Questions et exercices préparatoires du cours de

h
P(h) = ρgh
P ρ g
h
0 K = −273,15◦C
V= : Pf= 2Pi⇒Tf= 2Ti
T= : Pf= 2Pi⇒Vf=Vi
2
P= : Tf= 2Ti⇒Vf= 2Vi
1g
12g

P V =nRT
P V =NkBT
n=N
NA⇒N=nNA
P V =n NAkB
|{z}T R =NAkB
T=2
3
EC
kB
U≡NEC=3
2NkBT
U N EC
kBT
U≡NEC+NER+NEV=nd
2NkBT
EREV
nd
CV=nd
2R
CVR
H=kTS∆T
L
H kTS
∆T=Tf−TiL
∆U Q W
∆U=Q−W

Cp> Cv
W=ZVf
Vi
P(V)dV
W=P∆V
P V γ=
γ= 1 + 2
nd
W=NkB
γ−1(Ti−Tf)
Wcycle QH
r≡Wcycle
QH
= 1 −|QB|
QH
r= 1 −TB
TH
∆S≡Z
A→B
dQ
T

dS =dQ
T
∆S= 0
∆S=nd
2NkBln(Tf Ti
)
R
L L0=L+ ∆x
V=S×L⇒∆V=S×∆L S =πR2∆L= ∆x
⇒∆V=πR2∆x
m= 80
S= 400 2
P=mg
S=80 9,81 −2
0,04 2= 19 620
P= 1
P=P+P= 101 325 + 19 620 = 120 945
W=nRT ln(Vf
Vi)
W=RP(V)dV P (V) = nRT
V
W=RnRT
VdV =nRT RdV
V=nRT ln(Vf Vi)
pV γ=γ= 1 + 2
nd
Q= 0 ⇒∆U=−W⇒dU =−dW ⇒nd
2NkBdT =−P(V)dV P (V) = NkBT V
dT
T=−2 nd
dV
V
ZT
Ti
dT
T=−2 ndZV
Vi
dV
V

⇒ln T
Ti=−2 nd
ln V
Vi⇒ln T Ti= ln V Vi2
nd!⇒T
Ti
=V
Vi2
nd⇒T V
2
nd=TiV
2
nd
i
T V
2
nd
T V
2
nd=
T=P V
nRT
P V
nR V
2
nd=nR
γ= 1 + 2
nd
P V V
2
nd=P V 1+ 2 nd=P V γ=
Cp
Cp
Q≡Cpn∆T
Cpn= 1
∆T= 1 Q
∆T
Q
∆T=∆U ∆T+P∆V ∆T
U=nd
2NAkBT⇒U=nd 2RT ⇒∆U=nd 2R∆T
P V =RT ⇒P∆V=R∆T
Cp≡Q
∆T=nd 2R+R= (nd 2+ 1)R
q1q2
k0
1C= 1A×1s