Question

A polyatomic gas with \( n \) degrees of freedom has a mean kinetic energy per molecule given by (if \( N \) is Avogadro’s number):

• A. \( \frac{n k T}{N} \)
• B. \( \frac{n k T}{2N} \)
• C. \( \frac{n k T}{2} \)
• D. \( \frac{3 k T}{2} \)

Hint:

The mean kinetic energy of a gas molecule depends on the degrees of freedom. For any gas, it is given by \( E = \frac{f}{2} k T \), where \( f \) is the degrees of freedom.

Answer 1

The Correct Option is C

To determine the mean kinetic energy per molecule of a polyatomic gas with \( n \) degrees of freedom, we need to use the principle of equipartition of energy. According to this principle, each degree of freedom contributes \( \frac{1}{2} k T \) to the kinetic energy, where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature.

For a molecule with \( n \) degrees of freedom, the total mean kinetic energy per molecule is given by multiplying the contribution per degree of freedom by the number of degrees of freedom:

\( \text{Mean Kinetic Energy} = \frac{n}{2} k T \)

This result does not depend on the number of Avogadro's number \( N \) directly, because it pertains to individual molecules rather than a mole of molecules. Therefore, the mean kinetic energy per molecule is:

\( \frac{n k T}{2} \)

Thus, the correct answer is the option:

\( \frac{n k T}{2} \)

Answer 2

Step 1: Understanding Kinetic Energy of a Gas Molecule
The mean kinetic energy of a molecule of an ideal gas is given by: \[ E = \frac{f}{2} k T \] where: - \( f \) is the degrees of freedom,
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature.
Step 2: Applying for a Polyatomic Gas
For a polyatomic gas with \( n \) degrees of freedom, the kinetic energy per molecule becomes: \[ E = \frac{n k T}{2}. \] Step 3: Conclusion
Thus, the mean kinetic energy per molecule of a polyatomic gas is: \[ \mathbf{\frac{n k T}{2}}. \]