278
4 Thermodynamics and Statistical Physics
4.39 By Maxwell’s first equation
∂ S
∂ V
T
=
∂ P
∂ T
V
(1)
dS =
dU + PdV
T
(2)
using (2) in (1)
∂U
∂ V
T
= T
∂ P
∂ T
− P
For perfect gases,
P =
RT
V
∂U
∂ V
T
=
RT
V
− P = 0
Thus, temperature remaining constant, the internal energy of an ideal gas
is independent of the volume.
4.40
dP
dT
=
L
T (ν 2 − ν 1 )
dT =
T
L
(ν 2 − ν 1 )dP
=
373(1677 − 1)(2 × 10
6 )
546 × 4.2 × 10 7
= 55.1
◦ C
4.41 ν 1 = 1 cm
3 ; v 2 =
1
0.091
= 10.981 cm
3
dP =
LdT
T (ν 2 − ν 1 )
=
80 × 4.2 × 10
7
× 1
(−1 + 273)(10.981 − 1.0)
= 1.238 ×
10
6 dynes
cm 2 = 1.24 atm
P 2 = P 1 + dP = 1.0 + 1.24 = 2.24 atm
4.42 ν 1 =
1
ρ 1
=
1
1.145
= 0.873 cm
3
/g
ν 2 =
1
ρ 2
=
1
0.981
= 1.019 cm
3
/g
dT =
T (ν 2 − ν 1 )dP
L
=
(80 + 273)(1.019 − 0.873)(1.0 × 10
6 )
35.5 × 4.2 × 10 7
= 0.0346
◦ C
4 Thermodynamics and Statistical Physics
4.39 By Maxwell’s first equation
∂ S
∂ V
T
=
∂ P
∂ T
V
(1)
dS =
dU + PdV
T
(2)
using (2) in (1)
∂U
∂ V
T
= T
∂ P
∂ T
− P
For perfect gases,
P =
RT
V
∂U
∂ V
T
=
RT
V
− P = 0
Thus, temperature remaining constant, the internal energy of an ideal gas
is independent of the volume.
4.40
dP
dT
=
L
T (ν 2 − ν 1 )
dT =
T
L
(ν 2 − ν 1 )dP
=
373(1677 − 1)(2 × 10
6 )
546 × 4.2 × 10 7
= 55.1
◦ C
4.41 ν 1 = 1 cm
3 ; v 2 =
1
0.091
= 10.981 cm
3
dP =
LdT
T (ν 2 − ν 1 )
=
80 × 4.2 × 10
7
× 1
(−1 + 273)(10.981 − 1.0)
= 1.238 ×
10
6 dynes
cm 2 = 1.24 atm
P 2 = P 1 + dP = 1.0 + 1.24 = 2.24 atm
4.42 ν 1 =
1
ρ 1
=
1
1.145
= 0.873 cm
3
/g
ν 2 =
1
ρ 2
=
1
0.981
= 1.019 cm
3
/g
dT =
T (ν 2 − ν 1 )dP
L
=
(80 + 273)(1.019 − 0.873)(1.0 × 10
6 )
35.5 × 4.2 × 10 7
= 0.0346
◦ C