TITLE.PM5

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