Kinetic Theory of Gases

Kinetic Theory of Gases Chapter 3 P. J. Grandinetti Chem. 4300 Aug. 28, 2017 P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 1 / 45

History of ideal gas law 1662: Robert Boyle discovered with changing pressure at constant temperature that the product of the pressure and volume of a gas at equilibrium is constant, pv = constant at constant T. 1780 s Jacques Charles found that the ratio of volume to temperature was also invariant when temperature was changed with the pressure kept constant, V /T = constant at constant p 1811 Amedeo Avogadro found the ratio of volume to amount remained constant with changing amount at fixed pressure and temperature, V /n = constant at constant p and T. 1834 Emile Clapeyron combined the gas laws of Boyle, Charles, and Avogadro into the ideal gas equation of state, where R is the gas constant. pv = nrt P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 2 / 45

Bernoulli s derivation of Boyle s law, pv = constant. As early as 1738 Daniel Bernoulli proposed a microscopic kinetic explanation of Boyle s law, but only after Clapeyron s work did Bernoulli s kinetic theory gain widespread acceptance. Bernoulli s derivation z x y m Remembering that pressure is defined as force per unit area, we ask what is the force of one gas molecule hitting a wall? P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 3 / 45

Force on wall is momentum change when a molecule hits it. v y -v y The force along y is given by the ratio of the change in momentum to the time between collisions. F y = p t. Assuming collisions with the walls are elastic we write p y = p y,final p y,initial = ( mv y ) (mv y ) = 2mv y P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 4 / 45

Force of one molecule hitting one wall of the box. We take t = 2l/v y as the time to travel the length of the box, hit wall, and travel back. The average force of one molecule hitting one wall of the box is F y = p y t = 2mv y 2l/v y = mv 2 y l. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 5 / 45

Force of N molecules hitting all walls of the box. The sum over N molecules hitting one wall is N F yn = m /l. If we add together the magnitude of all forces on the walls (i.e., ignore signs) we get j=1 v 2 y j F total = 2 m l N ) (v 2 xj + v 2 yj + v 2 zj = 2 m l j=1 N j=1 v 2 j, where v 2 j = v 2 x j + v 2 y j + v 2 z j. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 6 / 45

Force of N molecules hitting all walls of the box. F total = 2 m l N j=1 v 2 j, We can define the mean square velocity as v 2 = 1 N N j=1 v 2 j, and get the total force on all walls of the box as F total = 2 m l Nv 2. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 7 / 45

Pressure from N molecules inside the box. The pressure is the force per unit area. The total area of the box walls is 6 times the area of one wall: 6A wall. Combining p = F total /(6A wall ) and F total = 2 m l Nv 2 we obtain p = 2 m l Nv 2 /(6A wall ) = Nmv 2 3A wall l = Nmv 2 3V where V is the volume of the box. Rearranging we obtain we obtain Boyle s law pv = Nmv 2 3 P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 8 / 45

Molecular translational energy and temperature. With further rearrangement we find pv = Nmv 2 3 = 2 3 N ( 1 2 mv 2 ) = 2 3 Nϵ k, where ϵ k = 1 2 mv 2 is the average kinetic energy per molecule. Set N = N A for 1 mole of gas molecules pv = 2 3 N Aϵ k = 2 3 E k, and define E k as the kinetic energy of a mole of gas. Connecting this expression to the ideal gas law we find pv = 2 3 E k = RT and discover E k = 3 2 RT. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 9 / 45

Temperature is a quantity derived from energy The kinetic energy of 1 mole of an ideal gas E k = 3 2 RT This equation reveal the absolute nature of temperature. You can t have negative temperatures because you can t have negative kinetic energy. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 10 / 45

Molecular translational energy and temperature. E k = 3 2 RT. If you raise the temperature of the gas then you increase the total energy of the gas. Dividing by N A we obtain the relationship on a per molecule basis ϵ k = 3 R T = 3 2 N A 2 k BT. The Boltzmann constant is defined as k B = R/N A. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 11 / 45

Average molecular speed Since 1 2 mv 2 = 3 2 k BT, then the root mean square speed, c rms = v 2, is related to temperature and molecular mass, m, or molar mass, M, according to c rms = 3kB T 3RT m = M. As temperature increases the average kinetic energy of the molecules increases and thus the average molecular speed increases. We also see that as the molecular mass increases for a fixed average kinetic energy the average molecular speed must decrease. P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 12 / 45

Average molecular speed Example Calculate the root mean square speed for a mole of vanillin molecules at room temperature. Solution Since Vanillin has the chemical formula C 8 H 8 O 3 with a molecular weight of 152.1 g/mol we obtain 3R(300 K) c rms = 221 m/s. 152.1 g/mol If vanillin has such a high speed why does it take so long for the scent to travel across a room? P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 13 / 45

Maxwell Distribution Laws In 1859 James Clerk Maxwell worked out the probability distribution of molecular velocities, f ( v), for gas molecules as perfectly elastic spheres. Maxwell assumed that the distribution of velocities in each direction were uncorrelated, that is, f ( v) can be written as the product of three independent distributions f ( v) = f (v x )f (v y )f (v z ). He also reasoned that the distribution of velocities is independent of direction, implying that f ( v) should only depend on the magnitude of the velocities, f (v x )f (v y )f (v z ) = ϕ(v 2 x + v 2 y + v 2 z ). This is an example of a functional equation: an equation in which the unknowns are functions. P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 14 / 45

How do we solve this functional equation? f (v x )f (v y )f (v z ) = ϕ(v 2 x + v 2 y + v 2 z ). Product of the functions on left gives the sum of their variables as the function argument on the right. A function, f (v i ), that satisfies this functional equation is f (v i ) = ae bv 2 i. Putting this function into our functional equation gives ϕ(v 2 x + v 2 y + v 2 z ) = a 3 e b(v 2 x +v 2 y +v 2 z ) P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 15 / 45

Normalizing Maxwell s distribution for molecular velocities As the f (v i ) are probability distributions we require them to be normalized, f (v i )dv i = 1. This leads to a = b/π and f (v i ) = b π e bv 2 i. Taken together Maxwell s probability distribution becomes f ( v) = ( ) b 3/2 e b(v x 2 +vy 2 +vz 2). π P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 16 / 45

Maxwell s distribution law for molecular velocities From Bernoulli s kinetic theory we learned that v 2 = 3k B T /m. This mean square speed should also be obtained from our probability distribution according to v 2 = (v 2 x + v 2 y + v 2 z )f (v x )f (v y )f (v z )dv x dv y dv z. Substituting our normalized solutions for f (v i ) we obtain ( ) b 3/2 v 2 = (vx 2 + vy 2 + vz 2 ) e b(v x 2 +vy 2 +vz 2) dv x dv y dv z. π Evaluating this integral and setting it equal to 3k B T /m yields b = m/(2k B T ). Thus we find f ( v) = ( ) 1 m 3/2 e 1 2 (v x 2 +vy 2 +vz 2 )/(k B T /m). (2π) 3 k B T This is Maxwell s distribution law for molecular velocities. P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 17 / 45

Maxwell s distribution law for molecular velocities Maxwell s distribution law is a three dimensional Gaussian distribution centered on v = 0. The distribution for one component of the velocity vector for N 2 gas at three different temperatures. 0.0025 0.0020 100 K N 2 gas / s/m 0.0015 0.0010 0.0005 0.0000 300 K 1000 K -1500-1000 -500 0 500 1000 1500 velocity/ m/s P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 18 / 45

Maxwell s distribution law for molecular speeds P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 19 / 45

Maxwell s distribution law for molecular speeds Speed is the magnitude of a velocity vector. To get the speed distribution we transform Maxwell s velocity distribution into spherical coordinates, c = vx 2 + vy 2 + vz 2, With this change of variables we find f ( v) = f (c, θ, ϕ) = 1 (2π) 3 cos θ = v z c, tan ϕ = v y v x. ( ) m 3/2 e 1 2 c2 /(k B T /m). k B T This is independent of θ and ϕ we we put it into the normalization π 2π 0 0 0 f (c, θ, ϕ)c 2 dc sin θ dθ dϕ = 1 and integrate away the θ and ϕ to obtain ( ) 2 m 3/2 f (c) = c 2 e 1 2 c2 /(k B T /m). π k B T This is Maxwell s distribution law for molecular speeds. P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 20 / 45

Maxwell s distribution law for molecular speeds f (c) = ( ) 2 m 3/2 c 2 e 1 2 c2 /(k B T /m). π k B T 0.0035 0.0030 0.0025 100 K N 2 gas / s/m 0.0020 0.0015 0.0010 300 K 0.0005 1000 K 0.0000 0 500 1000 1500 2000 speed/ m/s P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 21 / 45

Maxwell s distribution law for molecular speeds f (c) = ( ) 2 m 3/2 c 2 e 1 2 c2 /(k B T /m). π k B T 0.0025 0.0020 Ar 300 K / s/m 0.0015 0.0010 Ne 0.0005 He 0.0000 0 500 1000 1500 2000 2500 speed/ m/s P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 22 / 45

Kinetic Theory of Gases Simulation P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 23 / 45

Mean speed With Maxwell s distribution law for molecular speeds we can calculate the mean speed ( ) 2 m 3/2 c = cf (c)dc = 1 ( ) 2kB T 2, π k B T 2 m 0 and obtain c = 8kB T 8RT πm = πm. Note that the mean speed, c, is smaller than the root mean square speed, c rms. P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 24 / 45

Most probable speed To calculate the most probable speed we need to find the value of c where f (c) is a maximum. So we set df (c)/dc = 0 and solve for c to obtain 2kB T 2RT c mode = m = M. 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 speed / P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 25 / 45

Fraction of molecules with speeds between c 1 and c 2. The fraction of molecules with speeds between c 1 and c 2 is obtained by integrating Maxwell speed distributions between these two limits, δn N = c2 c 1 f (c)dc. 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 speed / P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 26 / 45

Miller and Kusch experiments In the 1955 Miller and Kusch published the first convincing measurements of the speed distribution for K and Tl atoms in the gas phase. For each fixed rotation speed only molecules with a small range of speeds can travel from the furnace to the detector. With the dimensions given in the instrument diagram the selected speed is v 0 = ωl/ϕ. P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 27 / 45

Miller and Kusch experiments 20 Run 31 Run 60 Run 57 20 Run 99 Run 97 15 15 Intensity 10 Intensity 10 5 5 0 0.2 0 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 speed / speed / P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 28 / 45

Distribution of kinetic energies Homework Derive the distribution of kinetic energies, f (ϵ k ) = 2 ( ) 1 3/2 ϵ 1/2 k e ϵ k/(k B T ), π k B T from Maxwell s speed distribution. P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 29 / 45

Collisions and Mean Free Path P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 30 / 45

Collisions and Mean Free Path It is through molecular collisions that molecules react. Let s use the kinetic theory of gases to get the collision rate of molecules. How close the molecules have to come to be considered a collision? Let s assume a spherical molecule of diameter d. d d 2d effective collision area = collision cross section = π d 2 The effective collision area is called the collision cross section, σ cs, given by σ cs = πd 2. P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 31 / 45

Collision Rate How many collisions occur in a given time t? A molecule moving at c will sweep out a collision volume given be Collision Volume = collision area length = (σ cs ) (ct) How many target molecules inside this volume? We multiply the number density, N/V, by the volume to find the number of molecules inside the collision volume. # collision in time t inside collision volume = (N/V )(σ cs )(ct) P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 32 / 45

Collision Rate How many collisions occur in a given time t? WAIT! We incorrectly assumes the target molecules are not moving, that is, v all others = 0. A proper treatment uses the relative speed of the colliding molecules, c rel not c. One can show (see notes) that c rel = 2 c. Thus we add a factor of 2 to correct our previous expression, # collisions in time t inside collision volume = (N/V ) (σ }{{} cs 2 ct). }{{} number density collision volume P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 33 / 45

Z 1 : Collision rate for single molecule inside the collision volume Finally, we can calculate Z 1, the collision rate for a single molecule inside the collision volume, by dividing by time, t, to get Z 1 = σ cs 2c ( N V ). Bu what is the total collision rate for all N molecules in a gas? P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 34 / 45

Z N : Total collisions per unit time per unit volume For all N molecules inside a total volume, V, the total collision rate is [ ( )] [ ] [ ] N 1 1 Z N = σ cs 2c V 2 N. V }{{}}{{}}{{} 1 2 3 1 is the single molecule collision rate. 2 is a scaling up by N molecules. Use 1/2 to avoid double counting. 3 is to get collisions occurring inside the total volume (not collision volume). Altogether we obtain Z N = 1 ( 2 σ N cs 2c V ) 2, as the collisions per unit time per unit volume. P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 35 / 45

Mean Free Path P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 36 / 45

Mean free path: λ Definition Mean free path, λ, is average distance a particle travels between collisions in the gas phase. We calculate λ according to λ = c t where c average speed, and t time between collisions. t is the inverse of the collision rate, that is, 1/Z 1. Thus we obtain λ = c Z 1 = 1 σ cs 2(N/V ). P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 37 / 45

Mean free path in terms of p and T We can eliminate (N/V ) from λ = 1 σ cs 2(N/V ). by rearranging the ideal gas law, pv = nrt, N V = pn A RT = p k B T, and expressing the mean free path in terms of pressure and temperature as λ = k BT 2σcs p. P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 38 / 45

Example Calculate the mean free path for a vanillin molecules at room temperature and pressure. Take σ cs = 0.5 nm 2 for vanillin. λ = k BT 2σcs p = k B (300 K) 2(0.5 nm 2 = 57.81 nm. )(1 atm) For vanillin we find a mean free path of λ 60 nm, which explains why it takes so long for the scent to travel across a room. P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 39 / 45

Effusion P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 40 / 45

Effusion Effusion The rate at which gas escapes through a pin hole into vacuum or low pressure region. Pin hole diameter must be smaller than mean free path. High pressure side Vacuum side P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 41 / 45

Graham s law of effusion Determined empirically by Scottish physical chemist Thomas Graham in 1848. Definition Graham s law of effusion: Rate of effusion varies inversely with square root of the molecular weight, rate1 rate2 = M2 M 1 where M 1 and M 2 are the molar masses of the two gases. P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 42 / 45

Effusion Rate How many molecules pass through pin hole in a given time interval, t? There is a range of velocities in the gas and those with fast (high) v x values can be further away and still pass through the pin hole while those with a slower (low) v x value will need to be closer to pass through the pin hole during the same t. Only molecules with A = Area of pin hole y x z v x > 0 can move towards the pin hole, and if it is within v x t from the pin hole during t, and if it is inside a volume v x ta during t. P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 43 / 45

Effusion Rate Average number of molecules passing through a pin hole of area A in t is ( ) [ N ] v x f (v x )dv x t A, V 0 where f (v x ) is Maxwell s velocity distribution for v x. The rate of effusion, Z A, is the number of molecules escaping per unit area per unit time (i.e, divide above Eq by t A), ( ) N Z A = v x f (v x )dv x. V Using the Maxwell-Boltzmann distribution we obtain ( ) ( ) N m 1/2 Z A = v x e mv x 2/(2kBT dv x V 2πk B T which becomes Z A = 0 0 ( ) ( ) N kb T 1/2 V 2πm P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 44 / 45

Effusion Rate Using pv = nrt = Nk B T gives the effusion rate Z A = p (2πmk B T ) 1/2. Taking the ratio at a constant temperature, Z 1 Z 2 = m2 m 1 we find agreement with Graham s law of effusion. P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 45 / 45