276 ENGINEERING THERMODYNAMICS

dharm
/M-therm/th5-4.pm5

(i)Change in enthalpy, H 2 – H 1 :

We know that, H dH Vdp

H

p

p

1

2

1

2

zz= ...(i)

Also pV 11 n=pVn

∴ V =

pV

p

n n

11

F^1

HG

I

KJ

/

Substituting this value of V in eqn. (i), we get

dH pVp dp

n n

p

p

H

H

=

F

HG

I

KJ

L

N

M

M

O

Q

P

zz P

11

1

1

2

1

2

/

=()pVnnzp p− ndp

p

11 1/ 1/

1

2

=

−+

L

N

M

M

M

O

Q

P

P

P

−+

()pV p

n

nn n

p

p

11 1/

(^11)
(^11)
1
2

−
−
L
N
M
M
M
O
Q
P
P
P
HFG− IKJ FHG− IKJ
[(pV)]
pp
n
nn nn
11
1/ 21
1
1
1 1
1 1
=×− −
L
N
M
M
M
O
Q
P
P
P
FH− IK FH− IK
()pV 11 n^1 /n ()nn ()pp 2 nn()
1 1
1
1 1
1

npV pV
n
()
()
22 11
1
−
−
[Q p 1 V 1 n = p 2 V 2 n]

−×

1. 
   


2. 
   

   25
   ()25 1 103
   [20.87 × 10^5 × 0.08 – 4 × 10^5 × 0.3] kJ

   
   

   ×

   
   


3. 
   


4. 
   

   25
   25 10^3
   × 10^5 (20.87 × 0.08 – 4 × 0.3) kJ = 234.8 kJ.
   Hence, change in enthalpy = 234.8 kJ. (Ans.)
   (ii)Change in internal energy, U 2 – U 1 :
   H 2 – H 1 = (U 2 + p 2 V 2 ) – (U 1 + p 1 V 1 )
   = (U 2 – U 1 ) + (p 2 V 2 – p 1 V 1 )
   ∴ U 2 – U 1 = (H 2 – H 1 ) – (p 2 V 2 – p 1 V 1 )
   = 234.8 – 20.87 10 08 4 10^3
   10
   55
   3
   F ×× −××
   HG
   I
   KJ

   
   


5. 
   
   
   
   0. kJ
   
   
   
   

= 234.8 – 46.96 = 187.84 kJ.

Hence, change in internal energy = 187.84 kJ. (Ans.)

(iii)Change in entropy, S 2 – S 1 = 0. (Ans.)

(iv)Heat transfer, Q1–2 = 0. (Ans.)