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The Boltzmann distribution for energy can be leveraged to find a distribution of the speeds of the molecules. This is the Maxwell-Boltzmann ...
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Lana Sheridan
De Anza College
May 8, 2020
For an adiabatic process (Q = 0):
PV γ^ = const.
and:
TV γ−^1 = const.
(Given the first one is true, the second follows immediately from the ideal gas equation, P = nRTV .)
On the eastern side of the Rocky Mountains there is a phenomenon called chinooks.
These eastward moving wind patterns cause distinctive cloud patterns (chinook arches) and sudden increases in temperature.
In a gas at temperature T , we know the average translational KE of the molecules.
However, not all of the molecules have the same energy, that’s just the average.
How is the total energy of the gas distributed amongst the molecules?
Ludwig Boltzmann first found the distribution of the number of particles at a given energy given a thermodynamic system at a fixed temperature.
Assuming that energy takes continuous values we can say that the number of molecules per unit volume with energies in the range E to E + dE is:
N[E ,E +dE] =
∫ (^) E +dE
E
nV (E ) dE
Where nV (E ) = n 0 e−E^ /kB^ T
and n 0 is a constant setting the scale: when E = 0, nV (E ) = n 0.
This particular frequency distribution:
nV (E ) ∝ e−E^ /kB^ T
is called the Boltzmann distribution or sometimes the Gibbs distribution (after Josiah Willard Gibbs, who studied the behavior of this distribution in-depth).
This distribution is even easier to understand for discrete energy levels.
The probability for a given particle to be found in a state with energy Ei drawn from a sample at temperature T :
p(Ei ) =
e−Ei^ /kB^ T
where Z is simply a normalization constant to allow the total probability to be 1. (The partition function.)
p(Ei ) =
e−Ei^ /kB^ T
If we know the energies of two states E 1 and E 2 , E 2 > E 1 , we can find the ratio of the number of particles in each:
nV (E 2 ) nV (E 1 )
= e−(E^2 −E^1 )/kB^ T
States with lower energies have more particles occupying them.
Suppose a type of atom has only 2 energy states, separated in energy by 12.0 eV.^1 For a very large sample of these atoms, at what temperature would 1% of the atoms in the sample be in the excited (higher energy) state?
∆E = E 2 − E 1 = 12 eV
Suppose a type of atom has only 2 energy states, separated in energy by 12.0 eV.^1 For a very large sample of these atoms, at what temperature would 1% of the atoms in the sample be in the excited (higher energy) state?
nV (E 2 ) nV (E 1 )
(^1) This does not describe any real atom.
Suppose a type of atom has only 2 energy states, separated in energy by 12.0 eV.^2 For a very large sample of these atoms,
∆E = E 2 − E 1 = 12 eV
At what temperature would the number of atoms in each state be equal?
(^2) This does not describe any real atom.
Suppose a type of atom has only 2 energy states, separated in energy by 12.0 eV.^2 For a very large sample of these atoms,
∆E = E 2 − E 1 = 12 eV
At what temperature would the number of atoms in each state be equal?
nV (E 2 ) nV (E 1 )
(^2) This does not describe any real atom.
Lasers emit coherent light. One photon interacts with an atom and causes another to be emitted with the same phase.
This starts a cascade.
Inside a laser cavity there are atoms that are in a very strange state: a higher energy level is more populated than a lower one. This is called a “population inversion”.
This is necessary for the photon cascade. Since:
nV (E 2 ) nV (E 1 )
= e−(E^2 −E^1 )/kB^ T^ , E 2 > E 1
we can associate a “negative temperature”, T , to these two energy states in the atoms.