PHYS 172: Modern Mechanics

Fall 2010

Lecture 25 - Kinetic theory of gases; pressure

Chapter 12.8, bits of Chapter 13

CLICKER QUESTION #1 Reading Question (Sections 12.7–12.8)

There is only a trace amount of helium in the air we breathe. Why? A. There never was much helium on Earth to begin with. B. At ordinary temperatures, most helium atoms have speeds greater than the escape velocity. C. Most of the helium can be found at very high altitudes, not at the surface. D. The low mass helium atoms are more likely to gradually escape from the Earth. E. Since helium does not bond to other elements, it is less likely to be trapped on Earth.

CLICKER QUESTION #1 Reading Question (Sections 12.7–12.8)

There is only a trace amount of helium in the air we breathe. Why? B. At ordinary temperatures, most helium atoms have speeds greater than the escape velocity. NO!! But A FEW do. D. The low mass helium atoms are more likely to gradually escape from the Earth. The rare tail of the Maxwell Boltzmann distribution.

Summary of Statistical Mechanics (Chapter 12 Entropy) Ω=

(q + N − 1)! q!(N − 1)1

S = kB ln Ω

∂S 1 ≡ T ∂Eint ΔEatom 1 ΔEsystem C = = ΔT N ΔT

P(ΔE) ∝ e

−

ΔE kT

Counting states

Entropy

Temperature

Heat capacity

Boltzmann distribution

Rhetorical Question How do we know that the definition of temperature that we got from statistics & entropy agrees with standard definitions?

∂S 1 ≡ T ∂Eint The most common definition of temperature comes from the ideal gas law: PV=(#ofMoles)RT, or P = nkT. (n = N/V = number density) Where R is the ideal gas constant 8.3 J/K/mole and k is Boltzmann’s constant We can start with our statistical mechanics ideas, extend them to the kinetic theory of gases, and show that we get the result: P=nkT. This will prove that our definition, obtained from fundamental physics, is in agreement with the gas law result.

The Kinetic Theory of Gases Assume there are no interactions between molecules, except when they collide. Assume these molecules have the Maxwell-Boltzmann velocity distribution. Apply geometrical scattering arguments & the momentum principle, etc.

Elastic scattering of molecules off wall of box

Boltzmann Distribution Boltzmann Distribution 1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

6

Quanta on Atom A

From an earlier example. The ground state is always the most likely state, for any finite temperature.

7

Application: The Boltzmann distribution in a gas Ideal gas: molecules do not interact with each other (low density) Energy of a molecule of mass M:

Ktrans + Evib + Erot + Mgycm Notes: Omit rest energy, nuclear and electronic energies Omit cm for brevity Measure Evib and Erot relative to ground state

Probability to find molecule at energy E:

P(E) ∝ e

−

Ktrans + Evib + Erot + Mgycm kT

Translational and gravitational energies are quantized, but if kT>>quantum, (~classical limit) probability to find a molecule at certain location with certain velocity is:

P(E) ∝ e

−

Ktrans + Evib + Erot + Mgycm kT

dxdydzdvx dv y dvz

Application: The Boltzmann distribution in a gas

P(E) ∝ e

−

Ktrans + Evib + Erot + Mgycm kT

dxdydzdvx dv y dvz

E E trans cm − vib ⎞⎛ − rot ⎞ ⎛ − KkT ⎞⎛ − Mgy ⎞⎛ P(E) ∝ ⎜e dvx dv y dvz ⎟⎜ e kT dxdydz ⎟⎜ e kT ⎟⎜ e kT ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

Velocity distribution

Height distribution

Vib. and Rot. Energy distribution

Ktrans = ½ Mv2, also = sum of x, y, and z terms; P is product of terms For the total translational form, go to spherical polar coordinates for the differentials: dvxdvydvz = 4πv2dv where the factor 4π is the solid angle and we have assumed isotropy

Distribution of speeds in a gas See 12.8 for details

He at RT

Maxwell-Boltzmann speed distribution (low-density gas) 3 2

⎛ M ⎞ 2 − P ( v ) = 4π ⎜ ⎟ ve ⎝ 2π kT ⎠ He

1 Mv 2 2

kT

e

−

1 Mv 2 x 2

kT

The Maxwell-Boltzmann Speed Distribution: P(v) ∝ v e 2

−

This is determined by looking only at a molecule’s kinetic energy, summed over the three orthogonal components of its velocity, and then averaging over all directions (see page 499-500). Use this “distribution function” to calculate averages:

K trans

3 = kT 2

vrms

3kT = m

2 1 mv 2

kT

The Kinetic Theory of Gases Using these results, now apply geometrical scattering arguments & the momentum principle, etc. to rediscover the relationship PV = NkT where N is the number of gas molecules and k is Boltzmann’s constant.

Elastic scattering of molecules off wall of box

Use Kinetic Theory to determine PRESSURE: A. Determine the number of molecules striking a surface per second. B. Calculate the momentum transfer per collision. C. Determine Pressure = force per unit area

1 p P= n 3 m

2

from Kinetic Theory (p. 520). Also, note that K=p2/2m, so:

From Boltzmann, statistical mechanics, & entropy we have: Ergo,

P = nkT

K trans

2 P = nK 3 3 = kT 2

Thermodynamics agrees with statistics!

NOTE: n = N / V = number of MOLECULES/m3

it is NOT the number of moles.

Application: The Boltzmann distribution in a gas

P(E) ∝ e

−

Ktrans + Evib + Erot + Mgycm kT

dxdydzdvx dv y dvz

E E trans cm − vib ⎞⎛ − rot ⎞ ⎛ − KkT ⎞⎛ − Mgy ⎞⎛ P(E) ∝ ⎜e dvx dv y dvz ⎟⎜ e kT dxdydz ⎟⎜ e kT ⎟⎜ e kT ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

Velocity distribution

Height distribution

Vib. and Rot. Energy distribution

We have dealt with the momentum-space part of the distribution. Now let’s deal with the position (coordinate-space) part. Generally the only important coordinate is height (due to gravitational potential energy varying with the height, y )

Height distribution in a gas E E trans − vib ⎞⎛ − rot ⎞ ⎛ − KkT ⎞ ⎛ − Mgy ⎛ ⎞ P(E) ∝ ⎜e dvx dv y dvz ⎟ ⎜ e kT dxdydz ⎟ ⎜ e kT ⎟⎜ e kT ⎟ ⎠⎝ ⎝ ⎠⎝ ⎠⎝ ⎠

P ( y) ∝ e

−

Mgy kT

dxdydz Volume

∝density

Important factor: kT At height 0: At height

kT y= : Mg

P ( 0 ) ∝ 1⋅ dxdydz P ( y ) ∝ e−1 ⋅ dxdydz = 0.38P(0)

Example exercises: 12.X.15 & 12.X.16

kT Mg

Height distribution in a gas P ( y) ∝ e

−

Mgy kT

dxdydz

Amusingly, a simple classical treatment gives essentially the same result – an exponential drop-off in pressure (and density) with height – without any reference to the Boltzmann distribution!! NOTE: either derivation assumes Temperature is uniform with height. Also either derivation neglects the variation of g with altitude (which isn’t too inaccurate since the scale height of the atmosphere is a lot less than the radius of the Earth.)

kT Mg

dP = Mg/A = ρA(dy)g/A = ρg(dy) A But ρ is proportional to P, call it cP. Then dP = Pcgdy. exponential in y

This is the D.E. for an

h

How does pressure change with height? dP ΔP = Δy dy

P = nkT

The density n is also governed by the Boltzmann distribution:

n( y) = n0 e

−

mgy kT

⎛ mg ⎞ dn( y) dP = kT = n0 kT ⎜ − dy dy ⎝ kT ⎟⎠ Note: the exponential term is very close to 1, since here we are interested in small departures, h, from y = 0.

ΔP = −n0 mgh

What is the weight of a box of gas?

Weighing in vacuum allows us to ignore buoyancy forces.

Fnet , y = Ptop A − Pbottom A = AΔP

What is the weight of a box of gas?

Fnet , y

N = AΔP = −n0 mg(hA) = − mgV = −Mg!!! V

For any container of height h, there is always a difference in pressure between the top and the bottom such that the force on the top is less than the force on the bottom by an amount exactly equal to the weight of the enclosed gas molecules.

Clicker Question Ice

water

steam

Three identical rigid containers rest on a table. One contains a block of ice, one contains water, and one has just steam (water vapor). All three contain the same number of H2O molecules. Which is the most correct statement? A. Ice system weighs most - all the force is directed down. B. Water system weighs less than ice - some of the force is directed to the side walls. C. Steam system weighs least - most molecules don’t touch the bottom. D. They have equal weights.

Clicker Question A large, standard gas cylinder (tank) has a volume of about 100 liters and is filled with nitrogen gas to a pressure of 2500 psi. Somebody opens the valve and all the nitrogen escapes. Choose the correct statement: A. The tank weighed less after the nitrogen escaped. B. The weight of the tank did not change, because the nitrogen gas molecules don’t rest on the bottom of the tank. C. The weight of the tank did not change, because any nitrogen gas molecules striking the bottom of the tank are balanced by nitrogen gas molecules striking the top of the tank.