Question

The average thermal energy for a mono-atomic gasis : (kBis Boltzmann constant and T, absolute temperature)

• A. \(\frac12K_BT\)
• B. \(\frac32K_BT\)
• C. \(\frac52K_BT\)
• D. \(\frac72K_BT\)

Answer 1

The Correct Option is B

To solve the given problem about the average thermal energy of a mono-atomic gas, we will use the concept of the kinetic theory of gases.

The average thermal energy for a mono-atomic ideal gas can be derived from the equipartition theorem, which states that each degree of freedom contributes \(\frac{1}{2}k_B T\) to the energy. Here, \(k_B\) is the Boltzmann constant, and \(T\) is the absolute temperature.

For a mono-atomic gas, the molecules are in three-dimensional translational motion, which means it has three degrees of freedom. Therefore, the average thermal energy \(E\) for a mono-atomic gas is calculated as follows:

The energy contribution from each degree of freedom is:

\(E = \frac{1}{2}k_B T \times \text{number of degrees of freedom}\)

Substituting for three degrees of freedom, we have:

\(E = \frac{1}{2}k_B T \times 3 = \frac{3}{2}k_B T\)

Thus, the average thermal energy for a mono-atomic gas is \(\frac{3}{2}k_BT\).

Now, we'll verify and rule out the other options:

• \(\frac{1}{2}k_BT\): This is the energy contribution per degree of freedom, not the total energy for a mono-atomic gas.
• \(\frac{5}{2}k_BT\) and \(\frac{7}{2}k_BT\): These values apply to diatomic or polyatomic gases, which have additional degrees of freedom due to rotational and vibrational motions.

Hence, the correct answer is \(\frac{3}{2}k_BT\), corresponding to option 2, which represents the average thermal energy for a mono-atomic gas.