TITLE.PM5

440 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-3.pm5

∴ p(CO ) 22 = 0.7 ×^00909

03766

.

.

= 0.17 bar

Similarly, p(N ) 22 = 0.7 ×

0 2857

0 3766

.

.

= 0.53 bar

∴ Change in entropy, ∆S

= mc

T

T

mR p

p eep

log log

CO

2

1

2

(^12)
−
L
N
M
O
Q
P +mc
T
T
mR p
p eep
log log
N
2
1
2
(^12)
−
L
N
M
O
Q
P
= 4 085 398 2
313
8 314
44
017
14
.log ..log.
ee.
L −
NM
O
QP

• 8 104 398 2
  433
  8 314
  28
  053
  10
  .log ..log.
  ee.
  L −
  NM
  O
  QP
  = 4(0.2046 + 0.3984) + 8(– 0.0871 + 0.1885) = 3.2232 kJ/K
  i.e., Change in entropy = 3.2232 kJ/K. (Ans.)
  +Example 9.20. An insulated vessel containing 1 mole of oxygen at a pressure of 2.5 bar
  and a temperature of 293 K is connected through a valve to a second insulated rigid vessel
  containing 2 mole nitrogen at a pressure of 1.5 bar and a temperature of 301 K. The valve is
  opened and adiabatic mixing takes place. Assuming that oxygen and nitrogen are perfect gases
  calculate the entropy change in the mixing process.
  Assume the following specific heats at constant volume :
  cv(O 2 ) = 0.39 kJ/kg K
  cv(N 2 ) = 0.446 kJ/kg K.
  Solution. Consider the system within the boundary of Fig. 9.11.
  ×
  Valve
  closed
  O 2 N 2
  Boundary
  Stage 1
  ×
  Valve
  open
  N+ O 22
  Boundary
  Stage 2 N+ O^22
  Fig. 9.11
  In the process : Q = 0, W = 0, hence by the first law for a non-flow process ∆U = 0. Let T 2
  be the final temperature of the mixture.
  ∴ mc T mc TOON N 2 vv(O ) 2 22 + (N ) 2 2 = mcON 2 vv(O ) 2 +mc 2 (N ) 2 T 2
  or [(1 × 32) × 0.39 × 293 + (2 × 28) × 0.446 × 301] = (1 × 32 × 0.39 + (2 × 28) × 0.446] T 2
  or T 2 =
  32 0 39 293 56 0 446 301
  32 0 39 56 0 446
  ××+× ×
  ×+×
  ..
  .. =
  11174 4
  37 45
  .
  .
  = 298.4 K