TITLE.PM5

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