82. To model the “uniform rates” described in the problem statement, we have expressed
the volume and the temperature functions as follows:
V
=
V
i
+
©
¨
§
¹
¸
·
V
f
–
V
i
τ
f
t
and
T
=
T
i
+
©
¨
§
¹
¸
·
T
f
–
T
i
τ
f
t
where
V
i
= 0.616 m
3
,
V
f
= 0.308 m
3
,
τ
f
= 7200 s,
T
i
= 300 K and
T
f
= 723 K.
(a) We can take the derivative of
V
with respect to
t
and use that to evaluate the
cumulative work done (from
t =
0 until
t
=
):
W
=
´
¶
p dV
=
´
¶
©
¨
§
¹
¸
·
nRT
V
©
¨
§
¹
¸
·
dV
dt
dt
= 12.2
+
238113 ln(14400
−
)
−
2.28 × 10
6
with SI units understood.
With
=
f
our result is
W
=
−
77169 J
≈−
77.2 kJ, or 
W

≈
77.2 kJ.
The graph of cumulative work is shown below. The graph for work done is purely
negative because the gas is being compressed (work is being done
on
the gas).
(b) With
C
V
=
3
2
R
(since it’s a monatomic ideal gas) then the (infinitesimal) change in
internal energy is
nC
V
dT
=
3
2
nR
©
¨
§
¹
¸
·
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 Spring '08
 Any
 Physics

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