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Since the initial gauge pressure is 1.03
×
10
5
Pa,
p
i
= 1.03
×
10
5
Pa + 1.013
×
10
5
Pa = 2.04
×
10
5
Pa.
The final pressure is atmospheric pressure:
p
f
= 1.013
×
10
5
Pa. Thus
()
(
)
5
53
4
5
2.04
10 Pa
2.04
10 Pa 0.14m
ln
2.00
10 J.
1.013
10 Pa
W
§·
×
=×
=
×
¨¸
×
©¹
During the constant pressure portion of the process the work done by the gas is
W
=
p
f
(
V
i
–
V
f
). The gas starts in a state with pressure
p
f
, so this is the pressure throughout this
portion of the process. We also note that the volume decreases from
V
f
to
V
i
. Now
V
f
=
p
i
V
i
/
p
f
, so
(
)
55
3
4
1.013 10 Pa
2.04 10 Pa 0.14m
1.44 10 J.
ii
fi
f
i
i
f
pV
Wp
V
pp
V
p
¡¬
¢
¢
=
.
=
.
.
×
¢
¢
£®
=
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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, Work

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