Question 47: Thermodynamics & KTG

Question

The mean energy per molecule for a diatomic gas is:

Accepted Answer

The correct choice is Option A:
A diatomic molecule (neglecting vibrational modes at ordinary temperatures) has 5 degrees of freedom (3 translational + 2 rotational). By the equipartition theorem, each degree contributes ÃÂ½Ã¢ÂÂkBT, so the mean energy per molecule is
$$\frac{5}{2}\,k_{B}T\,. $$

Answer

$\left(\frac{5}{2}\right) \mathrm{K}_{\mathrm{B}} \mathrm{T}$

Answer

$\left(\frac{3}{2}\right) \mathrm{K}_{\mathrm{B}} \mathrm{T}$

Answer

$\left(\frac{3}{2 N}\right) \mathrm{K}_{\mathrm{B}} \mathrm{T}$

Answer

$\left(\frac{5}{2 N}\right) \mathrm{K}_{\mathrm{B}} \mathrm{T}$

Thermodynamics & KTG

The mean energy per molecule for a diatomic gas is:

Choose the correct answer: