360 ENGINEERING THERMODYNAMICSdharm
\M-therm\Th7-1.pm5
Solution. The general equation for finding dh is given bydh = cp dT + vTv
− T p
∂
∂F
HI
KL
NM
MO
QP
P^ dpdh
12
z = 0 + vTv
T p
T− ∂
∂F
HGI
KJR
S
|
T|U
V
|
W|L
N
M
MO
Q
P
z 1 P2as dT = 0 for isothermal change.
From the given equation of state, we have
∂
∂F
HI
Kv
T p
= R
pC
T+^34 ...(i)∴ h 2 – h 1 =
RT
pC
TRT
pC
Tdp
TF −
HGI
KJ−−
R
S
|
T|U
V
|
W|L
NM
MO
QP
z (^133) P
(^23)
= −
F
HG
I
KJ
L
N
M
O
Q
z P =− −
44
1 3
2
3 21
C
T
dp C
T
pp
T
[( )]T
The general equation for finding ds is given by
ds = cp dTT Tv
p
−FH∂∂ IK dp
ds
1
2
z = −
∂
∂
F
HG
I
KJ
L
N
M
M
O
Q
P
z P
v
T
dp
(^1) p T
2
as dT = 0 for isothermal change.
Substituting the value from eqn. (i), we get
(s 2 – s 1 ) = −+
F
HG
I
KJ
L
N
M
O
Q
z P
R
p
C
T
dp
T
3
1 4
2
= – R loge
p
p
2
1
F
HG
I
KJ –
3C
T^4
F
H
I
K^ (p^2 – p^1 ) (Ans.)
Example 7.7. For a perfect gas obeying pv = RT, show that cv and cp are independent of
pressure.
Solution. Let s = f(T, v)
Then ds =
∂
∂
F
H
I
K
s
T v^ dT +
∂
∂
F
H
I
K
s
vT^ dv
Also u = f(T, v)
Then du =
∂
∂
F
H
I
K
u
T v^ dT +
∂
∂
F
H
I
K
u
vT^ dv = cv dT +
∂
∂
F
H
I
K
u
vT^ dv
Also, du = Tds – pdv
Tds – pdv = cv dT + FH∂∂uvIK
T
dv
ds = cv dT
TT
u
- vT pdv
∂
∂
F
H
I
K +
L
N
M
O
Q
P
1