Kinetic Theory
Problem:
A 5liter vessel contains 0.125 moles of an ideal gas at a pressure of 1.5
atm. What is the average translational kinetic energy of a single molecule?
Solution:
 Concepts:
Kinetic theory, the ideal gas law
 Reasoning:
PV = nRT.
The temperature is a direct measure of the average translational molecular
kinetic energy.
½m<v^{2}> = (3/2)k_{B}T.
 Details of the calculation:
PV = nRT yields T.
T = PV/(nR)
= (1.5*1.01*10^{5 }Pa)(5000 cm^{3} * 1 m^{3}/10^{6
}cm^{3})/(0.125*8.31 J/K) = 729 K.
½m<v^{2}> = (3/2)k_{B}T = (3/2)*1.38*10^{23}*729
J = 1.51*10^{20} J.
The average translational kinetic energy of a single molecule is 1.51*10^{20
}J.
Problem:
A cylinder contains a mixture of helium and argon gas in equilibrium at 150
^{o}C.
(a) What is the average kinetic energy of each gas
molecule?
(b) What is the rootmeansquare speed of each type of
molecule?
m_{He }= 4 u, m_{Ar }= 39.9 u.
1 u = 1 atomic
mass unit = 1.66*10^{27 }kg.
Solution:
 Concepts:
Kinetic
theory
 Reasoning:
The temperature is a direct measure of the
average translational molecular kinetic energy,
½m<v^{2}> =
(3/2)k_{B}T
 Details of the calculation:
(a) The average kinetic energy of each
molecule is
(3/2)k_{B}T = (3/2)1.38*10^{23 }J/K(423 K)
= 8.76*10^{21 }J.
The average kinetic energy is the same for
both types of atoms.
(b) v_{rms}^{2 }= 2*8.76*10^{21
}J/m.
v_{rms}(He) = 1.62*10^{3 }m/s.
v_{rms}(Ar)
= 514 m/s.
The more massive molecules have a lower average speed.
Problem:
(a) For a gas of electrons in which quantum
mechanical effects can be neglected, above what temperature would you expect
special relativistic effects to become important?
(b) For a gas of
nonrelativistic electrons, above what electron number density would you expect
quantum mechanical behavior of the gas to become important if its temperature is
10^{7} K?
Solution:

Concepts:
Limitations of Newtonian Physics

Reasoning:
(a) We expect relativistic effects
to become when kT ~ mc^{2},
(b) We expect quantum
mechanical effects to become important when the particles become
indistinguishable for practical purposes, i.e. when their wave packets start
overlapping. This happens when the average spacing between the
particles becomes comparable to their deBroglie wavelength.

Details of the calculation:
(a) T ~ mc^{2}/k
= 0.511*10^{6} eV/(8.617*105 eV/K) ~ 5.9*10^{9} K.
(b)
deBroglie wavelength: λ∫ = h/p = (h^{2}/p^{2})^{½}
~ (h^{2}/(2mkT))^{½} = hc/(2mc^{2}kT)^{½}.
Average spacing between the particles: d ~ (V/N)^{⅓} = n^{⅓}.
Therefore n ~ 1/λ^{3}.
For T = 10^{7} K, λ =
(1240 eV nm)/((2*0.511*10^{6} eV)(8.62*10^{5} eV/K)(10^{7}
K))^{½}~ 4.2*10^{2} nm.
Therefore n ~ 1.4*10^{4}/nm^{3}
= 1.4*10^{25}/cm^{3}.
Problem:
An apparent limit on the lowest
temperature achievable by laser cooling is reached when an atom's recoil energy
upon absorbing or emitting a single photon is approximately equal to its total
kinetic energy (3/2 kT). Calculate this “recoil temperature” for the
Rubidium atom.
(m = 85 u, u = 1.66 * 10^{27}kg) if the wavelength of
the photon in resonance with Rubidium is 780 nm.
Solution:
 Concepts:
Kinetic theory, momentum conservation: h/λ = p_{recoil}
 Reasoning:
We want E_{recoil} = p^{2}_{recol}/2m
= h^{2}/(2λ^{2}m) = (3/2)kT.
 Details of the calculation:
T = h^{2}/(3kλ^{2}m) =
1.24*10^{7} K.
Problem:
What is the rootmeansquare speed of a thermal (room temperature)
neutron? Can you outrun it? (Show work!)
Solution:
 Concepts:
Kinetic theory
 Reasoning:
The temperature is a direct measure of the average translational
kinetic energy.
 Details of the calculation:
½mv^{2} = (3/2)kT, v = (3kT/m)^{½}
= (3 * 1.38
* 10^{23} J/K
* 293 K /1 .675
* 10^{27} kg)^{½} = 2.7
* 10^{3} m/s = 2.7 km/s
No, you cannot outrun the neutron.