TITLE.PM5

GASES AND VAPOUR MIXTURES 443

dharm

\M-therm\Th9-3.pm5

∴∆S = () ()∆∆SSHN 22 + = 0.00626 + 0.424 = 0.43026 kJ/K

i.e., Total change in entropy = 0.43026 kJ/K. (Ans.)
Example 9.22. A perfect gas mixture consists of 4 kg of N 2 and 6 kg of CO 2 at a pressure
of 4 bar and a temperature of 25°C. Calculate cv and cp of the mixture.
If the mixture is heated at constant volume to 50°C, find the change in internal energy,
enthalpy and entropy of the mixture.

Take : cv(N 2 )= 0.745 kJ/kg K, cv(CO 2 ) = 0.653 kJ/kg K

cp(N 2 ) = 1.041 kJ/kg K, cp(CO 2 ) = 0.842 kJ/kg K.

Solution. mN 2 = 4 kg, mCO 2 = 6 kg, pmix = 4 bar

T 1 = 25 + 273 = 298 K, T 2 = 50 + 273 = 323 K

cv(mix) = ?, cp(mix) =?

Using the relation,

()mmcNCO(mix) 22 + v = mcN(N) 22 v + mcCO 22 v(CO )

(4 + 6) cv(mix) = 4 × 0.745 + 6 × 0.653

∴ cv(mix) =

4 0.745 + 6 0.653

46

××

+

= 0.6898 kJ/kg K. (Ans.)

Similarly, cp(mix) =

4 1.041 + 6 0.842

46

××

+

= 0.9216 kJ/kg K. (Ans.)

Change in internal energy, ∆∆∆∆∆U :

∆∆∆∆∆U = [mcv(T 2 – T 1 )]mix

= (4 + 6) × 0.6898(323 – 298) = 172.45 kJ. (Ans.)

Change in enthalpy, ∆H :

∆H = [mcp(T 2 – T 1 )]mix

= (4 + 6) × 0.9216(323 – 298) = 230.4 kJ. (Ans.)

Change in entropy, ∆S :

(∆s) = c

T

T

R v

ve ev

log^2 log

1

2

1

+

F

HG

I

KJ

= c

T

veT

log^2

1

F

HG

I

KJ

()Qvv 21 =

∴ (^) ()∆sN 2 = c
T
veT
log
N
2
(^12)
F
HG
I
KJ
and ()∆sCO 2 = c
T
veT
log
CO
2
(^12)
F
HG
I
KJ
Hence, ∆S = mc
T
veT
log
N
2
(^12)
F
HG
I
KJ + mc
T
veT
log
CO
2
(^12)
F
HG
I
KJ