Calculate the ratio of the average kinetic energies of "8 g" "H"_2 compared to "8 g" of "O"_2 at 27^@ "C"?

1 Answer
Jun 4, 2017

The average kinetic energy at a high enough temperature is given by:

$n \left\langle\kappa\right\rangle \equiv \left\langleK\right\rangle \approx \frac{5}{2} n R T$ in $\text{J}$,

and we got that:

$\frac{{\left\langleK\right\rangle}_{{H}_{2}}}{{\left\langleK\right\rangle}_{{O}_{2}}} \approx \frac{9.922 {n}_{{H}_{2}} \cdot R T}{0.625 {n}_{{O}_{2}} \cdot R T} \approx 15.87$

How to use the equipartition theorem is given below.

DISCLAIMER: LONG ANSWER!

The equipartition theorem can be best used to approximate the average kinetic energy in the high temperature limit (also called the "classical" limit), where rotational and vibrational states are not accessed.

$\left\langle\kappa\right\rangle = \frac{\left\langleK\right\rangle}{n} = \frac{N}{2} R T$,

where:

• $\left\langle\kappa\right\rangle$ is the average kinetic energy in $\text{J/mol}$, while $\left\langleK\right\rangle$ is in $\text{J}$.
• $N$ is the number of degrees of freedom, which we can consider the number of dimensions of each type of motion that should be considered.
• $R = \text{8.314472 J/mol"cdotK}$ is the universal gas constant
• $n$ is the mols of gas.
• $T$ is temperature in $\text{K}$.

In considering whether we are at a high enough temperature, it turns out that:

• For pretty much all molecules, the high temperature limit applies for translational motion ($x , y , z$) and rotational motion. This includes ${\text{O}}_{2}$ and ${\text{H}}_{2}$.

So, we can assume that equipartition applies for the translational and rotational kinetic energy, ${\left\langleK\right\rangle}_{t r}$ and ${\left\langleK\right\rangle}_{r o t}$.

• For $T = 27 + 273.15 = \text{300.15 K}$, both ${\text{O}}_{2}$ and ${\text{H}}_{2}$ have ${\Theta}_{v i b} \equiv \frac{{N}_{A} h \nu}{R}$ $\text{>>}$ $T$, where:
• $\nu$ is the fundamental vibrational frequency of ${\text{O}}_{2}$ or ${\text{H}}_{2}$.
• ${N}_{A}$ is Avogadro's number.
• $h$ is Planck's constant.
• ${\Theta}_{v i b}$ is the vibrational temperature, the temperature above which vibrational motions become important.

We actually have that:

${\Theta}_{v i b , {O}_{2}} = \text{2256 K}$

${\Theta}_{v i b , {H}_{2}} \approx \text{6323 K}$

So, ${\Theta}_{v i b}$ is much greater than $\text{300.15 K}$ for both molecules. This allows us to assume these molecules are in their ground vibrational state.

In other words, for ${\text{O}}_{2}$ and ${\text{H}}_{2}$, it is better to just not invoke equipartition for vibrational motion, so that we don't overestimate $\left\langleK\right\rangle$.

In considering the degrees of freedom, $N$, we say that:

• For each dimension of translational motion ($x , y , z$), we assign $1$ to $N$.
• For each dimension of rotational motion ($\theta , \phi$), we assign $1$ to $N$.
• We don't assign a value for $N$ for vibrational motion because the high temperature limit doesn't apply for the vibrations of ${\text{O}}_{2}$ and ${\text{H}}_{2}$.

So far, we then have from equipartition, a general expression for ${\text{O}}_{2}$ and ${\text{H}}_{2}$:

$\textcolor{g r e e n}{\left\langle\kappa\right\rangle} \approx {\left\langle\kappa\right\rangle}_{t r} + {\left\langle\kappa\right\rangle}_{r o t} + {\cancel{{\left\langle\kappa\right\rangle}_{v i b}}}^{\text{small}}$

$= \frac{3}{2} R T + \frac{2}{2} R T$

$= \textcolor{g r e e n}{\frac{5}{2} R T}$

So, their average kinetic energy is predicted to be:

${\left\langleK\right\rangle}_{{O}_{2}} = n {\left\langle\kappa\right\rangle}_{{O}_{2}} \approx \frac{5}{2} n R T$

${\left\langleK\right\rangle}_{{H}_{2}} = n {\left\langle\kappa\right\rangle}_{{H}_{2}} \approx \frac{5}{2} n R T$

Thus, with $\text{8 g}$ of each gas:

${\left\langleK\right\rangle}_{{O}_{2}} \approx \frac{5}{2} \left(8 \cancel{\text{g") xx "1 mol O"_2/(31.998 cancel"g}}\right) R T \approx 0.625 R T$

$\approx$$\text{1559.8 J}$

${\left\langleK\right\rangle}_{{H}_{2}} \approx \frac{5}{2} \left(8 \cancel{\text{g") xx "1 mol H"_2/(2.0158 cancel"g}}\right) R T \approx 9.922 R T$

$\approx$ $\text{24760.3 J}$

Therefore, their kinetic energy ratio at $\text{300.15 K}$ is:

$\textcolor{b l u e}{\frac{{\left\langleK\right\rangle}_{{H}_{2}}}{{\left\langleK\right\rangle}_{{O}_{2}}}} \approx \frac{9.922}{0.625} = \frac{24760.3}{1559.8}$

$\approx \textcolor{b l u e}{15.87}$

Thus, $\text{8 g}$ of ${\text{H}}_{2}$ is expected to have a greater average kinetic energy, mainly because there are more $\text{mol}$s of it than the $\text{mol}$s of ${\text{O}}_{2}$.