10.3 Solutions
445
whence γ = 1 +
log 1.32
log 2.0
= 1.4
∴
f =
2
1.4 − 1
= 5
10.47 The efficiency of the engine is
e = 1 −
T 2
T 1
= 1 −
300
400
= 0.25 or 25%
Work done by the engine
W = e × Q 1 = 0.25 × 10
8 cal
= 0.25 × 10
8
× 4.18 J
= 1.05 × 10
8 J
10.48 e = 1 −
T 2
T 1
=
1
6
(1)
∴
5T 1 − 6T 2 = 0
( 2 )
When the temperature of the sink is reduced by 62 ◦ C, the efficiency becomes
e
= 2e
(3)
e
= 2e = 1 −
(T 2 − 62)
T 1
=
1
3
(4)
∴ 2T 1 − 3T 2 − 186 = 0
( 5 )
Solving (2) and (5), T 1 = 372 K = 99 ◦ C, T 2 = 310 K = 37 ◦ C.
10.49 (a) PV = P 0 V 0
(isothermal conditions)
(1)
Dividing by m, the mass of air
PV
m
=
P 0 V 0
m
or
P
ρ
=
P 0
ρ 0
∴
ρ =
Pρ 0
P 0
(2)
where the density of air on earth’s surface is ρ 0 and pressure is P 0 , the
corresponding quantities at height y being ρ and P.
445
whence γ = 1 +
log 1.32
log 2.0
= 1.4
∴
f =
2
1.4 − 1
= 5
10.47 The efficiency of the engine is
e = 1 −
T 2
T 1
= 1 −
300
400
= 0.25 or 25%
Work done by the engine
W = e × Q 1 = 0.25 × 10
8 cal
= 0.25 × 10
8
× 4.18 J
= 1.05 × 10
8 J
10.48 e = 1 −
T 2
T 1
=
1
6
(1)
∴
5T 1 − 6T 2 = 0
( 2 )
When the temperature of the sink is reduced by 62 ◦ C, the efficiency becomes
e
= 2e
(3)
e
= 2e = 1 −
(T 2 − 62)
T 1
=
1
3
(4)
∴ 2T 1 − 3T 2 − 186 = 0
( 5 )
Solving (2) and (5), T 1 = 372 K = 99 ◦ C, T 2 = 310 K = 37 ◦ C.
10.49 (a) PV = P 0 V 0
(isothermal conditions)
(1)
Dividing by m, the mass of air
PV
m
=
P 0 V 0
m
or
P
ρ
=
P 0
ρ 0
∴
ρ =
Pρ 0
P 0
(2)
where the density of air on earth’s surface is ρ 0 and pressure is P 0 , the
corresponding quantities at height y being ρ and P.