Chapter 16. Kinetic Theory of Gases Summary molecular interpretation of the pressure and pv = nrt the importance of molecular motions elocities and speeds of gas molecules distribution functions for molecular speeds and elocities application to effusion (gas leak through a pin hole) molecular collision rates important for chemical kinetics
Section 16.1 Kinetic Theory of Gas Pressure and Energy What is Kinetic Theory? gases are treated as a collection of moing particles kinetic theory applies if the aerage distance between molecules is ery large compared to the molecular size useful for gases at low pressures many applications to physical and chemical processes one of the most successful theories of fluids
Key assumptions of the Kinetic Theory of Gases the gas consists of particles of mass m in ceaseless random motion* the size of the particles is negligible compared to the aerage distance between collisions the particles do not interact, ecept for perfectly elastic collisions** *random motion not organized motion, as in bulk flow through a tube **translation kinetic energy unchanged by collisions no internal energy modes (such as ibrations or rotations) are ecited
Eample: Argon gas at 300 K and 1 bar (No internal rotations or ibrations. Why?) olume per mole: V m 5 L mol 1 = 0.05 m 3 mol 1 Diide V m by Aogradro s number to get olume per argon atom: 4.1 10 6 m 3 = 41 nm 3 Use the argon atom diameter ( 0.9 nm) to calculate olume occupied by one Ar atom: (4/3)r 3 = 0.013 nm 3 Conclusion Only about 0.03 % of the container olume is occupied by argon atoms. The system is mostly empty space.
Kinetic Theory of Gas Pressure Eperiment results for gases at low pressures pressure at fied n, V is proportional to T pressure at fied n, T is inersely proportional to V pressure at fied T, V is proportional to n combine to gie the ideal gas law: p nrt V But is there a theoretical interpretation of the ideal gas law?
A particle of mass m with -component elocity collides with a wall of area A. The collision changes the -component of the elocity from to. Momentum change per collision: m m( ) = m
Net, calculate the number of particles colliding with the wall in the time interal t. All particles within the distance t from the wall will hit it in the time interal t if they are moing toward the wall. One half of the particles the olume A t (the ones moing right) hit the wall in the time interal t.
For a container of olume V filled with n moles of particles, the number of particles in olume A t is nn A At V One half of these molecules hit the wall, so the momentum change in time interal t is (number of collisions) (momentum change per collision) 1 At nna m V nn A ma V t
The pressure on the wall is the force (rate of change of momentum) diided by the wall area. Diide by t and A to get p nn ma V t A 1 At nnam V nm V using M = N A m for the particle molar mass. This kinetic theory result is starting to resemble the ideal gas law! p nrt V
Not all of the molecules trael with the same elocity. As a result, the pressure is the aerage alue of nm /V. Using angular brackets < > to denote the aerage alue of gies p nm V Because molecules are moing randomly, < > = < > = < > and the aerage particle speed squared c is c y z 3 Important kinetic theory result: p 1 3 nmc V
Comparison of the Kinetic Theory and Ideal Gas Pressure Equations p 1 3 nmc V nrt V Mean (Aerage) Particle Speed Squared: c 3RT M Root Mean Squared (rms) Particle Speed: c 3RT M
Comparison of the Kinetic Theory and Ideal Gas Pressure Equations Mean (Aerage) Particle Translational Kinetic Energy (per mole): Equipartition Theorem for the Mean Translational Kinetic Energy for Each of the, y, and z Dimensions (per mole): RT M Mc z y 3 ) ( 1 1 RT M M M z y 1 1 1 1 Important interpretation of the temperature T a measure of aerage molecular kinetic energy
Comparison of the Kinetic Theory and Ideal Gas Equations Mean (Aerage) Translational Kinetic Energy (per mole): 1 Mc 3 RT Pressure: p nrt V 3 3 RT V 3 translational kinetic energy olume A new interpretation of the pressure of a gas! /3 of the translational kinetic energy per unit olume
Mean argon atom speed squared: c 1 1 3 3 (8.314 J K mol )(300 K) 187,300 m s 1 M RT Eample: Argon gas at 300 K 0.03995 kg mol Mean molar translational kinetic energy: 1 3 1 Mc RT 3741 J mol Root-mean-squared (rms) argon atom speed: c 187,300 m s 43.8 m s 1 1558 km/hr 968.1 miles/hr
from Chem 31: p = gas pressure C Vm = molar heat capacity at constant olume = gas density C pm = molar heat capacity at constant pressure adiabatic compressibility S isothermal compressibility T (at constant entropy) (at constant temperature) T Vm p S S C C p c 1 1 m sound The Speed of Sound c sound (another important speed) S S S p V V p 1 1 T T T p V V p 1 1
Eample: The Speed of Sound for Argon assuming ideal-gas behaior T = 1/p and: p = nrt/v = nm/v C Vm = 3R/ (for monatomic gases) C pm = 5R/ (for monatomic gases) Eercise: Show c sound 5 3 RT M for monatomic gases For argon at 300 K the speed of sound is c sound = 3.6 m s 1 Points to ponder: the speed of sound is independent of pressure proportional to T 1/ proportional to M 1/
Section 16. Velocity Distribution in One Dimension Detailed mechanical description of macroscopic systems. Need to specify: i, y i, z i coordinates for the positions of 10 3 molecules? i, yi, zi elocity components for 10 3 molecules (constantly changing due to molecular collisions)? Mission Impossible!
Impossible to Calculate the Positions and Velocities of Molecules in Macroscopic Systems. What to do? Gie up? No! Plan B: Huge numbers ( 10 3 ) of molecules are moing in random directions with different elocities use statistics. Calculate the elocity distribution function giing the probability of different molecular elocities, from to + Aerage elocities, aerage speeds, and other useful information can then be obtained.
Distribution Functions What are they? Why are they important? How can we use them?
An Eample of a Distribution Function histogram plot of final-eam marks for Chem 3 N i (number of students with mark i ) number N i of students with mark i plotted against mark i a clear, concise isual representation of the data 5 4 3 1 0 87 88 89 90 91 9 93 94 95 96 97 mark i / % a discrete distribution Why? N i and i are not continuous N i = 0, 1,, 3, 4 i = 89, 91, 9, 93, 94
What is the aerage mark? i i N i N i i (1)(89) (3)(91) (4)(9) (1)(93) 1 3 4 1 ()(94) 1101 11 91.90 %
another way to calculate the aerage mark: the area under the histogram plot is Ni 11 Diiding by Ni 11 gies P i = the probability of mark i the normalized histogram is the mark probability distribution P i ( probability of mark i ) 0.5 0.4 P i Ni N i Ni 11 Normalized Plot 0.3 0. 0.1 0.0 87 88 89 90 91 9 93 94 95 96 97 mark i / % Area = 1 under P i plotted s. N i Why? The area represents the sum of all possible probabilities.
Calculating aerage alues for a discrete probability distribution P i Ni N i i P i i P 11 P P3 3 P4 4 P5 5 P6 6 P7 7 P8 8 P9 9 P1010 P1111 1 89 11 0 90 11 3 91 11 4 9 11 1 93 11 94 11 91.90 aerage mark
Calculating aerage alues from a discrete probability distribution P i Ni N i 0.5 i P i i The probability P i of mark i = 0, 1,,, 97, 98, 99, 100 is the area of a rectangle of height P i and width = 1. P i ( probability of mark i ) 0.4 0.3 0. 0.1 0.0 87 88 89 90 91 9 93 94 95 96 97 mark i / %
Aerage of discrete 0, 1,, alues: i P i i What about aerages of continuous alues, such as elocities, speeds, translational kinetic energies,? 1. Use probability distribution function P() to gie the probability is between and + d. Instead of adding discrete i P i alues, integrate P()d alues: P ( ) d
Eample: Molecular Speeds Distribution function: F( ) 4 m 3 / e m / gies the probability F()d of molecules with speeds in the range from to + d. Aerage speed: 0 F( )d 8 m
Distribution Functions applications in eery branch of science and technology Important Eample: Electron Probability Distributions * Soling Schrodinger s equation Hˆ E gies quantum mechanical wae functions (, y, z) describing 1s, s, p, 3s, 3p, 3d, electron orbitals. The probability of finding an electron at coordinates to + d, y to y + dy, z to z + dz in olume element ddydz is: (, y, z)ddydz or * (, y, z)(, y, z)ddydz [ if (, y, z) is real ] [ if (, y, z) is comple ]
Probability distribution functions * (, y, z)(, y, z) for electrons in 1s, s, p, 3s, 3p, 3d orbitals (higher probability density in the brighter regions)
Another important eample of a distribution function: Normal (Gaussian) Distribution of Random Errors f ( ) 1 e / = the ariance (aerage alue of ) narrow distribution: small (high precision) wide distribution: large (low precision)
Normal (Gaussian) Distribution of Random Errors f ( ) 1 e / f() is normalized: 1 f ( ) d Why? aerage alue of is zero: 0 f ( ) d Why? aerage alue of : (the ariance ) f ( ) d
Normal (Gaussian) Distribution of Random Errors 0.954 f ( )d error 95.5 % probability 0.68 f ( ) d 3 0.998 f ( )d 3 error 68. % probability error 3 99.8 % probability
Normal (Gaussian) Distribution of Random Errors 0 0.500 f ( )d negatie error: 50 % probability 0 0.500 f ( )d positie error: 50 % probability
Another Eample: Boltzmann Distribution oer Molecular Energy Leels The probability that a molecule is in energy leel E i is proportional to the Boltzmann factor ep(e i /) f ( E i ) proportional to e E i / f ( E i ) Ae E i / A is a normalization factor
Barometric Equation: Boltzmann Distribution of Molecules in a Graity Field A molecule at eleation h aboe sea leel has graitational potential energy mgh The probability that a molecule is at eleation h is proportional to the Boltzmann factor ep(mgh /) find: f ( h) mg e mgh/ 1 h e h / h aerage eleation: h mg 0 hf ( h)dh
f (h) / km 1 0.14 0.1 0.10 0.08 0.06 0.04 0.0 0.00 Barometric Distribution of He, N and O O N He 0 0 40 60 80 100 h / km He N O T = 300 K g = 9.80 m s <h> = 63.6 km <h> = 9.09 km <h> = 7.95 km (height aboe sea leel) 0 1 f ( h)dh H and He are by far the most abundant elements in the Unierse. Why is there no H and no He in our atmosphere?
Back to Section 16.: Velocity Distribution f ( ) in One Dimension A molecule of mass m moing in the -direction with elocity has translational kinetic energy: E 1 m The Boltzmann distribution gies f ( ) Ae E / Ae m / with normalization constant A
Velocity Probability Distribution f ( ) in One Dimension The pre-eponential normalization factor A is ealuated using: 1 Find: f ( f )d ( ) A e Ae m E / d / A m use a table of definite integrals to ealuate this term A m and f ( ) m e m /
Velocity Distribution Function in One Dimension f ( ) m e m / What is f( )? f( )d is the probability molecular elocity is in the range to + d m = molecular mass k = R/N Aogadro = Boltzmann constant T = temperature
Velocity Distribution Function in One Dimension f ( ) m e m / Most probable alue of? Why? (hint: there is no bulk flow) Why is f( ) symmetrical? f( ) = f( ) f( ) 0 as Why? Same distribution for y and z?
Velocity Distribution Function in One Dimension f ( ) m e m / Compare f ( ) with the normal distribution of random errors: f ( ) 1 e / Notice (it s no accident!): f ( ) is a normal distribution with ariance = < > =/m
Velocity Distribution Functions for Neon, Argon and Krypton T = 98 K As the molecular mass increases, the elocity distributions get: narrower taller Does this make sense?
Velocity Distribution for Ar at 98 K and 1000 K Temperature Effects As the temperature decreases, elocity distributions get taller and narrower. Does this make sense?
Section 16.3 Distribution of Molecular Speeds A molecular elocity is the ector sum of the elocity components in the, y and z directions. z y y z magnitude and direction
Section 16.3 Distribution of Molecular Speeds A molecular speed is the magnitude (no direction) of the elocity ector. z y z y 1 1 1 00 m s 350 m s 100 m s z y Notice that different elocity ectors can gie the same speed (important later). Eample: and (Pythagoras Theorem) 1 1 1 100 m s 00 m s 350 m s z y
Probability of a elocity ector with, y and z components to + d y to y + d y z to z + d z z y z y f f f d d )d ( ) ( ) ( z y m m m m m m z y d d d e e e / / / z y m m z y d d d e )/ ( 3/ z y m m d d d e / 3/
d d y d z can be pictured as a tiny cube in three-dimensional elocity space with edges d and d y and d z. To calculate the probability a gien molecular speed, add up the olumes d d y d z of cubes in a spherical shell that gie the same speed in any direction A speed shell from to + d has area 4, thickness d and olume 4 d. The sum of d d y d z alues that gies speeds from to + d is 4 d.
Show the olume of a speed-shell of radius and differential thickness d is 4 d shell olume = olume of sphere(radius + d) olume of sphere(radius ) = (4/3) ( + d) 3 (4/3) () 3 = (4/3) [ ( + d) 3 () 3 ] negligible nd -order and 3 rd -order differential terms = (4/3) [ 3 + 3 d + 3(d) + (d) 3 3 ] = (4/3) [ 3 d ] = 4 d
Mawell Speed Distribution Function F() Important result: the probability of a molecular speed in the range from to + d is in terms of molecular mass m: F( )d 4 m 3/ e m / d in terms of molar mass M: F( )d 4 M RT 3/ e M / RT d
What do speed distribution functions look like? temperature effects F() for argon at 98 K and 1000 K
mass effects m Ne < m Ar < m Kr Speed distribution functions at 98 K for Ne, Ar and Kr
Mawell Speed Distribution Function F() F( )d 4 m 3/ e m / d The probability of a molecule with a speed in the range 1 to is the area under the F() cure plotted from 1 to F() 0 as 0 and as 0. This means ery few molecules hae ery low or ery high speeds The fraction of molecules with speeds larger than 10 times the most probable speed (calculated net) is 9 10 4. No molecules hae speeds this high!
Section 16.4 Most Probable, Aerage and Root-Mean-Square (rms) Molecular Speeds Most Probable Molecular Speed ( mp ) Gas molecules moe with a wide range of different speeds. What is the most probable molecular speed? Find mp that maimizes F(). Sole: F( ) d d m 4 d 3 / d m / e 0 4 m 3 / e m / m 3 0 Find ( try it! ): mp m RT M
What is the aerage molecular speed? Sole: Aerage Molecular Speed ( ae ) 0 / 3 / 0 ae d e 4 )d ( m F m M RT m 8 8 ae Find ( try it! ): 0 / 3 3 / d e 4 m m 3 / 1 4 m m use a table of definite integrals to ealuate this term
Root-Mean-Squared Speed ( rms ) Root-mean-squared speed sounds complicated, but it s just the square root of the aerage squared speed* (the ariance): rms 3 m 3RT M This might look familiar! rms is identical to the square root of the aerage squared speed calculated from equipartition theory: 1 1 m m( y z ) 3
Root-Mean-Squared Speed ( rms ) *Note that the root-mean-squared speed rms (first square, then aerage, then take the square root) rms 3 m 3RT M differs from the root-squared-mean speed (first aerage, then square, then take the square root) which is just the aerage speed rsm 8 m 8RT M
Most Probable, Aerage, and Root-Mean-Square Speeds Argon at 5 o C The most probable speed is smaller than the aerage speed. Also, the aerage speed is smaller than the rms speed. Why? Hint: Notice that F() is not symmetrical.
Theory is fine, but can speed distributions be measured? One approach: Use a molecular beam and elocity selector An oen with a pinhole and collimating slits emits a narrow beam of gas molecules. The molecules pass through a pair of slots offset by angle in rotating metal discs. Discs separated by distance L rotating at angular elocity proide L = t and = t which selects molecules with speed = L/. Molecules with other speeds hit the discs and do not reach the detector.
Figure 16.8 Measured () and predicted ( ) speed distributions for potassium apor at 466 K. Miller and Kusch. Phys. Re. 1955, 99, 1314. Why does this eperiment gie F() for the distribution of molecular speeds, not f( ) for elocities in the direction of the molecular beam from the oen?
Distribution of Translational Kinetic Energies E For some applications it is important to know the probability distribution of molecular energies. The probability F(E)dE of a molecule with translational kinetic energy in the range E to E + de is calculated from the probability F()d of molecular speeds The kinetic energy of a molecule is E = m /, which gies = (E/m) 1/ d = de/(me) 1/ for the molecular speed and its differential.
Distribution of Translational Kinetic Energies E Substitution of = (E/m) 1/ and d = de/(me) 1/ into m m / F( )d 4 e d for the probability of molecular speeds gies (try it!) 3/ F( E)dE E 3 / e E / de F(E)dE is independent of the molecular mass. Why? The aerage translational kinetic energy is E 0 EF( E)dE 3
Distribution of Translational Kinetic Energies E Probability of molecular translational energies: F( E)dE E 3 / e E / de Probability of molar translational energies (use E m = N A E and R = N A k): F E RT m Em / RT ( Em)dEm e de 3 / m F(E)dE and F(E m )de m are independent of molecular mass. Why?
0.0007 0.0006 Translational Kinetic Energy Distributions F(E m ) / (J mol 1 ) 1 0.0005 0.0004 0.0003 0.000 T = 100 K 0.0001 0.0000 0 000 4000 6000 8000 10000 1000 E m / (J mol 1 ) T = 300 K
Lots and lots of theoretical equations for molecular elocities and speeds. Another academic eercise. Another Zzzzzz No! Lots and lots of practical applications too! Predict: rates of molecular collisions (important for reaction rates reactant molecules need to collide to react) rates of molecular collision with surfaces (important for gas/solid reactions, catalysis, crystallization, etc.) leak rates isotope separation (e.g., 35 UF6 and 38 UF 6 ) transport properties, such as iscosity, diffusion and heat conduction (net Chapter)
Section 16.5 Molecular Collisions with Surfaces and Gas Effusion How many gas molecules hit the wall of a container per unit area per second? What does the collision rate depend on? If there is a hole in the wall of a container, how rapidly do the gas molecules leak out (effusion)? use rates of effusion to measure apor pressures?
Calculating wall collision rates illustrates a useful application of elocity distribution functions. Analysis: Consider molecules hitting a wall of area A perpendicular to the ais. In time interal dt, a molecule moes distance dt in the -direction A molecule hits the wall in the time interal dt if the molecule is moing toward the wall ( > 0) and within distance dt of the wall. The number of molecules in olume A dt with elocities from to + d is N V ( A dt) f ( )d
Collision rate per unit surface area per unit time is calculated from by: N V ( A dt) f ( )d diiding by wall area A diiding by time interal dt integrating oer all molecules moing toward the wall Z N p m m f e / ( )d V d c 0 0 Molecular collision rate with a wall per unit area per unit time. Z c p N m V m Z c is proportional to the number of molecules per unit olume, T 1/ and m 1/.
Eample 1 A container holds argon at 1 atm and 98 K. How many argon atoms hit the container wall per square centimeter per second? Collision rate per unit area per unit time: Z c p m (0.0400 kg mol / 6.010 10135 Pa mol ) (1.38110 1 3 1 3 1 J K ) (98 K).44 10 7 m s 1 Collision rate per unit time for area A = 1.00 cm = (0.0100 m) : AZ c (0.000100 m ) (.44 10 7 m s 1 ).44 10 3 s 1 Wow! About Aogadro s number per second per square centimeter!
Eample A container is filled with an equimolar miture of argon at 0.500 bar and helium at 0.500 bar at 98 K. Calculate the leak rate of argon and helium into a acuum through a circular pinhole of area 0.0100 m. Also calculate the composition of the leaked gas. Collision rate per unit area per unit time: p Zc m Z c for argon: (0.0400 kg mol / 6.010 50,000 Pa mol ) (1.38110 1 3 1 3 1 J K ) (98 K) 1.0 10 7 m s 1 Leak rate for argon through hole of area A = 0.0100 m : AZ c 1.0 10 (0.010010 13 1 m ) (1.0 argon atoms per second 10 7 m s 1 )
Eample cont. Collision rate per unit area per unit time for helium: Z c (0.00400 kg mol 1 / 6.010 50,000 Pa 3 mol 1 ) (1.38110 3 J K 1 ) (98 K) 3.79 10 7 m s 1 Leak rate for helium through hole of area A = 0.0100 m : AZ c 3.79 10 (0.010010 13 1 m ) (3.79 helium atoms per second Mole fraction of He in the leaked gas miture: Z c Zc (He) (He) Z (Ar) c 10 7 13 3.7910 13 3.7910 1.010 13 m s 1 ) 0.760 Application: isotope separation (compared to He mole fraction 0.500 in the container)
Eample 3 Inert the procedure: use measured leak rates to calculate the apor pressure of solids and liquids. An organic compound (molar mass M = 56 g mol 1 ) leaks out of a small hole (area A = 1.35 10 10 m ) in the wall of a Knudsen cell into a acuum chamber. A leak rate of 0.13 mg in 6 minutes is measured at 500 K. Calculate the apor pressure of the compound. Knudsen Vapor Pressure Cell The weight of the cell is measured before and after gas effusion for a timed interal.
Eample 3 (cont.) Measured leak rate in units of molecules per square meter per second: Z c N At 1 3 1 ( 0.00013 g / 54 g mol ) (6.010 mol ) 4 1 6.56 10 m s 10 1 (1.310 m ) (6 min 60 smin ) p Sole Zc for the apor pressure p. m 6.58 10 4 m s 1 1 0.56 kg mol 3 6.010 mol 1 p (1.38110 3 J K 1 ) (500 K) apor pressure p = 894 Pa = 0.00894 bar
Applications of Effusion apor pressure measurements molecular weight determination: leak rate M 1/ isotope separation ( e.g., 35 UF 6 / 38 UF 6 ) Graham s Law leak detection (Why is helium frequently used?)
Eample Gas molecules at initial pressure p(t = 0) leak out of a container (olume V) through a hole of area A into a acuum. Derie an epression for the gas pressure p(t) as a function of time. leak rate: (molecules per second) dn dt Z c A p m A substitute N = pv/: d( pv/ ) dt p m A rearrange: integrate from t = 0 to t: 1 p dp dt dlnp dt A m lnp( t) lnp(0) A m ( t 0) p( t) p(0)e At / m
Section 16.6 Molecular Collisions Kinetic theory treating gas molecules as point masses used in Sections 16.1 16.5 to analyze: pressure eerted on a wall translational kinetic energy molecular elocities molecular speeds rates of molecular collisions with a wall rates of molecular effusion through a hole Molecular collisions analyzed in this Section are different: depend on molecular sizes and cross-sectional areas Rates of molecular collisions are important for understanding rates of chemical reactions, mean free paths, iscosity, thermal conductiity, and miing by molecular diffusion
Collisions of Gas Molecules Analyzing molecular collisions is tricky! Depends on: molecular masses, sizes and shapes intermolecular potential energies relatie elocities direction of approach
Collisions of Gas Molecules Approimation used here: molecules are treated as hard spheres intermolecular potential energy is zero at distances between centers beyond r 1 + r potential energy is infinite at shorter distances molecules do not interact unless the distance between their centers is r 1 + r collisions are elastic (no change in total kinetic translational kinetic energy)
Collisions of Gas Molecules If molecules of type are stationary: molecules of type 1 collide in time dt with all molecules of type in the cylinder of olume (r 1 +r ) ae dt = ae dt collision cross sectional area = (r 1 + r )
But wait! All molecules are moing! Use aerage relatie speeds to calculate collision rates. Angles of approach range from 0 o to 180 o. 90 o is the aerage angle.
Mean Relatie Speed < 1 > of Molecules 1 and using the Pythagorean theorem: 1 aerage approach angle reduced mass: m1m m m 1 1 1 8 m 1 8 8 m
Eample Calculate the aerage relatie speed < 1 > of H (1) and O () molecules at 98 K. molecular masses m m 1 1 3 3.016 10 kg mol 3 1 6.0 10 mol 3.00 10 kg mol 3 1 6.0 10 mol 3.348 5.314 10 1 3 6 10 kg kg reduced mass m m m m 1 3.150 10 1 6 kg 3 1 8 8 (1.381 10 J K ) (98 K) 1 1 184 m s 6 (3.150 10 kg) Note: The mean relatie speed of H and O is closer to the mean speed of H (190 m s 1 ) than to the mean speed of O (48 m s 1 ). Why?
Collisions of Molecule 1 with Molecule In time dt molecule 1 sweeps out collision olume < 1 >dt. Number of collisions per second for a single molecule of type 1 with molecules of type : z N V 1 dt dt p 1 8 (units: s 1 ) Total number of collisions of molecules 1 and molecules per unit olume per second (called the collision density): N1 Z1 z1 V p1 p 8 (units: m 3 s 1 ) Collision cross sectional area = (r 1 + r ). Reduced mass = m 1 m /(m 1 + m ).
Collisions of Molecule 1 with Molecule 1 Number of collisions per second for a single molecule of type 1 with other molecules of type 1: z 11 N V 1 11 dt dt p1 8 m 1 Total number of collisions of molecules 1 and molecules 1 per unit olume per second (note: a factor of ½ is included so each collision is only counted once): Z 11 1 N V 1 z 11 1 p1 z 11 p1 8 m 1 Collision cross sectional area = (r 1 + r 1 ). Reduced mass = m 1 m 1 /(m 1 + m 1 ) = m 1 /.
Eample Calculate the collision frequency z 11 and the collision density Z 11 for O gas at 98 K and 1 bar. p1 8 z11 m p 1 1 8RT M (1.381 10135 Pa 10 J K )(98 K) (0.40 10 m ) 1 1 8 (8.314 J K mol ) (98 K) 1 (0.0300 kg mol ) 18 3 1 z 11 = 6.18 10 9 s 1 (collisions per second for one O molecule) Z 11 1 p1 9 1 z11 (6.1810 s 1 1 (1.381 10135 Pa 3 10 J K )(98 K) ) Z 11 = 7.61 10 34 m 3 s 1 Z 11 = 1.6 10 8 mol L 1 s 1 collisions per cubic meter per second moles of collisions per liter per second
Section 16.7 Mean Free Path mean free path the aerage distance a molecule traels between collisions with other molecules Useful for analyzing transport of heat, momentum, and mass. Calculated by diiding the aerage distance traelled per unit time by the number of collisions per unit time. For a gas consisting of one type of molecules: z N V 11 N V p is proportional to T, p 1 and 1
Eample Calculate the mean free path for argon at 1 atm and 98 K. Is large compared to the size of an Ar atom? Data: = 0.36 10 18 m Ar atom diameter = 0.9 10 9 m p (1.38110 3 (10135 Pa) J K 1 ) (0.3610 (98 K) 18 m ) mean free path = 79.8 10 9 m = 79.8 nm An Ar atom traels on aerage 79.8 nm/0.9 nm = 75 diameters between collisions.
! Warnings! The flow properties of a gas change significantly at pressures so low the mean free path is comparable to the dimensions of the container holding the gas ballistic flow. The hard-sphere billiard ball model used in this Chapter for molecular collisions is an approimation. More realistic (and more complicated!) models hae been deeloped, such as the Lennard-Jones intermolecular potential energy V ( r) r 1 r 6
Collision Trajectories for Lennard-Jones Interaction Potentials Low Speeds High Speeds