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Kinetic Theory - Kinetic Theory of an Ideal Gas

Grade 11CBSEPhysics

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

🔑Concepts

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Postulates of Kinetic Theory: A gas consists of a large number of identical, tiny, spherical particles (atoms or molecules) in constant random motion. The volume of the molecules is negligible compared to the volume of the gas container.

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Elastic Collisions: Molecular collisions among themselves and with the walls of the container are perfectly elastic. No energy is lost; only momentum is transferred.

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Pressure Interpretation: Pressure PP is exerted by the gas due to the continuous bombardment of molecules against the walls of the container. It is defined as P=13ρvrms2P = \frac{1}{3} \rho v_{rms}^2.

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Kinetic Interpretation of Temperature: The average kinetic energy of a molecule is directly proportional to the absolute temperature TT of the gas. At 0K0 K, the molecular motion ceases.

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Degrees of Freedom (ff): The number of independent ways in which a molecule can possess energy. For Monatomic gases f=3f=3, for Diatomic gases f=5f=5 (at room temperature), and for Polyatomic gases f=6f=6.

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Law of Equipartition of Energy: The total energy of a system in thermal equilibrium is equally divided among all its degrees of freedom, and the energy associated with each degree of freedom per molecule is 12kBT\frac{1}{2} k_B T.

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Mean Free Path (λ\lambda): The average distance traveled by a molecule between two successive collisions. It is inversely proportional to the number density and the square of the molecular diameter.

📐Formulae

P=13Nmvˉ2V=13ρvrms2P = \frac{1}{3} \frac{Nm\bar{v}^2}{V} = \frac{1}{3} \rho v_{rms}^2

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

vav=8RTπMv_{av} = \sqrt{\frac{8RT}{\pi M}}

vmp=2RTMv_{mp} = \sqrt{\frac{2RT}{M}}

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

Cv=f2R,Cp=(f2+1)RC_v = \frac{f}{2}R, \quad C_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 Helium atoms at 27∘C27^\circ C. (Given R=8.31 J mol−1K−1R = 8.31 \, J \, mol^{-1} K^{-1} and atomic mass of He=4 uHe = 4 \, u)

Solution:

Step 1: Convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300 \, K. Step 2: Molar mass of Helium M=4×10−3 kg/molM = 4 \times 10^{-3} \, kg/mol. Step 3: Use the formula vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}. vrms=3×8.31×3004×10−3v_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{4 \times 10^{-3}}} vrms=74790.004=1869750≈1367.39 m/sv_{rms} = \sqrt{\frac{7479}{0.004}} = \sqrt{1869750} \approx 1367.39 \, m/s.

Explanation:

The RMS speed is determined by the absolute temperature and the molar mass. Since Helium is light, its RMS speed is quite high even at room temperature.

Problem 2:

Determine the total internal energy of 22 moles of an ideal diatomic gas at 300 K300 \, K.

Solution:

Step 1: For a diatomic gas at room temperature, degrees of freedom f=5f = 5. Step 2: Total internal energy U=nf2RTU = n \frac{f}{2} RT. Step 3: Substitute n=2n = 2, f=5f = 5, R=8.31R = 8.31, and T=300T = 300. U=2×52×8.31×300U = 2 \times \frac{5}{2} \times 8.31 \times 300 U=5×8.31×300U = 5 \times 8.31 \times 300 41.55×30012465\begin{array}{r} 41.55 \\ \times 300 \\ \hline 12465 \end{array} U=12465 JU = 12465 \, J.

Explanation:

Internal energy of an ideal gas depends only on its temperature and degrees of freedom. For diatomic molecules, we consider 3 translational and 2 rotational degrees of freedom.