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Q
net
=
1
2
(2.00)(50) = 50 J
.
(d) Since we are dealing with an ideal gas (so that
ǻ
E
int
= 0 in an isothermal process),
then
W
1
ĺ
2
=
Q
1
ĺ
2
= 700 J
.
(e) Using Eq. 1914 for the isothermal work, we have
W
1
ĺ
2
= nRT
ln
V
2
V
1
.
where
T
= 350 K.
Thus, if
V
1
= 0.200 m
3
, then we obtain
V
2
=
V
1
exp (
W
/
nRT
)
= (0.200)
e
0.12
= 0.226 m
3
.
(f) Process 2
ĺ
3 is adiabatic; Eq. 1956 applies with
Ȗ
= 5/3 (since only translational
degrees of freedom are relevant, here).
T
2
V
2
Ȗ
1
=
T
3
V
3
Ȗ
1
This yields
V
3
= 0.284 m
3
.
(g) As remarked in part (d),
ǻ
E
int
= 0 for process 1
ĺ
2.
(h) We find the change in internal energy from Eq. 1945 (with
C
V
=
3
2
R
):
ǻ
E
int
=
nC
V
(
T
3
–
T
2
) = –1.25
×
10
3
J
.
(i) Clearly, the net change of internal energy for the entire cycle is zero.
This feature of a
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 Spring '08
 Any
 Physics

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