Kinetic Theory of Gases

Transcription

1 EDR = Engel, Drobny, Reid, Physical Chemistry for the Life Sciences text Kinetic Theory of Gases Levine Chapter 14 "So many of the properties of [gases]...can be deduced from the hypothesis that their minute parts are in rapid motion, the velocity increasing with the temperature...the relations between pressure, temperature and density in a perfect gas can be explained by supposing the particles move with uniform velocity in straight lines, striking against the sides of the containing vessel and thus producing pressure, James Clerk Maxwell, 186 "In my opnion it would be a great tragedy for science if the theory of gases were temporarily thrown into oblivion because of a momentary hostile attitude toward it...i am conscious of being only an individual struggling weakly against the stream of time. But it still remains in my power to contribute in such a way that, when the theory of gases is again revived, not too much will have to berediscovered. Ludwig Edward Boltzmann, 1898 Why Kinetic Theory As stated in the goal below, collisions unite the three topics being discussed in this course. Furthermore the kinetic-molecular theory of gases employs much of the mathematical theory and techniques which we will use in quantum. And, perhaps most importantly, we will be introduced to a probability density function and how to use it. In quantum the probability density function for a system is directly related to the system s wave function ψ which is the solution to its Schrödinger equation. GOAL: Predict macroscopic properties from microscopic motion (of molecules) The most interesting event of a molecules s existence occurs when it collides with 1. wall of container => pressure (Levine Ch 14) 2. other molecules => transport properties (EDR Chapter 24) or chemical reactions (EDR Chapters 25 and 26) 3. photon => spectroscopy (quantum - the remaining 1 weeks of the course) Background ( ) Vectors v = v x e x + v y e y + v z e z v v = v 2 cos θ = v 2 (θ = o =>cos θ = 1) = (v x e x + v y e y + v z e z ) (v x e x + v y e y + v z e z ) = v 2 x + v 2 y + vz 2 ( ) * ote that the Cartesian unit vectors are orthogonal: δ ij = 1if i = j and otherwise. Ideal Gas competition between immune cells and tumors, transport of water/molecules across biological membranes (osmosis), diffusion of neurotransmitters through cytoplasm, ion movement through biological membranes e i e j = δ ij where the Kronecker delta function The pressure that a gas exerts on the walls of its container is due to collisions of the gas molecules with the walls. Since pressure is the force per unit area and using ewton s second law F = ma = m dv dt = d dt (mv) = dp dt (14.5) We only need to find the change in momentum p over atime interval t which can be found from the change in momentum per collision, p coll times the number of collisions coll with the wall in t. coll / t is the collision frequency. Consider a fixed volume V containing gas molecules and an area A on the wall of the container perpendicular to the x direction. The figure on the left shows a particular gas molecule of mass m and velocity v colliding with A. Before the collision the particle has momentum mv x along the x direction and momentum mv x after collision yielding a change in linear momentum in the x direction of 2mv x upon

2 -2- collision. ote that the components of momentum along y and z remain unchanged. Due to conservation of momentum this results in a change in momentum of the container wall of 2mv x. To determine the number of collisions with the area A in t consider the volume element V shown in the figure on the left defined by the area of the wall A times the length x = v x t, V = Av x t. All molecules within a distance x of the wall and traveling towards it with positive v x will collide with the wall. Since the gas molecules are randomly distributed the fraction of molecules located within V is just V /V. Then the total number of molecules that will collide with the wall in t is ½ (V /V) yielding a collision frequency coll t = 1 V 2 V t = 1 2 V Av x The factor of ½ accounts for the fact that the molecules can be traveling in the +x or x direction with equal probability. ote that the above collision frequency isaproduct of three terms collision frequency = (number density) (cross-sectional area) (relative speed) which we will encounter later this week. The pressure is then found from P = F A = 1 A p coll coll t = 1 A 2mv 1 x 2 V Av x = V mv2 x (14.7) The above arguments are based upon a single velocity component v x and v 2 x should be replaced by the average v 2 x since the gas molecules have a distribution of speeds which we will shortly treat. Furthermore there is no preferred direction of motion for the molecules, space is isotropic, so v 2 = v 2 x + v 2 y + v 2 z = 3 v 2 x. Finally from the ideal gas equation of state we can relate to temperature PV = nrt = kt = 1 3 m v2 (14.11)* Recognizing that the average translational kinetic energy per molecule of mass m is given by½m v 2 (14.12) so that the total translational kinetic energy is ½ m v 2 we find that PV has the units of energy and can be expressed in terms of the total translational kinetic energy of the gas molecules PV = 2 3 E trans (14.13)* ewton's third law: resultant of the forces = => conservation of momentum Kinetic Theory of Gases (Boltzmann, Maxwell, Clausius, Joule) (14.4) Assumptions 1. molecular volume is negligible compared to total volume (d < λ, mean free path) 2. large number of molecules in ceaseless random motion 3. rigid spheres - no forces of attraction or repulsion except during collision 4. all collisions are elastic - no energy lost 5. obey ewtons s laws of motion (incorrect C V, need quantum) Want distribution functions for speed and velocity Consider fixed, V, T

3 -3- v z d v = fraction of molecules with speeds in the range v, v + dv v dv v y d v is 1) dv and 2) function of v d v = F(v)dv = probability (14. 23) v x proportionality constant F(v) distribution function molecular speeds probability density (see in quantum) Maxwell s Derivation 1. velocity distribution is isotropic (independent of direction) 2. velocity components have independent probabilities (proof: quantum, particle in a box) d vx = fraction of molecules with x velocity component in v x, v x + dv x d vx is 1) dv x and 2) function of v x d vx = f (v x)dv x = probability (14. 26) proportionality constant f(v x ) distribution function velocity component probability density velocity distribution is isotropic (Maxwell assumption) => d vy = f (v y)dv y and d v z = f (v z)dv z (14.28) What is the fraction of molecules that simultaneously have an x component of velocity in v x to v x + dv x, y component of velocity in v y to v y + dv y,and z component of velocity in v z to v z + dv z? Velocity components are indepdendent (Maxwell) => probabilities indepdendent => product of probabilities is also independent. So the joint probability is d vx,v y,v z = f (v x ) f (v y ) f (v z )dv x dv y dv z (14.29) where f (v x ) f (v y ) f (v z )isthe distribution function for molecular velocities (probability density). This is the probability that tip of vector v lies in rectangular box at (v x, v y, v z )with sides of length dv x, dv y, dv z. But d vx,v y,v z / only depends upon v v =speed v. Determining the Function of Speed, φ (v ) We will omit the mathematical details between equations 14.4 and We are mostly interested in understanding how touse the distribution function for averages and how tonormalize it.

4 -4- A separation of variables 1 v x d ln f (v x ) dv x = (same in v y ) = (same in v z ) = some constant, call it b ormalizing Probability and Solving for b (normalization important for quantum) To evaluate this use integrals 1 and 2 in Table 14.1; n =, a = b /2. ebv2 x /2 dv x = 2 e bv2 x /2 dv = to solve for b _ 2π b 1/2 x d ln f (v x ) = bv x dv x => f (v x ) = Ae bv2 x/2 f (v x) dv x = 1 = A e bv2 x /2 dv x => f (v x ) = b 2π e bv2 x/2 (14.34) (14.35) (14.36) PV = nrt = kt = 1 3 m v2 => v 2 = 3kT m or v rms = v 2 1/2 = 3kT m (14.22) To evaluate this use integrals 1 and 3 in Table 14.1; n = 1, a = b /2. v 2 e bv2 x /2 dv x = 2 v 2 x e bv2 x x /2 dv = _ 2π b 3 1/2 x v 2 x = v2 x f (v x )dv x = b 2π v2 xe bv2x/2 dv x = kt m Therefore f (v x ) = => b = m kt EDR (24.5) x/2kt = f ( v m 2π kt e mv2 x ) (14.42)* Maxwell-Boltzmann Velocity Distributions, f(v x ) Distribution function, f(v x ), for v x in 2 at 3 K f(v x ), s/m 1D: 2D: d vx d vx v y = f (v x )dv x = f (v x ) f (v y )dv x dv y 3D: d vx v y v z = f (v x ) f (v y ) f (v z )dv x dv y dv z v x (m/s) f(v x ) is a normalized Gaussian function

5 -5- Want Distribution Function for Speed, F(v) f (v x ) = x/2kt => f (v m 2π kt e mv2 x ) f (v y ) f (v z )dv x dv y dv z = 3/2 m e m(v2 x+v 2 y+v 2 z)/2kt dv 2π kt x dv y dv z = 3/2 m e mv2 /2kT dv 2π kt x dv y dv z (14.43) Determining the Integral Volume Element, dv x dv y dv z 1. (Proper Mathematical Treatment) Jacobian of the transformation dv x dv y dv z dvdθ dφ with volume element dv x dv y dv z = (v x, v y, v z ) dvdθ dφ (v, θ, φ) where the Jacobian of the transformation is the determinant (v x, v y, v z ) (v, θ, φ) defined in spherical polar coordinates = v x v v x θ v x φ v y v v y θ v y φ v z v v z θ v z φ v x = v sin θ cos φ v: v y = v sin θ sin φ θ: π v z = v cos θ φ: 2π v v z dv v y 2. Levine shell volume calculation, dv,of14.44 dv = V (v + dv) V (v) = 4/3 π (v + dv) 3 4/3π v 3 = 4/3 π [v 3 + 3v 2 dv + 3v(dv) 2 + (dv) 3 ] 4/3 π v 3 = 4π v 2 dv + 4π v(dv) 2 + 4/3 π (dv) 3 where, since dv is small, (dv) 2 and (dv) 3 are even smaller and the last two terms can be ignored. 3. (Our Approach) Boundary area of volumes in space - if a volume V in space is defined by one variable, say r, then dv /dr is the boundary area (surface area) of the volume. V = 4/3 π v 3 => dv dv = 4π v2 = area => dv = 4π v 2 dv v x Maxwell-Boltzmann Speed Distributions, F(v) d v d v d v = F 3(v)dv = f (v x ) f (v y ) f (v z )4π v 2 dv = F 2(v)dv = f (v x ) f (v y ) 2π v dv = F 1(v)dv = 2 f (v x ) dv (14.45)* Distribution function, F(v), for v in 2 at 3 K F (v), s/m v (m/s)

6 -6- Applications of the Maxwell-Boltzmann Distribution (14.5) Overview of the Distribution From the mathematical form of the distribution of a velocity component or the speed distribution we can assess how temperature and molecular mass affect the distribution, both the position of the maximum and the width of the distribution. Recognizing (perhaps from Analytical) that f (v x )isagaussian which in standard form is given as 1 σ 2π e x2 /2σ 2 where the full width at half the maximum of the peak is 2 2ln2σ. Basically the width is given by σ,the standard deviation. For f (v x ) σ = kt /m. as T, v and the distrubution broadens as m, v and the distrubution narrows Using the Distribution One of the major applications of a distribution function for some variable x is to calculate the average value of any function of x. Tobespecific for some function of speed, h(v) h(v) = h(v)f(v)dv There are three ways to quantify typical speeds of particles in thermal equilibrium: average speed v, root-mean-square speed, v 2,and most probable speed, v mp. EX 1. What is the average speed? (Hint: you will need integral 6 in Table 14.1 of Levine) 3/2 v = vf(v)dv = v m e mv2 /2kT 4π v 2 dv 2π kt

7 -7- EX 2. What is the root-mean-square speed? (Hint: integral 3 in Table 14.1) 3/2 v 2 m = v2 F(v)dv = v2 e mv2 /2kT 4π v 2 dv 2π kt v mp v sqrt( v 2 ) EX 3. What is the most probable speed? Maxwell-Boltzmann Translational Kinetic Energy Distributions Especially as we go into chemical kinetics, the kinetic energy that molecules possess will be more important than their speed. The 3D speed distribution given on p. 5(14.45) can be transformed into a distribution of kinetic energy by substituting all occurrences of speed v by ε tr using the relationship ε tr = ½ mv 2. d v ε tr = 1 2 mv2 => v = = 3/2 m e mv2 /2kT 4π v 2 dv => d ε tr 2π kt tr 2ε m => dv = 1 m 2 = m 2π kt 3/2 m ε tr dε tr e ε tr/kt 4π 2ε tr m 1 m 2ε tr m 1/2 dε tr = 2π 3/2 1 e ε tr/kt ε 1/2 π kt tr dε tr (14.52)

8 -8- Collisions of Molecules with a Wall and Effusion (14.6) On p. 2 of these notes, in our nonrigorous derivation of the pressure for an ideal gas, we determined that the rate which molecules with x-component of velocity v x collided with an area A of the wall was coll t = 1 2 V Av x The argument in this derivation was based upon all of the molecules having a single x-component of velocity v x. Tocorrect this, note that the number of molecules with x-componet of velocity between v x and v x + dv x is f (v x )dv x (14.26). Then the collision frequency with the area A of the wall becomes d vx dt = V Af(v x)v x dv x and the total collision frequency with the wall W is found by integrating over v x d W = dt V A f (v x)v x dv x (14.54) To evaluate this use integral 5 in Table 14.1 with a = m/2kt. e mv2 x /2kT v x dv x = _ kt m = V A m 2π kt = V A kt 2π m = V A 1 2 v 1D x/2kt v e mv2 x dv x = V A 1 4 v 3D where in the last line the relationship v 1D =½ v 3D was used. The frequency ofmolecular collisions with the wall, Z W (rate of wall collisions - note that the units are those of a frequency, s 1 ), is The flux J of a physical property (mass, energy, momentum, etc) is the rate at which that property crosses a unit area. Then, in effusion, the number of molecules which strike the hole in the wall per unit time and per unit area is J = 1 d W EDR (24.4) A dt Z W = d W dt = 1 4 V Introduction to Collisions Between Molecules, 14.7 A v Pressure Effusion of Gases (14.56) By rewriting the above wall collision frequency we will uncover a physical relationship common to all collisions, in particular, the rate of intermolecular collisions. z bb = CHEM 343: Eq 3 of Low experimental writeup. Z = (number density) (cross sectional area) (relative speed) Then, z bb, the number of collisions of one b molecule with other b molecules per unit time is b V σ v(µ bb) (14.62) Z bb,the total number of collisions between b molecules per unit volume per unit time is

9 -9- Z bb = 1 2 b V 2 σ v(µ bb ) (14.64) (where the factor of one-half is necessary so as not to count every collision twice), z bc,the number of collisions of one b molecule with c molecules per unit time is z bc = c V σ bc v(µ bc ) (14.61) and Z bc,the total number of collisions between b and c molecules per unit volume per unit time is Z bc = b V c V σ bc v(µ bc ) (14.63) where σ is the collisional cross sectional area and µ (which we will encounter again in quantum) is the reduced mass For particles 1/(reduced mass) is 1 1 = µ m 1 m 2 m σ bc = π (r b + r c ) 2 = π [(d b + d c )/2] 2 = π d 2 bc 1 µ bc = 1 m b + 1 m c Collisions Between Molecules The most interesting events in the life of a molecule occur when it makes a collision with another molecule. Chemical reactions between molecules depend upon such collisions. Important transport properties, through which kinetic energy (by thermal conductivity), mass (by diffusion), momentum (by viscosity), and charge (by ionic conductivity) are transferred from one point to another involve collisions between molecules (EDR Chapter 24). A brief simplified account of molecular collisions will be given to rationalize the above formula for Z bc, neglecting the distribution of molecular velocities. Consider two kinds of molecules, b and c, which interact as rigid spheres with radii r b and r c and diameters d b and d c. The contact between two such molecules is shown below. A collision occurs whenever the distance between centers becomes as small as d bc = (d b + d c )/2. Imagine a sphere of radius d bc circumscribed about the center of b. Whenever the center of a c molecule comes within this sphere, c can be said to make a rigid-sphere collision with b. r b r c b c b Hy bc I c c c db d c hard sphere model Hy bc I dt Suppose that all the c molecules are at rest and the b molecule is moving through the volume of stationary c molecules with an average speed v b. Wecan imagine that the moving b molecule in unit time sweeps out a cylindrical volume π d 2 bc v b. If c /V is the number of c molecules per unit volume, there will be π d 2 bc v b c /V centers of c molecules encountered per unit time in the sweep by the b molecule. We can thus estimate the collision frequency for one b molecule as z bc (est ) = c V σ bc v b where σ bc is the cross sectional area, π d 2 bc. Ifthere are ( b /V )molecules of b per unit volume, the total collision frequency between b and c molecules would be

10 -1- Z bc (est ) = b V c V σ bc v b This estimate neglects the effect of the excluded volumes of the molecules (similar to the van der Waals correction to the equation of state of an ideal gas), but at low pressures this correction would be small. Amuch more important error in this estimate comes from the assumption that the c molecules are stationary while the b molecules sweep through the volume. Actually, itisthe speed of b relative to c that determines the frequency ofcollisions. This relative speed v bc is the magnitude of the vector difference between the velocities of b and c. Asshown below, the magnitude v bc depends on the angle between v b and v c (as given bythe law ofcosines from trigonometry). v bc = (v 2 b + v 2 c 2v b v c cos θ ) 1/2 Thus, the expressions for the collision frequencies should be corrected to employ the average relative speed, v bc z bc = c V σ bc v bc Z bc = b V c V σ bc v bc (14.63) In the rigorous derivation of the molecular collision frequency (collision frequency handout) one finds that v bc = v(µ bc ) = 8kT π µ bc where µ bc is the reduced mass. ote that if we were to consider a volume of a single kind of gas where m b = m c = m in the reduced mass formula, the average relative speed becomes 8kT v bb = v(µ bb ) = 2 π m = 2 v An important quantity in kinetic theory is the average distance a molecule travels between two successive intermolecular collisions. This is called the mean free path (we will encounter this in EDR Chapter 24.2). The average number of collisions experienced by one molecule in unit time is z bb. The distance the molecule has traveled in this unit time is v b. Therefore, recognizing that distance = velocity time and time is just the reciprocal of the collision f requency, the mean free path λ is λ = v b z bb = 1 2σ (/V ) (14.67)