TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 279

dharm

/M-therm/th5-4.pm5

Example 5.32. A rigid cylinder containing 0.004 m^3 of nitrogen at 1 bar and 300 K is

heated reversibly until temperature becomes 400 K. Determine :

(i)The heat supplied. (ii)The entropy change.

Assume nitrogen to be perfect gas (molecular mass = 28) and take γ = 1.4.

Solution. Given : V 1 = 0.004 m^3 ; p 1 = 1 bar ; T 1 = 300 K ; T 2 = 400 K ; M for N 2 = 28 ;

γ = 1.4.

(i)The heat supplied :

Gas constant R = R

M

0 8 314

28

(

(

Universal gas constant).

Molecular mass)

= = 0.297 kg/kg K

Mass, m = pV

RT

11

1

1 10^5 0 004

0 297 1000 300

= ××

××

().

(. )

= 0.00449 kg

cv = R

γ−

=

1 −

0 297

14 1

.

.

= 0.742 kJ/kg K

∴ Heat supplied = m cv(T 2 – T 1 )

= 0.00449 × 0.742(400 – 300) = 0.333 kJ. (Ans.)

(ii)The entropy change :

The entropy change, S 2 – S 1 = m cv loge T

T

2

1

F

HG

I

KJ

= 0.00449 × 0.742 × loge 400

300

F

HG

I

KJ = 9.584 × 10

–4 kJ/kg K. (Ans.)

Example 5.33. A piston-cylinder arrangement contains 0.05 m^3 of nitrogen at 1 bar and
280 K. The piston moves inwards and the gas is compressed isothermally and reversibly until the
pressure becomes 5 bar. Determine :
(i)Change in entropy. (ii)Work done.
Assume nitrogen to be a perfect gas.
Solution. Given : V 1 = 0.05 m^3 ; p 1 = 1 bar ; T 1 = 280 K ; p 2 = 5 bar.
(i)Change in entropy, (S 2 – S 1 ) :
Gas constant, R = R
M

0 8 314

28

=. = 0.297 kJ/kg K

Mass of the gas, m = pV

RT

11

1

110 005^5

0 297 1000 280

= ××

××

().

(. )

= 0.06 kg

∴ Change in entropy S 2 – S 1 = mR loge

p

p

1

2

F

HG

I

KJ

= 0.06 × 0.297 loge

1

5

F

HG

I

KJ = – 0.0287 kJ/K. (Ans.)

Heat interaction, Q = T(S 2 – S 1 )

= 280 × (– 0.0287) = – 8.036 kJ

∴ Work done, W = Q = – 8.036 kJ. (Ans.) (Q In its other process, W = Q)

Alternatively W p V V

V

pV p

eep

e

:

kJ = 8.04 kJ

=

F

HG

I

KJ

=

F

HG

I

KJ

=× × × F

HG

I

KJ×

L

N

M

M

M

M

O

Q

P

P

P

P

−

11 2

1

11 1

2

110 005^531

5

10

log log

.log