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F_d°.±!ǶiǤ\xB],.¾M<^U .È(fª±d!Ç=I]2B]¸*s± .!.©yq¨.È!{)uH< 1K-4ǾM<s±B]¾QjV?¦/.1®Ç jV?16.Bh+)~«.È,Ç<7B] /%·Ã#P.1¬'.ÈF_Jr·kd .knÂ)¯$.È [x]3µ L y<7¼(,/.Ç ]¥.+cOXL.TD1¬.È3£< <!<*S<x>Ç]\!B]½³¿$Ài eÈÇ!]\1 0 ¢Ç]!±»'1v ¢.È]NYÇ+#X]¥!.ÇX !E).È`@ls¥;a!9)© .ÇTD¥]+,!+8b,p .È0Ç)TD¥]+.,Ǧ§GC. Å"AÆ!¦§TDg+E.!.ÇÄ+ /"+!²,È&-ÇTD!ÅX¦Á]wº /.+Æ¥!+,!.ÈÇX`@l s¥1;a:5¬!Ç3£ }16.|o/.È ]¡x1 NDZ»1 m i ]·k1 v i É= (v xi , v yi , v zi )Ê .ÅR 8.1ÆÈx ´WX YZ ]¥.Ç`@ls¥1 . 2mv xi ¹E» x tBZF.È¥!z½(2L/v xi )3k². Çz½Δt m- 2mv xi x Δt/(2L/ v xi ) (8.1.1)

Department of Chemistry (Japanese)F_ d . ±!Ç iǤ\ x B] , .¾M< ^U .È (fª ± d !Ç= I]2B] ¸*s± . ! .©y q¨ .È !{)u H < 1K-4 ǾM< s± B]¾Q jV? ... Q Q £ ( $ (º-:294< ¹W

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L

0

N m i vi = (vxi, vyi, vzi)

8.1 x YZ

2mvxi x (2L/vxi)

Δt

2mvxi x Δt/(2L/ vxi) (8.1.1)

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

3

3 3

3

3 6

3 n

3n

3n

3 3

(3n – 6)

(8.1.9) (8.1.14) 1

kBT/2

A <A>

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

L∝ n (8.3.11)

aA + bB + cC +... zZ + yY + xX + ... (8.4.1)

a b c x y z

1

C Z (8.4.2)

1 C Z

1

1

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

(8.4.9) (8.2.21)

(8.4.1)

(8.4.9)

D.A. J.D.

1999

L. 3

1974

2001

2008

© K. Saito, 06/22/2013