Page 80 - DJJ20063- Thermodynamics 1
P. 80
DJJ20063- Thermodynamics 1
Feedback To Activity 2F
2.14 From the question,
m = 2 kg
Q = 200 kJ
o
(T2 – T1) = 100 C = 373 K
Q = mCv(T2 – T1)
Q 200
C = = = . 0 268 kJ/kgK
v
m (T −T ) ( 2 373 )
2 1
2.15 From the question,
P1 = 3 bar
o
T1 = 315 C = 588 K
P2 = 1.5 bar
M = 26 kg/kmol
= 1.26
From equation,
R 8314
R = o = = 319 8 . J/kg K
M 26
From equation 2.26,
R 319 8 .
C = = = 1230 J/kg K = 1.230 kJ/kg K
v
( − ) 1 . 1 26 − 1
During the process, the volume remains constant (i.e. rigid vessel) for the mass
of gas present, and from equation 3.4,
P 1 V 1 = P 2 V 2
T 1 T 2
71 | P a g e
