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22. (a) The final pressure is
()
(
)
33 3
1.00 m
2.00 m
1.00 m
= 5.00 kPa
= 5.00 kPa
1.84 kPa .
i
f
f
VV a
pe
e
−
−
=
(b) We use the ratio form of the gas law (see Sample Problem 191) to find the final
temperature of the gas:
3
3
(1.84 kPa)(2.00 m )
600 K
441 K .
(5.00 kPa)(1.00 m )
ff
fi
ii
pV
TT
§·
==
=
¨¸
©¹
For later purposes, we note that this result can be written “exactly” as
T
f
= T
i
(2
e
–1
). In
our solution, we are avoiding using the “one mole” datum since it is not clear how precise
it is.
(c) The work done by the gas is
(
)
/
/
1.00
3
1.00
2.00
(5.00 kPa)
5.00 kPa
5.00 kPa
1.00 m
3.16 kJ .
f
f
i
i
i
i
fV
V
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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