TITLE.PM5

142 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-3.pm5

pV = mRT

∴ m

pV

RT

==××

××

11

1

5

3

02 10 0

0 10 295

1. .015
   .287 [Q R for air = 0.287 × 10

(^3) ]
= 0.01807 kg
∴ W=
×× −
−
0 01807 0 287 10 295 507 24)
41
..(.^3
()1.
= – 2751 J or – 2.751 kJ
i.e., Work done = 2.751 kJ. (Ans.)
(– ve sign indicates that work is done on the air).
Example 4.29. 0.44 kg of air at 180°C expands adiabatically to three times its original
volume and during the process, there is a fall in temperature to 15°C. The work done during the
process is 52.5 kJ. Calculate cp and cv.
Solution. Refer Fig. 4.23.
Reversible
adiabatic
2
1
p (Pressure)
V (Volume)
Fig. 4.23
Mass of air, m = 0.44 kg
Initial temperature, T 1 = 180 + 273 = 453 K
Ratio = VV^2
1
= 3
Final temperature, T 2 = 15 + 273 = 288 K
Work done during the process, W1–2 = 52.5 kJ
cp = ?, cv =?
For adiabatic process, we have
T
T
V
V
2
1
1
2
1
=F
HG
I
KJ
−γ
288
453
1
3
1
=FH IK
−γ
or 0.6357 = (0.333)γ–1