SECTION XII – ENTROPY CHANGE FOR IDEAL GASES (Exact treatment)
dh
dP
v
T
T
2
2
2 dP
v R
dT
   ds   c p
 R
For an ideal gas, dh  c p dT,
1
1
1 P
T P
T
Recall Gibb’s eqn.: ds 
2
s 2  s1   c p
1

P 
dT
 Rn 2 
T
 P1 
dT
, which depends upon temperature alone.
T
Note that s is zero at absolute temperature of zero.
T
Define an entropy function s   c p
0
2
c
1
p
P 
dT 2 dT 1 dT
 c p
  cp
 s 2  s1  s 02  s10  Rn 2 
0
T 0
T
T
 P1 



Unlike the change in internal energy and the change in enthalpy, the change in entropy of an
ideal gas does not depend upon the temperature alone.

Gibbs eqns.:
du P
ds 
 dv
T T
ds 
(1)
dh v
 dP
T T

For ideal gases: Pv = RT, du = cvdT, dh = cpdT

From (1)  ds  c v
dT
dv
R
T
v

and from (2) ds  c P
dT
dP
R
T
P

For constant cv, (3) gives: s = cvnT + Rnv + s1
(2)
(3)
( 4)
(5)
integration constant
and for constant cp, (4) gives: s = cpnT - RnP + s2

For constant v, (5) becomes
s  c v nT  c1
(7) where c1 = s1 + Rnv
and for constant p, (6) becomes
s  c p nT  c 2

(6)
(8) where c2 = s2 -RnP
From (7)  T  e ( s c1 ) / c v and from (8)  T  e (s c 2 ) / c P
constant
Section XII – Entropy Change for Ideal Gases (Exact treatment)
Page 53
constant pressure
constant volume
T

s
Cv
Cp
n=k
n =  (v = const)
T
n = 0 (P = const)

n=1
s
Isentropic Processes for ideal gases

Isentropic  constant entropy

An internally reversible, adiabatic process is an isentropic process but an isentropic process is
not necessarily reversible and adiabatic. Consider, for example, an isentropic process involving
an incompressible substance.
2
P 
P 
dT
Recall s 2  s1   c p
 Rn 2   s 2  s1  Rn 2 
1
T
 P1 
 P1 

For an isentropic process from
(1) T1 = 0 K, P1 = P0
Section XII – Entropy Change for Ideal Gases (Exact treatment)
to
s
P
Pr   
 Pr  e R
 P0  s const
relative
(2) P2 = P  0  s  Rn
Page 54
 P  s
P
 n  
P0
 P0  R
s  s(T)  Pr  Pr (T)
For an isentropic process between any two states (1) and (2)
 P2

 P1

P P


  2  0
 s const  P0 P1
 Pr 


 2 
 
 s const  Pr1 
isentropic
( s 2 s1 )
R
pressure  e
ratio
 v 
Relative specific volume: v r   
 v 0  s const
v0 = v at T0 = 0 K and P = P0
 RT P0 
1 T

with Pv  RT  v r  

   
 P RT0  s const  Pr T0  s const
but vr will be infinite since T0 = 0 K
 v
Hence define v r  
 v ref

T P

 v r    ref
 s const
 P Tref
 T Pref 1 
T
v r  

 
 vT  c
Pr
 Tref P0 Pr  s const

 T Pref P0 

 

 
T
P
P  s const
0
 s const  ref
C: arbitrary constant (Pref and Tref are fixed)
Pr  Pr (T)  v r  v r (T)
 v2

 v1

 T Pr 
T P P 

 RT
 v   vr
P 

  2  1 
  2  1  0 
 2  1 
  2    2
RT1  s const  T1 P0 P2  s const  T1 Pr2 
 s const  P2
 v1   v r1
s  const
For ideal gases with constant specific heats
Recall
T 
v 
s 2  s1  c v ve n 2   Rn 2  ,
 T1 
 v1 
T 
P 
s 2  s1  c Pve n 2   Rn 2 
 T1 
 P1 
– for an isentropic process from (1) to (2), s2 = s1
 c vave  T2 
v 
 n 2 

n 
R
 v1  s const
 T1  s const




Section XII – Entropy Change for Ideal Gases (Exact treatment)
 C v ave 

R 
 v2 
 T2  
 
  
 v1  s const  T1 

P
n 2
 P1
and
 v2

 v1

 v1

 v2
 C v ave 


R 




 s const
 k ave

k

T 
 T 

  2 
  2 
 s const  T1  s const
 T1 
 1




 T   k ave 1 

  2 
 s const  T1  s const

 R  cP  cv 


 k  cp



c
v


 1

 k 1 
ave

 T 
  1 
 T2 
cP

T

 ave n 2
R
 s const
 T1
Page 55


ave 1 
k ave
 P2 
v 
 
  1 
 P1  s const  v 2  s const
 T2 
v 
 
  1 
 T1  s const  v 2 
k ave
5.44 Nitrogen at 800 K, 2 MPa proceeds along an isentropic path until its temperature is reduced to
300 K. Assuming ideal-gas behavior, calculate the pressure at the final state for the following
conditions:
(a) The nitrogen has variable specific heats.
(b) The nitrogen has constant specific heats.
( s 2 s1 )
P 
P 
R
(a) s 2  s1  s 2  s1  Rn 2  Process is isentropic  s 2  s1   2   e
 P1 
 P1 
s 2  191.682 kJ / kg mol K
s 2  6.8426 kJ / kg K
s1  220.907 kJ / kg mol K
s  7.8859 kJ / kg K

1
P2  P1  e
( s 2 s1 )
R
P2  59.5 KPa
 2(e
3.5152
)  0.0595 MPa
R N 2  0.2968 kJ / kg K
M N 2  28.013 kg / kg mol
Section XII – Entropy Change for Ideal Gases (Exact treatment)
P
(b)  2
 P1
  T2
  
  T1



T 
P2  P1  2 
 T1 
k k 1
k k 1
, Tave 
 300 
 2

 800 
800  300
 550 K
2
1.388
0.388
Page 56
kave = 1.388
 0.0599 MPa
P2  59.9 KPa
5.31 Carbon dioxide changes state from 150 kPa, 30C to 300 kPa, 300C. Calculate the entropy
change of the CO2 during the process, assuming the following:
(a) The CO2 is an ideal gas with constant specific heats.
(b) The CO2 is an ideal gas with variable specific heats.
T 
P 
(a) s 2  s1  C Pave n 2   Rn 2 
 T1 
 P1 
Tave 
T1  T2 303  573

 438 K
2
2
C Pave  0.939 
(0.978  0.939)( 438  400)
 0.970 kJ / kg K
(450  400)
(see Table D-8, p. 723)
R CO 2  0.1889 kJ / kg K
 573 
 300 
s 2  s1  0.970 n
  0.1889 n 
  0.487 kJ / kg K
 393 
 150 
P 
(b) s 2  s1  s 2  s1  Rn 2 
 P1 
 (215.146  213.915)(303  300) 
  214.284 kJ / kg mol K
s1  213.915  
(310  300)


 (241.602  240.789)(573  570) 
  241.033 kJ / kg mol K
s 2  240.789  
(580  570)


M CO 2  44.011 kg / kg mol
s1 
214.282
241.033
 4.869 kJ / kg K , s 2 
 5.477 kJ / kg K
44.011
44.011
Section XII – Entropy Change for Ideal Gases (Exact treatment)
Page 57
 300 
s 2  s1  (5.477  4.869)  0.1889 n
  0.477 kJ / kg K
 150 
5.53 Air enters an air compressor at 27C, 1 atm pressure. The air follows a reversible adiabatic
process in the compressor and leaves a 2 MPa. The mass flow rate of the air through the
compressor is 2.5 kg/s. Neglecting changes in kinetic energy, calculate the power required by
the compressor and the temperature of the air as it leaves the compressor.

Ti  300 K  Tcrit air  133 K, Pi  101.3 KPa  Pcrit air  3.76 MPa
P e  2 MPa  P crit air  3.76 MPa  air can be treated as ideal

Process is reversible and adiabatic  Process is isentropic  se = si

Process in T-S diagram
Pe = 2 MPa
Te
e

e
Pi = 101.3 kPa
Ti


ie
Exit temperatrue
P 
s e  s i  s e  s i  R n e 
 Pi 
P
s e  s i  R n e
 Pi

 2000 
  6.7018  0.287 n
  7.5579 kJ / kg K
 101.3 

Te  690 7.5579  7.5572

 Te  690.45 K
700  690 7.5726  7.5572

Assume flow through compressor is steady and uniformly distributed at exit and inlet

Neglect changes in potential and kinetic energy of air. These will be small compared to the
change in enthalpy
Section XII – Entropy Change for Ideal Gases (Exact treatment)

Page 58
Conservation of energy applied to the compressor as an open system:
 W
 m
 m
 (h e  h i )  W
 (h i  h e )
Q
ie
ie
ie
 adiabatic 


 process 
h i  300.19 kJ / kg
h e  702.52
7.5579  7.5572

 h e  703.01kJ / kg
713.27  702.52 7.5726  7.5572
  2.5 (300.19  703.01)  1007.1 KW
W
ie
Power input
system boundary
ei
e

e
Air, at a temperature of 155.5C and at a pressure of 0.1 MPa, undergoes an isothermal, internally
reversible process. The final specific volume is 0.28 m3/kg. Find the change in entropy, heat
transferred and work done

Air can be treated as ideal since T1 = T2 = 428.5 K >
Tcrit ave  133 K and P1  0.1MPa  Pcrit ave  3.76 MPa

Initial specific volume: v1 

Final pressure: P2 
RT1 0.287  428.5

 1.23 m3 kg
P1
100
RT2 0.287  428.5

 439.2 KPa
v2
0.28
Section XII – Entropy Change for Ideal Gases (Exact treatment)

Page 59
Process on P – v and T – s diagrams
P
T
P2 = 0.439 MPa

P1 = 0.1 MPa
 


T1 = T2


v

s
Change in entropy
P
s 2  s1  s 2  s1  R n 2
 P1
Pr
ocess
is




isothermal



 439.2 
  0.287 n
   0.4247 kJ / kg K
 100 

alternatively
T 
v 
 0.28 
s 2  s1  C vave n 2   R n 2   0.237 n

 1.23 
 T1 
 v1 
(T1 =T2)
Q12
Air
T1 = T2 = 155.5C
= 628.4 K
P1 = 0.2 MPa
V2 = 0.23 m3/kg
s 2  s1  0.4248 kJ / kg K
system boundary

Heat transfer:
Process is internally reversible and isothermal 
q 12  T(s 2  s1 )  428.5 (0.4247)   182.0 kJ / kg

Work done:
First Law: q12 = W12 + du
W12  182.0 kJ / kg
(du = cvdT, dT = 0)
W12
Section XII – Entropy Change for Ideal Gases (Exact treatment)
Page 60
5.54 An ideal gas in a closed system undergoes an internally reversible constant pressure process
during which the temperature of the gas decreases as a result of heat transfer to the
environment. Determine the correct response to each of the following and briefly justify your
responses.
(a) The entropy change of the gas is (greater than, equal to, less than) zero.
(b) The entropy change of the environment is (greater than, equal to, less than) zero.
(c) The total entropy change is (greater than, equal to, less than) zero.
(a) Entropy change of the gas:
P 
s 2  s1  s 2  s1  R n 2 
 P1 
Environment
Tenv
(P1 = P2)
ideal
gas
Q12
since T2 < T1  s 2  s1  s 2 - s1  0
(b) Entropy change of the environment:
S env 
Q > 0
Tenv > 0
system
boundary
Q
Tenv
since environment receives heat,
environment is a heat sink
 Senv  0
(c) Total entropy change
Heat transfer through a finite temperature difference, from the gas to the environment, is
an irreversible process.
 (S) total > 0
(Principle of increase in entropy)