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 of determining the distribution of internal energy in an ideal diatomic gas at 300 K, we can follow these steps:

### Step 1: Understand the Degrees of Freedom
For a diatomic gas, the degrees of freedom (DOF) are:
- Translational: 3 (movement along x, y, and z axes)
- Rotational: 2 (rotation about two axes perpendicular to the bond axis)
- Total DOF = 5

**Hint:** Remember that each degree of freedom contributes \(\frac{1}{2} k_B T\) to the internal energy, where \(k_B\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.

### Step 2: Calculate the Total Internal Energy
The total internal energy \(U\) of the gas can be expressed as:
\[ U = \frac{f}{2} k_B T \]
where \(f\) is the total degrees of freedom. Given \(U = 100 \, \text{J}\) and \(f = 5\), we can set up the equation:
\[ 100 = \frac{5}{2} k_B T \]

**Hint:** Rearranging the equation will help you isolate \(k_B T\).

### Step 3: Solve for \(k_B T\)
Rearranging the equation gives:
\[ k_B T = \frac{2 \times 100}{5} = 40 \, \text{J} \]

**Hint:** Make sure to double-check your arithmetic when calculating \(k_B T\).

### Step 4: Calculate the Rotational Kinetic Energy
The rotational kinetic energy \(U_{rot}\) can be calculated using its degrees of freedom:
\[ U_{rot} = \frac{2}{2} k_B T = 1 \times k_B T = 40 \, \text{J} \]

**Hint:** Remember that you only consider the rotational degrees of freedom (2 in this case).

### Step 5: Calculate the Translational Kinetic Energy
The translational kinetic energy \(U_{trans}\) is given by:
\[ U_{trans} = \frac{3}{2} k_B T = \frac{3}{2} \times 40 = 60 \, \text{J} \]

**Hint:** Again, focus on the translational degrees of freedom (3 in this case) when calculating.

### Step 6: Determine Potential Energy
Since the problem states that the internal energy is entirely kinetic (for an ideal gas), the potential energy \(U_{pot}\) is:
\[ U_{pot} = 0 \, \text{J} \]

**Hint:** In an ideal gas, potential energy changes are negligible, so it remains zero.

### Summary of Results
- Total Internal Energy: \(100 \, \text{J}\)
- Rotational Kinetic Energy: \(40 \, \text{J}\)
- Translational Kinetic Energy: \(60 \, \text{J}\)
- Potential Energy: \(0 \, \text{J}\)

**Final Answer:**
- Potential Energy: \(0 \, \text{J}\)
- Rotational Kinetic Energy: \(40 \, \text{J}\)
- Translational Kinetic Energy: \(60 \, \text{J}\)