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Behavior of Perfect Gas and Kinetic Theory - Degrees of Freedom

Grade 11ICSEPhysics

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

🔑Concepts

Degrees of Freedom (ff) is defined as the total number of independent coordinates or independent ways in which a system can possess energy.

For a system containing NN particles with kk independent relations/constraints between them, the degrees of freedom is given by f=3Nkf = 3N - k.

Monoatomic gases (e.g., HeHe, NeNe, ArAr) have only 3 translational degrees of freedom (f=3f = 3).

Diatomic gases (e.g., O2O_2, H2H_2, COCO) at moderate temperatures have 3 translational and 2 rotational degrees of freedom, giving f=5f = 5. At very high temperatures, 2 additional vibrational degrees of freedom are added.

Polyatomic non-linear molecules (e.g., H2OH_2O, NH3NH_3) have 3 translational and 3 rotational degrees of freedom, totaling f=6f = 6.

The Law of Equipartition of Energy states that for any system in thermal equilibrium, the total energy is equally distributed among its various degrees of freedom, and each degree of freedom contributes an average energy of 12kBT\frac{1}{2} k_B T.

The internal energy (UU) of nn moles of an ideal gas is related to the degrees of freedom by the expression U=f2nRTU = \frac{f}{2} nRT.

📐Formulae

f=3Nkf = 3N - k

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

U=f2nRTU = \frac{f}{2} nRT

Cv=f2RC_v = \frac{f}{2} R

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

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

💡Examples

Problem 1:

Calculate the ratio of specific heats (γ\gamma) for a rigid diatomic gas molecule.

Solution:

For a rigid diatomic molecule, there are 3 translational and 2 rotational degrees of freedom. Thus, f=5f = 5. Using the formula 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 diatomic molecule has a fixed bond length, meaning it does not vibrate. Therefore, we only sum the translational (33) and rotational (22) degrees of freedom to find ff.

Problem 2:

Find the total internal energy of 22 moles of a monoatomic gas at temperature TT.

Solution:

For a monoatomic gas, the number of degrees of freedom is f=3f = 3. The formula for internal energy is U=f2nRTU = \frac{f}{2} nRT. Substituting f=3f = 3 and n=2n = 2: U=32×2×RTU = \frac{3}{2} \times 2 \times RT U=3RTU = 3RT

Explanation:

Each mole of a monoatomic gas has 32RT\frac{3}{2} RT internal energy due to its 3 translational degrees of freedom. For 2 moles, the energy is doubled.

Degrees of Freedom - Revision Notes & Key Formulas | ICSE Class 11 Physics