TITLE.PM5

364 ENGINEERING THERMODYNAMICS

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Also h = g + Ts = g – T FH∂∂TgIK

p

Hence h = g – T

∂

∂

F

H

I

K

g

T p. (Ans.)

(iii) From eqn. (7.23), we have

cv = T

∂

∂

F

H

I

K

s

T v ...(i)

Also

∂

∂

F

H

I

K

a

T v = – s

or

∂

∂

F

H

I

K

s

T v = –

∂

∂

F

HG

I

KJ

2

2

a

T v ...(ii)

From eqns. (i) and (ii), we get

cv = – T

∂

∂

F

HG

I

KJ

2

2

a

T v. (Ans.)

(iv) From eqn. (7.26), we have

cp = T

∂

∂

F

H

I

K

s

T p ...(i)

Also

∂

∂

F

H

I

K

g

T p = – s

or

∂

∂

F

H

I

K

s

T p = –

∂

∂

F

HG

I

KJ

2

2

g

T p ...(ii)

From eqns. (i) and (ii), we get

cp = – T

∂

∂

F

HG

I

KJ

2

2

g

T p. (Ans.)

Example 7.10. Find the expression for ds in terms of dT and dp.

Solution. Let s = f(T, p)

Then ds = HF∂∂TsIK

p

. dT + HF∂∂spIK
T

dp

As per Maxwell relation (7.21)

∂

∂

F

H

I

K

s

pT = –

∂

∂

F

H

I

K

v

T p

Substituting this in the above equation, we get

ds = FH∂∂TsIK

p

dT – HF∂∂TvIK

p

. dp ...(i)

The enthalpy is given by

dh = cpdT = Tds + vdp

Dividing by dT at constant pressure

∂

∂

F

H

I

K

h

T p = cp = T^

∂

∂

F

H

I

K

s

T p + 0 (as dp = 0 when pressure is constant)