1 Problème : Autour de l`eau

◦2
p T
p0T0
p=a T +b

p0=a T0+b
a b
a=p0−p
T0−T=−1,007.107−1b=p0T−T0p T−T0= 2,751.109
m= 75
l= 30
e= 4
p=mg
le
p=p0+p=p0+mg
le ∼7,14.105
◦T= 268,15
p=a T +b∼5,07.107
p

n V
pV =nRT
v M
n
v=V
nM
pv =RT
M
v=RT
pM
∂v
∂p T
=−RT
p2M=−v p
χT=−1
v∂v ∂p T=1 p
pv =RT
M
T=pv
T > T
pv p

f f : (p, T )−→ f(p, T ) = pv
f p
∂f
∂p T
>0
χT=−1
v∂v ∂p T
=−1
v
∂f
p
∂p 
T
=−1
v−f p2+1 p∂f ∂p T=1 pf pv −1 v∂f ∂p T
χT=1
pf(p, T ) = pv ∂f ∂p T>0
χT=χT1−1
v∂f ∂p T< χT
pv
∂f
∂p T= 0
χT=χT
[t, t+dt]
k k0dmk=dkdtdmk0=d0
kdt
dHk=hkdmk= (dkhk) dtdHk0= (dk0hk0) dt
dkhkdk0hk0
k k0

p= 0
p= 0
p+p= 0
d1=d2=dgd3=d4=de
dg(h2−h1) + de(h4−h3)=0
h2−h1=cg(T2−T1)h4−h3=ce(T4−T3)
dgcg(T2−T1) + dece(T4−T3)=0
de=dgcg(T2−T1)
ce(T3−T4)= 13,1−1
h s
X
k0∈
dk0hk0−X
k∈
dkhk
←→  X
k0∈
dk0sk0−X
k∈
dksk
p+p←→ 0