Chapter 03 Kinetic Theory of Gases

Chapter 03 Kinetic Theory of Gases

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases History of ideal gas law 1662: Robert Boyle discovered with changing pressure at constant temperature that product of pressure and volume of a gas at equilibrium is constant, pV = constant at constant T 1780s: Jacques Charles found that ratio of volume to temperature was also invariant when temperature was changed with pressure kept constant, V∕T = constant at constant p 1811: Amedeo Avogadro found 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 gas laws of Boyle, Charles, and Avogadro into ideal gas equation of state, pV = nRT where R is gas constant. P. J. Grandinetti Chapter 03: Kinetic Theory of Gases 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

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Remember pressure is defined as force per unit area. What is the force of one gas molecule hitting a wall? P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Force on wall is momentum change when a molecule hits it.

Force along y is given by ratio of change in momentum to time between collisions. vy Δpy F = y Δt -vy Linear momentum is conserved in collision with wall

Δpy = py,final − py,initial = (−mvy) − (mvy) = −2mvy

퓁 Time to travel length of box, hit wall, and travel back is Δt = 2 ∕vy

2 Δpy −2mvy mvy Average force of 1 molecule hitting 1 wall of box is Fy = = 퓁 = − 퓁 Δt 2 ∕vy P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Force of N molecules hitting all walls of the box. Sum over N molecules hitting one wall is ∑N m F = − v2 yN 퓁 y훼 훼=1 Add magnitude (i.e., ignore signs) of all forces on all 6 walls (top, bottom, left, right, front, back) ∑N ∑N ∑N ∑N ( ) ∑N m m m m m F = 2 v2 + 2 v2 + 2 v2 = 2 v2 + v2 + v2 = 2 v2 total 퓁 x훼 퓁 y훼 퓁 z훼 퓁 x훼 y훼 z훼 퓁 훼 훼=1 훼=1 훼=1 훼=1 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 훼=1 2 v훼 Define mean square velocity as ∑N ∑N 1 v2 = v2 or Nv2 = v2 N 훼 훼 훼=1 훼=1 and write total force on all 6 walls of box as m 2 Ftotal = 2 퓁 Nv

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Pressure from N molecules inside the box.

Pressure is force per unit area. Total area of box walls is 6 times area of 1 wall: Atotal = 6Awall. m Substituting previous result: 2 p = Ftotal∕Atotal = Ftotal∕(6Awall) Ftotal = 2 퓁 Nv gives m Nmv2 Nmv2 2 where 퓁 is volume of box p = 2 퓁 Nv ∕(6Awall) = 퓁 = V = Awall 3Awall 3V

Rearranging gives Boyle’s law (pV = constant) ( ) Nmv2 2 1 2 pV = = N mv2 = N휖 , where 휖 is mean kinetic energy of molecule 3 3 2 3 k k

Rewriting in terms of moles, i.e., N = nNA where NA is Avogradro constant, 2 2 pV = nN 휖 = nE where E is kinetic energy of 1 mole of gas 3 A k 3 k k

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Temperature is a quantity derived from energy Finally connect to Ideal gas law: 2 pV = nRT = nE 3 k and we discover 3 E = RT kinetic energy of 1 mole of ideal gas k 2

Equation reveals true nature of temperature—reflects kinetic energy of atoms and molecules. Can’t have negative temperatures because can’t have negative kinetic energy. Raising gas temperature increases kinetic energy of gas molecules and vice versa.

Dividing by NA we obtain relationship on per molecule basis E 휖 k 3 R 3 k = = T = kBT NA 2 NA 2 . −23 kB = R∕NA = 1 38064852 × 10 J/K is defined as Boltzmann constant. P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Average molecular speed

Given 1 3 mv2 = k T 2 2 B √ 2 define root mean square speed, crms = v , and obtain √ √ 3k T 3RT c = B = rms m M

crms is related to temperature and molecular mass, m, or molar mass, M

Molecular speeds increase with increasing temperature. Molecular speeds decrease with increasing molecular mass.

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Average molecular speed

Example Calculate the rms speed for a mole of vanillin molecules at room temperature.

Solution

Since Vanillin has chemical formula C8H8O3 with a molecular weight of 152.1 g/mol we obtain √ √ 3RT 3R(300 K) c = = ≈ 221 m/s ≈ 500 mph rms M 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 Chapter 03: Kinetic Theory of Gases Maxwell Distribution Laws

What is parent probability distribution function for molecular velocities, p(v⃗) molecular speeds, p(c) molecular energies, p(E)?

James Clerk Maxwell 1831-1879 P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell Distribution Laws

In 1859 James Clerk Maxwell worked out the probability distribution of molecular velocities, f (v⃗), for gas molecules as perfectly elastic spheres.

Maxwell assumed that distribution of velocities in each direction were uncorrelated, that is, f (v⃗) can be written as product of 3 independent distributions

⃗ f (v) = f (vx) f (vy) f (vz)

He also reasoned that distribution of velocities is independent of direction, implying that f (v⃗) should only depend on magnitude of velocities,

휙 2 2 2 f (vx) f (vy) f (vz) = (vx + vy + vz )

This is an example of a functional equation: an equation in which the unknowns are functions.

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases How do we solve this functional equation?

휙 2 2 2 f (vx) f (vy) f (vz) = (vx + vy + vz )

Product of functions on left must give sum of their variables as function argument on right.

A function, f (vi), that satisfies this functional equation is

−bv2 f (vi) = ae i

Putting this function into functional equation gives

b v2 v2 v2 휙 2 2 2 3 − ( x + y + z ) (vx + vy + vz ) = a e Need to determine a and b.

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Normalizing Maxwell’s distribution for molecular velocities

As the f (vi) are probability distributions we require them to be normalized, ∞ ∞ −bv2 ∫ f (vi)dvi = 1 so ∫ ae i dvi = 1 −∞ −∞

√ 휋 requires that a = b∕ and √ b −bv2 f (vi) = 휋 e i

Taken together Maxwell’s probability distribution then becomes

( )3∕2 b b v2 v2 v2 ⃗ − ( x + y + z ) f (v) = f (vx) f (vy) f (vz) = 휋 e

Still need to determine b.

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell’s distribution law for molecular velocities 2 From Bernoulli’s kinetic theory we learned that v = 3kBT∕m. This mean square speed should also be obtained from probability distribution

2 2 2 2 ⃗ ⃗ v = ∫ (vx + vy + vz ) f (v) dv = 3kBT∕m

Substituting our normalized solution for f (v⃗) we obtain

∞ ∞ ∞ ( )3∕2 b −b(v2+v2+v2) v2 = (v2 + v2 + v2) e x y z dv dv dv = 3k T∕m ∫ ∫ ∫ x y z 휋 x y z B −∞ −∞ −∞

Solving this 3D integral equation requires b = m∕(2kBT), and finally obtain ( ) 3∕2 1 m − 1 (v2+v2+v2)∕(k T∕m) f (v⃗) = √ e 2 x y z B (2휋)3 kBT

This is Maxwell’s distribution law for molecular√ velocities: A 3D Gaussian probability distribution with a standard deviation of kBT∕m. P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell’s distribution law for molecular velocities Maxwell’s distribution law is a 3D Gaussian distribution centered on v⃗ = 0.

Distribution for one component of velocity vector for N2 gas at 3 different temperatures. 0.0025 N2 gas 100 K 0.0020

0.0015 / s/m

0.0010 300 K

0.0005 1000 K 0.0000 -1500 -1000 -500 0 500 1000 1500 velocity/ m/s

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell’s distribution law for molecular speeds

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell’s distribution law for molecular speeds Speed is magnitude of velocity vector. To get speed distribution transform Maxwell’s velocity distribution into spherical coordinates, √ v v 2 2 2, 휃 z , 휙 y . c = vx + vy + vz cos = tan = c vx With this change of variables we find ( ) 3∕2 1 m − 1 c2∕(k T∕m) f (v⃗) = f (c, 휃, 휙) = √ e 2 B (2휋)3 kBT This is independent of 휃 and 휙 so if we put it into the normalization ∞ 휋 2휋 , 휃, 휙 2 휃 휃 휙 ∫ ∫ ∫ f (c )c dc sin d d = 1 0 0 0 we can integrate away 휃 and 휙 and obtain √ ( ) 3∕2 1 2 Maxwell’s distribution law 2 m 2 − c ∕(kBT∕m) f (c) = 휋 c e 2 for molecular speeds. kBT

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell’s distribution law for molecular speeds √ ( ) 3∕2 1 2 2 m 2 − c ∕(kBT∕m) f (c) = 휋 c e 2 kBT

0.0035 N gas 100 K 2 0.0030

0.0025

0.0020 / s/m

0.0015 300 K

0.0010

0.0005 1000 K

0.0000 0 500 1000 1500 2000 speed/ m/s P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Maxwell’s distribution law for molecular speeds √ ( ) 3∕2 1 2 2 m 2 − c ∕(kBT∕m) f (c) = 휋 c e 2 kBT

0.0025 300 K Ar 0.0020

0.0015 Ne / s/m

0.0010

0.0005 He

0.0000 0 500 1000 1500 2000 2500 speed/ m/s P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Mean speed With Maxwell’s distribution law for molecular speeds √ ( ) 3∕2 1 2 2 m 2 − c ∕(kBT∕m) f (c) = 휋 c e 2 kBT

we can calculate the mean speed √ ( ) ( ) ∞ 3∕2 2k T 2 2 m ⋅ 1 B c = ∫ cf (c)dc = 휋 0 kBT 2 m which simplifies to √ √ 8k T 8RT c = B = 휋m 휋M √ Note, mean speed, c, is smaller than root mean square speed, c = 3kBT . rms m

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Most probable speed

√ ( ) 3∕2 1 2 2 m 2 − c ∕(kBT∕m) f (c) = 휋 c e 2 0.6 kBT 0.5 To calculate most probable 0.4 speed need to find c value where f (c) is maximum. 0.3

Set df (c)∕dc = 0, solve for c to 0.2 obtain √ √ 0.1 2k T 2RT 0.0 c = B = mode m M 0 1 2 3 4 5 speed /

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Fraction of molecules with speeds between c1 and c2. Fraction of molecules with speeds between c1 and c2 is obtained by integrating Maxwell speed distributions between these two limits, 훿N c2 = f (c)dc N ∫ c1

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0.0 0 1 2 3 4 5 speed / P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Web Apps by Paul Falstad

Kinetic Theory of Gases Simulation

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases 1955 - Miller and Kusch experimentally verify Maxwell’s molecular speed distribution.

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Miller and Kusch experiments

In 1955 Miller and Kusch published first convincing measurements of speed distribution for K and Tl atoms in gas phase. For each fixed rotation speed only molecules with a small range of speeds can travel from the furnace to the detector. With dimensions given in instrument 휔 휙 diagram the selected speed is v0 = l∕ . Measuring intensity as a function of rotation speed gives the atomic speed distribution.

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Miller and Kusch experiments

Potassium vapor Thallium vapor 20 20 Run 31 Run 99 Run 60 Run 97 Run 57

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10 10 Intensity Intensity

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0 0 0.2 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 Chapter 03: Kinetic Theory of Gases Miller and Kusch experiments

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases Distribution of kinetic energies

Homework

0.35 100 K Given 0.30 √ ( ) 3∕2 0.25 2 m 2 − 1 c2∕(k T∕m) f (c) = c e 2 B 휋 k T J 0.20

B -21

10 0.15 derive distribution of kinetic energies, ( ) 0.10 300 K 3∕2 2 1 1∕2 −휖 ∕(k T) 0.05 f (휖 ) = √ 휖 e k B 1000 K k 휋 k T k B 0.00 0 5 10 15 20 25 30 Energy/ 10-21 J

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases