TITLE.PM5

260 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-3.pm5

5.17.3. Heating a gas at constant pressure

Refer Fig. 5.26. Let 1 kg of gas be heated at constant pressure, so that its absolute tempera-
ture changes from T 1 to T 2 and entropy s 1 to s 2.

s 1

T 1

T 2

s 2 s

Q

2

1

T

v = const.

p = const.

Fig. 5.26. T-s diagram : Constant pressure process.
Then, Q = cp(T 2 – T 1 ).
Differentiating to find small increase in heat, dQ of this gas when the temperature rise is
dT.
dQ = cp. dT
Dividing both sides by T, we get
dQ
T
= cp.
dT
T
or ds = cp. dT
T
Integrating both sides, we get

ds c

dT

s T

s

p T

T

1

2

1

2

zz=

s 2 – s 1 = cp loge

T

T

2

1

...(5.32)

5.17.4. Isothermal process

An isothermal expansion 1-2 at constant temperature T is shown in Fig. 5.27.

Entropy changes from s 1 to s 2 when gas absorbs heat during expansion. The heat taken by

the gas is given by the area under the line 1-2 which also represents the work done during expan-

sion. In other words, Q = W.

But QTdsTss

s

s

==−z

1

2

() 21

and W = p 1 v 1 loge

v

v

2

1

= RT 1 loge

v

v

2

1

per kg of gas [Q p 1 v 1 = RT 1 ]