Introduction to Thermodynamics and Statistical Physics phần 1 ppt
Introduction to Thermodynamics and Statistical Physics phần 1 ppt
... MMM
ee
++
2
p
…
1
2
p
q
1
1
p
q
M
1
1
+M
e
2
1
p
q
M
2
1
+M
e
21
MM
e
+
…
2
2
1
p
q
M +
2
21
p
q
MM +
1
211 M
qqqp +++=
211
12 MMM
qqp
++
++=
1
p
1
e
1
, ,,
21 M
eee
1
1
p
q
2
e
1
M
e
…
211
, ,
1 MMM
ee
++
2
p
…
1
2
p
q
1
1
p
q
M
1
1
+M
e
2
1
p
q
M
2
1
+M
e
21
MM
e
+
…
2
2
1
p
q
M ... Buks Thermodynamics and Statistical Physics 5...
Introduction to Thermodynamics and Statistical Physics phần 2 ppt
... p
r
)(logp
s
+1) . (1. 1 01)
Thus, using both expressions (1. 100) and (1. 1 01) yields
dσ
dt
=
1
2
X
r
X
s
Γ
sr
(p
s
− p
r
)(logp
s
− log p
r
) . (1. 102)
In general, since log x is a monotonic increasing ... (1. 69)
Z
c
=exp
µ
βε
2
¶
+exp
µ
−
βε
2
¶
=2cosh
µ
βε
2
¶
, (1. 75)
thus using Eqs. (1. 70) and (1. 71) one finds
hUi = −
ε
2
tanh
µ
βε
2
¶
, (1. 76)
and
D
(∆U)
2
E
=
³
ε
2
´
2...
Introduction to Thermodynamics and Statistical Physics phần 4 ppt
... occupation
Fermions 1 + λ exp (−βε)
1
exp[β(ε−µ)] +1
Bosons
1
1−λ exp(−βε)
1
exp[β(ε−µ)] 1
classical limit 1 + λ exp (−βε)exp[β (µ − ε)]
0
0.5
1
1.5
2
-2 246 810 1 214 1 618 20
energy
2.4 Ideal Gas in the ... Ideal Gas
…
orbital 1
N
1
=0
orbital 2
N
2
=1
orbital 3
N
3
=2
orbital 4
N
4
=0
3 1
2
…
orbital 1
N
1
=0
orbital 2
N
2
=1
orbital 3
N
3
=2
orbital 4
N
4
=0
33...
Introduction to Thermodynamics and Statistical Physics phần 7 pptx
... λ
0
=2πc/ω
0
one has
λ
0
µm
=5 .1
µ
T
10 00 K
¶
1
. (3.50)
0.2
0.4
0.6
0.8
1
1.2
1. 4
0246 810
x
The function x
3
/ (e
x
− 1) .
The total energy is found by integrating Eq. (3.48) and by employing the
variable ... (3 .14 ) one finds that
q∇
2
u =
1
c
2
u
d
2
q
dt
2
. (3 .17 )
Eyal Buks Thermodynamics and Statistical Physics 98
3 .1. Electromagnetic Radiation
Multiplying by...
Introduction to Thermodynamics and Statistical Physics phần 8 ppt
... leads
to
Eyal Buks Thermodynamics and Statistical Physics 12 1
3.6. Solutions Set 3
thus
µ = τ log
·
exp
µ
nπ~
2
m
e
τ
¶
− 1
¸
. (3 .17 9)
Eyal Buks Thermodynamics and Statistical Physics 12 5
Chapter ... here n ∈ {0, 1} ,andby
p
B
(n)= {1 − exp [(µ − ε) β]} exp [n (µ − ε) β] , (3 .11 8)
where n ∈ {0, 1, 2, }, for the case of Bosons.
Eyal Buks Thermodynamics and...
Introduction to Thermodynamics and Statistical Physics phần 9 pptx
... by
p =(n
1
(h)+n
2
(h)) τ
=
µ
M
1
N
1
exp (βM
1
gh) − 1
+
M
2
N
2
exp (βM
2
gh) − 1
¶
g
S
.
(4 .10 3)
Eyal Buks Thermodynamics and Statistical Physics 14 5
Chapter 4. Classical Limit of Statistical ... ex p (−βε)=
1
β
Γ (1) = 1
Γ
µ
1
2
¶
=
√
π
one finds
hvi
¿
1
v
À
=
µ
∞
R
0
dεεexp (−βε)
¶µ
∞
R
0
dε exp (−βε)
¶
µ
∞
R
0
dεε
1/ 2
exp (−βε)
¶
2
=
Γ (1) β
1
1
β
1...
Introduction to Thermodynamics and Statistical Physics phần 10 pptx
... function
f
FD
(ε)=
1
exp [β (ε − µ)] + 1
. (6.46)
Eyal Buks Thermodynamics and Statistical Physics 16 0
Chapter5. ExamWinter2 010 A
µ
V
2
V
1
¶µ
τ
2
τ
1
¶
c
V,A
+c
V,B
N
A
+N
B
=1. (5.55)
Alternatively, ... erage energy U and average number of particle N are calcu-
lated using Eqs. (1. 80) and (1. 81) respectively
Eyal Buks Thermodynamics and Statistical Physics...
Introduction to Thermodynamics and Statistical Physics phần 3 pdf
... 0.8
20−n
=5.6 × 10
−4
. (1. 144)
8. Using
N
+
+ N
−
= N, (1. 145a)
N
+
− N
−
=
M
m
, (1. 145b)
one has
N
+
=
N
2
µ
1+
M
mN
¶
, (1. 146a)
N
−
=
N
2
µ
1 −
M
mN
¶
, (1. 146b)
or
N
+
=
N
2
(1 + x) , (1. 147a)
N
−
=
N
2
(1 ... general
∞
X
N=0
Nx
N 1
=
d
dx
∞
X
N=0
x
N
=
d
dx
1
1 − x
=
1
(1 − x)
2
,
thus
¯
N =
1
6
∞
X
N=0
N
µ
5
6
¶
N 1
=
1
6
1
¡
1 −
5
6
¢
2
=6.
3. Let W (m)...
Introduction to Thermodynamics and Statistical Physics phần 5 pdf
... (2 .11 5)
Using Eqs. (2 .11 1) and (2 .11 3) one finds
η =1+
τ
l
log
V
d
V
c
τ
h
log
V
b
V
a
. (2 .11 6)
Employing Eq. (2 .10 8) for both isen tropic processes yields
τ
h
V
γ 1
b
= τ
l
V
γ 1
c
, (2 .11 7)
τ
h
V
γ 1
a
= ... has
log
τ
2
τ
1
=log
µ
V
2
V
1
¶
1 γ
. (2 .10 7)
Thus
τ
1
V
γ 1
1
= τ
2
V
γ 1
2
, (2 .10 8)
or using Eq. (2 .55)
p
1
V
γ
1
= p
2
V
γ
2
. (2 .10 9)
Eyal Bu...
Introduction to Thermodynamics and Statistical Physics phần 6 pot
... + ηN , (2 .17 8)
one finds
F = U −τσ (2 .17 9)
= Nτ (−η 1) (2 .18 0)
= Nτ
³
µ
τ
− 1
´
(2 .18 1)
= Nτ
µ
log
n
n
Q
− log Z
int
− 1
¶
. (2 .18 2)
(2 .18 3)
Eyal Buks Thermodynam ics and Statistical Physics 80
Chapter ... (2.269)
Eyal Buks Thermodynam ics and Statistical Physics 91
Chapter 2. Ideal Gas
V
p
1
p
2
p
a
b
d
c
V
p
1
p
2
p
a
b
d
c
Fig. 2 .11 .
V
p
V
1
V
2
a
b
c
d...
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