P20 060

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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 .


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