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During process
B
→
C
, the system is neither expanding nor contracting. Thus,
(c)
W
= 0.
(d) The sign of
Δ
E
int
must be the same (by the first law of thermodynamics) as that of
Q
which is given as positive. Thus,
Δ
E
int
> 0.
During process
C
→
A
, the system is contracting. The environment is doing work on the
system, which implies
W
<
0. Also,
Δ
E
int
<
0 because
¦
Δ
E
int
= 0 (for the whole cycle)
and the other values of
Δ
E
int
(for the other processes) were positive. Therefore,
Q
=
W
+
Δ
E
int
must also be negative.
(e)
Q
< 0.
(f)
W
< 0.
(g)
Δ
E
int
< 0.
(h) The area of a triangle is
1
2
(base)(height). Applying this to the figure, we find
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 Spring '08
 Any
 Physics, Work

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