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.