270
4 Thermodynamics and Statistical Physics
δ Q = Ldm
(4)
If ν 1 and ν 2 are the specific volumes (volumes per unit mass) of the liquid and
vapor respectively
δν = (ν 2 − ν 1 )dm
(5)
Using (4) and (5) in (3)
L
ν 2 − ν 1
= T
∂ P
∂ T
V
(6)
Here, various thermodynamic quantities refer to a mixture of the liquid and
vapor in equilibrium. In this case
∂ P
∂ T
V
=
∂ V
∂ T
sat
since the pressure is due to the saturated vapor and is therefore independent of
V , being only a function of T . Thus (6) can be written as
∂ P
∂ T
sat
=
L
T (ν 2 − ν 1 )
(Clapeyron’s equation)
(7)
4.25 L = T (ν 2 − ν 1 )
dP
dT
= 373.2(1,674 − 1) ×
2.71
76
× 1.013 × 10
6
= 2.255 × 10
10 erg g
−1
= 2.255 J/g
=
2.255
4.18
= 539.5 cal/g
4.26
∂ S
∂ V
T
=
∂ P
∂ T
V
(1)
Substitute
dS =
dU + PdV
T
(2)
in (1)
∂U
∂ V
T
= T
∂ P
∂ T
V
− P
(3)
If u is the energy density and P the total pressure,
∂U
∂ V
= u and the total
pressure P = u/3, since the radiation is diffuse. Hence (3) reduces to
4 Thermodynamics and Statistical Physics
δ Q = Ldm
(4)
If ν 1 and ν 2 are the specific volumes (volumes per unit mass) of the liquid and
vapor respectively
δν = (ν 2 − ν 1 )dm
(5)
Using (4) and (5) in (3)
L
ν 2 − ν 1
= T
∂ P
∂ T
V
(6)
Here, various thermodynamic quantities refer to a mixture of the liquid and
vapor in equilibrium. In this case
∂ P
∂ T
V
=
∂ V
∂ T
sat
since the pressure is due to the saturated vapor and is therefore independent of
V , being only a function of T . Thus (6) can be written as
∂ P
∂ T
sat
=
L
T (ν 2 − ν 1 )
(Clapeyron’s equation)
(7)
4.25 L = T (ν 2 − ν 1 )
dP
dT
= 373.2(1,674 − 1) ×
2.71
76
× 1.013 × 10
6
= 2.255 × 10
10 erg g
−1
= 2.255 J/g
=
2.255
4.18
= 539.5 cal/g
4.26
∂ S
∂ V
T
=
∂ P
∂ T
V
(1)
Substitute
dS =
dU + PdV
T
(2)
in (1)
∂U
∂ V
T
= T
∂ P
∂ T
V
− P
(3)
If u is the energy density and P the total pressure,
∂U
∂ V
= u and the total
pressure P = u/3, since the radiation is diffuse. Hence (3) reduces to