ME320 Thermodynamics
Homework set #4
Due: Wednesday 10/1/03 at the start of class
1. 2–65C (Textbook)
Solution:
A gas can be treated as an ideal gas when it is at a high temperature
or low pressure relative to the critical temperature and pressure.
2. 2–70 (Textbook)
Solution:
Initially, the absolute pressure in the tire is
P
1
=
P
g
+
P
atm
= 210 kPa + 100 kPa = 310 kPa
Treating the air as an ideal gas, it follows that
P
1
V
1
T
1
=
P
2
V
2
T
2
.
Thus,
P
2
=
P
1
V
1
T
2
V
2
T
1
and, assuming that the volume of the tire remains constant, the final
pressure in the tire is given by
P
2
=
P
1
T
2
T
1
=
(310 kPa)(323 K)
298 K
= 336 kPa
Thus, the pressure rise is
Δ
P
=
P
2

P
1
= (336

310) kPa = 26 kPa
.
Finally, using the gas constant of air given in Table A1, we find that
m
1
=
P
1
V
1
RT
1
=
(310 kPa)(0
.
025 m
3
)
(0
.
287 kPa m
3
/
kg K)(298 K)
= 0
.
0906 kg
and
m
2
=
P
2
V
2
RT
2
=
(310 kPa)(0
.
025 m
3
)
(0
.
287 kPa m
3
/
kg K)(323 K)
= 0
.
0906 kg
,
whereby the mass of air that needs to be bled off to restore pressure
to its original value is
Δ
m
=
m
2

m
1
= 0
.
0070 kg
.
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3. 2–76C (Textbook)
Solution:
The compressibility factor
Z
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 Spring '08
 Shen
 kPa, P2 V2 P1, P1 V1 T2

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