TITLE.PM5

GASES AND VAPOUR MIXTURES 441

dharm

\M-therm\Th9-3.pm5

Now entropy change in the mixing process is given by

∆S = mmON 2 ∆∆ss(O ) 2 + 2 (N ) 2

where ∆s = cv loge

T

T

2

1

+ R loge V

V

2

1

Now to find initial and final volumes of O 2 and N 2 using the relation,

V =

nR T

p

0

∴ The initial volume of O 2 =

1 8 314 293 10

25 10

3

5

×××

×

.

.

= 9.74 m^3

and Initial volume of N 2 =

2 8 314 301 10

15 10

3

5

×××

×

.

.

= 33.4 m^3

Final volume of the mixture = 9.74 + 33.4 = 43.14 m^3

Thus ∆S = ().log

..log.

.

132039 298 4

293

8 314

32

4314

974

×× +RS

T

U

V

W

L

N

M

O

Q

eeP

+ ().log

..log.

.

2 28 0 446 298 4

301

8 314

28

4314

33 4

×× +RS

T

U

V

W

L

N

M

O

Q

eeP
= 12.60 + 4.04 = 16.64 kJ
i.e., Entropy change in the mixing process = 16.64 kJ. (Ans.)
+Example 9.21. A tank of capacity 0.45 m^3 is insulated and is divided into two sections
through a partition. One section initially contains H 2 at 3 bar and 130°C and has a volume of
0.3 m^3 and the other section initially holds N 2 at 6 bar and 30°C. The gases are then allowed to
mix after removing the adiabatic partition. Determine :
(i)The temperature of the equilibrium mixture ;
(ii)The pressure of the mixture ;
(iii)The change in entropy for each component and total value.
Assume : cv(N 2 ) = 0.744 kJ/kg K, cv(H 2 ) = 10.352 kJ/kg K

cp(N 2 )= 1.041 kJ/kg K, cp(H 2 ) = 14.476 kJ/kg K.

Solution. Total capacity of the tank, V = 0.45 m^3

VH 2 = 0.3 m^3 ; TH 2 = 130 + 273 = 403 K

pH 2 = 3 bar ; VN 2 = 0.15 m^3 (i.e., 0.45 – 0.3 = 0.15 m^3 )

pN 2 = 6 bar ; TN 2 = 30 + 273 = 303 K.

(i)Temperature of equilibrium mixture, T 2 :

Now pVHH 22 = mRTHHH 222

∴ mH 2 =

310 03

8 314

2

403 10

5

3

××

F

HG

I

KJ

××

.

. = 0.0537 kg

QR R

M

L =

NM

O

QP

0