TITLE.PM5

442 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-3.pm5

and pVNN 22 = mRTNNN 222

∴ mN 2 = 610 015

8 314

28

303 10

5

3

××

F

HG

I

KJ××

.

.

= 1.0 kg

According to the first law for a closed system

∆U = 0

i.e., Uinitial = Ufinal

or mc TH 2 vv vv(H ) 2 (H) 21 +=+mc TN 2 (N ) 2 (N) 21 mc TH 2 (H ) 2 (H) 22 mc TN 2 (N ) 2 (N) 22

or mcH 2 v(H ) 2 TT(H ) 22 − (H ) 21 + mcN 2 v(N ) 2 TT(N ) 22 − (N ) 21 = 0

or 0.0537 × 10.352 T(H ) 22 − 403 + 1.0 × 0.744 T(N ) 22 − 303 = 0
TTT(H ) 22 ==(N ) 22 2
or 0.556(T 2 – 403) + 0.744(T 2 – 303) = 0
or 0.556T 2 – 224 + 0.744T 2 – 225.4 = 0
∴ T 2 = 345.7 K
i.e., Temperature of the mixture = 345.7 K. (Ans.)
(ii)Pressure of the mixture, p 2 :
Now p(H ) 22 V = mRHH 22 T 2

∴ p(H ) 22 =

0 0537 8 314

2

345 7 10

045 10

3

5

...
.

×F

HG

I

KJ

××

×

= 1.71 bar

Similarly p(N ) 22 =

10 8 314

28

345 7 10

045 10

3

5

...
.

×FHG IKJ××

×

= 2.28 bar

∴ p 2 = pp(H ) 22 + (N ) 22 = 1.71 + 2.28 = 3.99 bar. (Ans.)

(iii)Change in entropy :

Now ()∆SH 2 = m c

T

T

R p

pe ep

log^2 log

1

2

1

F −

HG

I

KJH 2

= 0.0537 14 476 345 7

403

8 314

2

171

3

L .logee..− log.

NM

O

QP

= 0.00626 kJ/K. (Ans.)

()∆SH 2 = m c T

T

R p

p eep

log^2 log

1

2

1

−

F

HG

I

KJ

= 1.0 1 041 345 7

303

8 314

28

228

6

L.logee..− log.

NM

O

QP

= 0.424 kJ/K. (Ans.)