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Kinetic Theory - Degrees of Freedom and Law of Equipartition of Energy

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Degrees of Freedom (ff): It is defined as the total number of independent coordinates or independent ways in which a system can possess energy. For a molecule with NN atoms and kk constraints, f=3N−kf = 3N - k.

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Law of Equipartition of Energy: In thermal equilibrium at temperature TT, the total energy of a dynamical system is equally distributed among its various degrees of freedom. The energy associated with each degree of freedom per molecule is 12kBT\frac{1}{2} k_B T, where kBk_B is the Boltzmann constant.

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Monoatomic Gases: These gases (e.g., HeHe, ArAr) have only 3 translational degrees of freedom, so f=3f = 3. Total internal energy per molecule is 32kBT\frac{3}{2} k_B T.

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Diatomic Gases: At room temperature, gases like O2O_2 or N2N_2 have 3 translational and 2 rotational degrees of freedom, so f=5f = 5. At very high temperatures, vibrational modes are also excited, making f=7f = 7.

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Polyatomic Gases: For a non-linear polyatomic molecule (e.g., H2OH_2O), there are 3 translational and 3 rotational degrees of freedom, hence f=6f = 6 (ignoring vibration).

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Specific Heat Capacities: The molar specific heat at constant volume (CvC_v) and constant pressure (CpC_p) are directly related to the degrees of freedom ff.

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Internal Energy (UU): For nn moles of an ideal gas, the internal energy is given by U=f2nRTU = \frac{f}{2} nRT.

📐Formulae

Eavg=12kBT (per degree of freedom per molecule)E_{avg} = \frac{1}{2} k_B T \text{ (per degree of freedom per molecule)}

U=f2nRT (Internal energy for n moles)U = \frac{f}{2} nRT \text{ (Internal energy for } n \text{ moles)}

Cv=dUdT=f2RC_v = \frac{dU}{dT} = \frac{f}{2} R

Cp=Cv+R=(f2+1)RC_p = C_v + R = \left( \frac{f}{2} + 1 \right) R

γ=CpCv=1+2f\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}

f=3 (Monoatomic), f=5 (Diatomic at room temp)f = 3 \text{ (Monoatomic), } f = 5 \text{ (Diatomic at room temp)}

💡Examples

Problem 1:

Calculate the internal energy of 2 moles2 \text{ moles} of Oxygen (O2O_2) gas at temperature 300K300 K. Assume the gas behaves ideally and neglect vibrational modes.

Solution:

Oxygen is a diatomic gas, so its degrees of freedom f=5f = 5. Using the formula for internal energy: U=f2nRTU = \frac{f}{2} nRT Substituting the values: U=52×2×8.314×300U = \frac{5}{2} \times 2 \times 8.314 \times 300 U=5×8.314×300U = 5 \times 8.314 \times 300 U=12471JU = 12471 J

Explanation:

Since vibrational modes are neglected and O2O_2 is diatomic, we use f=5f=5. The internal energy depends only on the temperature and degrees of freedom for an ideal gas.

Problem 2:

Find the ratio of specific heats (γ\gamma) for a rigid linear triatomic molecule.

Solution:

A rigid linear triatomic molecule (like CO2CO_2) has 3 translational and 2 rotational degrees of freedom, similar to a diatomic molecule, so f=5f = 5. Using the formula for γ\gamma: γ=1+2f\gamma = 1 + \frac{2}{f} γ=1+25\gamma = 1 + \frac{2}{5} γ=75=1.4\gamma = \frac{7}{5} = 1.4

Explanation:

A 'rigid' molecule implies we ignore vibrational degrees of freedom. A linear molecule only has 2 axes of rotation perpendicular to the internuclear axis, hence f=5f=5.

Degrees of Freedom and Law of Equipartition of Energy Revision - Class 11 Physics CBSE