Boltzmann Statistics

Question 1

Relationship between probability of microstates and multiplicity of atom at states 1 and 2.

Answer 1

P(s2)/ P(s1) = ΩR(s2) / ΩR(s1)

If we rewrite in terms of entropy:

P(s2) / P(s1) = e^SR(s2)/k / e^SR(s1)/k = e[SR(s2)-SR(s1)]/k

Since change in entropy is small, we can use dSR = 1/T (dUR + PdVR - udNR) to get:

SR(s2) - SR(s1) = 1/T[ UR(s2) - UR(s1)] = -1/T[E(s2) - E(s1)]

So plugging this in we get:
P(s2)/ P(s1) = e^-[E(s2) - E(s1)]/kT = e^-E(s2)/kT / e^-E(s1)/kT

Which is Boltzmann Factor! e^-E(s)/kT

Question 2

How is the Boltzmann Factor converted to a probability?

Answer 2

s1 and s2 are independent of each other when taking form P(s2)/e = P(s1)/e so both must be equal to a constant.

By constant of proportionality 1/Z

P(s) = 1/Z e^-E(s)/kT

Rewrite as:
P(s) = 1/Z e^-BE(s)

Where B is 1/kT (Boltzmann factor) and Z is partition function.

Question 3

Partition Function

Answer 3

The total probability of finding the atom in some state or other must be 1.

1 = sum of P(s) = sum of 1/Z e^-E(s)/kT

So Z = sum of e^-E(s) / kT (sum of all Boltzmann factors)

Question 4

Average value of Energy of System

Answer 4

N = atoms, N(s) = atoms in particular state S

Average Value:

Ē = sum to s E(s) N(s) / N = sum to s E(s)P(s)

thus Ē = 1/Z sum to s E(s)e^-BE(s)

so the average energy is the sum of all energies weighted by their probabilities.

Question 5

What about if the average values are additive? (Average total energy of two objects is sum of individual average energies)

Answer 5

You can compute total energy from average energy. (Since particles are identical and independent)

U = NĒ