91Ó°ÊÓ

The equipartition theorem says that any quadratic term in the expression for the total energy E of a particle contributes to the average energy $\stackrel{\mathbf{¯}}{\mathbf{E}}$by an amount$\stackrel{}{\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}$, where${\mathbf{k}}_{\mathbf{B}}$is the Boltzmann constant, and T is temperature.

First consider the 1-D harmonic oscillator. Its total energy writes:

$\mathrm{E}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{x}}^2+\frac{1}{2}{\mathrm{kx}}^2$

where$V_x$ is speed,$x$ is position, $m$is mass, and$k$ is the force constant.

The two quadratic terms pop up. Use the equipartition theorem, each degree of freedom namely$v_x^2$ and $X^2$contributes to the average energy by an amount $\frac{1}{2}{k}_{B}T$. Thus, the average energy in a harmonic oscillator must be:

$\stackrel{¯}{E}=\frac{1}{2}{k}_{B}T+\frac{1}{2}{k}_{B}T=k_{B}T$

Now for an atom in a monatomic ideal gas, the atom possesses three degrees of freedom in its velocity as it freely moves in 3-D space. Since it is not acted on by a conservative force, then its total energy must simply be its kinetic energy in 3-D:

$\mathrm{E}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{x}}^2+\frac{1}{2}{\mathrm{mv}}_{\mathrm{y}}^2+\frac{1}{2}{\mathrm{mv}}_{\mathrm{z}}^2$

Since the atom has three degrees of freedom, then its average energy from to the equipartition theorem must be:

$\begin{array}{c}\stackrel{¯}{\mathrm{E}}=\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}+\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}+\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\\ =\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\end{array}$