### Boltzmann statistics

#### Problem:

Neutrons are released into an evacuated cubical chamber and lowered to a very cold T = 1 mK temperature.  What is their mean height above the floor of the chamber, once their gravitational energy has reached an equilibrium distribution?  (Show work!  Derive a formula and quote a number.)

Solution:

• Concepts:
Boltzmann distribution
• Reasoning:
We have a macroscopic-size chamber.
• Details of the calculation:
The probability per unit length of finding the neutron at height z is
P(z) = P(0)exp(-mgz/(kBT)).
1 = P(0)∫0exp(-mgz/(kBT))dz = P(0)kBT/(gm),   P(0) = gm/(kBT).
<z> = gm/(kBT) ∫0z exp(-mgz/(kBT)) dz = kBT/(gm).
kB = 1.380658 *10-23 J/K.
mn = 1.6749286 * 10-27 kg.
<z> = 0.841 m.

#### Problem:

Consider ideal gas particles of mass m at temperature T.  What is the average speed <v> (not the rms <v2>½ ) of the particle in terms of m, T and the Boltzmann constant kB?

0x2n exp(-x2)dx = π½ (1*3*5* ... *(2n -1))/2n+1

Solution:

• Concepts:
Boltzmann statistics
• Reasoning:
The Boltzmann distribution gives the probability of finding particles with energy E at a given temperature T.
• Details of the calculation:
Boltzmann statistics:  Ni/N = giexp(-Ei/kT)/∑i(giexp(-Ei/kT))
Here gi is the degeneracy.
For this problem:
n(v)dv/N = P(v)dv = g(v)exp(-mv2/(2kT))dv/∫0g(v')exp(-mv'2/(2kT))dv'
= v2exp(-mv2/(2kT))dv/∫0v'2exp(-mv'2/(2kT))dv',
since g(v) is proportional to v2.
1/∫0v'2exp(-mv'2/(2kT))dv' = (m/(2kT))3/2(4/π½).
So <v> = ∫0 vP(v)dv = (m/(2kT))3/2(4/π½) 0v3exp(-mv2/(2kT))dv
= (m/(2kT))3/2(2/π½) 0v2exp(-mv2/(2kT))dv2 = (8kT/(mπ))½.

The Boltzmann distribution describes the distribution of energy among classical (distinguishable) particles.
P(E) = A exp(-E/(kT))
It can be used to evaluate the average energy per particle in the circumstance where there is no energy-dependent density of states to skew the distribution.
To represent the probability for a given energy, it must be normalized to a probability of 1.
0 Aexp(-E/(kT))dE = 1 ==> A = 1/(kT).
This normalized distribution function can then be used to evaluate the average energy.
<E> = (kT)-10 Eexp(-E/(kT))dE = kT.
This is the average energy if the energy is randomly distributed among the available energy states and there is no energy-dependent density of states.
Note that this average energy for randomly distributed energy is not the same as the average kinetic energy.
The average kinetic energy is found by evaluating
<v2> = ∫0 v2P(v)dv = (m/(2kT))3/2(4/π½) 0v4exp(-mv2/(2kT))dv = 3kT/m.
½m<v2> = (3/2)kT.
We have an energy dependent density of states.

We can use E instead of v as the variable we integrate over.
g(v)dv = g(E)dE = g(E)mvdv, g(E) = g(v)/(mv).
Since g(v) is proportional to v2, g(E) is proportional to E½.
We can then rewrite all the integrals in terms of E instead of v.

#### Problem:

Suppose a small meteorite makes a hole of area A = 1 mm2 in the wall of a spaceship.  The habitable volume of the spaceship is V = 10 m3.  The temperature of the air in the spaceship is T = 27o C and the pressure P = 105 kPa.  The molar mass of air is M = 29 g /mole.  Estimate, how much time will be available to astronauts to put on spacesuit, if the pressure should not drop by more than to 50% of its initial value.

Maxwell-Boltzmann speed distribution:  f(v) = (m/(2πkT))3/2 4πv2exp(-mv2/(2kT)

Solution:

• Concepts:
The Maxwell Boltzmann distribution, ideal gas law
• Reasoning:
We need to find the escape rate Resc = -dN/dt, then we can integrate to find N(t).  The rate with which the molecules escape from the hole
depends on their average speed.
• Details of the calculation:

Consider an area dA and gas molecules with speeds between v and v + dv in a ring at a distance between r and r + dr from dA.  Assume that the gas molecules travel in straight lines.
The probability that gas molecules from the ring will reach the area dA is
dA cosθ/(4πr2).  This is the solid angle subtended by dA at any dV in the ring.
The number of molecules from the ring reaching dA with speeds between v and v + dv is
nf(v)dv2πr2sinθ dr dθ dA cosθ/(4πr2) = nf(v)dv sinθ dr dθ dA cosθ/2.
Here n = N/V is the density of the gas.
The total number of molecules with speeds between v and v + dv reaching dA in a time interval dt comes from rings with radii between r = 0 and dr = vdt.
It is therefore given by
0π/2 dθ nf(v)dv dA vdt sinθ cosθ/2 = ¼ n f(v) dv dA vdt.
Integrating over all speeds we find Resc = ¼ n dA <v>,
where Resc denotes rate at which gas molecules reach dA and <v> = ∫0 vf(v) dv.
For the Maxwell Boltzmann speed distribution we find
<v> = (8kT/(mπ))½, where m is the mass of a gas molecule.
Resc = (P/kT) dA (kT/(2mπ))½.

Resc  = -dN/dt is the rate at which molecules are escaping from the chamber.
dN/dt = -(N/V) dA (kT/(2mπ))½.
N(t) = N0exp(-Ct), where C = (dA/V) (kT/(2mπ))½ = (dA/V) (RT/(2Mπ))½
If N/N0 = ½, then  t = ln(2)/C.
C = (10-6 m2/10 m3)*[(8.314 J/K mole * 300 K)/(2* 29*10-3 kg/mole * π)]½ = 1.71*10-3/s.
t = 592 s ~ 10 minutes.

Quantized energies