Question

What is the average KE of N2 at 1 atm and 298K?

Answer 1

`K_(avg) ~~ 1.03 xx 10^20 "J/N"_2` `"molecule"`

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The average kinetic energy at `"298 K"` and `"1 atm"` uses the `N/V` standard state (i.e. one molecule per unit volume), and for many molecules at `"298 K"`, we can assume the high-temperature limit to invoke the equipartition theorem:

`<< kappa >> = K_(avg)/N`

`= << kappa >>_(tr) + << kappa >>_(rot) + cancel(<< kappa >>_(vib))^"really small" + cancel(<< kappa >>_(el ec))^"practically zero"`

`~~ N_(tr)/2 k_B T + N_(rot)/2 k_B T`

where:

• `N_k` is the degrees of freedom for the `k`th type of motion (translation, rotation, vibration, electronic). To keep it simple I am ignoring vibrational and electronic energy states.
• The `N` in `K_(avg)/N` is a different `N` and is just the number of particles. Here we have `N = 1` for one `"N"_2` molecule.
• `k_B = 1.38065 xx 10^(-23) "J/K"` is the Boltzmann constant.
• `T` is the temperature in `"K"`.

For translational motion, `N_(tr) = 3`, because we are obviously in a three-dimensional world (`x,y,z`).

For rotational motion, a linear molecule has `N_(rot) = 2`, and a nonlinear molecule has `N_(rot) = 3`. Each degree of freedom corresponds to a rotation angle, either `theta`, `phi`, or some third angle `gamma`.

We ignore vibrational and electronic degrees of freedom, because that would wayyyyy overestimate `<< kappa >>` at ordinary temperatures to use equipartition on them.

Therefore:

`color(blue)(<< kappa >>) = K_(avg)/1`

`~~ 3/2k_BT + 2/2 k_B T`

`~~ 5/2 k_BT`

`= 5/2 cdot 1.38065 xx 10^(-23) "J/K" cdot "298 K"`

`= color(blue)(1.03 xx 10^(-20) "J/N"_2)` `color(blue)("molecule")`