TITLE.PM5

284 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

= 0.0209 × (0.718 + 0.287) × loge 300

1500

F

HG

I

KJ

= – 0.0338 kJ/K. (Ans.)

lConstant temperature (isothermal) process 3-1 :

p 3 = p 2 = 5 bar

Change in entropy,

S 1 – S 3 = mR loge

p

p

3

1

F

HG

I

KJ

= 0.0209 × 0.287 × loge

5

1

F

HG

I
KJ = 0.00965 kJ/K. (Ans.)
Example 5.38. Derive expressions for entropy change with variable specific heat.
Solution. Let us assume that the specific heats of a gas vary with temperature according
to the linear relations :
cp = a + kT, and cv = b + kT
where, a, b and k = Constants, and T = Temperature, K.
For unit mass of gas, Q = T ds = du + pdv
= cvdT + pdv

or, ds = c dT

T

pdv

T

c dT

T

Rdv

vvv

+= + (Q pv = RT)

Now, R = cp – cv = (a + kT) – (b + kT) = a – b

∴ ds = (b + kT) dT

T

abdv

v

+−()

This is the differential form of entropy change.

Integrating both sides between limits 1 and 2, we get

s 2 – s 1 = b loge

T

T

2

1

F

HG

I

KJ

+ k(T 2 – T 1 ) + (a – b) loge

v

v

2

1

F

HG

I

KJ

...(1)

For the entropy change the following expressions can be obtained by suitable manipulations
to eqn. ( 1 ) :

1. Expression for entropy change in terms of temperature only :

T

T

v

v

n

2

1

1

2

1

=

F

HG

I

KJ

−

or, logeeT ( ) log ( ) loge

T

n v

v

n v

v

2

1

1

2

2

1

=− 11

F

HG

I

KJ

=− −

F

HG

I

KJ

∴ s 2 – s 1 = b T

T

kT T ab

n

T

eeT

log^2 ( ) log

1

21 2

(^11)
F
HG
I
KJ
+−−−
−
F
HG
I
KJ
F
HG
I
KJ
...[From eqn. (1)]
or, s 2 – s 1 = b
ab
n
T
T
− − e kT T
−
F
HG
I
KJ
F
HG
I
KJ
+−
1
2
1
log ( 21 ) ...(i)