Equipartition Theorem for a Monoatomic Ideal Gas

Correct Answer

(B) ⟨½mv²⟩ = 3⁄2 k_BT

Understanding Thermal Equilibrium

Thermal equilibrium is the state in which all parts of a system have the same temperature and there is no net transfer of heat between them. For an ideal gas in thermal equilibrium, molecules move randomly in every direction with different speeds. Although individual molecules possess different kinetic energies, the average kinetic energy depends only on the absolute temperature of the gas.

This remarkable result is independent of the type of gas, the mass of its molecules, or the pressure of the gas. Only the absolute temperature determines the average translational kinetic energy.

What is the Equipartition Theorem?

The Equipartition Theorem states that every independent quadratic degree of freedom contributes an average energy of

½k_BT

to each molecule in thermal equilibrium.

A monoatomic gas molecule can move independently along three mutually perpendicular directions:

• x-direction
• y-direction
• z-direction

Therefore, it possesses three translational degrees of freedom.

Each degree of freedom contributes

½k_BT

Hence, the total average translational kinetic energy becomes

Average Kinetic Energy = 3 × ½k_BT = 3⁄2 k_BT

Derivation of the Average Kinetic Energy

The kinetic energy of one gas molecule is

K = ½mv²

Since the molecular speed changes continuously because of random collisions, we consider the average value.

Thus,

⟨K⟩ = ⟨½mv²⟩

According to the Equipartition Theorem,

⟨½mv²⟩ = 3⁄2 k_BT

This equation is one of the most important results in kinetic theory and should be remembered for all competitive examinations.

Relation Between Velocity Components

The total velocity of a molecule is related to its three components by

v² = v_x² + v_y² + v_z²

Because the gas is isotropic in thermal equilibrium, the average kinetic energy is equally distributed among the three directions.

Therefore,

⟨½mv_x²⟩ = ½k_BT

Similarly,

⟨½mv_y²⟩ = ½k_BT

and

⟨½mv_z²⟩ = ½k_BT

Adding all three contributions gives

⟨½mv²⟩ = 3⁄2 k_BT

Detailed Option-Wise Analysis

Option (A): ⟨½mv²⟩ = ½k_BT

This equation assigns only one degree of freedom to the molecule.

A monoatomic gas molecule has three translational degrees of freedom, not one.

Therefore, the average total translational kinetic energy must be three times larger.

Option (A) is Incorrect.

Option (B): ⟨½mv²⟩ = 3⁄2 k_BT

This is the standard result obtained from the Equipartition Theorem.

It correctly accounts for all three translational degrees of freedom of a monoatomic gas molecule.

This equation is universally valid for an ideal monoatomic gas in thermal equilibrium.

Option (B) is Correct.

Option (C): ½mv² = 3⁄2 k_BT

This equation refers to the kinetic energy of an individual molecule without taking an average.

Individual molecules continuously change their speeds because of collisions. Therefore, their kinetic energies are not constant.

Only the average kinetic energy satisfies the Equipartition Theorem.

Option (C) is Incorrect.

Option (D): ⟨½mv_x²⟩ = 3⁄2 k_BT

The x-component represents only one translational degree of freedom.

According to the Equipartition Theorem, each translational degree of freedom contributes only

½k_BT

Therefore,

⟨½mv_x²⟩ = ½k_BT

not

3⁄2 k_BT.

Option (D) is Incorrect.

Physical Interpretation

The molecules of an ideal gas are always moving randomly in three-dimensional space. Although each molecule has a different speed at any instant, the average kinetic energy of all molecules depends only on temperature. Increasing the temperature increases the average molecular speed and therefore increases the average kinetic energy. This explains why hotter gases exert greater pressure and diffuse more rapidly.

Real-Life Applications

The Equipartition Theorem is used extensively in thermodynamics, statistical mechanics, atmospheric physics, chemical kinetics, astrophysics, and plasma physics. It helps explain the behavior of gases in engines, refrigeration systems, weather prediction, stellar interiors, and many industrial processes involving gases.

Exam-Oriented Key Concepts

Students should remember that a monoatomic ideal gas possesses three translational degrees of freedom. Each degree contributes an average energy of ½k_BT, making the total average kinetic energy equal to 3⁄2k_BT. Individual molecules do not all possess the same kinetic energy; only the average follows this relation. Another frequently tested result is that the average kinetic energy associated with any one velocity component is ½k_BT.

Final Answer

For a monoatomic ideal gas in thermal equilibrium, the average translational kinetic energy of one molecule is

⟨½mv²⟩ = 3⁄2 k_BT

Correct Option: (B)