TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 255

dharm

/M-therm/th5-3.pm5

Since entropy is a thermodynamic property, we can write

dS dSRIdS

zzzL M

=+ =() ()

() 2 ()

1

1

2

(^0) ...(5.20)
(Subscript I represents the irreversible process).
Now for a reversible process, from eqn. (5.19), we have
()
() ()
dS Q
(^1) L R L T R
2
1
2
zz=
F
HG
I
KJ
δ
...(5.21)
Substituting the value of ()
()
dSR
1 L
2
z in eqn. (5.20), we get
δQ
T
dS
R
L M I
F
HG
I
KJ
+=zz ()
() 2 ()
1
1
2
(^0) ...(5.22)
Again, since in eqn. (5.20) the processes 1-L-2 and 2-M-1 together form an irreversible cycle,
applying Clausius equality to this expression, we get
δδQ δ
T
Q
T
Q
zzzL RIM T
= GFH IKJ + FHG IKJ
2
1
1
2
() () < 0 ...(5.23)
Now subtracting eqn. (5.23) from eqn. (5.22), we get
()
() ()
dS Q
2 MMI T I
1
2
1
zz>
F
HG
I
KJ
δ
which for infinitesimal changes in states can be written as
()dS
Q
I T I

zFHG IKJ
δ
...(5.24)
Eqn. (5.24) states that the change in entropy in an irreversible process is greater than δQ
T
Combining eqns. (5.23) and (5.24), we can write the equation in the general form as
dS ≥ δQ
T
...(5.25)
where equality sign stands for the reversible process and inequality sign stands for the irrevers-
ible process.
It may be noted here that the effect of irreversibility is always to increase the entropy of the
system.
Let us now consider an isolated system. We know that in an isolated system, matter, work
or heat cannot cross the boundary of the system. Hence according to first law of thermodynamics,
the internal energy of the system will remain constant.
Since for an isolated system, δQ = 0, from eqn. (5.25), we get
(dS)isolated ≥ 0 ...(5.26)
Eqn. (5.26) states that the entropy of an isolated system either increases or remains con-
stant. This is a corollary of the second law. It explains the principle of increase in entropy.

5.14. Change in Entropy of the Universe

We know that the entropy of an isolated system either increase or remains constant, i.e.,

(dS)isolated (^) _≥ 0