p = −

F I T

R

F I H K 2 T

F I dv

v T H K G J

G J 2

2 v H K G J c v

v

F I a

F I L O p

N Q

358

But,


F p I L R RT a UO T

v T v −b v2

v −b

2

du c 2

dT 2 L

T F R I

−pOdv

1 1 1 = cv z2

dT z2 LT

Fv −b

I−

Rv −b

−a V

Odv

= cv

2

dT 2

Mv −b

−v −b

a P dv

2 2 c dT .dv

1 1

u2 – 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 + HT Kv

dT

(dp)T = H v KT

dv + 0 as dT = 0 ...(2)

From equation (1),

(dh)T = Nv −T HT K

pQ(dp)T.

Substituting the value of (dp) from eqn. (2), we get (dh)T = N

v −T HT Kp

OH

v KT

dv

= Nv H v K

T

−T HT Kp

H v KT Q dv ...(3)

Using the cyclic relation for p, v, T which is

F v I FT I Fp I T

p p

v v

T

F v I G pJ −G p J

p T v

dharm

\M-therm\Th7-1.pm 5

H K − S V T W N Q M P v

= = R

v b v z z z − H K G J N Q

M P M P

T W N Q M P

v

L O N Q z z

a

v z z

v v G J

R RT

v H K G J S U M P

2 1 1 RT RT

2 1 1

= v 2

H K

L O p T H K N Q M P

H K p H K T

F I F I p p

F I p

L O v F I M P M P T

v p L M M Q

P P F I F I

F I F I F I L O

p v p M P M P

H K H H G G K K J J − 1

F I F I H K H K H K T v T


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)

HT Kv

= T MH

RT −

a

K P v

= v −b

...(6)

Substituting the values of eqns. (5) and (6) in equation (1), we get (dh)T = Mv R−

(v −b)2

v

aUT H v

R

bK P dv

1

(dh)T = – RT 1 (v −b)

2 dv + 2a 1

dv + RT

1 (v −b)

(h2 – h1)T = – RT Mloge Fv −b

J −b Rv −b

−v −b

UP

– 2a F 1

−1 I

RT log F v2 −b I

2 1 1

= bRT M(v2 −b) −

(v1 −b) Q – 2a Mv2

−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

ds = cv

2

MdT

P R 2

(v

dv

b)

s2 – s

1 = cv log

e 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-therm\Th7-1.pm 5

F I

F I

L O T T v H K H K N Q

F I H K v H K = −

N Q M P RT 2a

F I F I L O − G J N Q M P

v b v 2 p R

RT S T

V W −

F I G J

L N M

O Q P 3

2

z z z z

2 v 2

v 2 2 dv 2

v b 2 1 2 1

1 1 − H G I

K S V T W L N

O Q

v v v b e − H K H K G J G J

L O L O 1 1 N P

1 1

N Q

T T

H K

dT R

z L O z z

1

2

T N Q − 1 1

L O L O T T

2

N Q 1 v −b M P

p T 3


Solution. The general equation for finding dh is given by

dh = cp dT +

Lv −T

FT

Ip

O dp

2dh = 0 + z 2 MR

v −T FT

Ip

UPT

as dT = 0 for isothermal change.

From the given equation of state, we have

HTK = R

3C ...(i)

h2 – h1 = z 2

MRF RT

−C I

−RT

−3C U

dpPT

= M1

G − 4C J dpPT

−4C

[( p2 − p )]

The general equation for finding ds is given by

ds = cp

dT −H

v

Kp

dp

z2ds =

L−

2FT

Ip

dpO

T

as dT = 0 for isothermal change.

Substituting the value from eqn. (i), we get (s2 – s1) = M

2

−G R

3C Idp

O

= – R loge

F p2 I

– F 3C I

(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 = HT Kv

dT + Hv KT

dv

Also u = f(T, v)

Then du = HT Kv

dT + H v KT

dv = cv dT + H v KT

dv

Also, du = Tds – pdv

Tds – pdv = cv dT + Hu

KT

dv

ds = cv T

T MF

v

IT

pP dv

dharm

\M-therm\Th7-1.pm 5

v H K N Q

M P M P

z L O

1 v

H G K J S | T |

V | W | N Q M P 1

F I v

p p T 4

p p T T H G K J S | T |

V | W |

L O N Q M P 3 3

1

F I L O z

H K N Q 2

3 3 1 T T

T

F I T T

1 H G K J N M M Q

P P z

v

1

F L z H K J N Q P p T T

4 1 p 1 H K G J

T 4 H K v p

F I F I s s

F I F I F I u u u

F I v

L O dT u H K N Q

1


Equating the co-efficients of dT in the two equations of ds, we have

cv s T T v

cv

= T Hs

K Fcv I

2s

v T Tv

From eqn. (7.20), F s I F p I v

T T

v

2s

2 p

vT T2

Fcv I F 2p I

v T T2

v

Also p = RT

...(Given)

F p I R

T v

FT

p Iv

0 or G v JT

0

This shows that cv is a function of T alone, or cv is independent of pressure.

Also, cp = T Hs

Kp

H p K T Tp

From eqn. (7.21), F sI

−F v I

T p

2s 2v

pT T2 p

Gcp 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-therm\Th7-2.pm 5

p

H K G J

H G K J

F I H G K J

F I T

v

H K G J T

H K H K F I

H G K J

v

H K G J H G K J T

v

H K v

H G K J

2 2

F I H K

c v

F I T

F I G J c s p

T

2

H K H K p T

F I −

H K G J

F I F I H K H K p

v T

p

F I G J c

p


Example 7.8. Using the first Maxwell equation, derive the remaining three.

Solution. The first Maxwell relation is as follows :

H v Ks

−H s Kv

(1) Using the cyclic relation

H v Ks

. H s KT

. HT Kv

= – 1

Hv KT

= – H v Ks

. HT Kv

Substituting the value from eqn. (i) in eqn. (ii), we get

Hv KT

= H s Kv

. HT Kv

Using the chain rule,

H s Kv

. HT Kv

. H p Kv

= 1

Substituting the value of eqn. (iv) in eqn. (iii), we get

Hv KT

= HT Kv

This is Maxwell Third relation.

(2) Again using the cyclic relation

vI

p v

v s

s p

v F pI F v I s

p s v

p s

Substituting the value from eqn. (i) into eqn. (v)

HsKp

= H v Ks

. HpKs

Again using the chain rule,

H v Ks

. HpKs

. HT Ks

= 1

Substituting the value of (vi) into (v), we get

HsKp

H p Ks

This is Maxwell second relation.

(3) HT Kp

. 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

F I F I T p

T v s

s T s

s p s

p s T

F I F I s p

F I F I F H K H K H K s p

. . = – 1

H K = – H K H K .

v T v

T v p

F I F I v T

v T p

\M-therm\Th7-2.pm 5


F v I F p I F v I T

p T v

p T

= – H s Kv

HT Kv

HpKT

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

T s T

= TH v K s

. H sKT

. HTKvW H pK

T

= – HpKT

Hv

Kp

= – HpKT

This is Maxwell fourth relation.

Example 7.9. Derive the following relations :

(i) u = a – T FT

I (ii) h = g – T

F g Ip

(iii) cv = – T G2a

J (iv) cp = – T F 2 g J

v p

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 + HT Kv

dT

Also da = – pdv – sdT

Comparing the co-efficients of dT, we get

HT Kv

= – s

Also or

Hence

(ii) Let

Then

Also

a = u – Ts

u = a + Ts = a – T HT Kv

u = a – T HT Kv

. (Ans.)

g = f(p, T)

dg = HpKT

dp + HTKp

dT

dg = vdp – sdT

Comparing the co-efficients of dT, we get

HT Kp

= – s

dharm

\M-therm\Th7-2.pm 5

F I

F I

F I F I

H K = – H K H K .

F I F I F I F I p s s v

H K T = H G K J H H H G G G K K K J J J R U F I F I F I F I F I S V T v s s s

F I

T F I s

H K a

v H K T

F I I H K 2 T H G K 2

T

F I F I a a

F I a

a

a

g g

F I g


Also h = g + Ts = g – T HTKp

Hence h = g – T HTKp

. (Ans.)

(iii) From eqn. (7.23), we have

cv

= T Hs

K

Also F a I

= – s

v

or HT Kv

= – HT2 Kv

From eqns. (i) and (ii), we get

cv

= – T GT2 J

v

. (Ans.)

(iv) From eqn. (7.26), we have

cp = T Hs

Kp

Also HTK = – s

or FT

Ip

= – H2g

Kp

From eqns. (i) and (ii), we get

cp = – T HT2 Kp

. (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 = Hs

Kp

. dT + HpKT

dp

As per Maxwell relation (7.21)

HpKT

= – HTKp

Substituting this in the above equation, we get

ds = Hs

Kp

dT – Hv

Kp

. dp ...(i)

The enthalpy is given by

dh = cpdT = Tds + vdp

Dividing by dT at constant pressure

HTKp

= cp = T HT Kp

+ 0 (as dp = 0 when pressure is constant)

dharm

\M-therm\Th7-2.pm 5

F I

H K T

F I F I

F I

F I

F I p

F I

F I

F I g

F I g

T v

s G J 2 a

H K 2 a

T

g

H K s G J 2 T

G J 2 g

F I F I T

s

F I F I s v

F I F I T T

F I F I h s


Now substituting this in eqn. (i), we get

But

ds = cp

dT −H

s

K . dp ...(ii)

1 F v I

v T p

Substituting this in eqn. (ii), we get

ds = cp

dT – vdp (Ans.)


(i) H p K = Tv

(ii) H v K = – c K

.

where = Co-efficient of cubical expansion, and

K = Isothermal compressibility.

Solution. (i) Using the Maxwell relation (7.19), we have F T I FvI F v I FT I s s

p T p

s p

Also cp = T HTK

From eqn. (7.34), = v

FT

Ip

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 HT K

K = – 1 F v I

F T I T F p I v c T

Also FpI F v I FTI

= – 1 T p v

i.e., Hp

Kv

= – HvKT

HTKp

= – H −vK K v = K

(Eqn. 7.23)

(Eqn. 7.36)

dharm

\M-therm\Th7-2.pm 5

p G J

T T

= H K

H K p H K = = H K H K F I

F I

F I

F I

v p H K

s v v H K H K

F I F I F I F I

F I

T

F I F I c p

T T

s s v T

s

p

1 v H K

T c p

T Tv p

H K v = – = – H K H K H K s

v

T

Then = –

H K H K H K v T p

T p v 1


HT

Ks

= −T

. (Ans.)

+Example 7.12. Derive the third Tds equation

Tds = cv H p Kv

dp + cp Fv

Ip

dv

and also show that this may be written as :

Solution. Let

Then

or

But

Hence

Also

and

Tds = cv

Kdp + v

dv.

s = f(p, v) ds = Fp

I dp + H

s

K dv

Tds = T F s I

dp + T Hs

K dv

= T Hs

Kv

HT

Kv

+ T F s I

p

FT Ip

dv

Hs

Kv

= T and Hs

Kp

= T

Tds = cvHp K dp cp Hv K dv ...Proved.

Fp

I = F pI

−

F v I = – H v K Fp

I =

K

v T

T p

H v Kp

= 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 HTKp

dp. (U.P.S.C. 1988)

Solution.

where cp = T H TKp

Also,

s = f (T, p)

Tds = T HT Kp

dT + T HpKT

dp

Hs

KT

= – F v I

p

...(i) ......Maxwell relation

Substituting these in eqn. (i), we get

Tds = cp dT – T HTKp

dp. (Ans.)

dharm

p

H K v

F I p

H K v

F I v p

H K H K F I

v p

F I F I

F I

F I

F I F I

F I v c K v

F I T H G K J T

c

s v

p G J

F I F I T p T v

F I T

c v T

c p

T T

T

v H K H K

1 T

p H K H K v

T

T 1

c v p

F I v

F I G J s

s s

F I p H K T

F I v

\M-therm\Th7-2.pm 5


Example 7.14. Derive the following relations

F T I TG p J − p

v

Solution. F T I

can be expressed as follows : u

H v K = F vIKv = – F uI

u T

T v

Also Tds = du + pdv

or du = Tds – pdv

or Hu

K = T Hs

K – p H v K

or F uI

= T F s I

– p

T T

or Hu

Kv

= T Hs

K v

Dividing eqn. (i) by eqn. (ii), we get

F T I THvK −p

v u s

T v

Also cv = T HT K

and Fv

IT

= FT

Iv

... 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) Hh

KT

= v Hp

KT

+ T Hp

Kv

(ii) Hh

KT

= v – T HTKp

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 K v H K v

F I F I

T

F I

H F I

K F I

v

F I H K T T

F H

I K

F I F I F I F I F I

F I H K

u T

c v

v

H K = .

H K v

F I F I F I T

u

− H H K

H K H K

T u u v

T

v T v

T u

T

T T

H K =

s T

s

H G K J s H K p

T v v p v

RT a

v −b v

\M-therm\Th7-2.pm 5


Solution. We know that

ds = T

dT HT K dv

Also dh = Tds + vdp = T M v dT F p I

dvP + vdp v

i.e., dh = cvdT + T H

p

Kv

+ dv + vdp

Putting dT = 0, we get

G h J T G p J v G pJ ...Proved. T v T

(ii) Hh

KT

H h

KT

Hv

KT

= NT

F

p Iv

vF pI

T Q G

v J

i.e., Fp

IT

= T Hp

K F vI

+ v

Also F p I

v HvJ

T

= – Hv

Kp

Eqn. (i) becomes

HpKT

= v – T Hv

Kp

...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

(v −b )2

v2 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

= – cp

F T I (ii)

F uI = T H

p

K – p T h T

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\M-therm\Th7-2.pm 5

v

F I

L N

O Q

F I

F F F I I 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 K v = –

H K v =

N Q H K

= =

H K =

h = f(T) −

c p v

c T T H K

T

H H H K K K v T v

p v p

T T v p H K H K G J G J M P

H K

H K h F I G J G J T p

H K G J G K T p T

h T

T

H K v = M P v T

= – =

F I F I H K T

p H K H K H K p p v T v


With the aid of eqn. (ii) show that

H pKT

= – T HT K – p HpKT

The quantity cp

F T I 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 H3C

K −b .

Solution. We know that Fp

I = – cp

Also = cp

L HT K

p

−vO

F hI

= – T F v I

−vO

= 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 = HT Kv

dT + H v KT

dv

= cv

dT + H v KT

dv ...(i)

du = Tds – pdv

Substituting the value of Tds [from eqn. 7.24], we get

du = cv

dT + T F p I

v dv – pdv

= cv dT + MT Hp

Kv

− pP dv ...(ii)

From (i) and (ii), we get

Hu

K = T HT K – p ...Proved.

F uI F uI F v I p

T v

T p

T

or Fp

IT

= HpK T

T HT K v

− pQ

or H pKT

= T FT

Iv

Fp

IT

– p Gp

JT

...Proved.

dharm

\M-therm\Th7-2.pm 5

H K T

N Q M P L N Q M P

F I

F I F I

F I F I

F I

F I F I F I u v

p

v

H K p

F I p T 2 T 2

G J

h

T 1 v F I M P

H K p T H K M P H K T

T

h T

u u

u

H K G J T

F I L O T G J

N Q

F I F I v

T G J p

v

Also = H K H K H K G J G J

F I F I L O H K G J G J G J

N M P u v p

F I G J u p v H K H K G J G J F I

H K v

370

We know that

or

Also

and

Fp I F v I FT I v

T T

p p

v

p v v

T p T

Fp

IT

= – T FT

Ip

– p Fp

IT

= 1

M F u I

p

−vP


...Already proved.

...[Eqn. (7.46)]

Now v – b = RT

−C

...[Given]

HTKp

= R

2C

Substituting this value in the expression of above, we get

= 1 L

T HR

2C

K −vO

or cp = T G R

2C J – RT

C

– b = 3

2 −b

or cp H p Kh

= T 2

– 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 (p22 – p1

2)

= – 0.114 10

−3 8.610

−12

[(800 × 105)2 – (20 × 105)2]

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\M-therm\Th7-2.pm 5

F H G I

K J − F F H H G G I I

K K J J T v p

H K H K H K = – 1

H K v

H K H K G J G J u v

T c T p H K

L O M P N Q

p T 2

F I v

p T 3

3 c p T p

F I G J N M

Q P

F I p T H K 3 p T 2

C

T

F I G J T 3 C

–

z

v p H K

z z

2

2


= – 0.114 8.610

−15

× 1010 [(800)2 – (20)2]

= – 0.114 8.610

−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 :

Gp

JT

= – Hv

Kp

= – v

FT

Ip

= – vas 1

Hv

Kp

=

(ds)T

= – v (dp)T

Integrating the above equation, assuming v and remaining constant, we get

s2 – s

1 = – 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 :

cp – c

v =

2Tv ... [Eqn. (7.38)]

(5 10−5

)2

(15273) 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-therm\Th7-2.pm 5

2

2

F I H K s v v

H K T v T F I F I

K

F I dp 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 G T2 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

103

× loge

F 273I = 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 fg fg

dT v gT (vg −v )T

Since vf is neglected, therefore eqn. (i) becomes

dp fg

dT vgT

...(given)

...(i)

Now, log10

p = 7.0323 – 3276.6

– 0.652 log10

T

dharm

\M-therm\Th7-2.pm 5

= h v .

z z p = h

f T

F I h v fg

T 1 H K

F I 5 b g

263 H K

0 1 . 263 H K

= = h h

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 ×

T 2 – 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,

log10

p = 7.0323 – 3276.6

– 0.652 log10

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 × 0.110

5

– 0.652 × 0.110

5

(523)

563.17 = 275.75 – 12.46

g

i.e., vg = 2.139 m3/kg. (Ans.)

HIGHLIGHTS

1. Maxwell relations are given by

F T I = –

F pJ ; F T J =

Fs

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 =

vT2

; cv = T F s I

v

; cp = T GT

Jp

.

3. Joule-T hom son co-efficien t is expressed as

= H p Kh

.

dharm

\M-therm\Th7-2.pm 5

T

T

. = –

dp p p

h p p

T

T

T

2 523

v

H G K J v s H H G G I I

K K s p v s H G K J

v

p

H H G G K K J J − H H G G K K J J ; .

F I K H K G J

T H K s

F I G J T


4. Entropy equation s (Tds 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 :

G u J = T G p J – p T v

du = c dT + TF p I

−pV dv

v

Gp 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) c

p = T

F T Ip

(c) cp

= T FT

Ip

(d) cp

= T FT

Ip

.

2. The specific heat relation is

(a) (cp – cv) = vT2

(c) (cp – c

v) =

pTK

3. The relation of internal energy is

(a) du = F K

cv

I dp + H

cp −pK dv

(c) du = H cpJ dp + G v

−vJ dv

4. Tds equatio n is

(a) Tds = cvdT +

Tdv

(b) (cp – c

v) =

vTK

(d) (cp – c

v) =

v T.

(b) du = G K c

I dp + Hcp pK dv

(d) du = HK

cpK dp + F cv −p

I dv.

(b) Tds = cpdT – T

dv

(c) Tds = cvdT +

TK dv (d) Tds = c

vdT +

Tdp.

Answers

1. (a) 2. (a) 3. (a) 4. (a).

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\M-therm\Th7-2.pm 5

F I

F F H H

I I K K v T

v T H K R S T

U W

F H

I T

F I H K

p

v − F

H G I K J S V | |

K 2

2 K

H K

v H G K J T

G J v

G J J h

T

T W

p

H H G G K K J J T s

H H G G K K J J v v

G J v F I G J

K F G I K

c p F I H K

2

v F H K J v

F I G J

F I G J v H K G J

K K

K


1. Define the co-effic ien t of :

(i) Volum e expansion

EXERCISES

(ii ) Isothermal compressibil ity

(iii ) Adiabatic compressibility.

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.

4. A substance has the volume expansivity and isothermal compressibility :

= T

; K = p

Find the equation of state. NAns.

pv constant

Q 5. For a perfect gas, show that the difference in specific heats is

c

p – c

v =

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

G h J = 0. T

7. Using the cyclic equation, prove that

H T Kv

= KT

.

8. Prove that the change in entropy is given by

ds = cv

L KT . dp

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 -Tho m son 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 – c

v) for this change of state.

The followin g data may be assumed :

Volum e expansivity () = 5 × 10–5/K

Isothermal compressibility (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]

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\M-therm\Th7-2.pm 5

F I

v

1 1

L O T M P

R

H K p F H

I K p

F I G J p

T c v N M Q P

F H

I K T p H H H H G G G G K K K K J J J J

p p v T

R

Engineering

Cap 7 relaciones termodinamica