Potentiels thermodynamiques (PC*)

•
•
dU =δW +δQ dS =δScr +δQ
Text ≥δQ Text =dU−δW Text dU −δW −TextdS < 0
T0
dU −δW −TextdS =d(U−TextS)
∆ (U−TextS) = ∆ (U−T S)
dF =T dS −pdV −T dS −SdT =−pdV −SdT
T0
p0
dU −δW −TextdS =dU +p0dV −TextdS =d(U−TextS+p0V)
∆ (U−TextS+p0V) =
∆ (U−T S +pV )dF =T dS −pdV −T dS −SdT +pdV =V dp =
+V dp −SdT

2a S
M x n
T
ω=cste
φ
{gaz +piston}
φ
φ= Ψ0(T)+Ψ1(x, T )
Ψ1(x, T )
Cωω=cste
δQ =CωdT Cω
Tc=mω2a2
2nR
f=mω2x−→ uxδW =mω2xdx =
d1
2mω2x2dU −δW −TextdS =dU−TextS−1 2mω2x2
φ=U−T S −1
2mω2x2
φ x Ugaz
T Upiston Spiston
φ= Ψ(T)−T Sgaz −1
2mω2x2
n dU =T dS −pdV =ncvdT = 0 dS =p
TdV =nR dV VS(x) =
S(0) + nRln Vf
V0=S(0) + nRln S(a+x) Sa 
Sgaz = 2S(x= 0, T ) + nRln 1 + x
a+nRln 1−x a= 2S(x=
0, T ) + nRln 1−x2
a2φ= Ψ(T)−2T Sgaz (0) −nRT ln 1−x2 a2−1 2mω2x2
⇔∂xφ= 0 ⇔nRT 2x
a2−x2−mω2x= 0 ⇔x= 0 2nRT =mω2a2−x2
x=±qa2−2nRT
mω2=±aq1−T TcTc=mω2a2 2nR
δQ =CωdT δQ =T dS Cω=T∂S
∂T ω
Sgaz
T∂S
∂T ω=T∂S ∂T x,ω +T∂S ∂x T,ω ∂x ∂T ω= 2Cv−nRT 2x a2−x2∂x ∂T ω
•T≥Tcx= 0 Cω= 2Cv
•T < Tc,∂x
∂T ω=∓a Tc
1
2p1−T
Tc
2x
a2−x2∂x ∂T ω=−2ap1−T
Tc
a2−a2(1−T
Tc)
a
Tc
1
2p1−T
Tc
=−1
T
nRT 2x
a2−x2∂x ∂T ω=nR Cω= 2Cv+nR

ρ M
T P < PSPS
T
Pi
Pi
P T
PS
Pi
Pi=P+2γ
Rγ'9.10−4
R
333K PS= 4.5bars M = 1 kg.mol−1
ρ= 58 kg.m−3
gliq(P, T ) = ggaz (Pi, T )dG =V dP −SdT
dg =vmassiquedP vmassique =1
ρ=cste vmassique =1 Mvmolaire =1 M
V
n=RT MP
gliq(PS, T ) = ggaz(PS, T )gliq(PS, T ) =
1
ρ(PS−P) + gliq(P, T )ggaz (PS, T ) = ggaz(Pi, T ) + ´vmassiquedP =ggaz(Pi, T ) + RT Mln PS Pi
1
ρ(PS−P) = RT Mln PS PiPi=PSexp M RT ρ (P−Ps)
Pi= 4.486 R= 7.2

T−→
M
Tc
f(T, M) = u−T s =f0(T) + 1
2a(T−Tc)M2+1 4bM4
f0(T)f0=u−T s a b
f(T, M)MT
MST
T < Tc1−T
Tcβ
β
f(T, M)
T > TcT < Tc
−−→
Bext
−−→
dMδwmag =−−→ Bext.−−→ dM
g
dg =−MdB −sdT
T Bext