FIRST LAW OF THERMODYNAMICS 137

dharm

/M-therm/Th4-3.pm5

Also γ= =

c

c

p

v

1

0 714.

= 1.4

∴ V 2 = 0.2 ×

410

102 10

5

5

1

×^1

×

F

HG

I

. KJ

.4

= 0.53 m^3

Also, TT^2 pp

1

2

1

1

=F

HG

I

KJ

−γ

γ

∴ TT 21 pp^2

1

1

= FHG IKJ

−γ

γ

=

×

×

F

HG

I

KJ

−

403 02 10

410

5

5

1

1.^14

1.4

.

= 272.7 K

Mass of the gas,

m

pV

RT

=^11

1

[Q pV = mRT]

where, R = (cp – cv) = (1 – 0.714) kJ/kg K
= 0.286 kJ/kg K = 286 J/kg K or 286 Nm/kg K

∴ m =

410 02

286 403

××^5

×

.

= 0.694 kg.

Process 2-3. Constant pressure :

Q2–3 = mcp (T 3 – T 2 )

72.5 = 0.694 × 1 × (T 3 – 272.7)

∴ T 3 = 72.5

0.694

+ 272.7 = 377 K

Also, V

T

V

T

2

2

3

3

=

or 0.53
272.7
= V^3
377
∴ V 3
0 377
272.7

.53×
= 0.732 m^3
Work done by the path 1-2-3 is given by
W1–2–3 = W1–2 + W2–3

=

pV 11 p V2 2

1

−

−γ + p^2 (V^3 – V^2 )

=

4 10 0 02 10 0

41

××− ××^55

−

.2 1. .53

1. 
   
   
   
   • 1.02 × 10
   
   
   
   

(^5) (0.732 – 0.53)

10 4 0 02 0
04
(^5) ()
.
×− ×.2 1. .53

• 1.02 × 10^5 (0.732 – 0.53)
  = 64850 + 20604 = 85454 Nm or J
  Hence, total work done = 85454 Nm or J. (Ans.)