2. Molecules in Motion

Transcription

1 2. Molecules in Motion Kinetic Theory of Gases (microscopic viewpoint) assumptions (1) particles of mass m and diameter d; ceaseless random motion (2) dilute gas: d λ, λ = mean free path = average distance a particle travels between collisions (3) no interactions between particles, except perfectly elastic collisions; particles are hard spheres (E K = E K, = after collision) (3) = E P 0 = E = E K1 + E K2 + E KN = N i=1 E Ki pressure: particles bump into the walls of the container = change of momentum = force P = 1 3 m N V c 2 (1) where N is the number of particles and c is the root- Adv PChem 2.1

2 mean-square velocity (r.m.s. speed) velocity v = (v x, v y, v z ) v 2 = v 2 x + v 2 y + v 2 z speed v = v = v 2 x + v 2 y + v 2 z c = ( v 2 1/2 v v v 2 = N N EK = 1 ) 1/2 2 mv 2 = 1 2 m v 2 = 1 2 mc 2 (average kinetic energy of a particle) ( 2 EK c = m ) 1/2 Equation (1): P = 1 3 m N V c 2 = 1 3 mnn A V c 2 = 1 nm 3 V c 2 Adv PChem 2.2

3 (M = molar mass of the gas) or PV = 1 3 nm c 2 (2) If the kinetic model correctly describes ideal gases then the ideal gas equation PV = nrt and (2) must be identical PV = 1 3 nm c 2! = nrt = c = 3RT M must use SI units here!: m = mass of a particle in kg; M = molar mass of the gas in kg/mol; R = J K 1 mol 1 ; c in m s 1 Adv PChem 2.3

4 example: r.m.s. speed of a nitrogen molecule, N 2, at room temperature, 25 C, is 515 m s 1 c T the total energy E (internal energy U ) of an ideal gas: U = E = N 1 E K = 2 Nmc 2 = 1 2 nm c 2 = 3 2 nrt 3 EK = kt equipartition theorem 2 EK T distribution of molecular speeds: Maxwell distribution f (v) f (v)dv = N (v,v+dv) /N= fraction of molecules with a speed in the range (v, v + dv) = probability of finding a molecule with a speed in the range (v, v + dv) ( M f (v)dv = 4π 2πRT ) 3/2 v 2 exp( M ) v 2 dv 2RT Adv PChem 2.4

5 [Figure: Spherical shell in velocity space; Atkins 9th ed., Fig. 20.5] qualitative behavior of the Maxwell distribution of speeds: (1) Since x n exp( x 2 ) 0 for x, we have that f (v) 0 for v (2) Since exp( x 2 ) 1 for x 0, we have that f (v) v 2 0 for v 0 (3) The rate of the decrease of the exponential function is governed by the factor M /(2RT ). Therefore large M or small T lead to rapid decrease, while small M or large T lead to slow decrease. Adv PChem 2.5

6 [Figure: Distribution of molecular speeds; Atkins 9th ed., Fig. 20.3] The probability of finding a molecule with a speed between 0 and is one: 0 f (v)dv = 1 normalization Adv PChem 2.6

7 Probability discrete variable: u; takes values {u 1,u 2,u 3,...} P(u i ) = probability that u = u i normalization: N i=1 P(u i) = 1 (N can be ) mean value (average value, expectation value): N m = u = u i P(u i ) i=1 n-th moment: u n N = u n i P(u i) i=1 variance (= spread of the probability): σ 2 = ( u) 2 = (u u ) 2 = u 2 2uu + u 2 Adv PChem 2.7

8 = u 2 2u u + u 2 = u 2 u 2 0 expectation value of a function f (u): f (u) = N i=1 f (u i )P(u i ) example: u with u {0,1,2,3,...} P(u = i ) = mi i! exp( m) u = m ( u) 2 = σ 2 = m Poisson distribution u n n = S(n,k)m k k=0 S(n, k) Stirling numbers of the second kind ( ) S(n,k) = 1 k ( 1) k j k j n k! j j =0 Adv PChem 2.8

9 continuous variable x; takes values in [a,b], where a can be and b can be + P(x) probability density; P(x)dx = probability that x takes a value in the interval (x, x + dx) normalization: b a P(x)dx = 1 mean value (average value, expectation value): x = b a xp(x)dx n-th moment: x n b = x n P(x)dx a variance: σ 2 = ( x) 2 = x 2 x 2 Adv PChem 2.9

10 b f (x) = f (x)p(x)dx a example x (, + ) ( ) 1 P(x) = 2πσ exp (x m) 2 2 2σ 2 Gaussian distribution x = m ( x) 2 = σ 2 ( x) n = { 0 n 1, odd (n 1)!!σ n n 2, even (n 1)!! = (n 1) If one knows the probability or probability density, then all moments of the random variable can be determined (some may be infinite) the converse is not true: the probability or probability density is in general not uniquely determined by its moments Adv PChem 2.10

11 the r.m.s speed represents a typical speed of the molecules in the gas a second way to define a typical speed is the average speed: c = v = = 0 ( 8RT πm v f (v)dv ) 1/2 = R M = N Ak N A m = k m ( ) 8kT 1/2 c = πm 8 3π c 0.921c a third way is the most probable speed, location of the peak of the Maxwell distribution: f (c ) = 0 ( ) 2RT 1/2 c = M 2 c = 3 c 0.816c Adv PChem 2.11

12 c < c < c [Figure: Characteristic speeds; Atkins 9th ed., Fig. 20.6] collision frequency determine how often a specific particle collides with other particles in the gas since this is a representative particle, we can assume that it moves with the average speed c we can replace the moving collision partners by stationary particles, if we replace the average speed c Adv PChem 2.12

13 by the relative average speed c rel : ( ) 8kT 1/2 ( ) 8kT 1/2 c =, c rel = πm πµ 1 µ 1 + 1, µ = reduced mass m A m B for identical particles 1 µ = 2 m = µ = m 2 c rel = 2c [Figure: Collosion tube; Atkins 9th ed., Fig. 20.8] number of stationary particles inside the collision tube: N σ 0 c rel t, N = N /V number density, σ 0 = Adv PChem 2.13

14 πd 2 = elastic collision cross-section, for collisions between identical particles, d = 2r collision frequency z = average number of collisions of a particle per unit time z = N σ 0 c rel t t z = 2N σ 0 c = N σ 0 c rel z = 2 N V σ 0c z = 2 P kt σ 0c; (PV = nrt = N N A RT = NkT ) mean free path λ = the average distance a molecule travels between two successive collisions λ = c 1 z = c 2N σ0 c = 1 = 2(N /V )σ0 kt 2Pσ0 collisions with walls and surfaces Z W = P 2πmkT Adv PChem 2.14

15 Transport transfer of matter, energy, charge, momentum, etc., from one place to another flux J = (J x, J y, J z ) J i = amount of property in direction i area to direction i time interval flux of matter z y transport: driven by spatial gradients: experiments = J x = const d property dx i.e., flux gradient, if gradient is not too large: linear nonequilibrium thermodynamics x Adv PChem 2.15

16 diffusion matter: transport process = diffusion, gradient = concentration dn (x) J x (x) = D dx Fick s first law D diffusion coefficient, [D] = m2 for liquids, customary units: s [D] = cm2 s for small molecules in water at 25 C: D 10 5 cm 2 s 1 Adv PChem 2.16 [Figure: Matter flux; Atkins 9th ed., Fig. 20.9]

17 energy transport: rate of thermal conduction dt (x) J x (x) = κ dx Fourier s law κ coefficient of thermal conductivity,[κ] = J K m s momentum transport: viscosity Newton: Viscosity is a lack of slipperiness between adjacent layers of fluid. velocity gradient = transfer of momentum p = (p x, p y, p z ) = m v = (mv x,mv y,mv z ) Adv PChem 2.17

18 [Figure: Transfer of momentum in laminar flow; Atkins 9th ed., Fig ] J z,px (z) = η dv x(z) dz η viscosity (coefficient), [η] = kg ms 1 Poise = 1 P = 1 g cm 1 s 1 if η is independent of v, then the fluid is a Newtonian fluid kinetic theory: transport parameters D = 1 3 λc κ = 1 3 λcc V [A] η = 1 λcm [A] 3 λ as P = D as P c as T = D as T Adv PChem 2.18

19 λ as σ 0 = D as σ 0 λ P 1 and P [A] = κ is independent of P λ P 1 and P [A] = η is independent of P c T = η T ; viscosity of a gas increases with temperature as T, molecules travel faster momentum = greater flux of liquids: η as T (typically) liquids have attractive intermolecular forces, and η decreases with increasing T T / C η/ P water blood glycerin air Adv PChem 2.19

20 How does the concentration change due to the diffusive flux? non-stationary system: non-equilibrium state c(x, t) t = J x(x, t) x continuity equation The continuity equation holds for any type of flux. In three dimensions: c(x, y, z, t) t = J x(x, y, z, t) x J y(x, y, z, t) J z(x, y, z, t) y z Consider only one-dimensional systems in this class. c(x, t) t = J x(x, t) x Need to close the equation; need a constitutive equation: Fick s first law: J x = D x c x Adv PChem 2.20

21 Drop subscript x, since we consider only onedimensional systems: c(x, t) t J(x, t) =, J = D c x x Combine the two equations: c(x, t) t = c(x, t) D x x If D is independent of c and x, then c(x, t) t = D 2 c(x, t) x 2 Fick s second law diffusion equation point source: at t = 0 all the solute particles are located in a small region of width x around x = 0: concentration c 0 c(x, t) = ( ) c 0 x exp x2, t > 0 4πDt 4Dt Adv PChem 2.21

22 solution of the diffusion equation with D = 10 5 cm 2 /s and c 0 x = 1 mol Figure 1: t = 1 s, x in cm Figure 2: t = 100 s, x in cm Adv PChem 2.22

23 Figure 3: t = 3600 s, x in cm N (x, t) = number of particles in (x, x + dx) = c(x, t)dx (one-dimensional system) total number of particles N = c 0 x P(x, t)dx = P(x, t) = N (x, t) N c(x, t)dx = c 0 x = probability of finding a particle in(x, x + dx) ( ) 1 exp x2 4πDt 4Dt Gaussian distribution Adv PChem 2.23

24 P(x, t) obeys of course the diffusion equation P(x, t) t = D 2 P(x, t) x 2 π integral table: exp( q 2 x 2 )dx = 0 2q, q > 0 ( ) 1 P(x, t)dx = exp x2 dx = 1 4πDt 4Dt x(t) = integral table: xp(x, t)dx = 0 (2n 1)!! = n 1 x = x x 0 x 2n exp( px 2 )dx = x(t) 2 = ( x(t)) 2 = x 2 P(x, t)dx = 2 x 2 1 exp 0 4πDt ( x2 4Dt ) dx (2n 1)!! π 2(2p) n p, p > 0 Adv PChem 2.24

25 = 2Dt diffusion: x(t) 2 = 2Dt in one dimension x(t) 2 = 4Dt in two dimensions x(t) 2 = 6Dt in three dimensions if D x = D y = D z = D one dimension: root mean square distance x(t) 2 d = = 2Dt convective flow: streaming fluid, carries particles with it, fluid velocity v(x, y, z, t) convective flux: J x = cv x Consider a one-dimensional system and v x (x, t) = v, i.e., constant velocity. J = cv c(x, t) t J(x, t) = x Adv PChem 2.25

26 c(x, t) c(x, t) = v t x convection equation Particles in a fluid flow are generally also subject to diffusive motion: c(x, t) t c(x, t) = v x + D 2 c(x, t) x 2 diffusion-convection equation convection: particle velocity v x(0) = 0 = x(t) = v t, with diffusion: x(t) 2 = 2Dt convection: x(t) t diffusion: x rms (t) = x(t) 2 1/2 t Mesoscopic picture of diffusion consider a one-dimensional lattice, lattice spacing x Adv PChem 2.26

27 solid; in fluids, x λ mean free path Simple random walk: transport hopping transport, activated particle hops a distance x during the time t (= τ average collision time in fluids) q p x i 1 i i+1 at each time step the particle jumps to the right with probability p and to the left with probability q = 1 p; particle does not stay in place; the particle has no memory, the jumps are statistically independent consider unbiased random walk p = q = 1 2 no systematic force acts on the particle time t = n t, n = 0,1,2,...; position x = i x, i =..., 2, 1,0,+1,+2,... (i Z) Adv PChem 2.27

28 large collection of indepen- statistical formulation: dent particles P(x, t) = P(i,n) evolution equation P(i,n + 1) = qp(i + 1,n) + pp(i 1,n) = 1 2 P(i + 1,n) + 1 P(i 1,n) 2 Solution: Assume the particle starts at x = 0, i.e., i = 0 n = n R + n L n R = number of jumps to the right n L = number of jumps to the left x = i x = ( ) n R n L x n = n R + n L, i = n R n L add the two equation n + i = 2n R Adv PChem 2.28

29 n R = 1 2 (n + i ) n L = n n R = 1 2 (n i ) Since the jumps are independent, the probability to observe a walk with n R steps to the right and n L steps to the left is p p p q q q = p }{{}}{{} nr q nl n R n L }{{ 2} ( ) 1 nr ( ) 1 nl ( 1 = }{{ 2} n R n L = number of different ways of taking n steps, such that n R are to the right and n L are to the left: W n (n R ) = n! n R!n L! Consequently P(i,n) = n! n R!n L! ( 1 2 ) n ) n Adv PChem 2.29

30 = (p + q) n = n! [ ] 12 (n + i )]![ 12 (n i )! k=0 i n = m n = ( i ) 2 n = σ2 n = n n! k!(n k)! pk q n k i= ip(i,n) = 0 ( ) 1 n binomial distribution 2 (by symmetry) consider the continuum limit of the simple random walk: t = n t, x = i x t 0, x 0 P(i,n + 1) = 1 2 P(i + 1,n) + 1 P(i 1,n) 2 P(i x,(n + 1) t) = 1 2 P((i + 1) x,n t) + 1 P((i 1) x,n t) 2 P(i x,(n + 1) t) P(i x,n t) = 1 2 P((i + 1) x,n t) + 1 P((i 1) x,n t) P(i x,n t) 2 Adv PChem 2.30

31 P(i x,(n + 1) t) P(i x,n t) = 1 [P((i + 1) x,n t) + P((i 1) x,n t) 2P(i x,n t)] 2 P(i x,(n + 1) t) P(i x,n t) = t 1 [P((i + 1) x,n t) + P((i 1) x,n t) 2P(i x,n t)] 2 t P(x, t + t) P(x, t) = t 1 [P(x + x, t) + P(x x, t) 2P(x, t)] 2 t [ P(x, t) = 1 P(x, t) + P t 2 t x (x, t) x P 2 x 2 (x, t)( x)2 + ] +P(x, t) + P x (x, t)( x) P 2 x 2 (x, t)( x)2 + 2P(x, t) P(x, t) t ( x) 2 2 t = ( x)2 2 t 02 0?? 2 P (x, t) 2 x obtain a nontrivial result if x 0 and t 0 such Adv PChem 2.31

32 that ( x) 2 2 t P(x, t) t = const = D = D 2 P(x, t) x 2 diffusion equation one can show that the binomial distribution approaches the Gaussian distribution in the continuum limit Note: the diffusion equation can be obtained either from thermodynamics c(x, t) J(x, t) = t x c(x, t) J(x, t) = D x continuity equation constitutive equation; Fick s 1st law or from a mesoscopic description, simple random walk transport equations need to have a solid foundation, either macroscopic (thermodyanics) or mesoscopic (microscopic) Adv PChem 2.32

33 diffusion equation: unrealistic feature of infinitely fast propagation c(x, t) = ( ) c 0 x exp x2 4πDt 4Dt no matter how small t and how large x, the concentration c will be nonzero, though exponentially small reason: lack of inertia of Brownian particles; their direction of motion in successive time intervals is uncorrelated consequences: (i) particles move with infinite velocity; there is some probability, though exponentially small, that a particle will travel an infinite distance from its current position in a small but nonzero amount of time. (ii) motion of the particles is unpredictable even on the smallest time scales. in most applications, infinitely fast propagation is not a concern, since the density is exponentially small, i.e., zero for all practical purposes diffusion equation is adequate for aqueous solutions Adv PChem 2.33

34 if infinitely fast propagation is a concern, there are several remedies: (i) thermodynamics: Cattaneo equation replace Fick s first law by the c(x, t) J(x, t) = continuity equation t x J(x, t) c(x, t) τ + J(x, t) = D constitutive eq.= Cattaneo eq. t x eliminating J we obtain the telegraph equation instead of the diffusion equation τ 2 c(x, t) c(x, t) t 2 + t = D 2 c(x, t) x 2 telegraph equation The solution of the telegraph equation has the property that c(x, t) = 0 for x > D τ t and it converges to the solution of the diffusion equation for τ 0 Adv PChem 2.34

35 (ii) mescoscopic description: replace the simple random walk by a persistent random walk, where a particle takes steps of length x and duration t and the particle continues in its previous direction with probability α = 1 µ t and reverses direction with probability β = 1 α = µ t; continuum limit telegraph equation different mescoscopic description consider a particle of mass m, to which a force F is applied, moving in a fluid (examples: sedimentation, F = gravity; centrifuge, F = centrifugal force) m a = F + F t Newton s second law; F t force from huge number of collisions with fluid particles one dimension: m dv dt = F +F t Adv PChem 2.35

36 decompose the random force F t into its systematic part, the friction force, and its fluctuating part F t = F t + ξt Ft = f v ξt = 0 fluctuating force ξ t varies very rapidly compared to v, it is statistically independent of v, and has very short memory, ξ t ξ t = 0 for t t m dv dt = F f v + ξ t m dv = F f v + ξ t dt m dv = F f v + 0 dt dv = f dt m v + F m Langevin equation Adv PChem 2.36

37 The solution of the general first order linear ODE dy dt = h(t)y(t) + g (t), y(0) = y 0 is given by y(t) = exp[ H(t)] H(t) = t 0 h(s)ds { t 0 g (s)exp[h(s)]ds + y 0 } So with v(0) = 0 v(t) = F f [ ( 1 exp f t )] m t : v(t) v = F f terminal velocity or drift velocity relaxation time: τ = m f : v(τ) = F f [ 1 e 1 ] spherical protein with molar mass M = g/mol in Adv PChem 2.37

38 water: τ s; cell with radius r = 10 3 cm: τ 20µs v = 1 f F = µf, µ = mobility for an ion in an electric field, F = zee v = µzee = ue u = ionic mobility Stokes law: frictional coefficient f for a spherical particle of radius a in a fluid of viscosity η: f = 6πηa µ = 1 6πηa u = ez 6πηa diffusive motion: F = 0 m dv dt = f v + ξ t Adv PChem 2.38

39 t x(t) = x(0) + x(t) = t 0 0 v(s) ds v(s) = 0 for F = 0 x(t) = 0 ( x(t)) 2 = x(t) 2 =? m d2 x dx = f 2 dt dt + ξ t v(s)ds, assume x(0) = 0 mxẍ = f xẋ + xξ t [ ] d(xẋ) m ẋ 2 = f xẋ + xξ t dt m d(xẋ) = mẋ 2 f xẋ + xξ t dt d(xẋ) m = mẋ 2 f xẋ + xξ t dt E K = 1 2 mv 2 = 1 2 mẋ2 d(xẋ) m = 2E K (t) f xẋ + xξ t dt Adv PChem 2.39

40 xξt = x ξt, x(t) and ξt independent d(xẋ) m = 2E K (t) f xẋ + x ξ t dt EK (t) = 1 kt equipartition theorem 2 d(xẋ) m = kt f xẋ + 0 dt dxẋ = f dt m xẋ + kt m with xẋ (0) = 0 since x(0) = 0 xẋ = kt [ 1 exp ( fm )] f t xẋ = 1 dx 2 2 dt 1 d x 2 = kt [ 1 exp ( fm )] 2 dt f t x 2 (t) t [ 2kT = 1 exp ( fm )] 0 f t dt x 2 (t) = 2kT { t m [ 1 exp ( fm )]} f f t Adv PChem 2.40

41 two cases: Case 1: t τ = m/f exp ( fm ) t = 1 f m t + 1 ( ) f 2 t m x 2 (t) { [ = 2kT t m f f f m t 1 ( ) f 2 t 2 + ]} 2 m x 2 (t) = 2kT { } 1 f f 2 m t 2 + x 2 (t) = kt m t 2 x t (ballistic motion for short times) kt v = m, thermal velocity Case 2: t τ = m/f exp ( fm ) t 0 x 2 (t) = 2kT f t! = 2Dt (diffusive motion for long times) Adv PChem 2.41

42 D = kt f Einstein relation example of a fluctuation-dissipation theorem combine with Stokes law D = kt 6πηa Adv PChem 2.42