Page 81 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Therefore since V1 = V2,

P 5 . 1
T = T 2 = 588 x = 294 K
1
2
P 1 3
Then from equation 3.12,
Heat rejected per kg gas, Q = Cv(T2 – T1)

= 1.230(588 – 294)

= 361.6 kJ/kg

2.16 From the question

m = 0.18 kg

o
T = 15 C = 288 K
3
V = 0.17 m
Cv = 720 J/kg K = 0.720 kJ/kg K

i. From equation 2.14,

PV = mRT
PV 130 x . 0 17
R = = = . 0 426 kJ/kgK
mT . 0 18 x 288

ii. From equation 2.18,

R
R = o
M
R . 8 3144
M = o = = 19 . 52 kg/kmol
R . 0 426

iii. From equation 2.24,

R = Cp - Cv

Cp = R + Cv = 0.426 + 0.720 = 1.146 kJ/kg K

iv. From equation 2.25,

C . 1 146
 = p = = . 1 59
C v . 0 720

72 | P a g e