krit.club logo

Kinetic Theory - Molecular Nature of Matter

Grade 11CBSEPhysics

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

🔑Concepts

•

Molecular Nature of Matter: Matter is made up of atoms and molecules. In gases, the molecules are in constant, random motion. Dalton's atomic theory and Avogadro's law provide the foundation for this understanding.

•

Ideal Gas Postulates: (1) Gas consists of very small particles called molecules. (2) Molecules are in continuous random motion. (3) The volume of molecules is negligible compared to the volume of the container. (4) There are no intermolecular forces except during collisions. (5) Collisions are perfectly elastic.

•

Pressure of an Ideal Gas: Pressure is exerted due to the momentum transfer during collisions of gas molecules with the walls of the container. It is given by P=13ρv2‾P = \frac{1}{3} \rho \overline{v^2}.

•

Kinetic Interpretation of Temperature: The average kinetic energy of a molecule is directly proportional to the absolute temperature TT. The relation is E=32kBTE = \frac{3}{2} k_B T.

•

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

•

Degrees of Freedom: The number of independent ways in which a system can possess energy. For a monoatomic gas, f=3f = 3; for a diatomic gas (at room temperature), f=5f = 5.

•

Mean Free Path: The average distance traveled by a molecule between two successive collisions, denoted by λ\lambda.

📐Formulae

PV=nRTPV = nRT

P=13Nmv2‾V=13ρv2‾P = \frac{1}{3} \frac{N m \overline{v^2}}{V} = \frac{1}{3} \rho \overline{v^2}

vrms=3kBTm=3RTMv_{rms} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 RT}{M}}

Eavg=32kBTE_{avg} = \frac{3}{2} k_B T

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}

λ=12πd2n\lambda = \frac{1}{\sqrt{2} \pi d^2 n}

💡Examples

Problem 1:

Calculate the root mean square (rms) speed of Nitrogen molecules (M=28 g/molM = 28 \text{ g/mol}) at a temperature of 27∘C27^\circ C. Given R=8.31 J mol−1 K−1R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1}.

Solution:

First, convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300 \text{ K} Convert molar mass to kg/mol: M=28×10−3 kg/molM = 28 \times 10^{-3} \text{ kg/mol} Use the formula for RMS speed: vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}} vrms=3×8.31×30028×10−3v_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{28 \times 10^{-3}}} vrms=74790.028≈267107.14≈516.8 m/sv_{rms} = \sqrt{\frac{7479}{0.028}} \approx \sqrt{267107.14} \approx 516.8 \text{ m/s}

Explanation:

The RMS speed is determined by the absolute temperature and the molar mass of the gas. Nitrogen, being a diatomic molecule, follows the standard kinetic theory speed distribution.

Problem 2:

Determine the total internal energy of 22 moles of an ideal monoatomic gas at 300 K300 \text{ K}.

Solution:

For a monoatomic gas, degrees of freedom f=3f = 3. The internal energy UU is given by: U=nf2RTU = n \frac{f}{2} RT Substitute the values: n=2,f=3,R=8.31,T=300n = 2, f = 3, R = 8.31, T = 300 U=2×32×8.31×300U = 2 \times \frac{3}{2} \times 8.31 \times 300 U=3×8.31×300=7479 JU = 3 \times 8.31 \times 300 = 7479 \text{ J}

Explanation:

According to the law of equipartition of energy, each degree of freedom contributes 12RT\frac{1}{2} RT per mole. Since a monoatomic gas has 3 translational degrees of freedom, the total energy is 32nRT\frac{3}{2} nRT.