Law of Equipartition of Energy

A single atom is free to move in space along the X, Y and Z axis. However, each of these movements requires energy. This is derived from the energy held by the atom. The Law of Equipartition of Energy defines the allocation of energy to each motion of the atom (translational, rotational and vibrational). Before we understand this law, let’s understand a concept called ‘Degrees of Freedom’.

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Degrees of Freedom

Imagine a single atom. In a three dimensional space, it can move freely along the X, Y and Z axis. Motion from one point to another is also known as translation. Hence, this movement along the three axes is called translational movement. If you have to specify the location of this atom, then you need three coordinates (x, y, and z).

We can also say that a single atom has 3 Degrees of Freedom. Most monoatomic molecules (i.e. molecules having a single atom like Argon) have 3 translational degrees of freedom, provided their movement is unrestricted.

Let’s now imagine a diatomic molecule (a molecule having two atoms like O₂ or N₂). Apart from the three translational degrees of freedom, these molecules can also rotate around the centre of mass. Two such rotations are possible along the axis normal to the axis that joins the two atoms.

This adds two additional degrees of freedom (rotational) to the molecule. In simpler words, to specify the location of the molecule, you would need the X, Y and Z coordinates along with the rotational coordinates of the individual atoms.

It is important to note here that these diatomic molecules are not rigid rotators (where molecules do not vibrate) at all temperatures. Along with the translational and rotational movements, diatomic molecules also oscillate along the interatomic axis like a single dimensional oscillator. This adds a vibrational degree of freedom to such molecules.

Hence, to specify the location of a diatomic molecule, you would finally need the X, Y and Z coordinates along with the rotational and vibrational coordinates. So, in a nutshell, Degrees of Freedom is nothing but the number of ways in which a molecule can move. This forms the basis of the Law of Equipartition of Energy.

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Law of Equipartition of Energy

The law states that: “In thermal equilibrium, the total energy of the molecule is divided equally among all Degrees of Freedom of motion”. Before delving into the calculations, let’s understand the law better. If a molecule has 1000 units of energy and 5 degrees of freedom (which includes translational, rotational and vibrational movements), then the molecule allocates 200 units of energy to each motion.

Now, let us look at some equations!

Kinetic Energy of a single molecule: KE = 1/2 mv². A gas in thermal equilibrium at temperature T, the average Energy is:

E_avg = 1/2 mv_x² + 1/2 mv_y² + 1/2 mv_z²  = 1/2KT + 1/2 KT + 1/2 KT = 3/2 KT

where K = Boltzmann’s constant. In case of a monoatomic molecule, since there is only translational motion, the energy allotted to each motion is 1/2KT. This is calculated by dividing total energy by the degrees of freedom:

3/2 KT ÷ 3 = 1/2 KT

In case of a diatomic molecule, translational, rotational and vibrational movements are involved. Hence the Energy component of translational motion= 1/2 mv_x² + 1/2 mv_y² + 1/2 mv_z². Energy component of rotational motion= 1/2 I₁w₁² + 1/2 I₂w₂²  {I1 & I2 moments of inertia. w1 & w2 are angular speeds}

And, the energy component of vibrational motion= 1/2 m (dy/dt)²+ 1/2 ky². Where k is the force constant of the oscillator and y is the vibrational coordinate. It is important to note here that this has both kinetic and potential modes.

According to the Law of Equipartition of Energy, in thermal equilibrium, the total energy is distributed equally among all energy modes. While the translational and rotational motion contributes ½ KT to the total energy, vibrational motion contributes 2 x 1/2KT = KT since it has both kinetic and potential energy modes.

Some Questions for You:

Q: ‘N′ moles of a diatomic gas in a cylinder are at a temperature ′T′. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monatomic gas. What is the change in the total kinetic energy of the gas?

A) 5/2 nRT       B) 1/2 nRT      C) 0          D) 3/2 nRT

Solution: D) Initial K.E. = (3/2) nRT . Number of moles in the final sample = 2n

Since the gas is changed to monoatomic gas, we have: K.E. of the final sample = (3/2)×2nRT

Hence, the change in the K.E. = 3nRT – (3/2) nRT = 3/2 nRT.