1 Termodinamika és optika gyakorlat II. éves fizikushallgatók ...

§
∂ T ∂S
∂T p
∂p
T
= T ∂2 S
∂p∂T
= −T ∂V
∂T
∂Cp
∂p
T
= T
p
= −T V α 2 + ∂α
.
∂T
∂
∂S
∂p
T
∂T
α − T V ∂α
∂T
p
p
¨¤
u(x, y) v(x, y)
∂ (u, v)
∂ (x, y) =
§
§
−V κS = ∂V
∂p
= ∂S
∂T
∂u
∂x y
∂u
∂y
x
∂ (u, z)
∂ (x, z)
∂ (u, z)
∂ (z, x)
∂ (u, v)
∂ (x, y)
S
V
∂v
∂x y
∂v
∂y
x
∂u
=
∂x
∂u
= ,
∂x
z
y
∂u
= − ,
∂x
z
= ∂ (u, v)
∂ (p, q)
p
∂V ∂
∂T p
= −T
∂T
p
p
= −T
= −T V α 2 + ∂α
.
∂T
∂v
∂y
κS = CV
κT .
Cp
x
∂ (p, q)
∂ (x, y) .
− ∂u
∂y
x
∂v
.
∂x
= ∂ (V, S) ∂ (V, S) ∂ (V, T ) ∂ (p, T )
=
∂ (p, S) ∂ (V, T ) ∂ (p, T ) ∂ (p, S) =
∂V
∂p
1
∂S = Cv
T (−V κT ) T
Cp
= −V CV
Cp
T
∂T p
T V α2
Cp − CV = .
κT
y
κT .
∂ (V α)
∂T
p