TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 263

dharm

/M-therm/th5-3.pm5

= cv loge T

T

2

1

• R 1
  n− 1

F

HG

I

KJ

loge T

T

2

1

= cv loge T

T

2

1

• (cp – cv) ×^1
  n− 1

F

HG

I

KJ

loge T

T

2

1

[Q R = cp – cv]

= cv loge T

T

2

1

• (γ. cv – cv) ×^1
  n− 1

F

HG

I

KJ

loge T

T

2

1

[Q cp= γ. cv]

=−

−

−

F

HG

I

KJ

L

N

M

M

O

Q

P

P

c

v n

1

1

1

γ

loge

T

T

2

1

=

−− −

−

L

N

M

O

Q

c P

n

v n

()()

()

11

1

γ

loge

T

T

2

1

= −− +

−

F

HG

I

KJ

c n

v n

11

1

γ ) log

e^

T

T

2

1

= cv. n

n

−

−

F

HG

I

KJ

γ

1

loge T

T

2

1

per kg of gas

∴ s 2 – s 1 = cv n

n

−

−

F

HG

I

KJ

γ

1

loge

T

T

2

1

per kg of gas ...(5.35)

5.17.7. Approximation for heat absorbed

The curve LM shown in the Fig. 5.30 is obtained by heating 1 kg of gas from initial state L
to final state M. Let temperature during heating increases from T 1 to T 2. Then heat absorbed by
the gas will be given by the area (shown shaded) under curve LM.

s 1

T 1

T 2

s 2 s

Q

M

T

L

Fig. 5.30
As the curve on T-s diagram which represents the heating of the gas, usually has very
slight curvature, it can be assumed a straight line for a small temperature range. Then,
Heat absorbed = Area under the curve LM

= (s 2 – s 1 ) TT^12

2

F +

HG

I

KJ

...(5.36)