Question

Find the average energy per molecule of a triatomic gas at room temperature $T$.
A) $3{k_B}T$
B) $\dfrac{1}{2}{k_B}T$
C) $\dfrac{3}{2}{k_B}T$
D) $\dfrac{5}{2}{k_B}T$

Answer 1

A triatomic molecule can be linear or nonlinear. The law of equipartition energy gives the average energy per molecule of a gas (monatomic or diatomic or triatomic or polyatomic) to be dependent on the degrees of freedom of the gas. The degree of freedom of a gas molecule refers to the number of ways in which it can move, rotate or vibrate in space.

Formula used:
-The average energy per molecule of a gas is given by, $E = \dfrac{1}{2}f{k_B}T$ where $f$ is the degree of freedom of the molecule, ${k_B}$ is the Boltzmann constant and $T$ is the temperature.

Complete step by step answer.
Step 1: Mention the degrees of freedom of the triatomic gas.
A non-linear triatomic gas has $f = 6$ degrees of freedom. Three are from the translational motion of the molecule and the other three are from the rotational motion of the molecule. A linear triatomic gas has $f = 5$ degrees of freedom i.e., there translational and two rotational. At room temperature there exists no vibrational mode.
Step 2: Express the relation for the average energy per molecule of a gas.
The average energy per molecule of a gas can be expressed as $E = \dfrac{1}{2}f{k_B}T$ -------- (1) where $f$ is the degree of freedom of the molecule, $k$ is the Boltzmann constant and $T$ is the temperature.
Substituting for $f = 6$ in equation (1) we get, $E = \dfrac{1}{2} \times 6{k_B}T = 3{k_B}T$ .
Thus the average energy per molecule of a non-linear triatomic gas is $E = 3{k_B}T$ .
Now substituting for $f = 5$ in equation (1) we get, $E = \dfrac{1}{2} \times 5{k_B}T = \dfrac{5}{2}{k_B}T$
Thus the average energy per molecule of a linear triatomic gas is $E = \dfrac{5}{2}{k_B}T$ .

So the correct options are A and D.

Note: The vibrational mode exists only at high temperatures. The law of equipartition energy states that at equilibrium, the total energy will be distributed equally among the possible energy modes and each mode will have an average energy of $\dfrac{1}{2}{k_B}T$. But the energy modes vary with the type of the gas molecule. Each translational and rotational degree of freedom contributes $\dfrac{1}{2}{k_B}T$ to the total energy of the gas whereas a vibrational degree of freedom contributes $2 \times \dfrac{1}{2}{k_B}T$ as the vibrational mode has kinetic energy and potential energy.