TITLE.PM5

350 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th7-1.pm5

Now, from eqns. (7.34) and (7.37), we have

cp – cv =

β^2 Tv

K

...(7.38)

Thus at any state defined by T and v, cv can be found if cp, β and K are known for the

substance at that state. The values of T, v and K are always positive and, although β may some-

times be negative (e.g., between 0° and 4°C water contracts on heating at constant pressure), β^2 is

always positive. It follows that cp is always greater than cv.

The other expressions for cp and cv can be obtained by using the equation (7.14) as follows :

Since cv = ∂

∂

F

HG

I

KJ

u

T v

= ∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

u

s

s

vvT

We have cv = T ∂

∂

F

HG

I

KJ

s

T v

...(7.39)

Similarly, cp =

∂

∂

F

HG

I

KJ

h

T p =

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

h

s

s

p T p

Hence, cp = T ∂

∂

F

HG

I

KJ

s

T p

...(7.40)

Alternative Expressions for Internal Energy and Enthalpy

(i) Alternative expressions for equations (7.29) and (7.32) can be obtained as follows :

∂

∂

F

HG

I

KJ

u

v T = T^

∂

∂

F

HG

I

KJ

p

T v – p ...(7.29)

But ∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

p

T

T

v

v

v pT

= – 1

or

∂

∂

F

HG

I

KJ

p

T v = –

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

v

T

p

p vT

= +

βv

Kv =

β

K

Substituting in eqn. (7.29), we get

∂

∂

F

HG

I

KJ

u

v T = T^

β

K

• p ...(7.41)

Thus, du = cvdT + T

K

F β−p

HG

I

KJ

dv ...[7.28 (a)]

Similarly, ∂

∂

F

HG

I

KJ

h

pT

= v – T ∂

∂

F

HG

I

KJ

v

T p

...(7.32)

But by definition,

∂

∂

F

HG

I

KJ

u

T p = βv

Hence ∂

∂

F

HG

I

KJ

h

pT

= v(1 – βT) ...(7.42)