LAW OF EQUIPARTITION OF ENERGY
The kinetic energy of a single molecule is
For a gas in thermal equilibrium at temperature T the average value of energy denoted by `< epsi_t >` is
`color{purple} { (: epsi_t :) = (: 1/2 mv_x^2 :) + (: 1/2 mv_y^2 :) + (: 1/2 mv_z^2 :) = 3/2 k_B T} ` ............................13.23
Since there is no preferred direction, Eq. (13.23) implies
A molecule free to move in space needs three coordinates to specify its location. If it is constrained to move in a plane it needs two; and if constrained to move along a line, it needs just one coordinate to locate it. This can also be expressed in another way.
We say that it has one degree of freedom for motion in a line, two for motion in a plane and three for motion in space. Motion of a body as a whole from one point to another is called translation.
Thus, a molecule free to move in space has three translational degrees of freedom. Each translational degree of freedom contributes a term that contains square of some variable of motion, e.g., `color{green} {1/2 mv_x^2}` and similar terms in vy and vz. In, Eq. (13.24) we see that in thermal equilibrium, the average of each such term is `1/2 k_BT` .
Molecules of a monatomic gas like argon have only translational degrees of freedom. But what about a diatomic gas such as `O_2` or `N_2`? A molecule of `O_2` has three translational degrees of freedom. But in addition it can also rotate about its centre of mass.
Figure 13.6 shows the two independent axes of rotation 1 and 2, normal to the axis joining the two oxygen atoms about which the molecule can rotate.
The molecule thus has two rotational degrees of freedom, each of which contributes a term to the total energy consisting of translational energy `epsi_t` and rotational energy `epsi_r`
`color{blue} { epsi_t + epsi_r = 1/2 mv_x^2 + 1/2 mv_y^2 + 1/2m v_z^2 + 1/2I_1 omega_1^2 + 1/2 I_2omega_2^2}` ................13.25
where `ω_1` and `ω_2` are the angular speeds about the axes 1 and 2 and `I_1, I_2` are the corresponding moments of inertia. Note that each rotational degree of freedom contributes a term to the energy that contains square of a rotational variable of motion.
We have assumed above that the `O_2` molecule is a ‘rigid rotator’, i.e. the molecule does not vibrate. This assumption, though found to be true (at moderate temperatures) for `O_2`, is not always valid.
Molecules like `CO` even at moderate temperatures have a mode of vibration, i.e. its atoms oscillate along the interatomic axis like a one-dimensional oscillator, and contribute a vibrational energy term εv to the total energy:
`epsi = epsi_t + epsi_r + epsi_v` ...........(13.26
where k is the force constant of the oscillator and y the vibrational co-ordinate.
Once again the vibrational energy terms in Eq. (13.26) contain squared terms of vibrational variables of motion y and `dy//dt` .
At this point, notice an important feature in Eq.(13.26). While each translational and rotational degree of freedom has contributed only one ‘squared term’ in Eq.(13.26), one vibrational mode contributes two ‘squared terms’ : kinetic and potential energies.
Each quadratic term occurring in the expression for energy is a mode of absorption of energy by the molecule. We have seen that in thermal equilibrium at absolute temperature T, for each translational mode of motion, the average energy is `½ k_BT`.
A most elegant principle of classical statistical mechanics (first proved by Maxwell) states that this is so for each mode of energy: translational, rotational and vibrational. That is, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to `½ k_BT`.
This is known as the law of equipartition of energy. Accordingly, each translational and rotational degree of freedom of a molecule contributes `½ k_BT` to the energy while each vibrational frequency contributes `color{green} {2 × ½ k_BT = k_BT}` , since a vibrational mode has both kinetic and potential energy modes.
The proof of the law of equipartition of energy is beyond the scope of this book. Here we shall apply the law to predict the specific heats of gases theoretically. Later we shall also discuss briefly, the application to specific heat of solids.
`color{orange} {epsi_t} = 1/2 color{purple} {mv_x^2 }+ 1/2color{purple} {mv_y^2} + 1/2 color{purple} {mv_z^2}`
.................... 13.22For a gas in thermal equilibrium at temperature T the average value of energy denoted by `< epsi_t >` is
`color{purple} { (: epsi_t :) = (: 1/2 mv_x^2 :) + (: 1/2 mv_y^2 :) + (: 1/2 mv_z^2 :) = 3/2 k_B T} ` ............................13.23
Since there is no preferred direction, Eq. (13.23) implies
`color{orange} {(: 1/2 mv_x^2 :) = 1/2 k_B T ,
(: 1/2 mv_y^2 :) = 1/2 k_B T ,
(: 1/2 mv_x^2 :) =1/2 k_B T }`
A molecule free to move in space needs three coordinates to specify its location. If it is constrained to move in a plane it needs two; and if constrained to move along a line, it needs just one coordinate to locate it. This can also be expressed in another way.
We say that it has one degree of freedom for motion in a line, two for motion in a plane and three for motion in space. Motion of a body as a whole from one point to another is called translation.
Thus, a molecule free to move in space has three translational degrees of freedom. Each translational degree of freedom contributes a term that contains square of some variable of motion, e.g., `color{green} {1/2 mv_x^2}` and similar terms in vy and vz. In, Eq. (13.24) we see that in thermal equilibrium, the average of each such term is `1/2 k_BT` .
Molecules of a monatomic gas like argon have only translational degrees of freedom. But what about a diatomic gas such as `O_2` or `N_2`? A molecule of `O_2` has three translational degrees of freedom. But in addition it can also rotate about its centre of mass.
Figure 13.6 shows the two independent axes of rotation 1 and 2, normal to the axis joining the two oxygen atoms about which the molecule can rotate.
The molecule thus has two rotational degrees of freedom, each of which contributes a term to the total energy consisting of translational energy `epsi_t` and rotational energy `epsi_r`
`color{blue} { epsi_t + epsi_r = 1/2 mv_x^2 + 1/2 mv_y^2 + 1/2m v_z^2 + 1/2I_1 omega_1^2 + 1/2 I_2omega_2^2}` ................13.25
where `ω_1` and `ω_2` are the angular speeds about the axes 1 and 2 and `I_1, I_2` are the corresponding moments of inertia. Note that each rotational degree of freedom contributes a term to the energy that contains square of a rotational variable of motion.
We have assumed above that the `O_2` molecule is a ‘rigid rotator’, i.e. the molecule does not vibrate. This assumption, though found to be true (at moderate temperatures) for `O_2`, is not always valid.
Molecules like `CO` even at moderate temperatures have a mode of vibration, i.e. its atoms oscillate along the interatomic axis like a one-dimensional oscillator, and contribute a vibrational energy term εv to the total energy:
`color{ purple} { epsi_o = 1/2 m ((dy)/(dt))^2 + 1/2 ky^2}`
`epsi = epsi_t + epsi_r + epsi_v` ...........(13.26
where k is the force constant of the oscillator and y the vibrational co-ordinate.
Once again the vibrational energy terms in Eq. (13.26) contain squared terms of vibrational variables of motion y and `dy//dt` .
At this point, notice an important feature in Eq.(13.26). While each translational and rotational degree of freedom has contributed only one ‘squared term’ in Eq.(13.26), one vibrational mode contributes two ‘squared terms’ : kinetic and potential energies.
Each quadratic term occurring in the expression for energy is a mode of absorption of energy by the molecule. We have seen that in thermal equilibrium at absolute temperature T, for each translational mode of motion, the average energy is `½ k_BT`.
A most elegant principle of classical statistical mechanics (first proved by Maxwell) states that this is so for each mode of energy: translational, rotational and vibrational. That is, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to `½ k_BT`.
This is known as the law of equipartition of energy. Accordingly, each translational and rotational degree of freedom of a molecule contributes `½ k_BT` to the energy while each vibrational frequency contributes `color{green} {2 × ½ k_BT = k_BT}` , since a vibrational mode has both kinetic and potential energy modes.
The proof of the law of equipartition of energy is beyond the scope of this book. Here we shall apply the law to predict the specific heats of gases theoretically. Later we shall also discuss briefly, the application to specific heat of solids.
