27.

(a) Energy is added as heat during the portion of the process from a to b. This portion occurs at

constant volume (V

b

), so Q

in

= nC

V

∆T . The gas is a monatomic ideal gas, so C

V

=

3
2

R and the

ideal gas law gives ∆T = (1/nR)(p

b

V

b

− p

a

V

a

) = (1/nR)(p

b

− p

a

)V

b

. Thus, Q

in

=

3
2

(p

b

− p

a

)V

b

. V

b

and p

b

are given. We need to find p

a

. Now p

a

is the same as p

c

and points c and b are connected

by an adiabatic process. Thus, p

c

V

γ

c

= p

b

V

γ

b

and

p

a

= p

c

=

V

b

V

c



γ

p

b

=

1

8.00



5/3

(1.013

× 10

6

Pa) = 3.167

× 10

4

Pa .

The energy added as heat is

Q

in

=

3

2

(1.013

× 10

6

Pa

− 3.167 × 10

4

Pa)(1.00

× 10

−3

m

3

) = 1.47

× 10

3

J .

(b) Energy leaves the gas as heat during the portion of the process from c to a. This is a constant

pressure process, so

Q

out

=

nC

p

∆T =

5

2

(p

a

V

a

− p

c

V

c

) =

5

2

p

a

(V

a

− V

c

)

=

5

2

(3.167

× 10

4

Pa)(

−7.00)(1.00 × 10

−3

m

3

) =

−5.54 × 10

2

J .

The substitutions V

a

− V

c

= V

a

− 8.00V

a

=

−7.00V

a

and C

p

=

5
2

R were made.

(c) For a complete cycle, the change in the internal energy is zero and W = Q = 1.47

× 10

3

J

− 5.54 ×

10

2

J = 9.18

× 10

2

J.

(d) The efficiency is ε = W/Q

in

= (9.18

× 10

2

J)/(1.47

× 10

3

J) = 0.624.