Question

Find net kinetic energy (maximum possible) associated with 20 diatomic molecules (Here k_B is Boltzmann constant and T is absolute temperature of diatomic gas).

• A. 35 k_B T
• B. 70 k_B T
• C. 60 k_B T
• D. 30 k_B T

Answer 1

The Correct Option is B

To find the net kinetic energy associated with 20 diatomic molecules, we need to understand the degrees of freedom available to a diatomic gas molecule and the relation between kinetic energy and temperature.

Concept: Diatomic gas molecules have different degrees of freedom due to their ability to move and rotate. For an ideal diatomic gas, the degrees of freedom are as follows:

• 3 translational degrees of freedom.
• 2 rotational degrees of freedom (for most ideal cases where vibrational modes are not considered at lower temperatures).

This totals 5 degrees of freedom for a diatomic gas molecule.

Equipartition Theorem: According to the equipartition theorem, each degree of freedom contributes \(\frac{1}{2} k_B T\) to the kinetic energy per molecule, where \(k_B\) is the Boltzmann constant and \(T\) is the absolute temperature.

Thus, for each diatomic molecule, the kinetic energy associated is:

E = \frac{5}{2} k_B T

Given, there are 20 diatomic molecules. The net kinetic energy is:

E_{\text{total}} = 20 \times \frac{5}{2} k_B T = 50 k_B T

However, to find the maximum possible kinetic energy, we need to consider vibrational modes, which become active at higher temperatures. This adds 2 more degrees of freedom (1 for kinetic, 1 for potential energy), making it 7 degrees of freedom per molecule.

So, with vibrational modes, the energy becomes:

E = \frac{7}{2} k_B T

The net maximum kinetic energy for 20 molecules is:

E_{\text{max, total}} = 20 \times \frac{7}{2} k_B T = 70 k_B T

Therefore, the correct answer is 70 k_BT.