SECOND LAW OF THERMODYNAMICS AND ENTROPY 265

dharm

/M-therm/th5-3.pm5

If equation (5.37) is divided by dt, then it becomes a rate equation and is written as

dS

dt

≥

1

T 0

dQ

dt

. + s
dm
dt
∑. ...(5.38)

In a steady-state, steady flow process, the rate of change of entropy of the system

dS

dt

F

H

I

K

becomes zero.

∴ O ≥

1

T 0

dQ

dt

+ s dm

dt

∑.

or^1

T 0

Q

+ Σs. m& ≤ 0 ...(5.39)

where Q

dQ
dt
and m& =
dm
dt

For adiabatic steady flow process, Q

= 0

∑sm.&^ ≤^0 ...(5.40)

If the process is reversible adiabatic, then

∑sm.& = 0 ...(5.41)

5.19. The Third Law of Thermodynamics

l The third law of thermodynamics is stated as follow :
‘‘The entropy of all perfect crystalline solids is zero at absolute zero temperature’’.
l The third law of thermodynamics, often referred to as Nernst Law, provides the basis
for the calculation of absolute entropies of substances.
According to this law, if the entropy is zero at T = 0, the absolute entropy sab. of a substance
at any temperature T and pressure p is expressed by the expression.

sc

dT

T

h

T

c dT

T

h

ab ps T

TT sf

s

T pf

TT fg

g

sf

s

fg

=++ +

==

zz 0

(^12) c dT
T pg T
T
zg
...(5.42)
where Ts = Tf 1 = Tsf = Tsat ...... for fusion,
Tf 2 = Tg = Tfg = Tsat ......for vaporisation
cps, cpf , cpg = Constant pressure specific heats for solids, liquids and gas,
hsf , hfg = Latent heats of fusion and vaporisation.
Thus by putting s = 0 at T = 0, one may integrate zero kelvin and standard state of 278.15 K
and 1 atm., and find the entropy difference.
l Further, it can be shown that the entropy of a crystalline substance at T = 0 is not a
function of pressure, viz.,
∂
∂
s
p T
F
HG
I
KJ

= 0
0