444
10 Heat and Matter
∴
P 4 =
V 3
V 4
T 4
T 3
P 3 =
V 3
3V 3
T 3
T 3
P 2 =
P 2
3
=
3P 1
3
= P 1 = 10
5 N/m
2
V 4 =
nRT 4
P 4
=
2 × 8.31 × 361
10 5
= 0.06 m
3
Thus the P, V , T coordinates of the initial and final points on the indicator diagram are identical.
(c) Heat gained by N 2 in the first (isochoric) process (Fig. 10.10):
Fig. 10.10 The P–V diagram
C V = C P /γ = 29.12/1.4 = 20.8 J/(mol K)
Q 1 = nC v T = 2 × 20.8 × (1083 − 311) = 32115 J
W 12 = 0
( ∵ process 1 → 2 is isochoric)
∴ U 1 = Q 1 = 32115 J
Q 2 = nC p T = 2 × 29.12 × (361 − 1083) = −42049 J
W 23 = PV = P 2 (V 3 − V 2 ) = −3 × 10
5
×
2
3
× 0.06 = −12, 000 J
∴ U 2 = Q 3 − W 23 = −42049 + 12000 = −30049 J
U 3 = 0
( ∵ process 3 → 4 is isothermal)
(d) Net change in energy
U = U 1 + U 2 + U 3 = 32115 − 30049 + 0 = 2066 J
The expected value is zero.
10.46 Number of degrees of freedom, f =
2
γ − 1
we can find γ from the relation
T 2 V
γ −1
2
= T 1 V
γ −1
1
∴
V 2
V 1
γ −1
=
T 1
T 2
= 1.32
∴
2
γ −1
= 1.32
10 Heat and Matter
∴
P 4 =
V 3
V 4
T 4
T 3
P 3 =
V 3
3V 3
T 3
T 3
P 2 =
P 2
3
=
3P 1
3
= P 1 = 10
5 N/m
2
V 4 =
nRT 4
P 4
=
2 × 8.31 × 361
10 5
= 0.06 m
3
Thus the P, V , T coordinates of the initial and final points on the indicator diagram are identical.
(c) Heat gained by N 2 in the first (isochoric) process (Fig. 10.10):
Fig. 10.10 The P–V diagram
C V = C P /γ = 29.12/1.4 = 20.8 J/(mol K)
Q 1 = nC v T = 2 × 20.8 × (1083 − 311) = 32115 J
W 12 = 0
( ∵ process 1 → 2 is isochoric)
∴ U 1 = Q 1 = 32115 J
Q 2 = nC p T = 2 × 29.12 × (361 − 1083) = −42049 J
W 23 = PV = P 2 (V 3 − V 2 ) = −3 × 10
5
×
2
3
× 0.06 = −12, 000 J
∴ U 2 = Q 3 − W 23 = −42049 + 12000 = −30049 J
U 3 = 0
( ∵ process 3 → 4 is isothermal)
(d) Net change in energy
U = U 1 + U 2 + U 3 = 32115 − 30049 + 0 = 2066 J
The expected value is zero.
10.46 Number of degrees of freedom, f =
2
γ − 1
we can find γ from the relation
T 2 V
γ −1
2
= T 1 V
γ −1
1
∴
V 2
V 1
γ −1
=
T 1
T 2
= 1.32
∴
2
γ −1
= 1.32