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19.59:
a) From constant crosssection area, the volume is proportional to the length, and
Eq. (19.24) becomes
(
29
γ
p
p
L
L
/
1
2
1
1
2
=
and the distance the piston has moved is
(
29
Pa
10
5.21
Pa
10
01
.
1
1
m
250
.
0
1
400
.
1
/
1
5
5
/
1
2
1
1
2
1
×
×

=

=

γ
p
p
L
L
L
m.
0.173
=
b)
Raising both sides of Eq. (19.22) to the power
γ
and both sides of Eq. (19.24) to
the power
1

γ
, dividing to eliminate the terms
)
1
(
2
)
1
(
1
and


γ
γ
γ
γ
V
V
and solving for the ratio
of the temperatures,
(
29
(
29
(
29
C.
206
K
480
Pa
10
1.01
Pa
10
21
.
5
K
15
.
300
400
.
1
1
1
5
5
/
1
1
1
2
1
2
°
=
=
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