LAW OF EQUIPARTITION OF ENERGY

We have seen in Section 9.2.1 that the average kinetic energy of a molecule moving in the x-direction is \(\frac{1}{2}mv_x^2 = \frac{1}{2}kT_x\).

Similarly, when the motion is in the y-direction, \(\frac{1}{2}mv_y^2 = \frac{1}{2}kT_y\) and for the motion along the z-direction, \(\frac{1}{2}mv_z^2 = \frac{1}{2}kT_z\).

According to kinetic theory, the average kinetic energy of a system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x, y, or z directions of motion) so that each degree of freedom will get \(\frac{1}{2}kT\) of energy. This is called the law of equipartition of energy.

Average kinetic energy of a monatomic molecule (with f=3) \(= 3 \times \frac{1}{2}kT = \frac{3}{2}kT\).

Average kinetic energy of a diatomic molecule at low temperature (with f = 5) \(= 5 \times \frac{1}{2}kT = \frac{5}{2}kT\).

Average kinetic energy of a diatomic molecule at high temperature (with f =7) \(= 7 \times \frac{1}{2}kT = \frac{7}{2}kT\).

Average kinetic energy of a linear triatomic molecule (with f = 7) \(= 7 \times \frac{1}{2}kT = \frac{7}{2}kT\).

Average kinetic energy of a non-linear triatomic molecule (with f = 6) \(= 6 \times \frac{1}{2}kT = 3kT\).

Meyer’s relation C_P - C_V = R connects the two specific heats for one mole of an ideal gas.

Equipartition law of energy is used to calculate the value of C_P - C_V and the ratio between them \(\gamma = \frac{C_P}{C_V}\). Here \(\gamma\) is called the adiabatic exponent.

i). Monatomic Molecule

Average kinetic energy of a molecule $$= [\frac{3}{2}kT]$$

Total energy of a mole of gas $$= \frac{3}{2} \times 3 \times kT N_A R T_A = \frac{9}{2} N_A R T_A$$

For one mole, the molar specific heat at constant volume \(C_V = \frac{dU}{dT} = [\frac{3}{2}R]\)

$$C_V = [\frac{3}{2}R]$$

$$C_P = \frac{dU}{dT} + P\frac{dV}{dT} = \frac{5}{2}R$$

The ratio of specific heats, $$\gamma = \frac{C_p}{C_v} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} = 1.67 $$

ii). Diatomic Molecule

Average kinetic energy of a diatomic molecule at low temperature =\( \frac{5}{2}kT\)

Total energy of one mole of gas $$= \frac{5}{2} kT \times N_A = \frac{5}{2} RT $$ (Here, the total energy is purely kinetic)

For one mole Specific heat at constant volume $$C_v = \frac{dU}{dT} = \left[ \frac{5}{2} RT \right] = \frac{5}{2} R $$

But$$ C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$$

$$\therefore \gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7/5}{1} = 1.40$$.

Energy of a diatomic molecule at high temperature is equal to \(\frac{7}{2}\)RT $$C_v = \frac{dU}{dT} = \left[\frac{7}{2}RT\right] = \frac{7}{2}R$$.

$$\therefore C_p = C_v + R = \frac{7}{2}R + R$$

$$C_p = \frac{9}{2}R$$

Note that the CV and CP are higher for diatomic molecules than the mono atomic molecules. It implies that to increase the temperature of diatomic gas molecules by 1°C it require more heat energy than monoatomic molecules.

$$\gamma = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} = 1.28$$

iii) Triatomic molecule

a) Linear molecule

Energy of one mole= \(\frac{7}{2}kT \times N_A = \frac{7}{2}RT\)

$$C_v = \frac{dU}{dT} = \frac{d}{dT}\left[\frac{7}{2}RT\right]$$

$$C_v = \frac{7}{2}R$$

$$C_p = C_v + R = \frac{7}{2}R + R = \frac{9R}{2}$$

$$\gamma = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} = 1.28$$

b) Non-linear molecule

Energy of a mole =\(\frac{6}{2}kT \times N_A = \frac{6}{2}RT = 3RT\)

$$C_v = \frac{dU}{dT} = 3R$$

$$C_p = C_v + R = 3R + R = 4R$$

$$\gamma = \frac{C_p}{C_v} = \frac{4R}{3R} = \frac{4}{3} = 1.33$$

Note that according to kinetic theory model of gases the specific heat capacity at constant volume and constant pressure are independent of temperature. But in reality it is not sure. The specific heat capacity varies with the temperature.

EXAMPLE 9.5

Find the adiabatic exponent γ for mixture of μ1 moles of monoatomic gas and μ2 moles of a diatomic gas at normal temperature (27°C) .

Solution: The specific heat of one mole of a monoatomic gas \(C_V = \frac{3}{2}R\)

For \(\mu_1\) mole, \(C_V = \frac{3}{2} \mu_1 R\), \(C_P = \frac{5}{2} \mu_1 R\)

The specific heat of one mole of a diatomic gas $$C_V = \frac{5}{2}R$$

For \(\mu_2\) mole, \(C_V = \frac{5}{2} \mu_2 R\), \(C_P = \frac{7}{2} \mu_2 R\)

The specific heat of the mixture at constant volume $$C_V = \frac{3}{2} \mu_1 R + \frac{5}{2} \mu_2 R$$

The specific heat of the mixture at constant pressure $$C_P = \frac{5}{2} \mu_1 R + \frac{7}{2} \mu_2 R$$

The adiabatic exponent $$\gamma = \frac{C_P}{C_V} = \frac{\frac{5}{2} \mu_1 + \frac{7}{2} \mu_2}{\frac{3}{2} \mu_1 + \frac{5}{2} \mu_2}$$