32.

(a) Using Eq. 20-54 for process D

→ A gives

p

D

V

γ

D

=

p

A

V

γ

A

p

0

32

(8V

0

)

γ

=

p

0

V

γ

0

which leads to

8

γ

= 32

=

⇒ γ =

5

3

which (see

§20-9and §20-11) implies the gas is monatomic.

(b) The input heat is that absorbed during process A

→ B:

Q

H

= nC

p

∆T = n

5

2

R



T

A

T

B

T

A

− 1



= nRT

A

5

2



(2

− 1) = p

0

V

0

5

2



and the exhaust heat is that liberated during process C

→ D:

Q

L

= nC

p

∆T = n

5

2

R



T

D

1

−

T

L

T

D



= nRT

D

5

2



(1

− 2) = −

1

4

p

0

V

0

5

2



where in the last step we have used the fact that T

D

=

1
4

T

A

(from the gas law in ratio form – see

Sample Problem 20-1). Therefore, Eq. 21-10 leads to

ε = 1

−





Q

L

Q

H



 = 1 −

1

4

= 0.75 = 75% .