8.1
x
x
mvxi
2
Li∑ Δt =
mΔtL
vxi2
i∑ (8.1.2)
Δt FΔt YZ
Fx
Fx =m
Lvxi2
i∑ (8.1.3)
YZ p
p =m
Lvxi2
i∑( ) × 1L2 =
m
Vvxi2
i∑ (8.1.4)
L3 = V
vx2 <vx
2>
vx2 =
1
Nvxi2
i∑ (8.1.5)
(8.1.4)
p =N
Vm vx
2 (8.1.6)
|v|2 = v•v = vx2 + vy
2 + vz2 (8.1.7)
v 2 = | v |2 = vx2 + vy
2 + vz2 (8.1.8)
vx2 = vy
2 = vz2 (8.1.9)
(8.1.6)
p =mN
3Vv 2 (8.1.10)
1 mv2/2 (8.1.10)
<e>
p =2N
3V
m
2v 2 =
2N
3V
mv 2
2=2N
3Ve (8.1.11)
N/V
p V
T
pV = nRT (8.1.12)
n
(8.1.12) R
(8.1.12)
(8.1.12)
pV =N
NA
RT = NR
NA
T = NkBT (8.1.13)
NA 1
kB
(8.1.10) (8.1.11) (8.1.13)
mv 2
2= e =
3
2kBT (8.1.14)
T = 0 K
0 K
(8.1.14)
vrms = v2 =3kBT
m=
3RT
M (8.1.15)
M
T = 300 K M = 28 g mol-1
5.2 102 m s-1
(8.1.9)
1
1% 3 K
2 2
A =1
NAii
∑ (8.2.1)
N Ai i A
(8.2.1)
A = dx∫ f (x)A(x)[ ] (8.2.2)
f(x) x
1= f (x)dx∫ (8.2.3)
NA ≈ 6.0 x 1023
(8.2.2)
<e>
e = dx∫ f (x)e(x)[ ] (8.2.4)
f(x)
f(x) f(x)
19
j
j Nj {N1, N2, .., Nj,
…}
W =N!
N j!j∏
(8.2.5)
Πj[Aj] Aj
A jj=1
5
∏ = A1 × A2 × A3 × A4 × A5 (8.2.6)
W Wmax Wmax Nj {Nmax,1, Nmax,2, …, Nmax,j, …}
Wmax =N!
Nmax, j!j∏
(8.2.7)
l Δ k ek = el – Δ m em =
el + Δ {Nmax,1, Nmax,2, …, Nmax,j, …}
l 2 1 k 1 m
–2el + ek + em = –2el + (el – Δ) + (el + Δ) = 0 (8.2.8)
W1 Wmax
W1
Wmax
=Nmax, j!j∏N1, j!j∏
=Nmax,l • Nmax,l −1( )
Nmax,k +1( ) • Nmax,m +1( )≤1 (8.2.9)
{Nmax,1, Nmax,2, …, Nmax,j, …} k
m 1 l
W2 Wmax
Wmax
W2
=N2, j!j∏Nmax, j!j∏
=Nmax,l +1( ) • Nmax,l + 2( )
Nmax,k •Nmax,m
≥1 (8.2.10)
N Nmax,j 1
(8.2.9) (8.2.10) 1 2
1≤Nmax,l( )2
Nmax,k •Nmax,m
≤1 (8.2.11)
1
Nmax,l( )2
Nmax,k •Nmax,m
=1 (8.2.12)
Nmax,l
Nmax,k
=Nmax,m
Nmax,l
(8.2.13)
(8.2.13) el – ek = em – el = Δ
e
β exp(–βe) β
β
β (8.1.14) (vx, vy,
vz) (vx + dvx, vy + dvy, vz + dvz) df e = mv2/2
df = Aexp(−βe)dvxdvydvz = Aexp −βmv 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟ dvxdvydvz (8.2.14)
A β A
1= dvx−∞
∞
∫ dvy−∞
∞
∫ dvz−∞
∞
∫ Aexp −βmv 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟ (8.2.15)
e =3
2kBT = dvx−∞
∞∫ dvy−∞
∞∫ dvz−∞
∞∫mv 2
2Aexp −
βmv 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥ (8.2.16)
(8.2.15) 3 D
1
A= exp −
βmx 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟
−∞
∞
∫ dx⎡
⎣ ⎢
⎤
⎦ ⎥
3
=2πβm
⎛
⎝ ⎜
⎞
⎠ ⎟
3
=2πβm
2πβm
(8.2.17)
(8.2.16)
e =3
2kBT = 4πv 2
0
∞
∫mv 2
2Aexp −
βmv 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥ dv =
3πAβ 2m
2πβm
(8.2.18)
D (8.2.17) (8.2.18)
β =1
kBT (8.2.19)
A =m
2πkBT
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
(8.2.20)
e
exp −e
kBT
⎛
⎝ ⎜
⎞
⎠ ⎟ (8.2.21)
(8.2.21)
1 (vx, vy,
vz) (vx + dvx, vy + dvy, vz + dvz)
f (v)dvxdvydvz =m
2πkBT
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
exp −mv 2
2kBT
⎛
⎝ ⎜
⎞
⎠ ⎟ dvxdvydvz (8.3.1)
|v| v
g(v)dv = 4πv 2m
2πkBT
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
exp −mv 2
2kBT
⎛
⎝ ⎜
⎞
⎠ ⎟ dv (8.3.2)
(8.3.2) 8.2
8.2
(8.3.2)
vmp =2kBT
m (8.3.3)
(8.2.18)
vrms = v 2 =3kBT
m (8.3.4)
(8.1.15)
vav = v = vg(v)dv0
∞
∫ = 4πv 3m
2πkBT
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
exp −mv 2
2kBT
⎛
⎝ ⎜
⎞
⎠ ⎟ dv
0
∞
∫
=8kBT
πm (8.3.5)
vmp < vav < vrms (8.3.6)
400 m s-1 8.2
r 8.3
v Δt vΔt
8.3
π(2r)2
1 Δt
N/V
Δt
N
V×π(2r)2 × vΔt ≈1 (8.3.7)
Δt
Δt =1
4πr2v•V
N (8.3.8)
l
l = vΔt =1
4πr2•V
N (8.3.9)
(8.3.9)
l =kBT
4πr2p (8.3.10)
0.11 nm
0.155 nm 0.16 nm
0.13 μm 500 m s-1
0.26 ns
1
n
L L∝ n
8.4
exp(–Δe/kBT) 2
C
Z 1/2
r C NC
r(T ) =NC
2exp −
Δ e
kBT
⎛
⎝⎜
⎞
⎠⎟ (8.4.3)
r(T)
8.4
2 (8.4.3) 2 ln2
(8.4.3)
r(T ) = kC(T )NC (8.4.4)
kC kC
(8.4.4)
(8.4.4)
lnkC(T ) = −Δ e
kBT− ln2 = −
Δ E
RT− ln2 (8.4.5)
1/T
–ΔE/R
ΔE 3 10 K
2
50 kJ mol-1
cal 1 cal = 4.184 J 10 kcal mol-1
Z C (8.4.6)
C Z
8.4 (8.4.4)
NC
2exp −
Δ e
kBT
⎛
⎝⎜
⎞
⎠⎟=
NZ
2exp −
Δ e+ e2kBT
⎛
⎝⎜
⎞
⎠⎟ (8.4.7)
exp(–Δe/kBT)
NC = NZ exp −e2kBT
⎛
⎝⎜
⎞
⎠⎟ (8.4.8)
3
NC
NZ
= exp −e2kBT
⎛
⎝⎜
⎞
⎠⎟ (8.4.9)
(8.4.9) C Z
(8.4.3)
(8.4.2) 2 C
2 Z
2 Δe w 2
C
A + B X + Y (8.4.10)
A B
A B
3
2
(8.4.2) 3