Internal energy of an ideal diatomic gas at 300 K is 100 J. In this 100 J

Options

A. potential energy = 0
B. rotational kinetic energy = 40 J
C. translational kinetic energy = 60 J
D. translational kinetic energy is 100 J

Answer

To solve the problem regarding the internal energy of an ideal diatomic gas at 300 K, we will break down the total internal energy into its components: translational kinetic energy and rotational kinetic energy.

### Step-by-step Solution:

1. **Understand the Total Internal Energy**:
The total internal energy \( U \) of the ideal diatomic gas is given as 100 J. For an ideal gas, the internal energy consists of kinetic energy due to translational and rotational motion.

2. **Identify the Energy Distribution**:
For a diatomic gas, the degrees of freedom are:
- Translational motion: 3 degrees of freedom (x, y, z directions)
- Rotational motion: 2 degrees of freedom (rotation about two axes, as rotation about the axis of the molecule does not contribute significantly)

The total degrees of freedom for a diatomic gas is \( 5 \) (3 translational + 2 rotational).

3. **Calculate the Fraction of Energy**:
The translational kinetic energy contributes \( \frac{2}{5} \) of the total internal energy, while the rotational kinetic energy contributes \( \frac{3}{5} \) of the total internal energy.

4. **Calculate Translational Kinetic Energy**:
\[ U_{\text{translational}} = \frac{2}{5} \times U = \frac{2}{5} \times 100 \, \text{J} = 40 \, \text{J} \]

5. **Calculate Rotational Kinetic Energy**:
\[ U_{\text{rotational}} = \frac{3}{5} \times U = \frac{3}{5} \times 100 \, \text{J} = 60 \, \text{J} \]

6. **Summarize the Results**:
- Translational kinetic energy = 40 J
- Rotational kinetic energy = 60 J

### Final Answer:
- Translational Kinetic Energy: 40 J
- Rotational Kinetic Energy: 60 J