# Question #a1be8

Apr 11, 2017

${E}_{\text{kin,tot}} = 3.7326 \times {10}^{47} \cdot T$ [Joule]

#### Explanation:

The average speed of the $S {O}_{2}$ is not given, therefore we use the Equipartition principle.
The average kinetic energy can be estimated using the equipartition principle. This principle says that in every degree of freedom there is on average the same amount of kinetic energy.

In thermal equilibrium (the state variables (P, V, n, T) do not change) with high temperature, the average amount of kinetic energy in each degree of freedom is directly related to the temperature, therefore:

${E}_{\text{kin}} =$ $\frac{1}{2} {k}_{\text{B}} T$
With ${k}_{\text{B}}$ the Boltzmann constant ($1.38064 \times {10}^{23}$)and $T$ the temperature.

Since the temperature is not given, we cannot determine the average kinetic energy.

Degrees of freedom
A $S {O}_{\text{2}}$ molecule is not linear. For non-linear molecules, the total degrees of freedom is given by:
$3 \times N$, with N the number of atoms. In this case: 3, therefore:
$3 \times 3 = 9$ degrees of freedom per molecule.

We have 1 mol of $S {O}_{2}$, therefore we have
$6.022 140 857 \times {10}^{23}$ molecules (Avogadro's constant)
We have in total:
$6.022 140 857 \times {10}^{23} \cdot 9 =$
$5419926771300000000000000 = 5.41 \times {10}^{24}$ degrees of freedom

The total kinetic energy of 1 mol of $S {O}_{2}$ is therefore given by:
${E}_{\text{kin,tot}} = 5.41 \times {10}^{24} \cdot \frac{1}{2} \cdot 1.38064 \times {10}^{23} \cdot T$
$\to {E}_{\text{kin,tot}} = 3.7326 \times {10}^{47} \cdot T$ [Joule]

Now we only need the temperature to calculate the correct number.