Worked exercise — Equipartition and degrees of freedom

Problem

A diatomic gas (for example O$_2$) has a molar specific heat at constant volume equal to $(5/2)R$, compared with $(3/2)R$ for a monatomic gas (for example He). Where does the factor $5/2$ come from?

Solution

The equipartition theorem assigns to each quadratic degree of freedom in the energy a mean contribution $\tfrac{1}{2}k_B T$. For a monatomic molecule the degrees of freedom are the three translational ones (velocity along $x$, $y$, $z$): three squared terms in the kinetic energy $\tfrac{1}{2}m v_x^2 + \tfrac{1}{2}m v_y^2 + \tfrac{1}{2}m v_z^2$, hence $\bar E = \tfrac{3}{2}k_B T \qquad c_V = \tfrac{3}{2}R$ For a diatomic molecule, rotations must be added. A diatomic molecule has the shape of a small dumbbell: rotating about the molecular axis, the moment of inertia is practically zero (the mass is concentrated along the axis), so that degree of freedom does not activate. Two rotation axes perpendicular to the molecular axis remain, each with rotational energy $\tfrac{1}{2}I\omega^2$: two extra quadratic degrees. Total $3 + 2 = 5$ degrees of freedom, from which $\bar E = \tfrac{5}{2}k_B T \qquad c_V = \tfrac{5}{2}R$

The vibrations of the two atoms along the molecular axis are two further quadratic degrees (kinetic + elastic potential energy), but at room temperature quantum mechanics freezes them out: they only activate above roughly a thousand kelvin. This is one of the first historical clues, already visible to Maxwell, that classical physics is not enough to explain specific heats. $\ev{c_V = \tfrac{5}{2}R\ \text{from } 3\ \text{translations} + 2\ \text{rotations},\ \text{each } \tfrac{1}{2}k_B T}$

Links

Topics: Kinetic theory of gases Concepts: Internal energy · Specific heat and heat capacity Skills: Micro-macro interpretation