TITLE.PM5

374 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th7-2.pm5

4. Entropy equations (Tds equations) :

Tds = cvdT + T GFH∂∂TpIKJ

v

dv ...(1)

Tds = cpdT – T GFH∂∂TvIKJ

p

dp ...(2)

5. Equations for internal energy and enthalpy :
   ∂
   ∂

F

HG

I

KJ

u

vT = T^

∂

∂

F

HG

I

KJ

p

Tv – p ...(1)

du = cvdT + T

p

Tv p

∂

∂

F

HG

I

KJ −

R

S

T

U

V

W

dv ...[1 (a)]

∂

∂

F

HG

I

KJ

h

pT = v – T^

∂

∂

F

HG

I

KJ

v

T p ...(2)

dh = cpdT + vTTv

p

− FHG∂∂ IKJ

R

S|

T|

U

V|

W|

dp ...[2 (a)]

Objective Type Questions

Choose the Correct Answer :

1. The specific heat at constant pressure (cp) is given by

(a)cp = T GFH∂∂TsIKJ

p

(b)cp = T FHG∂∂TsIKJ

p

(c)cp = T GFH∂∂TvIKJ

p

(d)cp = T GFH∂∂TvIKJ

p

.

2. The specific heat relation is

(a)(cp – cv) = vTKβ

2

(b)(cp – cv) = vTK

β^2

(c)(cp – cv) = pTK

β^2

(d)(cp – cv) = vTK

(^2) β
.

3. The relation of internal energy is

(a)du = Kβcv

F

HG

I

KJ

dp +

c

v p

p

β−

F

HG

I

KJ^ dv (b)du =

Kc

β v

F

HG

I

KJdp +

c

v p

p

β+

F

HG

I

KJ

dv

(c)du = GHFKβcpIKJ dp +

c

v v

p

β−

F

HG

I

KJ

dv (d)du = FHGKβcpIKJ dp + GHFvcvβ−pIKJ dv.

4. Tds equation is
   (a)Tds = cvdT + TKβdv (b)Tds = cpdT – TKβ dv

(c)Tds = cvdT + TKβ dv (d)Tds = cvdT + TKβ dp.

Answers

1. (a) 2. (a) 3. (a) 4. (a).