Question

The translational kinetic energy of an ideal gas containing N molecules at temperature T is (k - Boltzmann constant) ________.

• A. $\frac{5}{2}NkT$
• B. $\frac{1}{2}NkT$
• C. $\frac{3}{2}NkT$
• D. $\frac{7}{2}NkT$
• E. $\frac{9}{2}NkT$

Hint:

Translational kinetic energy depends only on temperature, not on the type of gas (monatomic/diatomic).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This question is about the internal energy of an ideal gas, specifically the part of the energy associated with the translational motion of its molecules. The equipartition theorem states how this energy is distributed among the available degrees of freedom.
Step 2: Key Formula or Approach
The theorem of equipartition of energy states that for a system in thermal equilibrium at temperature T, the average energy associated with each quadratic degree of freedom is \(\frac{1}{2}kT\), where k is the Boltzmann constant.
- Translational motion always corresponds to 3 degrees of freedom, regardless of the type of gas (monatomic, diatomic, etc.). These correspond to motion along the x, y, and z axes.
- The total translational kinetic energy is the sum of the average energy for each molecule multiplied by the total number of molecules.
Step 3: Detailed Explanation
1. Determine the average translational kinetic energy per molecule.
- A molecule has 3 translational degrees of freedom.
- According to the equipartition theorem, each translational degree of freedom contributes \(\frac{1}{2}kT\) to the average energy of the molecule.
- Therefore, the total average translational kinetic energy per molecule is:
\[ E_{trans, avg} = 3 \times \left(\frac{1}{2}kT\right) = \frac{3}{2}kT \] 2. Calculate the total translational kinetic energy for N molecules.
The gas contains N molecules. To find the total translational kinetic energy of the gas (\(K_{trans, total}\)), we multiply the average energy per molecule by the number of molecules.
\[ K_{trans, total} = N \times E_{trans, avg} \] \[ K_{trans, total} = N \times \left(\frac{3}{2}kT\right) = \frac{3}{2}NkT \] This result is universal for any ideal gas because translational motion is always described by three degrees of freedom.
Step 4: Final Answer
The translational kinetic energy of the ideal gas is \(\frac{3}{2}NkT\).