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Kinetic Theory - Behaviour of Gases

Grade 11CBSEPhysics

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

🔑Concepts

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An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. It obeys the ideal gas law: PV=nRTPV = nRT.

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Pressure of an ideal gas is derived from the momentum transfer during collisions of gas molecules with the walls of the container, expressed as P=13ρvˉ2P = \frac{1}{3} \rho \bar{v}^2.

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The average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature TT of the gas, given by E=32kBTE = \frac{3}{2} k_B T, where kBk_B is the Boltzmann constant.

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The Law of Equipartition of Energy states that for any dynamic system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom, and the energy associated with each degree of freedom is 12kBT\frac{1}{2} k_B T.

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Degrees of freedom (ff) represent the number of independent ways a molecule can possess energy. For monoatomic gases f=3f=3, for diatomic gases f=5f=5 (at room temperature), and for non-linear polyatomic gases f=6f=6.

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Specific heat capacity at constant volume CvC_v and constant pressure CpC_p are related by Mayer's formula: Cp−Cv=RC_p - C_v = R.

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The mean free path λ\lambda is the average distance traveled by a molecule between two successive collisions, inversely proportional to the number density and the square of the molecular diameter.

📐Formulae

PV=nRT=NkBTPV = nRT = Nk_B T

P=13MVvrms2=13ρvrms2P = \frac{1}{3} \frac{M}{V} v_{rms}^2 = \frac{1}{3} \rho v_{rms}^2

vrms=3RTM=3kBTmv_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_B T}{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 speed of Nitrogen molecules at 27∘C27^{\circ}C. (Given: Molar mass of N2=28 g/molN_2 = 28 \text{ g/mol}, R=8.314 J mol−1 K−1R = 8.314 \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: vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}} vrms=3×8.314×30028×10−3v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{28 \times 10^{-3}}} vrms≈267235.7≈516.95 m/sv_{rms} \approx \sqrt{267235.7} \approx 516.95 \text{ m/s}

Explanation:

The vrmsv_{rms} formula relates the macroscopic temperature to the microscopic speed of the gas molecules. The units must be in SI (Kelvin for temperature and kg/mol for molar mass).

Problem 2:

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

Solution:

For a rigid diatomic gas, the degrees of freedom consist of 3 translational and 2 rotational modes: f=3+2=5f = 3 + 2 = 5 Using the formula for specific heat at constant volume: Cv=52RC_v = \frac{5}{2} R Using Mayer's relation: Cp=Cv+R=52R+R=72RC_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R The ratio is: γ=CpCv=7/2R5/2R=75=1.4\gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4

Explanation:

According to the law of equipartition of energy, each degree of freedom contributes 12R\frac{1}{2} R to CvC_v. A rigid diatomic molecule has 5 degrees of freedom at normal temperatures.