SECOND LAW OF THERMODYNAMICS AND ENTROPY 285
dharm
/M-therm/th5-4.pm5
- Expression for entropy change in terms of pressure, volume and temperature.
From eqn. (1), we have
s 2 – s 1 = b T
TkT T a v
vb v
eeev
log^2 ( ) log log
121 2
12
1F
HGI
KJ+−+
F
HGI
KJ−
F
HGI
KJ= av
vb
T
Tv
vlogee^2 log kT T( )
12
11
221F
HGI
KJ+×
F
HGI
KJ+−or, s 2 – s 1 = a
v
v
b p
p
logee^2 log kT T( )
12
1
21F
HGI
KJ+F
HGI
KJ+− ...(ii)- Expression for entropy change in terms of pressure and temperature only.
Again, from eqn. (1), we have
s 2 – s 1 = b T
TkT T a v
vb v
eeev
log^2 ( ) log log
121 2
12
1F
HGI
KJ+−+
F
HGI
KJ−
F
HGI
KJ= a
T
Tp
pb T
Tv
v
logee^2 log kT T( )
11
22
11
2× 21F
HGI
KJ
+×F
HGI
KJ
+−= a
T
Ta p
pb p
p
logeee^2 log log kT T( )
12
12
1
21F
HGI
KJ−F
HGI
KJ+F
HGI
KJ+−or, s 2 – s 1 = a T
T
ba p
p
logee^2 ( ) log kT T( )
1
2
121F
HGI
KJ+−
F
HGI
KJ+− ...(iii)lDerivation of the formula T b va-bekT = constant for the adiabatic expansion of gas :
We know that, ds = () ( )abdv
v
bkTdT
T
−++
s 2 – s 1 = avbvbTkT
alog v b T
vkTavb p
RkTav
p
ab
kTeee
eeeeeeblog log log
loglog loglog log−+ +
=+F
HGI
KJ
+=+F
HGI
KJ
+=+
−F
HGI
KJ
+U
V
|
|
|
|W
|
|
|
|=0 for adiabatic expansionThis gives : va pb ekT = constant
pva–b ekT = constant
Tb va–bekT = constant
The above expressions can be obtained by taking kT on right-side and taking the antilog of
the resulting expressions.
Example 5.39. Determine the entropy change of 4 kg of a perfect gas whose temperature
varies from 127°C to 227°C during a constant volume process. The specific heat varies linearly
with absolute temperature and is represented by the relation :
cv = (0.48 + 0.0096 T) kJ/kg K.
Solution. Given : m = 4 kg ; T 1 = 127 + 273 = 400 K ; T 2 = 227 + 273 = 500 K ;
cv = (0.48 + 0.0096 T) kJ/kg K.