# At 47^@"C", what is the average kinetic energy of "1 g" of "O"_2(g)?

Sep 12, 2017

Since we are at ordinary temperatures, oxygen gas might rotate and vibrate, but we should check. Thus, it has translational, rotational, and maybe vibrational degrees of freedom.

When we denote degrees of freedom as $N$, we can use the equipartition theorem at "high temperatures" (the so-called classical limit) to calculate the average kinetic energy:

${K}_{a v g} = \frac{N}{2} n R T$

where:

• $n$ is the mols of gas.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.
• $T = \text{47 + 273.15 K}$ is the temperature in $\text{K}$.

To check whether we are at high enough temperatures (the classical limit), you should ask yourself, is:

• ${\Theta}_{\text{vib" = tildeomega/k_B " << }} T$? This is usually NOT true at ordinary temperatures.
• ${\Theta}_{\text{rot" = tildeB/k_B " << }} T$? This is usually true at ordinary temperatures.

where $t i l \mathrm{de} \omega$ is the fundamental vibrational frequency and $t i l \mathrm{de} B$ is the rotational constant. ${k}_{b} = \text{0.695 cm"^(-1)//"K}$ is the Boltzmann constant.

For ${\text{O}}_{2}$, these are from NIST, and we have $t i l \mathrm{de} \omega = {\text{1580.19 cm}}^{- 1}$ and $t i l \mathrm{de} B = {\text{1.4377 cm}}^{- 1}$. So, we have:

Theta_"vib" = "1580.19 cm"^(-1)//"0.695 cm"^(-1)cdot"K" = ul("2273.7 K" " >> " T) $\textcolor{red}{\times}$
Theta_"rot" = "1.4377 cm"^(-1)//"0.695 cm"^(-1)cdot"K" = ul("2.0686 K" " << " T) color(green)(sqrt"")

In the classical limit (i.e. the temperature is much higher than ${\Theta}_{\text{rot}}$), we take linear diatomic molecules to have two degrees of rotational freedom.

In practice, it is better to ignore the vibrational degrees of freedom because we are NOT in the high temperature limit for vibration (meaning we do NOT have ${\Theta}_{\text{vib" " << }} T$).

This means the total number of degrees of freedom is:

$N = {\overbrace{3}}^{\text{Translational" + overbrace(2)^"Rotational" + overbrace(~~0)^"Vibrational}} = 5$

And so, the average kinetic energy for ${\text{O}}_{2}$ is:

$\textcolor{b l u e}{{K}_{a v g}} = \frac{5}{2} n R T$

$= 2.5 \left(\text{1 g" xx "1 mol"/"31.998 g")("8.314472 J/mol"cdot"K")("47 + 273.15 K}\right)$

$=$ $\textcolor{b l u e}{\text{207.97 J}}$