4-62
4-82 Air at a specified state contained in a piston-cylinder device undergoes an isothermal and constant volume process
until a final temperature. The process is to be sketched on the P-V diagram and the amount of heat transfer is to be
determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The
kinetic and potential energy changes are negligible, ke  pe  0 .
Properties The properties of air are R = 0.287 kJ/kgK and cv = 0.718 kJ/kgK (Table A-2a).
Analysis (a) The processes 1-2 (isothermal) and 2-3 (constant-volume) are
sketched on the P-V diagram as shown.
(b) We take air as the system. This is a closed system since no mass crosses
the boundaries of the system. The energy balance for this system fort he
process 1-3 can be expressed as
E E
inout


Net energy transfer
by heat, work, and mass
E system



Air
100 kPa
27°C
0.4 m3
Q
Change in internal, kinetic,
potential, etc. energies
 Wb,out ,1 2  Qin  U  mcv (T3  T1 )
P
(kPa)
The mass of the air is
m
P1V1
(600 kPa)(0.8 m 3 )

 1.394 kg
RT1 (0.287 kPa  m 3 /kg  K)(1200 K)
600
300
The work during process 1-2 is determined from boundary work relation
for an isothermal process to be
Wb,out ,1 2  mRT1 ln
V2
P
 mRT1 ln 1
V1
P2
 (1.394 kg )(0.287 kPa  m 3 /kg  K)(1200 K)ln
1
2
3
V (m3)
600 kPa
300 kPa
 332.8 kJ
since
V 2 P1

for an isothermal process.
V1 P2
Substituting these values into energy balance equation,
Qin  Wb,out,1-2  mcv (T3  T1 )
 332.8 kJ  (1.394 kg)(0.718 kJ/kg  K)(300  1200)K
 568 kJ
Thus,
Qout  568 kJ
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