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

∂V
∂T S
∂V
∂T
p
=
− ∂S
∂T | V
∂S
∂V | T
V α
= − CV
T V α
= − CV
T V α
1
− ∂V
∂T p
∂S
∂V
∂V
∂p
1
T
T
= − CV
T V α
= CV
T V α
¥
− CV κT CV
= −
T V α2 Cp − CV
¨¤
1
∂p
∂T
V
1
V α (−V κT ) = − CV κT
.
T V α2 = 1
1 − γ .
¥
M = f(H, T ),
dU = T dS + µ0HdM,
f ££
H M
M H M H £
¥
¥
M ←→ V,
µ0H ←→ −p.
§
CH
T
∂ (S, H) ∂ (S, H)
= =
∂ (T, H) ∂ (T, M)
= CM
T
+ µ0
∂H
∂T
M
CH − CM = µ0T χT V
χT = 1
V
∂ (T, M)
∂ (T, H) =
2 ∂M
∂H
T
,
∂S
∂M
∂H
∂M
∂H
∂S
∂T
= µ0
T
∂T M
∂H
∂T
T
M
∂H
∂T
.
M
M
∂H
∂M
2
,
T
− ∂S
∂M
T
∂H
∂T
M
∂M
∂H
¥ H χT > 0 χT < 0
T