368 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th7-2.pm5
Solution. We know that
ds = cTvdT Tp dv
v
+FH∂∂ IK [Eqn. (7.24)]
Also dh = Tds + vdp = T
c
TdT
p
T dv
v
v
L +FH∂∂ IK
N
M
O
Q
P + vdp
i.e., dh = cvdT + T
∂
∂
F
H
I
K
p
T v + dv + vdp
Putting dT = 0, we get
∂
∂
F
HG
I
KJ =
∂
∂
F
HG
I
KJ +
∂
∂
F
HG
I
KJ
h
v
T p
T
v p
TvTv ...Proved.
(ii) FH∂∂phIK =FH∂∂hvIK HF∂∂vpIK
TTT
= T p
T
v p
v
v
vTp T
∂
∂
F
HG
I
KJ
+ ∂
∂
F
HG
I
KJ
L
N
M
O
Q
P
∂
∂
F
HG
I
KJ
i.e.,
∂
∂
F
H
I
K
h
pT = T^
∂
∂
F
HG
I
KJ
∂
∂
F
HG
I
KJ
p
T
v
v pT + v ...(i)
Also ∂
∂
F
HG
I
KJ
∂
∂
F
HG
I
KJ
p
T
v
v pT
= – FH∂∂TvIK
p
∴ Eqn. (i) becomes
∂
∂
F
H
I
K
h
pT = v – T^
∂
∂
F
H
I
K
v
Tp ...Proved.
Now p =
RT
vb
a
− v
− 2
∂
∂
F
H
I
K
p
vT = –
RT
vb
a
()− v
23 +^2
and
∂
∂
F
H
I
K
p
T v =
R
vb−
∴
∂
∂
F
H
I
K
h
vT = v
RT
vb
a
v
− T vbR
−
+
L
NM
O
QP
+ FH − IK
()^23
2
= – RTv
vb
a
v
RT
()− vb
++
(^22) −
(^2) = −
− + − +
RTv
vb
RT
vb
a
()^22 v
2
= −+ −
−
RTv RT v b +
vb
a
v
()
()^22
(^2) = −+−
−
RTv RTv RTb+
vb
a
()^22 v
2
i.e., FH∂∂hvIK
T
= −
−
RTb+
vb
a
()^22 v
2
or h =
RTb
vb
2a
v
f(T)
−
−+ ...Proved.
This shows h depends on T and v.
Example 7.16. Derive the following relations :
(i)HF∂∂hpIK
T
= v – T
∂
∂
F
H
I
K
v
T p = – cp^
∂
∂
F
H
I
K
T
p h (ii)
∂
∂
F
H
I
K
u
vT = T^
∂
∂
F
H
I
K
p
T v – p