Kinetic Theory

Problem:

A 5-liter 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<v²> = (3/2)k_BT.
• Details of the calculation:
  PV = nRT yields T.
  T = PV/(nR)
  = (1.5*1.01*10⁵ Pa)(5000 cm³ * 1 m³/10⁶ cm³)/(0.125*8.31 J/K) = 729 K.
  ½m<v²> = (3/2)k_BT = (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 ^oC.
(a)  What is the average kinetic energy of each gas molecule?
(b)  What is the root-mean-square 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<v²> = (3/2)k_BT
• Details of the calculation:
  (a) The average kinetic energy of each molecule is 
  (3/2)k_BT = (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_rms² = 2*8.76*10^-21 J/m.
  v_rms(He) = 1.62*10³ 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 non-relativistic electrons, above what electron number density would you expect quantum mechanical behavior of the gas to become important if its temperature is 10⁷ K?

Solution:

• Concepts:
  Limitations of Newtonian Physics
• Reasoning:
  (a)  We expect relativistic effects to become when kT ~ mc²,
  (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 ~ mc²/k = 0.511*10⁶ eV/(8.617*10-5 eV/K) ~ 5.9*10⁹ K.
  (b)  deBroglie wavelength:  λ∫ = h/p = (h²/p²)^½ ~  (h²/(2mkT))^½ = hc/(2mc²kT)^½.
  Average spacing between the particles: d ~ (V/N)^(1/3) = n^-(1/3).
  Therefore n ~ 1/λ³.  
  For T = 10⁷ K, λ = (1240 eV nm)/((2*0.511*10⁶ eV)(8.62*10^-5 eV/K)(10⁷ K))^½
  ~ 4.2*10^-2 nm.
  Therefore n ~ 1.4*10⁴/nm³ = 1.4*10²⁵/cm³.

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^-27kg) 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 = p²_recol/2m = h²/(2λ²m) = (3/2)kT.
• Details of the calculation:
  T = h²/(3kλ²m) = 1.24*10^-7 K.

Problem:

What is the root-mean-square 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:
  ½mv² = (3/2)kT, v = (3kT/m)^½
  = (3 * 1.38 * 10^-23 J/K * 293 K /1 .675 * 10^-27 kg)^½ = 2.7 * 10³ m/s = 2.7 km/s
  No, you cannot outrun the neutron.