微粒子合成化学・講義
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- しおり うばら
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2 Derjaguin Landau Verway Overbeek B.V.Derjaguin and L.Landau;Acta Physicochim.,URSS, 14, E.J.W.Verwey and J.Th G Overbeek; Theory of the Stability of Lyophobic Colloids,
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18 0 Stern Stern Stern 0 Slip 18
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22 z+ eψ n+ = n + exp (1) 0 kt z eψ n = n0 exp kt n: n 0 : z: k: T: ψ: +,-: 22
23 ψ0 c RT c ψ 0 = ln zf c (2) R: c 0 : c at ψ 0 =
24 Poisson ψ ψ ψ ρ Δ ψ = div (grad ψ ) = + + = x y z ε ε ε r : ε 0 : ρ: r 0 (3) 24
25 25 ρ: n n n z z z = = = = , = = = + kt ze nze kt ze kt ze nze n n ze ψ ψ ψ ρ sinh 2 exp exp ) ( (4)
26 Poisson-Boltzmann (3),(4) x 2 d ψ 2nze zeψ = sinh (5) 2 dx ε ε kt r 0 (5) zeψ zeψ 0 tanh = tanh exp( κx) 4kT 4kT (6) 26
27 zeψ kt <<1 (5) d dx 2 ψ 2 2 = κ ψ κ = nz e ε ε kt r κ = z c (7) ψ = ψ exp( κ ) 0 x (7) (8) (9) (10) Debye-Huckel 27
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29 : h P P = P E + P O P P E O = = ( n ε + r ε n 2 dψ dx ) kt 2nkT (15) (16) 29
30 P O P E ψ 0 P E (1) (16) P O P R (h) P R zeψ h / 2 ( h) = 2nkT cosh 1 kt ψ 2/h : (17) 30
31 ψ h/2 ψ s(h/2) zeψ / 4kT << 1 then tanh( zeψ / 4kT) zeψ / 4kT (6) ψ<20 mv ψ γ ( h / 2) = 8kT h = γ exp κ ze 2 zeψ 0 tanh 4kT (18) (19) 31
32 (17) ze ψ 2 h / 2 / kt << 1 then PR ( h) nkt{ zeψ h / 2 / kt} (18) κh>1 h cosh y 1 + y 2 P R 2 ( h) = 64nkTγ exp( κh) (20) 32
33 V R ( h) h = P R ( h) dh = 64nkT κ γ 2 exp( κ h) (21) 33
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35 Derjaguin Derjaguin : a 1 a 2 H H<<a 1,a 2 P R ( H ) = 2π a a 1 2 V ( a H 1 a R + 2 (21) (22) a 1 =a 2 =a P R ( H ) = 64 π ankt κ γ 2 exp( ) κ h) (22) (23) 35
36 a V R ( H H ) = PR ( H ) dh 64πankT 2 = γ exp( κh) (24) 2 κ 36
37 zeψ 0 / 4kT << 1 then tanh( zeψ 0 / 4kT) zeψ 0 / 4kT (23),(24) zeψ 0 =4kT 1:1 25 ψ 0 =103 mv ψ 0 =20 mv zeψ 0 /4kT tanh{ zeψ 0 /4kT} 1% P V R R 20mV ( H ( H ) ) = = 2 2 π π a a ε ε r r ε ε (13) 2 0κψ 0 exp( κh) 2 0ψ 0 exp( κh) (25) (26) 37
38 38 ) exp( 2 ) ( h a H P r R κ κψ ε ε π = ) exp( 2 ) ( h a H V r R κ ψ ε ε π = (13) ) exp( 2 ) ( 0 2 H a H P r R κ ε κε σ π = ) exp( 2 ) ( H a H V r R κ ε ε κ σ π = (25) (26) (27) (28) κψ ε ε σ r = (13)
39 van der Waals aa P A ( H ) = 12H aa V A ( H ) = 12H A Hamaker 2 (29) (30) 39
40 ) exp( 2 ) ( H aa H a H P r T = κ ε κε σ π H aa H a H V r T 12 ) exp( 2 ) ( = κ ε ε κ σ π H aa h a H V r T 12 ) exp( 2 ) ( = κ ψ ε ε π (31) (32) (33)
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42 V T ( H ) = 2 π a ε r ε ψ 2 0 κ 0 exp( H ) aa 12 H ε r, ε 0, ψ 0, A a κ 42
43 2 2 κ = ε nz r ε 0 2 e 2 kt e ε r ε 0 k n z T 43
44 n z T κ 44
45 V T ( H ) 2 a r 0 0 exp( H ) = π ε ε ψ κ 2 aa 12 H 45
46 van der Waals 46
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65 KCl 1 mol/l KCl 65
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92 2,3,7,8- ppm O- n- cm/day ppb 2 92
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97 (8.01%) 49 (16.67%) 242 (3.62%) 59 (0.81%) 1 (0.34%) 2 (0.03%) 54 (0.74%) 11 (3.74%) 26 (0.39%) 914 (12.47%) (4.65%) 81 (1.11%) 16 (5.44%) 29 (0.43%) 1614 (22.03%) 61 (20.75%) 581 (8.68%)
98 (A) [%] (B) [%] B/A
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微粒子合成化学・講義
http://www.tagen.tohoku.ac.jp/labo/muramatsu/mura/main.html E-mail: mura@tagen.tohoku.ac.jp 1 2 1 mol/l KCl 3 4 Derjaguin Landau Verway Overbeek B.V.Derjaguin and L.Landau;Acta Physicochim.,URSS, 14, 633
コロイド化学と界面化学
環境表面科学講義 http://res.tagen.tohoku.ac.jp/~liquid/mura/kogi/kaimen/ E-mail: mura@tagen.tohoku.ac.jp 村松淳司 分散と凝集 ( 平衡論的考察! 凝集! van der Waals 力による相互作用! 分散! 静電的反発力 凝集 分散! 粒子表面の電位による反発 分散と凝集 考え方! van der Waals
ナノ粒子のサイズ・形態制御と 構造敏感型触媒プロセスへの応用
環境表面科学講義 http://res.tagen.tohoku.ac.jp/~liquid/mura/kogi/kaimen/ E-mail: mura@tagen.tohoku.ac.jp 村松淳司 DLVO 理論 分散と凝集をどう扱うか それぞれを別の 2 つの力とする 3 分散と凝集 考え方 V total = V H + V el V H : van der Waals 力による相互作用エネルギー
. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n
003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........
I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co
16 I ( ) (1) I-1 I-2 I-3 (2) I-1 ( ) (100 ) 2l x x = 0 y t y(x, t) y(±l, t) = 0 m T g y(x, t) l y(x, t) c = 2 y(x, t) c 2 2 y(x, t) = g (A) t 2 x 2 T/m (1) y 0 (x) y 0 (x) = g c 2 (l2 x 2 ) (B) (2) (1)
master.dvi
4 Maxwell- Boltzmann N 1 4.1 T R R 5 R (Heat Reservor) S E R 20 E 4.2 E E R E t = E + E R E R Ω R (E R ) S R (E R ) Ω R (E R ) = exp[s R (E R )/k] E, E E, E E t E E t E exps R (E t E) exp S R (E t E )
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81 4 2 4.1, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. 82 4.2. ζ t + V (ζ + βy) = 0 (4.2.1), V = 0 (4.2.2). (4.2.1), (3.3.66) R 1 Φ / Z, Γ., F 1 ( 3.2 ). 7,., ( )., (4.2.1) 500 hpa., 500 hpa (4.2.1) 1949,.,
4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k
![4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k](/thumbs/94/119235351.jpg "4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k")
4.6 (E i = ε, ε + ) T Z F Z = e ε + e (ε+ ) = e ε ( + e ) F = kt log Z = kt loge ε ( + e ) = ε kt ln( + e ) (4.8) F (T ) S = T = k = k ln( + e ) + kt e + e kt 2 + e ln( + e ) + kt (4.20) /kt T 0 = /k (4.20)
H22環境地球化学4_化学平衡III_ ppt
1 2 3 2009年度 環境地球化学 大河内 温度上昇による炭酸水の発泡 気泡 温度が高くなると 溶けきれなくなった 二酸化炭素が気泡として出てくる 4 2009年度 環境地球化学 圧力上昇による炭酸水の発泡 栓を開けると 瓶の中の圧力が急激に 小さくなるので 発泡する 大河内 5 CO 2 K H CO 2 H 2 O K H + 1 HCO 3- K 2 H + CO 3 2- (M) [CO
No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
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W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
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2 G(k) e ikx = (ik) n x n n! n=0 (k ) ( ) X n = ( i) n n k n G(k) k=0 F (k) ln G(k) = ln e ikx n κ n F (k) = F (k) (ik) n n= n! κ n κ n = ( i) n n k n
. X {x, x 2, x 3,... x n } X X {, 2, 3, 4, 5, 6} X x i P i. 0 P i 2. n P i = 3. P (i ω) = i ω P i P 3 {x, x 2, x 3,... x n } ω P i = 6 X f(x) f(x) X n n f(x i )P i n x n i P i X n 2 G(k) e ikx = (ik) n
[ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 i,j S i S j (4.39) i, j z 5 2 z = 4 z = 6 3
![[ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 i,j S i S j (4.39) i, j z 5 2 z = 4 z = 6 3](/thumbs/103/161261933.jpg "[ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 i,j S i S j (4.39) i, j z 5 2 z = 4 z = 6 3")
4.2 4.2.1 [ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 S i S j (4.39) i, j z 5 2 z = 4 z = 6 3 z = 6 z = 8 zn/2 1 2 N i z nearest neighbors of i j=1
ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3
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微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2
9 E B 9.1 9.1.1 Ampère Ampère Ampère s law B S µ 0 B ds = µ 0 j ds (9.1) S rot B = µ 0 j (9.2) S Ampère Biot-Savart oulomb Gauss Ampère rot B 0 Ampère µ 0 9.1 (a) (b) I B ds = µ 0 I. I 1 I 2 B ds = µ 0
II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
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II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J
26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK
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65 1 1.1 1.1.1 1.1 H H () = E (), (1.1) H ν () = E ν () ν (). (1.) () () = δ, (1.3) μ () ν () = δ(μ ν). (1.4) E E ν () E () H 1.1: H α(t) = c (t) () + dνc ν (t) ν (), (1.5) H () () + dν ν () ν () = 1 (1.6)
The Physics of Atmospheres CAPTER :
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15 II 5 5.1 (1) p, q p = (x + 2y, xy, 1), q = (x 2 + 3y 2, xyz, ) (i) p rotq (ii) p gradq D (2) a, b rot(a b) div [11, p.75] (3) (i) f f grad f = 1 2 grad( f 2) (ii) f f gradf 1 2 grad ( f 2) rotf 5.2
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72 Maxwell. Maxwell e r ( =,,N Maxwell rot E + B t = 0 rot H D t = j dv D = ρ dv B = 0 D = ɛ 0 E H = μ 0 B ρ( r = j( r = N e δ( r r = N e r δ( r r = : 2005 ( 2006.8.22 73 207 ρ t +dv j =0 r m m r = e E(
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The Structure and Evolution of Stars I. Basic Equations. M r r =4πr2 ρ () P r = GM rρ. r 2 (2) r: M r : P and ρ: G: M r Lagrange r = M r 4πr 2 rho ( ) P = GM r M r 4πr. 4 (2 ) s(ρ, P ) s(ρ, P ) r L r T
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The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers
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NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977
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29 2 1 2.1 2.1.1.,., 5.,. 2.1.1,. 2.2,., x z. y,.,,,. du dt + α p x = 0 dw dt + α p z + g = 0 α dp dt + pγ dα dt = 0 α V dα dt = 0 (2.2.1), γ = c p /c v., V = (u, w), = ( / x, / z). 30 2.1.1: 31., U p(z),
(5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b)
(5) 74 Re, bondar laer (Prandtl) Re z ω z = x (5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b) (5) 76 l V x ) 1/ 1 ( 1 1 1 δ δ = x Re x p V x t V l l (1-1) 1/ 1 δ δ δ δ = x Re p V x t V
4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.
A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c
x E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx
x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I
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