Question

How would the equipartition theorem be used to estimate the average kinetic energy of molecules?

Answer 1

At high enough temperatures,

`<< kappa >> -= K_(avg)/(nN_A) ~~ N/2 k_B T`

where `<< kappa >>` is the average molecular kinetic energy in `"J/molecule"cdot"K"`, `n` is the mols, `N_A` is Avogadro's number in `"mol"^(-1)`, `N` is the number of degrees of freedom, `k_B` is Boltzmann's constant in `"J/K"`, and `T` is temperature in `"K"`.

At room temperature, this holds for simple molecules such as `"N"_2` and `"O"_2`, but overestimates for more complicated molecules like `"CH"_4` and `"NH"_3`.

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THE EQUIPARTITION THEOREM AT THE CLASSICAL LIMIT

Well, the equipartition theorem for the free particle at high enough temperatures is:

`bb(E_(avg) ~~ K_(avg) = N/2nRT)`,

or

`K_(avg) = N/2 cdot stackrel("Number of Molecules")overbrace(nN_A) cdot k_B T`

where:

• `N` is the number of degrees of freedom in the molecule.
• `n` is the `"mols"` of substance.
• `N_A = 6.0221413 xx 10^(23) "mol"^(-1)` is Avogadro's number.
• `k_B = 1.38065 xx 10^(-23) "J/K"` is the Boltzmann constant.
• `T` is the temperature in `"K"`.

If we define `<< kappa >> = K_(avg)/(nN_A)` as the average molecular kinetic energy, then:

`color(blue)(<< kappa >> = N/2 k_B T)`

EXAMPLE USING NITROGEN MOLECULE

For instance, let's say we were looking at `"N"_2`, a linear molecule. Here is how we could determine its degrees of freedom:

• `3` translational dimensions (`x,y,z`)
  `->` `3` translational degrees of freedom
• `2` rotational angles (`theta, phi` in spherical coordinates) for linear molecules
  `->` `2` rotational degrees of freedom

(it would have been `3` for nonlinear molecules)

• Hardly any vibrational degrees of freedom, because of a very stiff triple bond.

So, for `"N"_2`, we expect its average kinetic energy to be:

`color(blue)(<< kappa >>) ~~ (3+2+0)/2 k_B T`

`= color(blue)(5/2k_B T)` `color(blue)("J/molecule"cdot"K")`

or,

`= color(blue)(5/2R T)` `color(blue)("J/mol"cdot"K")`

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CHECKING LITERATURE VALUES

We can't really look up the average molecular kinetic energy, but we can check by looking at the constant-pressure molecular heat capacity:

`CC_P -= C_P/(nN_A)`

`~~ (N+2)/2k_B` `"J/molecule"cdot"K"`,

or the constant-pressure molar heat capacity:

`barC_P -= C_P/n`

`~~ (N+2)/2 R` `"J/mol"cdot"K"`.

From the `5` degrees of freedom we found, we can approximate:

`barC_P ~~ (5+2)/2 R = 7/2 R ~~ ul("29.10 J/mol"cdot "K")`

From NIST, we have the graph for `barC_P`:

which shows `ul("29.124 J/mol"cdot"K")` at `"299 K"`, around `0.08%` error.