Statistical Mechanics

  1. Assuming a uniform gas, the Boltzmann distribution gives the fraction of particles which have energy E:
    f(E) ~ e-E / (k T).
    Note: this assumes that the particles in the gas are distinguishable. If they are not, the distribution function depends on the spin of the particles. For particles of integer spin, we have the Bose-Einstein distribution
    f(E) ~ 1 / (A eE / (k T) - 1),
    and for particles of half-integer spin, we have the Fermi-Dirac distribution
    f(E) ~ 1 / (A eE / (k T) + 1).
    If all of the energy is kinetic, and the gas is uniform, f(E) = e-m v2 / (2 k T) = f(v).
    Note that this is a normal distribution with standard deviation √ (k T / m). For air at room temperature (with molecular weight 28.94 g, temperature 20 C and mean speed 463 m/s), σ = 290.244 m/s:

    In three dimensions we normalize this with respect to position and velocity as
    ∫ f(v) d3r d3v = N,
    where N is the total number of particles in the gas. Now d3r = dV; assuming that the velocity distribution is isotropic (does not depend on direction, i.e., is spherically symmetric), we have
    d3v = 4 π v2 dv.
    Integrating with respect to v from 0 to ∞, we find
    f(v) = (N / V) (m / (2 π k T))3/2 e-m v2 / (2 k T).

    (Here we used ∫0 x2 e-a x2 dx = (√ π) / (4 a3/2).)

    The graph above was plotted using this normalization and standard pressure, for a mole of air occupying 22.4L. The right side of the plot is the relevant part, considering we are now treating v as a speed. For the 3-dimensional isotropic normal distribution, almost 87% of the molecules will have speeds within one standard deviation of the mean.
  2. The average of a function g(v) weighted by a distribution f(v) is
    <g(v)> = (V / N) ∫ g(v) f(v) d3v.
  3. The pressure due to collisions of gas particles (taking z normal to the wall) is:
    P = Fz / A = dpz/dt / A
    = (1 / A) * the number of particles colliding per unit time * average momentum transfer
    <P> = (1 / A) ∫ f(v) * (A dz / 2) / dt * 2 m vz d3v
    (V = A dz, and half of the particles are moving away from the wall during the time dt)
    = m ∫ f(v) vz2 d3v

    = m N <v2> / (3 V)

    (<vz2> = <v2>/3 since v is isotropic)
    = 2 N <E> / (3 V)
  4. An ideal gas is a dilute gas whose constituents can be assumed to be non-interacting (except for perfect collisions). The equation of state for an ideal gas is
    P V = n R T,
    where P is the pressure, V is the volume, n is the number of moles, and T is the temperature.
    R = NA kB is the molar gas constant, equal to 8.314 J / (mol K).
    Equations of state are usually thought of as relating pressure and density; in those terms, the ideal gas equation of state is
    P = (R T / (molecular weight)) ρ.
  5. So for an ideal gas, we find <E> = 3 P V / (2 N) = 3 n R T / (2 N) = 3 k T / 2.

    In general, the Equipartition theorem tells us that each quadratic degree of freedom in <E> contributes k T / 2.

  6. The first law of thermodynamics is a statement of conservation of energy:
    dE = dQ + dW
    where dE is the change in energy of the system, dQ is the heat added to it and dW is the work done on it:
    W = -∫ P dV
  7. For an ideal gas with χ*N quadratic degrees of freedom,
    cv = dQ/dT / n = dE/dT / n
    (constant volume implies that W = 0)
    = d(χ N k T / 2)/dT / n

    = χ R / 2

    cp = dQ/dT / n = (dE/dT + P dV/dT) / n
    = cv + R.
    (since P is constant, P dV/dT = d(P V)/dT = n R )
    χ is the sum of translational, rotational and vibrational modes. All atoms and molecules possess 3 translational degrees of freedom; linear molecules have 2 more rotational modes, whereas nonlinear molecules have 3 more rotational modes. Vibrational degrees of freedom require sufficient energy to rise to an excited quantum state. For gases, room temperature is mostly insufficient to the task. Note that vibrational modes have both a kinetic and a potential degree of freedom.
  8. Define γ ≡ cp / cv. For an ideal gas undergoing an adiabatic process (in which ΔQ = 0),
    dE = n cv dT = - P dV.
    d (P V) = P dV + dP V = n R dT
    = - (R / cv) P dV

    = - (γ - 1) P dV

    we have
    dP V = - γ P dV.
    Integrating we find
    P Vγ = constant.
    Note that for an ideal gas, γ = 1 + R / cv = 1 + 2 / χ.
  9. Note that the ideal gas is not the only equation of state!


©2011, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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