CHEM-3340 Physical Chemistry II Kinetic Theory of Gases PDF

CHEM-3340 Physical Chemistry II Kinetic Theory of Gases Kinetic Model Gases: simplest phase of matter Can be pictured as ◊ collection of particles of mass m in ceaseless, random motion ◊ The size of the molecules in negligible (diameters are much smaller than the average distance travelled between c...

CHEM-3340 Physical Chemistry II Kinetic Theory of Gases Kinetic Model Gases: simplest phase of matter Can be pictured as ◊ collection of particles of mass m in ceaseless, random motion ◊ The size of the molecules in negligible (diameters are much smaller than the average distance travelled between collisions) ◊ The molecules interact only through brief, infrequent, and elastic collisions Elastic collision: the total translational kinetic energy of the molecules is preserved Change of momentum in elastic collisions Pressure of gas is due to the change of momentum of a molecule as it collides with the wall of a container For x component: ◊ before collision: pi = mvx ◊ after collision: pf = -mvx Momentum change: pf - pi = 2mvx Note: the other (y,z) components are unchanged) Change of momentum: the number of colliding particles We have many molecules in the container, but within a timeframe of Δt seconds, not all of them will collide with the wall Only those will collide that are within a distance of |vx Δt| This distance corresponds to a volume of |vx Δt| A Concentration of the gas in number of atoms /L: c = n NA/ V n: number of particles in mol NA: Avogadro number V: volume of the container The number of particles (n2NA) in volume V2: n2 NA = c V2 = nNA/V V2 The number of particles that can collide with the wall within Δt seconds: nNA/V |vx Δt| A This quantity needs to be modified: half of the particles move from left to right, the other half from right to left -> dived by 2. Final equation: nNA vx Δt A / (2V) Change of momentum Change of momentum for one collision 2mvx Number of collision in Δt seconds nNA vx Δt A / (2V) Overall change of momentum: 2mvx nNA vx Δt A / (2V) mNA vx2Δt A n/ V mNA = molar weight: M Force on the wall: rate of change of momentum (change of momentum per time) M vx2Δt A n/ V F = M vx2A n/ V Pressure on the wall Pressure: P = F/A P = n M vx2/ V F = M vx2A n/ V Because the molecules ceaselessly batter in the wall, the pressure measures mean squared velocity () of the molecules P = n M / V The overall velocity, c, comes from three components (x,y,z) c2 = = + + Since there is no special direction, all these should equal c2 = + + = 3 = 1/3 c2 Perfect gas equation P = n M / V = 1/3 c2 P V= 1/3 n M c2 microscopic perfect gas equation c: root mean square speed of the molecules c= 0.5 Perfect gas equation from empirical measurements P V = nRT R=8.314 J/(molK): gas constant nRT = 1/3 n M c2 c= 3RT M Temperature and molecular speed c= 3RT M c, the root mean square speed is ◊ Proportional to the square root of temperature Higher temperature: larger c ◊ Inversely proportional to the square root of molar mass Heavier molecules travel slower than light molecules (H2 is very fast!) Example: N2, M=28 g/mol = 0.028 kg/mol R = 8.314 J/(molK), T=298 K (25 0C) c = 515 m/s (1150 mile/h) Distribution of speeds We got an expression for root mean square speed, however, it is also important what is the distribution of the speed of the molecules? ◊ How many molecules go faster than average, how many go slower? ◊ What is the mean speed of the molecules? ◊ What is fraction of molecules that have kinetic energies larger than a certain threshold? Distribution of speeds f(v): The fraction of molecules that have speeds in a range of [v, v+dv] is f(v)dv The fraction of molecules with speeds between [v1,v2] is v2 v1 f (v)dv Distribution of of speed Maxwell-Boltzmann distribution Number of molecules at energy level E at temperature T is proportional to exp(-E/kT) k: Boltzmann constant (R/NA) E = 1/2 mvx2 + 1/2 mvy2 + 1/2 mvz2 = 1/2 m v2 N(E) = N x K x exp(-mv2/(2kT)) We can rewrite this with molar mass M and R N(E) = N x K x exp(-Mvx2/(2RT)) exp(-Mvy2/(2RT)) exp(-Mvz2/ (2RT)) The speed distribution function is N(E)/N: f(v) = K x exp(-Mv2/(2RT)) K: normalization constant Normalization constant ⇥ 1= f (v)dvx dvy dvz = ⇥ =K ⇥ ⇥ K exp( M vx2 /2RT )dvx ⇥ exp( ⇥ 2 RT M M vy2 /2RT )dvy 2 RT M x ⇥ ⇥ K= ⇥ exp( M vz2 /2RT )dvz 2 RT M x exp( ax )dx = 2 ⇥ ⇥ M 2 RT /a ⇥3/2 = =1 Maxwell-Boltzmann Distribution Fraction of molecules with speeds within a range of [vx, vx+dvx], [vy, vy+dvy], [vz, vz+dvz] M 2 RT ⇥3/2 exp( M (vx2 + vy2 + vz2 )/(2RT ))dvx dvy dvz Not so useful quantity. Maximum probability: vx=vy=vz = 0 What we really want is the range [v,v+dv] independent of direction! Maxwell Distribution The fraction of molecules that have speeds between [v,v+dv]are important that are in a volume of 4πv2 dv f (v)dv = f (v) = 4 M 2 RT ⇥3/2 M 2 RT exp( M (v 2 )/(2RT )) ⇥ 4 v 2 dv ⇥3/2 v 2 exp( M v 2 /(2RT )) Maxwell distribution Maxwell distribution: properties f (v) = 4 M 2 RT ⇥3/2 v 2 exp( M v 2 /(2RT )) Exponential decay: probability of finding a molecule with very high speed is very small If M is large or T is small: fast decay Heavy molecules are unlikely to be found at large speeds Temperature effect: greater fraction of molecules is expected to be found at high speed at higher temperature Pre-exponential factor, v2: small fraction of molecules are expected to be found at speeds close to 0 The remaining part of pre-exponential term (M/[2πRT]) not so important Mean speed We can use Maxwell distribution to describe many important properties of speed Mean speed c̄ = ⇤ M 2 RT vf (v)dv = 4 0 1 x exp( ax ) = 2 2a 3 0 c̄ = 4 ⇥3/2 ⇤ M 2 RT 2 ⇥3/2 v 3 exp( M v 2 /(2RT )) 0 1/2 (2RT/M)2 a=M/(2RT) 1 (2RT /M )2 = 2 ⇤ 8RT M Most probable speed, c* Maximum of speed distribution df (v) =0 dv c = for v=c* 2RT M Comparison of speed Smallest: c*, most probable Largest: c, root mean square Always use which is most appropriate for a certain property (e.g., temperature, root mean square) For N2: c=515 m/s c=475 m/s c*=421 m/s Relative speed Mean speed with which two molecules approach each other Larger than mean speed, smaller than twice as mean speed (because most collisions have an angle) When two molecules are identical c̄rel = 2c̄ When two molecules are non identical Use mean speed with reduced mass! c̄rel = 8RT ⇥µ MA MB µ= MA + M B Collision Frequency How many times do the molecules collide with each other per second? The number of molecular encounters is the number of molecules a preselected molecule approaches within a distance of 2xr =d The area σ = πd2 is called collisional cross section In a timeframe Δt seconds the molecule travels a length of The collision tube has a volume of area x length = c̄rel t c̄rel t Collision Frequency,z Number of collisions: number of molecules in the collisional tube concentration=n/V n = concentration x V n= N c̄rel t N = number of molecules / volume Collision frequency: number of collisions per time z = n/Δt = N Perfect gas law: pV = nRT n/V = p/(RT) nNA/V = N = pNA/(RT) = p/(kT) z= c̄rel c̄rel p kT Collisional cross section Simplest to calculate σ =d2 π diameter of the molecule But molecules are not hard spheres -> they determined these cross sections in molecular beam experiments (study of colliding molecules) Collision frequency z = N c̄rel z= c̄rel p kT At constant volume ◊ with increasing temperature z increases because of increased relative speed ◊ with increase of pressure there are more collisions Example: N2, σ = 0.43 nm2= 0.43x10-18 m2, T=298K, p=100000Pa, crel = 21/2 475 m/s z = 7x109 1/s It takes about 1/z =1.43x10-10 s (140 ps) between two collisions Mean free path, λ Mean free path: average distance a molecule travels between collisions Average time between collision: 1/z Average speed: c̄ Mean free path= time x speed c̄ kT = = z 2⇥p Typical value (N2) = 70 nm (1000 molecular diameter) At constant volume: when T is increased, p=nRT/V, p is increased as well -> λ does not depend on T (less time between collisions, but the molecule is faster) At extremely low pressures (10-7 Pa), the molecules hit the walls before they typically collide with each other: ultra high vacuum, UHV (λ = 40 km) Molecular Picture of Gas N2 at 1atm, and 298 K ◊Collection of molecules traveling with a mean speed of about 500 m/s. ◊Each molecule makes a collision within about 1 ns ◊Between collisions it travels about 1000 molecular diameters