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# phys documents (dragged) 36 - if kT ≈ 6 B 7.2 The energy...

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Chapter 7 Statistical physics 7.1 Degrees of freedom A molecule consisting of n atoms has s = 3 n degrees of freedom. There are 3 translational degrees of freedom, a linear molecule has s = 3 n - 5 vibrational degrees of freedom and a non-linear molecule s = 3 n - 6 . A linear molecule has 2 rotational degrees of freedom and a non-linear molecule 3. Because vibrational degrees of freedom account for both kinetic and potential energy they count double. So, for linear molecules this results in a total of s = 6 n - 5 . For non-linear molecules this gives s = 6 n - 6 . The average energy of a molecule in thermodynamic equilibrium is E tot = 1 2 skT . Each degree of freedom of a molecule has in principle the same energy: the principle of equipartition . The rotational and vibrational energy of a molecule are: W rot = ¯ h 2 2 I l ( l + 1) = Bl ( l + 1) , W vib = ( v + 1 2 h ω 0 The vibrational levels are excited if kT ¯ h ω , the rotational levels of a hetronuclear molecule are excited if kT 2 B . For homonuclear molecules additional selection rules apply so the rotational levels are well coupled
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Unformatted text preview: if kT ≈ 6 B . 7.2 The energy distribution function The general form of the equilibrium velocity distribution function is P ( v x ,v y ,v z ) dv x dv y dv z = P ( v x ) dv x · P ( v y ) dv y · P ( v z ) dv z with P ( v i ) dv i = 1 α √ π exp ±-v 2 i α 2 ² dv i where α = ³ 2 kT/m is the most probable velocity of a particle. The average velocity is given by ± v ² = 2 α / √ π , and ´ v 2 µ = 3 2 α 2 . The distribution as a function of the absolute value of the velocity is given by: dN dv = 4 N α 3 √ π v 2 exp ±-mv 2 2 kT ² The general form of the energy distribution function then becomes: P ( E ) dE = c ( s ) kT ± E kT ² 1 2 s-1 exp ±-E kT ² dE where c ( s ) is a normalization constant, given by: 1. Even s : s = 2 l : c ( s ) = 1 ( l-1)! 2. Odd s : s = 2 l + 1 : c ( s ) = 2 l √ π (2 l-1)!!...
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