TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 259

dharm

/M-therm/th5-3.pm5

s 1

T 1

T 2

s 2 s

Q

2

v = Const.

T

1

= cv loge

p

p

2

1

+ (cv + R) loge

v

v

2

1

= cv loge p

p

2

1

+ cp loge v

v

2

1

∴ s 2 – s 1 = cv loge p

p

2

1

+ cp loge v

v

2

1

...(5.29)

Again, from gas equation,

pv

T

pv

T

11

1

22

2

= or v

v

p

p

T

T

2

1

1

2

2

1

=×

Putting the value of

v

v

2

1

in eqn. (5.28), we get

(s 2 – s 1 ) = cv loge

T

T

2

1

+ R loge

p

p

T

T

1

2

2

1

×

= cv loge

T

T

2

1

+ R loge

p

p

1

2

+ R loge

T

T

2

1

= (cv + R) loge

T

T

2

1

• R loge
  p
  p

2

1

= cp loge

T

T

2

1

• R loge
  p
  p

2

1

∴ s 2 – s 1 = cp loge T

T

2

1

• R loge p
  p

2

1

. ...(5.30)

5.17.2. Heating a gas at constant volume

Refer Fig. 5.25. Let 1 kg of gas be heated at
constant volume and let the change in entropy and
absolute temperature be from s 1 to s 2 and T 1 to T 2
respectively.
Then Q = cv(T 2 – T 1 )
Differentiating to find small increment of heat
dQ corresponding to small rise in temperature dT.
dQ = cvdT
Dividing both sides by T, we get
dQ
T = cv.

dT

T

or ds = cv. dT

T

Integrating both sides, we get

ds c

dT

s T

s

v T

T

1

2

1

2

zz=

or s 2 – s 1 = cv loge

T

T

2

1

...(5.31) Fig. 5.25. T-s diagram : Constant

volume process