Physical Chemistry week 10 Monday March 11, 2013 page 1
dP
Mg
=- ( ) dz
P
RT
Consider the Earth being a sphere. The atmosphere starts at the surface of the Earth and tapers off with
increasing altitude z. The altitude of interest is z’. At the Earth’s surface, P=P0 and z=0. At the altitude
of interest, z=z’ and P=P’.
P'
integrate ∫
P0
z'
dP
Mg
=- ∫
dz
P
0 RT
The radius of the Earth is roughly 6300km. About 99.9% of the atmosphere is below 42km. Since
6300km>>42km, we can approximate g as being constant. Although T varies widely with altitude, we’re
going to assume T is constant anyway.
P'
∫
P0
dP -Mg z'
≈
∫ dz
P
RT 0
ln
≈instead of=due to assumptions
P' -Mg '
≈
z take the inverse log:
P0 RT
P' -Mgz'
=e RT
P0
-Mg
P≈P0 e RT z barometric formula, approximate due to assumptions
That’s a special form of the Boltzmann distribution law.
The Boltzmann distribution law
P=nRT
1
V
ideal gas law
Substitute for the n:
P=
N
1
RT
NA
V
-Mgz
RT
Substitute that into P≈P0 e
N(z)
1 N(0)
1 -Mgz
RT =
RT e RT
NA
V
NA
V
N(z) is the number of molecules at altitude z. The R, T, NA, and 1/V all cancel.
∆εP
N(z)=N(0)e kT
εP is potential energy
ΔεP=εP(z)-εP(0)=mgz
Earth’s surface is reference so εP(0)=0. Potential energy is always relative to an arbitrary reference.
Mgz NA mgz
=
RT
kNA T
Mgz mgz ∆εP
=
=
RT
kT
kT
N(z) -∆εP
≈e kT
N(0)
NA cancels
k is Boltzmann constant
special case of Boltzmann distribution law
Consider two energy levels, with state 2 above state 1. The number of molecules in state 1 is N1. The
number of molecules in state 2 is N2. State 1 has energy level ε1. State 2 has energy level ε2. The
difference in energy between the two states is ΔεP. If ε2>ε1 then ΔG>0 and N2<N1. The farther apart the
energy levels are, the less molecules there will be in the higher stand and more molecules in the lower
state. For a gas at 5000K:
-5*10-19 J
N2 -∆G 1.38*10-23 J⁄ *5000K
K
=e RT =e
=7.13*10-4
N1
N1
1400
=1.40*103 ≈
N2
1
That means even at 5000K, for every molecule in the higher energy electronic state there will be about
1400 molecules in the lower electronic state. This is the justification for ignoring electronic energy while
doing calculations.
Molecular collision and mean free path
Assume molecules are hard spheres, every though they’re more like sponges. Assume molecules are
spherical, even though many of them aren’t.
Consider two molecules: b and c, which b being the smaller one. The radius of b is rb and the radius of c
is rc. Consider an imaginary cylinder with diameter d=db+dc=2(rb+2c). The molecules and this cylinder
are inside a container with volume V. Also assume collisions are elastic, which means kinetic energy is
conserved.
Assumptions:
1. Molecules are hard spheres
2. Collisions are elastic
3. There are no intermolecular forces between molecules except at the time of collision
Lower case z(b)c = the collision rate for one particular b molecule with c molecules
Upper case Zbc = the total collision rate for b molecules colliding with c molecules per unit volume
Upper case Zbb = the total collision rate for b molecules colliding with other b molecules per unit volume
Lower case zbb = the collision rate for one particular b molecule with other b molecules
Nb is the number of b molecules.
Nc is the number of c molecules.
Calculation of Zbc
<vbc> is average speed of b molecules relative to c
Collision occurs when distance between b and c reaches rb+rc, which = (1/2)(db+dc).
d is diameter r is radius
find volume of cylinder
height(length) of cylinder = <vbc>dt
volume of cylinder = π(rb+rc)2*<vbc>dt
Notice π(rb+rc)2 is the area of the end of the cylinder and <vbc>dt is the length of the cylinder.
Vcyl
number of c molecules with their center in the cylinder= ( ) Nc
V
V is container volume
number of collisions which a specific b molecule makes with c molecules = lower case z(b)c
lower case z(b)c =
lower case z(b)c =
Vcyl Nc
V dt
Nc
dt
cancels
(π(rb +rc )2 ⟨vbc ⟩) equation I since
V
dt