Thermodynamics: Lecture 8, Kinetic Theory

Transcription

1 Thermodynamics: Lecture 8, Kinetic Theory Chris Glosser April 15, 1 1 OUTLINE I. Assumptions of Kinetic Theory (A) Molecular Flux (B) Pressure and the Ideal Gas Law II. The Maxwell-Boltzmann Distributuion (A) Equipartion of Energy (B) Specific Heat Capacity (C) Speed Distribution III. Mean Free Path and Effusion Assumptions of Kinetic Theory The fundamental assumptions of Kinetic Theory are as follows; 1. A reasonably sized system contains a vary large number of molecules;. The seperation of the molecules is large in comparison with the characteristic distance of the intermplecular forces; 3. The intermolecular forces do not significantly determine the dynamics of the molecules; 4. No energy is lost during collisions; That is the collision s are elastic; 5. The molecules are uniformly distributed throughout the volume; 1

2 6. The directions of motion of the molecules are evenly distributed: That is, the center of mass of the gas remains motionless. To substantiate item 3, bit, we consider the Lennard-Jones Potential, [ (d ) 1 ( ) ] d 6 V (r) = 4 ( E). (1) r r One may derive a potential of this form in quantum mechanics. Typically, the parameter E is of the order of 1 electron Volts, and the parameter d is of the order of angstroms. Therefore, if the average kinetic energy of the molecules is much larger than the characteristic energy, and the avergage spacing between molecules (that is the cube root of V/n) is much larger than the characteristic distaance, d, then we may apply kinetic theory. We assume the existance of a distribution of speeds, f(v), which tells us the fraction of molecules with speeds between v and v + dv. The fraction of molecules over a large ranges of speeds is the integral of this speed distribution, v v 1 dv f(v). () f(v) therefore has dimensions of inverse velocity, and satisfies the following normalization condition; dv f(v) = 1. (3) The mean speed v and the mean square speed v are thus given as; v = v = dv v f(v), (4) dv v f(v), (5) and, the RMS speed is defined through the usual relation, v rms =.1 Molecular Flux v. Let s define the molecular flux as the number of molecules striking a surfance per unit time. Consider an infintessimal area da. The number of molecules with a speed between v and v + dv that strike the infinessimal area is is just the number of molecules in a cylender of length v dt. That is; dn = ( ρ m f(v)dv ) (cos(θ)davdt) ( ) sin(θ) dθ dφ 4π (6)

3 Therefore, the differential flux is; dφ = dn da dt = ρ cos(θ) sin(θ) dθ dφ f(v)vdv. (7) m 4π In the preceding equations, m is the molecular mass, and ρ is the density of the gas. We can integrate this to obtain the average flux (remember, you are only integrating over half the unit sphere);. Pressure and the Ideal Gas Law Φ = 1 ρ 4 m v (8) We may extend the molecular flux idea to derive an equation for the differential momentum; and therefore, the pressure is; dp = mv cos(θ) dn = 1 3 ρv da dt, (9) P = dp da dt = 1 3 ρv. (1) Since ρ is related to the total number of particles and the volume, we may rewrite this expression as; P = 3V ( 1 Nmv ). (11) The term in parenthesis is just the total energy, U; therefore, This also serves as a definition of temperatre: P V = U. (1) 3 3 kt = 1 mv. (13) At room temperature, you can easily check that the average kinetic energies of the molecules are about an order of magnitude larger than the average potential energies. 3

4 3 The Maxwell-Boltzmann Distributuion Up until now, we have avoided any specifying any details about f(v). As you can imagine, it might be advantageous to have an explicit expression for the distribution. To do so, we will make a number of assumptions, which will be supported later when we get into statistical mechanics. 3.1 Equipartion of Energy The first thing that we will assume is that the energy is evenly distributed over each of teh degrees of freedom. This turns out to be a decent assumption for rotational and translational degrees of freedom, but not vibrational ones. Hence, we have: v = v x + v y + v z, (14) and therefore; v = 3v x. (15) We can therefore assign a set amount of energy per degree of freedom; ɛ = 1 kt. (16) Again this works reasonably well for rotational and translational degrees of freedom, but not vibrational ones. We therefore take the following expression as our model of an ideal gas with f translational and rotational degrees of freedom. U = f NkT (17) The specific heat capacity at constant volume is easily derived; c v = By Mayer s equation, we have c p ; ( ) u = f T v R (18) c p = f R + R = f + R (19) The ratio γ = c p /c v is; γ = f +. () f 4

5 3. Speed Distribution Now, let s try to derive an expression for the speed distribution f(v). We will assume that three particle collisions are extremely improbable. Two particle collisions, of course satisfy momentum conservation; v 1 + v = v 3 + v 4 (1) Now the rate at which collisions occur is directly proportional to the number of particles in a given volume. Thus, F (v 1 )F (v ) = F (v 3 )F (v 4 ). () By inspection, the following solves both of these conditions; F (v) = A exp( αv v). (3) This gives the velocity distribution. To get the speed distribution, integrate thsi over the solid angle, as we did for the flux; f(v)dv = (4πv ) F (v ) dv = (4πv ) A exp( αv ) dv. (4) The constants A, α can now be determined through the normalization condition and the the defintion of temperature to be; α = m/(kt ), (5) A = ( ) 3 m N. πkt (6) We thus have the Maxwell-Boltzmann speed distribution, f(x)dx = 4 ( Nx exp x ) dx (7) π where x = αv. We may now write down explicit expressions for the mean velocity (and the remaining modes) in terms of Gamma functions and the parameter α: v n = ( ) 3 + n π α n Γ. (8) 5

6 4 Mean Free Path and Effusion Now, Let s discuss some applications of Kinetic Theory. First, we write down an order of magnitude expression for the mean free path, that is, the average distance that a molecule travels before it collides with another. If we assume that the molecules can be modeled with a characteristic crosssectional area of exclusion σ, then the molecule excudes a cylinder of volume σvt. If a molecule has a center in this volume, then it collides whith moving one. Therefore, the mean free path is just l = mvt ρvtσ = m ρσ. (9) The collision frequency is the mean velocity divided by this expression; f = v ρσ m. (3) Effusion is the process by which gas under pressure escapes from a small hole in a container. Let s try and calculate the average velocity of such molecules. Since we are interested in the average velocity of molecules incident on an area, we use the expression for the differential flux to calculate the average velocity of the escaping gas: dφ = 1 xf(x)dx (31) 4α Therefore, the average velocity of the gas incident on the hole is; v e = 1 α xdφ Φ = 3 π 4α. (3) Note that this is slightly larger than the average velocity of the gas. 6