FIRST LAW OF THERMODYNAMICS 143
dharm
/M-therm/Th4-3.pm5
or Taking log on both sides, we get
loge (0.6357) = (γ – 1) loge (0.333)
- 0.453 = (γ – 1) × (– 1.0996)
∴γ =
0
0996
.453
1.
+ 1 = 1.41
Also,
c
c
p
v = γ = 1.41
Work done during adiabatic process,
W
mR T T
12 − =^121
−
−
()
γ
∴ 52.5
0 453 288
41 1
=
−
−
.44
1.
R()
()
∴ R= −
−
52.5 41 1
44 453 288
()
()
1.
0.
= 0.296
∴ cp – cv = 0.296 [Q R = cp – cv]
Also
c
c
p
v
= 1.41 or cp = 1.41 cv
∴ 1.41 cv – cp = 0.296
or cv = 0.722 kJ/kg K. (Ans.)
and cp = 1.018 kJ/kg K. (Ans.)
- Example 4.30. 1 kg of ethane (perfect) gas is compressed from 1.1 bar, 27°C according
to a law pV1.3 = constant, until the pressure is 6.6 bar. Calculate the heat flow to or from the
cylinder walls.
Given : Molecular weight of ethane = 30, cp = 1.75 kJ/kg K.
Solution. Mass of ethane gas, m = 1 kg
Initial pressure, p 1 = 1.1 bar
Initial temperature, T 1 = 27 + 273 = 300 K
Final pressure, p 2 = 6.6 bar
Law of compression, pV1.3 = C
Quantity of heat transferred, Q :
Now, characteristic gas constant,
R R
M
=Universal gas constart ( )
Molecular weight ( )
0
=^831430 = 277.13 N-m/kg K = 277.31 J/kg K
= 0.277 kJ/kg K
Also cp – cv = R
∴ cv = cp– R = 1.75 – 0.277 = 1.473 kJ/kg K
γ =
c
c
p
v
=
1.
1.
75
473
= 1.188