What will be the average value of energy along one degree of freedom for an ideal gas in thermal equ

Q: Why isn't the answer $\frac{3}{2} k_B T$?

$\frac{3}{2} k_B T$ is the *total* translational kinetic energy for a molecule with 3 degrees of freedom (like a monoatomic gas). The question specifically asks for the energy along *one* degree of freedom, which is always $\frac{1}{2} k_B T$.

Q: What exactly is the Boltzmann constant $k_B$?

The Boltzmann constant is a fundamental physical constant that relates the average relative kinetic energy of particles in a gas with the thermodynamic temperature of the gas. It acts as a bridge between macroscopic and microscopic physics.

Q: Does this law apply to rotational and vibrational motion as well?

Yes! The Law of Equipartition of Energy applies to all active degrees of freedom. Each rotational degree of freedom gets $\frac{1}{2} k_B T$. However, a vibrational mode has both kinetic and potential energy, so it actually contributes $2 \times \frac{1}{2} k_B T = k_B T$ in total.

Q: What if the gas is diatomic? Does the energy per degree of freedom change?

No, the energy *per degree of freedom* is a universal constant at a given temperature, always equal to $\frac{1}{2} k_B T$. A diatomic gas just has *more* degrees of freedom ($f=5$), so its *total* energy is higher ($\frac{5}{2} k_B T$).

Q: Why do we use absolute temperature $T$ in Kelvin?

Kinetic energy is directly proportional to the absolute temperature. At $0$ Kelvin (absolute zero), all classical molecular motion stops, meaning the kinetic energy becomes zero. Celsius or Fahrenheit scales don't start at absolute zero, so the proportionality wouldn't work.

Question

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature $T$? ($k_B$ is Boltzmann constant)

Options:

(A) $\frac{1}{2} k_B T$

(B) $\frac{2}{3} k_B T$

(C) $\frac{3}{2} k_B T$

(D) $k_B T$

Sigma Percentile · Accepted Answer

The Law of Equipartition of Energy is one of the most elegant and fundamental principles in the kinetic theory of gases. It bridges the gap between the microscopic motion of individual molecules and the macroscopic temperature we measure with a thermometer.

### The Concept of Degrees of Freedom

Imagine a single gas molecule floating in a room. How many independent ways can it move? It can travel left or right along the x-axis, up or down along the y-axis, and forward or backward along the z-axis. Each of these independent modes of motion is called a **degree of freedom**.

For a monoatomic gas like Helium, there are exactly 3 translational degrees of freedom. If the molecule is diatomic, like Oxygen, it can also rotate, adding rotational degrees of freedom. The total number of degrees of freedom is denoted by the variable $f$.

### The Law of Equipartition of Energy

When a gas is in thermal equilibrium at an absolute temperature $T$, its total internal energy isn't hoarded by just one type of motion. Instead, nature is perfectly fair. The **Law of Equipartition of Energy** states that the total energy is distributed equally among all active degrees of freedom.

But how much energy does each degree of freedom get? The law provides a beautifully simple answer: exactly $\frac{1}{2} k_B T$, where $k_B$ is the Boltzmann constant.

$$E_{	ext{per degree}} = \frac{1}{2} k_B T$$

This means that whether the molecule is translating along the x-axis or rotating around its center of mass, that specific independent motion contributes exactly $\frac{1}{2} k_B T$ to the average kinetic energy.

### Answering the Specific Question

This is where many students fall into a classic trap. When asked about the energy of a gas, the immediate reflex is to write down the total energy formula, $E = \frac{f}{2} k_B T$, and start plugging in values for $f$.

However, we must read the question carefully. It specifically asks for the average value of energy along **one degree of freedom**. It does not ask for the total energy of a monoatomic or diatomic gas.

Since we are only looking at a single degree of freedom ($f = 1$), we don't need to multiply by the total number of modes. The energy associated with just one degree of freedom is simply:

$$E = \frac{1}{2} k_B T$$

This directly matches option (a). Always remember to distinguish between the energy per degree of freedom and the total internal energy of the molecule!

Animated Solution for Physics - Thermodynamics: What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature $T$ ? ( $k_{B}$ is Boltzmann constant)

Select Answer:

Visualized Solution

Degree of Freedom $f$

Law of Equipartition

Energy per DOF

Calculation for $f = 1$

Final Answer

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

The Concept of Degrees of Freedom

The Law of Equipartition of Energy

Answering the Specific Question

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