TITLE.PM5

FIRST LAW OF THERMODYNAMICS 141

dharm

/M-therm/Th4-3.pm5

(iv) Q = ∆ U + W

Now, W

pV pV

n

mR T T

= n

−

− =

−

−

11 2 2 1 2

11

()

= ×−

−

0 433 294.2 353 509 7

296 1

.(.)

1. = – 67438 N-m or – 67438 J = – 67.44 kJ
   ∴ Q = 49.9 + (– 67.44) = – 17.54 kJ
   ∴ Heat rejected = 17.54 kJ. (Ans.)
   Example 4.28. Air at 1.02 bar, 22°C, initially occupying a cylinder volume of 0.015 m^3 , is
   compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate :
   (i)The final temperature ;
   (ii)The final volume ;
   (iii)The work done.
   Solution. Initial pressure, p 1 = 1.02 bar
   Initial temperature, T 1 = 22 + 273 = 295 K
   Initial volume, V 1 = 0.015 m^3
   Final pressure, p 2 = 6.8 bar
   Law of compression : pvγ = C
   (i)Final temperature :
   Using the relation,

TT^2 pP

1

2

1

1

=FHG IKJ

−γ

γ

T^2

1

1

295

68

02

=F

HG

I

KJ

−

.

1.

1.4

.4 [

Q γ for air = 1.4]

∴ T 2 = 295

68

02

1

. 1 .4

1. 
   

1.4

F

HG

I

KJ

−

= 507.24 K

i.e., Final temperature = 507.24 – 273 = 234.24°C. (Ans.)

(ii)Final volume :

Using the relation,

p 1 V 1 γ = p 2 V 2 γ

p

p

V

V

1

2

2

1

=F

HG

I

KJ

γ

or VV^2 pp

1

1

2

1

=FHG IKJ

γ

∴ V 2 = V 1 × pp^1

2

1

F

HG

I

KJ

γ

= 0.015 ×

1.02

6.8

1

F 1.4

HG

I

KJ = 0.00387 m

3

i.e., Final volume = 0.00387 m^3. (Ans.)

Now, work done on the air,

W=mR T()γ()−^121 −T ...(i)

where m is the mass of air and is found by the following relation,