P20 071

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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

.


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