Topic Covered



The kinetic energy of a single molecule is

`color{orange} {epsi_t} = 1/2 color{purple} {mv_x^2 }+ 1/2color{purple} {mv_y^2} + 1/2 color{purple} {mv_z^2}`

.................... 13.22

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

`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.


`color{green} bbul "Monatomic Gases"`

The molecule of a monatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature T is `(3/2)k_BT.`

The total internal energy of a mole of such a gas is

`color{purple} {U = 3/2 k_B T xx N_A = 3/2 RT}` .................13.27

The molar specific heat at constant volume, `C_v`, is

`C_v` (monatomic gas) `= (dU)/(dT) = 3/2 RT` ................13.28

For an ideal gas, `color{green} {C_P-C_v = R}` ....................13.29

where `C_p` is the molar specific heat at constant pressure. Thus,

`C_P = 5/2 R` ..............13.30

The ratio of specific heats

` color{green} {gamma = C_P/C_v = 5/3}`


`color{green}bb ul "Diatomic Gases"`

As explained earlier, a diatomic molecule treated as a rigid rotator like a dumbbell has 5 degrees of freedom : 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a mole of such a gas is

`color{orange} { U = 5/2 k_B T xx N_A = 5/2 RT}` ................13.32

The molar specific heats are then given by

`Cv` (rigid diatomic) `= 5/2 R, C_P =7/2 R ` ...................13.33

`gamma ` (rigid diatomic) ` =7/5` ..................13.34

If the diatomic molecule is not rigid but has in addition a vibrational mode

`color{purple} { U = ( 5/2 k_B T + K_BT ) N_A = 7/2 RT}`

`color{green} { C_v =7/2 R , C_p = 9/2 R , gamma = 9/7 R}`


`color{green} ul bb"Polyatomic Gases"`

In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number ( f ) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has

`color{purple} {U = (3/2 k_B T + 3/2 k_BT + f k_B T ) N_A}`

i.e. `C_v = (3 + f ) R, C_p = (4 + f ) R`,

`color{orange} { gamma = (4 + f)/(3 + f)}` .....................13.36

Note that `C_p – C_v = R` is true for any ideal gas, whether mono, di or polyatomic.

Table 13.1 summarises the theoretical predictions for specific heats of gases ignoring any vibrational modes of motion.

The values are in good agreement with experimental values of specific heats of several gases given in Table 13.2. Of course, there are discrepancies between predicted and actual values of specific heats of several other gases (not shown in the table), such as `Cl_2, C_2H_6` and many other polyatomic gases.

Usually, the experimental values for specific heats of these gases are greater than the predicted values given in Table13.1 suggesting that the agreement can be improved by including vibrational modes of motion in the calculation.

The law of equipartition of energy is thus well verified experimentally at ordinary temperatures.

Q 3240701613

A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by `15.0 °C` ? (`R = 8.31 J mol^-1 K^-1`).
Class 11 Chapter 13 Example 8

Using the gas law `PV = μRT`, you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure `(1 atm = 1.01 × 10^5 Pa)` occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, `C_v = (3/2) R`, and molar specific heat at constant pressure, `C_p = (3/2) R + R = (5/2) R`. Since the volume of the cylinder is fixed, the heat required is determined by `C_v`. Therefore,

Heat required = no. of moles × molar specific heat × rise in temperature

`= 2 × 1.5 R × 15.0 = 45 R`

`= 45 × 8.31 = 374` J.

Specific Heat Capacity of Solids

We can use the law of equipartition of energy to determine specific heats of solids. Consider a solid of N atoms, each vibrating about its mean position. An oscillation in one dimension has average energy of `2 × 1/2 k_BT = k_BT`.

In three dimensions, the average energy is 3 kBT. For a mole of solid, `N = N_A`, and the total energy is

`U = 3 k_BT × N_A = 3 RT`

Now at constant pressure `ΔQ = ΔU + PΔV`

`= ΔU`, since for a solid `ΔV` is negligible. Hence,

`C= (Delta Q)/(DeltaT) = (Delta U)/(Delta T) = 3 R` .......................13.37

As Table 13.3 shows the prediction generally agrees with experimental values at ordinary temperature (Carbon is an exception).

Specific Heat Capacity of Water

We treat water like a solid. For each atom average energy is `3k_BT`. Water molecule has three atoms, two hydrogen and one oxygen. So it has

`color{orange} {U = 3 × 3 k_BT × N_A = 9 RT}`

and `C = ΔQ// ΔT =Δ U //ΔT = 9R` .

This is the value observed and the agreement is very good. In the calorie, gram, degree units, water is defined to have unit specific heat. As 1 calorie = 4.179 joules and one mole of water is 18 grams, the heat capacity per mole is ~ `75 J mol^-1 K^-1 ~ 9R`.

However with more complex molecules like alcohol or acetone the arguments, based on degrees of freedom, become more complicated.

Lastly, we should note an important aspect of the predictions of specific heats, based on the classical law of equipartition of energy. The predicted specific heats are independent of temperature. As we go to low temperatures, however, there is a marked departure from this prediction.

Specific heats of all substances approach zero as `T -> 0`. This is related to the fact that degrees of freedom get frozen and ineffective at low temperatures. According to classical physics degrees of freedom must remain unchanged at all times.

The behaviour of specific heats at low temperatures shows the inadequacy of classical physics and can be explained only by invoking quantum considerations, as was first shown by Einstein. Quantum mechanics requires a minimum, nonzero amount of energy before a degree of freedom comes into play. This is also the reason why vibrational degrees of freedom come into play only in some cases.


Molecules in a gas have rather large speeds of the order of the speed of sound. Yet a gas leaking from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room.

The top of a cloud of smoke holds together for hours. This happens because molecules in a gas have a finite though small size, so they are bound to undergo collisions. As a result, they cannot move straight unhindered; their paths keep getting incessantly deflected.

Suppose the molecules of a gas are spheres of diameter d. Focus on a single molecule with the average speed . It will suffer collision with any molecule that comes within a distance d between the centres. In time `Δt`, it sweeps a volume `πd^2 < v > Δt` wherein any other molecule will collide with it (see Fig. 13.7).

If n is the number of molecules per unit volume, the molecule suffers `nπd^2 < v > Δt` collisions in time Δt. Thus the rate of collisions is `nπd^2 < v >` or the time between two successive collisions is on the average,

`color{orange} { τ = 1/((nπ < v > d^2 ))} ` .......................(13.38)

The average distance between two successive collisions, called the mean free path l, is :

`color{green} {l = < v > τ = 1/(nπd^2)}` ........................ (13.39)

In this derivation, we imagined the other molecules to be at rest. But actually all molecules are moving and the collision rate is determined by the average relative velocity of the molecules.

Thus we need to replace by `< v_r >` in Eq. (13.38). A more exact treatment gives

`color{purple} {l =1/((sqrt2 n pi d^2))}` ...............................13.40

Let us estimate `l` and `τ` for air molecules with average speeds `< v > =` ( 485m/s). At STP

`color{green} {n = (0.02 xx 10^23)/(22.4 xx 10^-3)}`

`= 2.7 × 10^25 m^-3`.

Taking, `d = 2 × 10^-10` m,

`τ = 6.1 × 10^-10` s

and `l = 2.9 × 10^-7 m ≈ 1500 d`

As expected, the mean free path given by Eq. (13.40) depends inversely on the number density and the size of the molecules. In a highly evacuated tube n is rather small and the mean free path can be as large as the length of the tube.
Q 3230001812

Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises 13.1 and Eq. (13.41) above.
Class 11 Chapter 13 Example 9

The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.

So `n = 2.7 xx 10^25 xx 273/373 = 2 xx 10^25m^-3`

Hence, mean free path `l = 4 xx 10^-7m`

Note that the mean free path is 100 times the interatomic distance `~ 40 Å = 4 ×10^-9 m` calculated earlier. It is this large value of mean free path that leads to the typical gaseous behaviour. Gases can not be confined without a container.

Using, the kinetic theory of gases, the bulk measurable properties like viscosity, heat conductivity and diffusion can be related to the microscopic parameters like molecular size. It is through such relations that the molecular sizes were first estimated.