71.
(a) Since an ideal gas is involved, then ∆E
int
= 0 implies T
1
= T
0
(see Eq. 20-62). Consequently, the
ideal gas law leads to
p
1
= p
0
V
0
V
1
=
p
0
5
for the pressure at the end of the sudden expansion. Now, the (slower) adiabatic process is described
by Eq. 20-54:
p
2
= p
1
V
1
V
2
γ
= p
1
(5
γ
)
as a result of the fact that V
2
= V
0
. Therefore,
p
2
=
p
0
5
(5
γ
) =
5
γ
−1
p
0
which is compared with the problem requirement that p
2
= 5
0.4
p
0
. Thus, we find that γ = 1.4 =
7
5
.
Since γ = C
p
/C
V
, we see from Table 20-3 that this is a diatomic gas with rotation of the molecules.
(b) The direct connection between E
int
and K
avg
is explained at the beginning of
§20-8. Since ∆E
int
= 0
in the free expansion, then K
1
= K
0
.
(c) In the (slower) adiabatic process, Eq. 20-56 indicates
T
2
= T
1
V
1
V
2
γ
−1
= 5
0.4
T
0
=
⇒
(E
int
)
2
(E
int
)
0
=
T
2
T
0
= 5
0.4
≈ 1.9 .
Therefore, K
2
= 1.9K
0
.