60.
(a) In the free expansion from state 0 to state 1 we have Q = W = 0, so ∆E
int
= 0, which means that
the temperature of the ideal gas has to remain unchanged. Thus the final pressure is
p
1
=
p
0
V
0
V
1
=
p
0
V
0
3V
0
=
1
3
p
0
.
(b) For the adiabatic process from state 1 to 2 we have p
1
V
γ
1
= p
2
V
γ
2
, i.e.,
1
3
p
0
(3V
0
)
γ
= (3.00)
1
3
p
0
V
γ
0
which gives γ = 4/3. The gas is therefore polyatomic.
(c) From T = pV /nR we get
¯
K
2
¯
K
1
=
T
2
T
1
=
p
2
p
1
= (3.00)
1
3
= 1.44 .