entropy change for an ideal gas - IndiaStudyChannel.com

THERMODYNAMICS

THERMO
-

HEAT DYNAMICS
-
MOTION

Conversion of heat into mechanical work & vice versa

Thermodynamics is the branch which deals with quantitative relationship
between heat & other forms of energy.

Limitation of Thermodynamics

1.

Deals with
prop. Like temperature, pressure deals with

macroscopic
quantities of the system not with macroscopic quantities.

2.

Help to predict feasibility of a process but does not tell anything about time
taken for the process to complete or the rate at which process
would
proceed.

3.

It concerns with the initial and final state of system & not concern with the
path by which change is brought about. Throws no ligh
t on the mechanism
of a process i.e. it tells us where the system is going but does not tells how
to get to th
e equilibrium or even if we can get to the equilibrium.

State of simple homogenous system may be completely defined by specifying only
two of the properties out of P
, V

& T

P & T are independent variables

V=
dependent variable, depends upon P &T.

Thermodyna
mic Equilibrium

A system is said to be in thermodynamic equilibrium if macroscopic properties of
system in various phases do not undergo any change with time

T
HERMODYNAMIC

EQUILIBRIUM
ARE OF THREE TYPES:
-

1.

Thermal equilibrium
-
No flow of heat from one portio
n to another

2.

Mechanical equilibrium
-
No work is done by one part of system over
another (if P is constant)

3.

Chemical equilibrium
-
No change in composition should take place.

EXTENSIVE AND INTENSIVE PROPERTIES

Extensive depends upon quantity of matter, mass,
V
olume, energy heat capacity

Intensive properties
-
Depends upon nature of substances and are independen
t of
amount of substances,

density
,

temperature, pressure, refractive
index,
viscosity.

Thermodynamic Processes
-

When system
changes from one initial to final stage:
-

1)

Isothermal
-

temperature constant

2)

Adiabatic
-
no heat flow

3)

Isochoric
-
volume constant

4)

Isobaric
-
pressure constant

Reversible Processes

1)

Arrived infinitesimally slow

2)

Changes occurring in the direct process can be reversed

and system
remains in equilibrium.

3)

Work done in a reversible process is the maximum work done by a system.

Explanation



Work done by the system W exp=
-
P∆V



Work done on the system (cont.)=+
P∆V



If gas expand freely against vacuum P ext=0



No work is done by t
he gas



If work done by the system = P ext. X∆V for a given increase in volume (∆V),
the work done can be maximum if P.ext is maximum.



For this external pressure should always be less than internal pressure of
gas

These are the conditions of a reversible pr
ocess. Hence maximum work is done
if carried out under reversible condition.

In reversible processes W is replaced by W max

Contraction of gas under reversible conditions
-

The external pressure should
be infinitesimally greater than internal pressure of th
e gas. This is maximum
pressure to compress the gas under reversible conditions for given decrease in
volume (∆
V
) work

done under reversible conditions is maximum work.

Heat:
-

mode of energy exchanged between system and surrounding as a
result of the diff
erence of temperature between them q.

-
q = heat is given by the system

+q= heat absorbed by system

Unit
-
Calories.

SI
-
joules

1 calorie=4.184 J

1 J=2390 cal.

State
Function
:
-

Value of thermodynamic quantity depends upon its initial
state and in final state
does not depend on path. E.g. mass pressure, volume,
temperature, internal energy

Thermodynamic quantity, the change in the value of which during a process
depends upon path followed
-

is some time called path function e.g. heat &
work.



When
a process occur
s

slowly,

the total change in the value of a
thermodynamic quantity of a thermodynamic quantity can be obtained
by integration only if it is a state function.



A thermodynamic

quantity

can be differentiated only if it is a state
function



Internal energy ca
n be differentiated and integrated and represented by
dE



Heat & work are not state function. Cannot be differentiated

so
represented as §q and §w or dq and dw for small change of heat and
work



Internal energy is a definite property and is an exact
differential whereas
work and heat are not.

Zeroth law
:
-

When two bodies have equalit
y of temperature with a third
body they in turn have equality of temperature with each other.

This Law is the basis of temperature measurement. Every time a body has
equa
lity of temperature with thermometer. We can say that the body has
the temperature we read on the thermometer
.

First law:
-

Energy can neither be created nor destroyed

Or

Total energy of an isolated system remain constant, although it may
undergo
transformation from one form to the other

Limitations of the first law:
-

1.

Only states the exact equivalence of various forms of energy involved
but provides no information regarding the feasibility of the process.

2.

It does not say whether a gas can diffuse f
rom low pressure to high
pressure.

Mathematical relationship:
-

∆
E =q
-
w

+q=Heat gained by the system stem

-
q=Heat lost by the system

+W=work is done by the system

-
W=work is done on the system

For a cyclic process

E1=E2 or ∆
E=0

q
-
w=0

q=w Work done =
Heat absorbed

It is impossible to construct a perpetual
motion machine
i.e.

a machine that
would produce work continuously without consuming energy

Suppose

-

Q joules heat converted in to mechanical work

-

When same amount of work converted in to
heat, heat
produced is Q1
joules


If Q≠q1


Q>Q1


Q<Q1

This

indicates though initial state has regained, a certain amount of heat
has been created or destroy

which is against the first law.

Hen
ce there must exist equivalence between heat and mechanical work
(another definition of first law of thermodynamics).

Einstein showed that energy can be created by the destruction of mass
.


E=mc
2

(law of conservation of energy)

It shoul
d be called as the law of conservation of mass and energy.

Total mass and energy of system remain constant (relationship used in
nuclear chemistry not in thermodynamics) because only in that case
with very small amount of mass large amount of energy is cre
ated.


Heat capacity
-
amount of heat required to increase temperature by 1
0
C

Molar heat capacity at constant volume

Cv=(dE/dT)
v





Molar heat capacity at constant

pressure

Cp=(dH/dT)
p






Relation between Cp and Cv


H=E+PV

∆
H=∆
E+∆(PV)


When temperature
is changed

(∆H/∆T)=(∆E/∆T)+∆(PV)/∆T

Cp=Cv + ∆(PV)/∆T


For 1 mole of an ideal gas

P2V2=RT2

P1V1=RT1


∆
(PV)=R
0
(T2
-
T1)=R∆T

We know that

Cp=Cv+∆(PV)/∆T

Cp=Cv + R∆T/∆T

Cp = Cv + R


For 1 mole of gas for solids and liquids

∆
H=∆E

∆
PV=0

Cp=Cv


As entropy is state function a small change in its value can be represented by ds


dS=q
rev
/T

Units
-

cal K
-
1

Mol
-
1


-
JK
-
1

Mol
-
1



Second law of thermodynamics

1.

It is impossible to construct a machine working in cycles which will transfer
heat from

lower
to
a higher temperature without the aid of an external
agency.

2.

It is impossible to convert heat from a source into work by a cyclic process,
without transferring it into a colder sink.

3.

Heat cannot be completely converted into work without changes either in
th
e surroundings or in the system.

4.

Heat cannot pass itself from a colder to a hotter body.

5.

All spontaneous processes tend to take place in one direction and they
cannot be reversed.


Entropy change in a reversible process:
-

Suppose a process occur
s

under completely reversible condition i.e. heat
absorbed and lost reversibly.

1.

Entropy change of system will be given by

∆
S system= q
rev
/T

2.

Entropy change of surrounding



∆
S=
-
q
rev
/T

Total entropy changes=∆S system + ∆S surrounding




= q
rev
/T
–

q
r
ev
/T=0

In a reversible
process,

the net entropy change for the combined system &
surroundings is zero i.e. there is no net change in entropy

Entropy changes in a
n

irreversible process
:
-

∆
S=∆S system + ∆S surrounding


-
q/T
1
+q/T
2


q(T
1
-
T
2
)/T
1
T
2

T
1
>T
2

; T
1
-
T
2
=+ve

∆
S surrounding>0

Entropy change should be positive



ENTROPY CHANGE FOR AN IDEAL GAS

Entropy is a state function and varies with the state of the system. Its value
depends on any two of the three variables

T,P and V

The temperature T of gas

is usually taken as one of the variable

Second variable to be considered is either V or P


CASE 1 WHEN T &V ARE VARIABLES:
-

Suppose n moles of an ideal gas occupy a value V at temperature T and pressure
P.

If the system absorbs an infinitesimally small a
mount of heat dq reversibly at
temperature T, the increase in entropy is given by:
-

ds=dq
rev
/T (1)

According to 1
st

law of thermodynamics


dq
rev
=dE
-
dw (2)

If the work involved due to expansion of gas


dw= +pdv

dv= Infinitesimal increase in
volume against a pressure P. Further for n moles of
an ideal gas



Cv=(dE/dT)
v


dE=nCvdt

Substituting this value in equation (2)


dq
rev
=nCvdt + pdv



=nCvdt + nRTdv/v (PV=nRT)

Put value of dq in equation (1)


ds=

(
nCvdt + nRTdv/v
)/
T


ds=nCvdt/T + nRdv/v

(3)

For finite change of state of system, the entropy change ∆S can be obtained by
integrating equation (3) between limits of initial state (1) and final state (2) i.e.
between limits S
1
,S
2
,T
1
,T
2
,V
1
,V
2
. Assuming Cv remain const
ant over the
temperature range

S2 T2 V2


∫ds = nCv ∫(dT/T) + nR ∫(dV)/V

S1 T1 V1

S2
-
S1 = nCv[InT2
-
InT1] + nR[InV2
-
InV1]

∆
S=2.303nCv log
10
(T2/T1)

+ 2.303nR log
10
(V2/V1) (5)


CASE 2 WHEN T & P ARE THE VARIABLES


Let P1 be the pressure of an ideal gas in
the initial state.

P2 in the final state.

P1V1=nRT1 and P2V2=nRT2

P1V1/T1=P2V2/T2

V2/V1=P1T2/P2T1

Subtitling the value in equation
(4)

∆
S=nCv In(T2/T1) + nR

In(P1T2/P2T1)


=nCv In(T2/T1) + nR In(T2/T1) + nR In(P1/P2)

(As Cp
-
Cv=R)


=(nCp
-
R) In(T2/T1) + nR In(T2/t1) + nR In(P1/P2)


=nCp In(T2/T1) + nR In(P1/P2)

∆
S=nCp In(T2/T1)
–

nR In(P2/P1)

∆
S=2.303 nCp log
10
(T2/T1)
-
2.303
nRlog
10
(P2/P1) (6)


FREE ENERGY OF SYSTEM
-

Free energy of system (G) is thermodynamic state function, which is related to
enthalpy and entropy as




G=H
-
TS

Free energy change (∆G) of the system (or reaction) a measure of an energy
available for doing us
eful work. Thus energy available as useful work= Total
Energy available
-
Non Available form of energy



i.e. G=H
-
TS

The change in free energy
between two states of system


∆
G=∆H
-
T∆S (1)


PHYSICAL SIGNIFICANCE OF FREE ENERGY :
-

From the law of conservation of energy, the total heat supplied to the system.


q=∆E + Wexpansion + Wnonexpansion




=∆
E+P∆V +Wnonexpansion


q=∆H + Wnonexpansion


(2)

Now in a reversible change at constant temperature T, we have


∆
S=q/T


q=T∆S

From equatio
n (1) and (2)


T∆S=∆H + Wnonexpansion


∆
H
-
T∆S=
-
Wnonexpansion


-
Wnonexpansion=(∆G)
TP

i
i.e. is a measure of non expansion or useful

work obtainable in a reversible
reaction at constant temperature, in other words

Decrease in free energy during a process is
equal to useful work obtainable from
the process.


VARIATION OF FREE ENERGY WITH TEMPERATURE AND PRESSURE :
-

The mode of variation of free energy with temperature and pressure can be
obtained as follows

By definition



G=H
-
TS

H=E+PV

G=E+PV
-
TS

On different
iation we have:
-

dG=dE+Pdv
-
Vdp
-
Tds
-
SdT


(1)

For an infinitesimal change First Law of Thermodynamics can be written as:
-

dq=dE
-
dW

if work is done only due to expansion

-
dW
-
Pdv

dq=dE+Pdv

For an infinitesimal stage of a reversible process

dS=dq/T

TdS=dq=dE+Pd
v

Equation (1) becomes

dG=TdS+VdP
-
TdS
-
SdT

dG=VdP
-
SdT

(2)

Equation (2) gives variation of free energy when a system undergoes reversibly a
change of pressure as well as change of temperature if pressure remains constant

dp=0

equation (2) becomes

dG=SdT

(dG
/dT)
P
=
-
S


(3)

Equation (3) gives variation of free energy with temperature.


FREE ENERGY CHANGE OF AN IDEAL GAS IN ISOTHERMAL REVERSBLE CHANGE
(AT CONSTANT

TEMPURATURE):
-

G=H
-
TS

H=E+PV

G=E+PV
-
TS

Upon differentiation we get:

dG=dE+PdV+VdP
-
TdS
-
SdT

(1)

From 1
st

law of thermodynamics

dq=dE
-
dW


=dE+PdV


(2)

Equation (1) becomes

dG=dq+VdP
-
TdS
-
SdT

(3)

For a reversible process

dS=dq/T

or TdS=dq

or dq
-
TdS=0


(4)

From equation (3) and (4) we get

dG=VdP
-
SdT

(5)

If temperature is kept constant (isothermal change)

dT=0, then the equation (5)
takes the form:
-

dG=VdP


(6)

Free energy change from state

(1) to (2) at constant temperature is given by



P2


P2

∆
G=G1
-
G2= ∫VdP= ∫PVdP/P


(7)




P1

P1

For n moles of an ideal gas PV=nRT



P2

∆
G=nRT ∫
dP/P



P1

∆
G
=nRTInP2/P1

∆
G
=2.303nRTlogV1/V2


GIBB’S HELMHOLTZ EQUATION:
-

J.W Gibbs andHelmhaltz

derived important
equation, which can be applied to both physical and chemical changes occurring
in a system




GIBB’S HELMOLTZ EQUATION
INVOLVING FREE ENERGY FUNCATION:
-

Gibb’s free energy G is given by

G=H
-
TS

(1)

H=E+PV

G=E+PV
-
TS

ON DIFFERENTIATING THE EQUATION

dG=dE+PdV+VdP
-
TdS
-
SdT

(2)

According to fi
r
st law of thermodynamics

dq=dE
-
dW


=dE+PdV



Equation (2) becomes

dG=dq+VdP
-
TdS
-
SdT


(3)

For a reversible process

dS=dq/T

or TdS=dq

or dq
-
TdS=0


(4)

From equation (3) and (4) we get

dG=VdP
-
SdT


on differentiation with respect to temperature at constant pressure

[dG/dT]
P
-
S

(4)

If the initial and final states of an isothermal process are represented by S
1

& S
2

We know

∆
G=∆H
-
T∆S


(5)

∆
S=S
2
-
S
1

-
∆
S=S
1
-
S
2

From equation (4)

-
S=(dG/dT)
P

-
∆
S=
-
(dG
1
/dT)
P
+(dG
2
/dT)
p

-
∆
S=[d(G
2
-
G
1
)/dT]
P

-
∆
S=[d(∆G)/dT]
P

∆
G=∆H+T[d(∆G)/dT]
P

(6)


SIGNIFICANCE OF

GIBB’S HELMHOLTZ EQUATION:
-

This equation helps in better understanding of the force that causes the chemical
reaction to proceed forward.

If ∆G=
-
ve=reaction spontaneous


∆
G=0 reaction in equilibrium


∆
G=+ive reaction non spontaneous

From the relation

∆G=∆H
-
T∆S


∆
H

∆
S

∆
G=∆H
-
T∆S

RNM䅒O

Nxam灬e

1

†
-

+

-

S灯p
W
a湥潵o aW all
Wemp

H
2
(g)+Cl
2
(g) 2HCl(g)

C(s)+O
2
(g) CO
2
(g)

2


-

-

(A)
-

(B)+

Spont. At low temp

Non spont At high temp

H
2
O(g) H
2
O(l)

2SO
2
(g)+O
2
(g)
2SO
3
(g)

3


+

+

(A)+

(B)
-

Non spont. At low temp

Spont. At high temp

NH
4
Cl(s) NH
3
(g)+HCl(g)

N
2
(g)+O
2
(g)



2NO(g)

4


+

-

+

Non spont. At all temp.

3O
2
(g) 2O
3
(g)

2H
2
O(l)+O
2
(g) 2H
2
O(l)


APPLICATION OF GIBB’S HELMHOLTZ
EQUATIONS:
-

Calculation of emf of a reversible cell:
-


Suppose at constant P & constant T emf of a reversible cell=E

Faraday=n

i.e. nF coulumbs of electricity is yielded by the cell.

Electrical work obtained = nFE

Net electrical work=decrease in free
energy

nFE=
-
∆G

Substituting in Gibb’s Helmhaltz equation we have


-
nFE=∆H
-
T(d(nFE)/dT)
P


-
nFE=∆H
-
nFT[dE/dT]
P


E=
-
(∆H/nF) +T[dE/dT]
P

By knowing E(emf of cell) and dE/dT

(Variation of emf with t
emperature at
constant P) value of heat of reaction (H)
occurring in the cell can be calculated
.


HELMHOLTZ ENERGY OR WORK FUNCATION (A)

Helmholtz energy or work function is represented by A and is defined as

A=E
-
TS


A=state function




E=internal energy






T=Temperature




S=Entropy of system

Suppose a syste
m goes from states (1) to (2) at constant T. Corresponding to
these two states

A1=E1
-
TS1

(1)

A2=E2
-
TS2

(2)

Therefore

∆
A=A2
-
A1=(E2
-
TS2)
-
(E1
-
TS1)

∆
A=(E2
-
E1)
-
T(S2
-
S1)

∆
A=∆E
-
T∆S


(3)


SIGNIFICANCE OF WORK FUNCATION

A=E
-
TS

(1)

Differentiating we get

dA=dE
-
TdS
-
SdT

(2)

From first law of thermodynamics

dq=dE+Wrev

from definition of entropy

dS=dq/T

TdS=dE+Wrev

dE=TdS
-
Wrev

(3)

Substituting the value of dE from equation (3) to (2) we get

dA=TdS
-
Wrev
-
TdS
-
SdT

=
-
[Wrev+SdT]

At constant temperature, dT=0

Therefore
–
(dA)
T
=
Wrev

1.

A decrease in work function at constant temperature gives the maximum
reversible work done by the system, which includes both mechanical and
non
-
mechanical work.

2.

The spontaneous change at constant volume have dA<0

3.

The concept of ∆A
is

commonly used i
n discussion of solid & condensed
phase processes, where the change in volume can be negligibly small.

4.

Helmholtz free energy is commonly used in consideration of geochemical

problems where huge pressure changes can be involved but volume does
not change si
gnificantly.

GIBB’S HELMHOLTZ EQUATION INVOLVING WORK FUNCTION:
-

Work function is given by

A=E
-
TS

On differentiation we get

dA=dE
-
TdS
-
SdT

dE=dq+W


=dq
-
PdV

dq=TdS

dE=TdS
-
PdV

dA=TdS
-
PdV
-
TdS
-
SdT

dA=
-
PdV
-
SdT

(2)

On differentiation with respect to
temperature keeping volume constant we get

(dA/dT)
V
=
-
S

We know ∆A=∆E
-
T∆S

(1)

∆
S=S2
-
S1

-
∆
S=
-
(S2
-
S1)


=(dA2/dT)
V
-
(dA1/dT)
V

-
∆
S=[d(A2
-
A1)/dT]
v
=[d(∆A)/dT]
V

∆
A=∆E+T[d(∆A)/dT]
V

This equation represents another form of Gibb’s Helmholtz equation involving
work function. Gibb’s Helmholtz equation is of general nature and can be applied
to any physical chemical process. Since all ∆E & ∆A are extensive properties and
depend upon the q
uantity of matter, these equations are applicable only to those
process which occur in a closed system.


SPONTANIETY AND DRIVING FORCE FOR SPONTANEOUS PROCESSES:
-

Tendency of a process to occur spontaneously depends on some energy to drive it
called the dr
iving force

For a reaction to be spontaneous there have to be two driving forces

1.

∆
H to be
–
ve (exothermic)

2.

∆
S to be positive (entropy increase)

Overall of reaction

∆G=
-
ve

∆
H is not the sole criteria of reaction to be spontaneous

Evidences in its favors are
:
-

1.

Feasibility of Endothermic reaction
-
Occurrence of endothermic reaction
result with absorption of heat, system moves from lower energy state to
higher energy states.

e
.
g.

NH
4
Cl(s+aq)


NH
3
(aq)+HCl(aq) ∆H=15.1KJ


CaCO
3
(s) CaO(s
)+CO
2
(g) ∆H=
177.8KJ


HgO(s)

Hg(l)+1/2O
2
(g) ∆H=90.8KJ

2.

Attainment of equilibrium
-
If decrease in enthalpy is only criteria an
exothermic reaction should always get completed. Many exothermic
reactions when carried out in a state of equilibrium cannot go
to
completion.

3.

Reversible Nature of Reaction
-
There are many reactions in which forward
and backward reaction occur simultaneously.

H
2
+I
2
2HI

This forward reaction involves decrease in enthalpy where as backward
reaction involves increase in
enthalpy. From the above discussion it is clear
that decrease in enthalpy help in occurrence of chemical reaction but
cannot be regarded as only criterion of spontaneity.


CLAUSIUS CLAPEYRON EQUATION

This is important equation is applicable to the
equilibrium between two phases of
the same substance. Helpful in study of change taking place between two phases
in equilibrium of a heterogeneous one component system.

This
equation can be derived from second law of thermodynamics.

Let the phase 1 &2 are
of one component system in equilibrium with each other
and the system be a closed one.

G1 molar Gibb’s free energy of system in phase 1 & G2 in phase 2.

Since the system is in state of equilibrium the change in free energy i.e. ∆G should
be zero.

∆
G=G2
-
G1

∆
G=0

G1=G2 (At equilibrium)

(1)

Gibb’s free energy is same in both the phases in equilibrium.

A change in temperature and pressure of system will result in a change in the free
energies of phases under consideration.

Suppose the pressure and temperature of

system are changed

by very small
amounts dP and dT

respectively in such a way that the system still remain in
equilibrium.

Let dG1 ad dG2 be the corresponding changes in the free energy of two.
Phases are still in equilibrium, the changed value of Gibb’s
free energy in
two phases i.e. G1+dG1 & G2+dG2 respectively should again be equal
therefore

dG1+G1=dG2+G2


(2)

on comparing (1) and (2) we get

dG1=dG
2

(3)

Gibb’s free energy is defined by the equation

G=H
-
TS

Since H=E+PV
-
TS

G=E+PV
-
TS

On differentiation w
e get

dG=dE+PdV+VdP
-
TdS
-
SdT

also dq=dE+PdV

dG=dq+VdP
-
TdS
-
SdT

dS=dq/T

dG=TdS
-
VdP
-
TdS
-
SdT

=VdP
-
SdT

(4)

Putting proper subscripts for initial and final equation (4) become

dG1=V1dP
-
S1dT

dG2=V2dP
-
S2dT

Putting the value of dG1 and dG2 in equation (3) we get

V1dP
-
S1dT=V2dP
-
S2dT

(S2
-
S1)dT=(V2
-
V1)dP

dP/dT=(S2
-
S1)/(V2
-
V1)=∆S/(V2
-
V1)

(5)


If ∆H is the enthalpy change for the reversible transfer of one mole of substance


from phase 1 to phase 2 at constant T we have

∆
S=∆H/T

Equation (5) take the form

dP/dT=∆H/T(V2
-
V1)

(6)

This equation is called the CLAPEYRON equation

It is special case of Clapeyron equation for liquid vapour equilibrium.

Suppose we have a one component system having liquid & vapour phase in

Equilibrium.

Let ∆Hvap be the enthalpy of vaporization, V
1 &V2 be volumes of liquid & vapour


phase, T be the temperature of vaporization.

Except at critical temperature V1<<V2, volume of liquid is much less as compared

to that of its vapours at all other temperature V2
-
V1 may be replaced by V2. Thus


the Clapey
ron equation may be written as

dP/dT=∆H/T(V2
-
V1)

(1)

dP/dT=∆Hvap/TV2


(2)

At temperature much less than critical temperature, the vapour pressure is


relatively small. Hence vapour may be treated to be an ideal gas equation can be

applied

PV2=RT

V2=RT/P

Putting the value of V2 in equation (2) we have

dP/dT=∆HvapP/RT
2

d In p/dT=∆Hvap/RT
2

(3)

This expression represented by e
qua
tion (3) is known as Clapeyron C
lausius

equation

(symbol p is generally used to represent Vapor pressure in
thermodynamics, while P
is used to denote total external pressure. Hence P
is replaced by p)

Equation (3) can also be integrated between limits of pressure P1 & P2


Corresponding to temperature T1 &T2

P2

T2

∫d In p=(∆Hvap/R)∫dT/T
2

P1

T1

In(P2/P1)=(∆Hvap/R)[(
-
1/T2)+(1/T1)]

Log
10
(P2/P1)=(∆Hvap/2.303R)[(1/T1)
-
(1/T2)]

(4)

Equation (4) represent
integrated form of Clapeyron Clau
sius

equation


APPLICATION OF CLAPEYRON CLAUSIUS EQUATION

1.

In determination of mo
lar heat of vapourization. If vapour pressures P1
&P2 of a given liquid are known at two temperature T1 &T2. The moles of
heat of vapourizat
i
on

(∆Hvap) can be calculated.

2.

In determination of vapour pressure of a liquid at different temperature. If
the vapo
ur pressure P1 of a liquid at temperature T1 is known then it
boiling point T2 at some pressure P2 can be calculated.

3.

In determination of vapour pressure of a liquid at different temperature. If
the vapour pressure

P
1 of a liquid at temperature T1 is known
, its vapour
p
ressure at some other pressure P
2 can be calculated with the help of th
e

equation
.