140 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-3.pm5

The right hand side is a positive figure indicating the increase in energy of the system with
time. As the energy is increasing the temperature of air increases and hence the effect of coolness
gradually begins to wear off.
It may be pointed out here that in this case the energy rise manifests itself in a rise in
temperature.

+Example 4.27. A cylinder contains 0.45 m^3 of a gas at 1 × 10^5 N/m^2 and 80°C. The gas
is compressed to a volume of 0.13 m^3 , the final pressure being 5 × 10^5 N/m^2. Determine :
(i)The mass of gas ;
(ii)The value of index ‘n’ for compression ;
(iii)The increase in internal energy of the gas ;
(iv)The heat received or rejected by the gas during compression.
Take γ = 1.4, R = 294.2 J/kg°C.
Solution. Initial volume of gas, V 1 = 0.45 m^3
Initial pressure of gas, p 1 = 1 × 10^5 N/m^2
Initial temperature, T 1 = 80 + 273 = 353 K
Final volume after compression, V 2 = 0.13 m^3
The final pressure, p 2 = 5 × 10^5 N/m^2.
(i) To find mass ‘m’ using the relation

m =

pV

RT

11

1

110 045^5

294 353

= ××

×

.

.2 = 0.433 kg. (Ans.)

(ii) To find index ‘n’ using the relation

p 1 V 1 n = p 2 V 2 n

or

V

V

p

p

n

1

2

2

1

F

HG

I

KJ =

0

0

510

110

5

5

.45

.13

F

HG

I

KJ =

×

×

F

HG

I

KJ

n

= 5

or (3.46)n = 5
Taking log on both sides, we get
n loge 3.46 = loge 5
n = loge 5/loge 3.46 = 1.296. (Ans.)
(iii) In a polytropic process,
T
T

V

V

n

2

1

1

2

(^1) 296 1
0
0

F
HG
I
KJ
=F
HG
I
KJ
− −
.45
.13

1. = 1.444
   ∴ T 2 = 353 × 1.444 = 509.7 K
   Now, increase in internal energy,
   ∆ U = mcv (T 2 – T 1 )
   = 0.433 ×
   R
   ()γ− 1 (T^2 – T^1 ) Qc
   R
   v= −
   L
   NM
   O
   ()γ (^1) QP
   = 0.433 ×
   294.2
   ()1.4 1 1000− (509.7 – 353)
   = 49.9 kJ. (Ans.)