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Kinetic Theory of Gases. Boltzmann's Law and Maxwell's Distribution

  • Number of collisions exercised by gas molecules on a unit area of the wall surface per unit time:
    $$\nu = \frac{1}{4} n \langle v \rangle,$$

    where $ n $ is the concentration of molecules, and $ \langle v \rangle $ is their mean velocity.

  • Equation of an ideal gas state:
    $$ p = nkT.$$
  • Mean energy of molecules:
    $$\langle\epsilon\rangle = \frac{i}{2} kT,$$

    where $ i $ is the sum of translational, rotational, and the double number of vibrational degrees of freedom. 

  • Maxwellian distribution:
    $$dN (v_x) = N \left(\frac{m}{2\pi k T}\right)^{1/2} e^{-m v_x^2/2kT}dv_x,$$
    $$dN(v) =  N \left(\frac{m}{2\pi k T}\right)^{3/2} e^{-m v^2/2kT} 4\pi v^2 dv.$$
  • Maxwellian distribution in a reduced form:
    $$dN(u) = N \frac{4}{\sqrt{\pi}} e^{-u^2} u^2 du,$$

    where $\displaystyle u = v/v_p $, $ v_p $ is the most probable velocity.

  • The most probable, the mean, and the root mean square velocities of molecules:
    $$v_p = \sqrt {2 \frac{kT}{m}}, \; \langle v \rangle = \sqrt{\frac{8}{\pi}\frac{kT}{m}}, \; v_{sq} = \sqrt{3 \frac{kT}{m}.$$
  • Boltzmann's formula:
    $$n = n_0 e^{-\left(U-U_0\right)/kT}, $$

    where $ U $ is the potential energy of a molecules.

 

Keywords: 
kinetic theory, Boltzmann, Maxwell, mean velocity, mean energy, degrees of freedom, Maxwellian distribution, root mean square
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