Kinetic Theory — Class 11 Physics

1. Microscopic View of Matter

Kinetic theory connects the observable properties of a gas—pressure, volume and temperature—with the microscopic motion of its molecules. A gas contains a very large number of particles that move continuously and randomly and collide with one another and with the walls of the container.

The theory explains why pressure exists, why temperature is related to molecular motion, and how the ideal-gas equation can be interpreted microscopically.

2. Assumptions of the Ideal-Gas Model

• A gas contains a very large number of molecules separated by distances much larger than their molecular size.
• Molecules are treated as point particles for the idealised model.
• Molecules move randomly in all directions.
• Except during collisions, intermolecular forces are neglected.
• Collisions between molecules and with the container walls are taken to be perfectly elastic.
• The time spent in a collision is negligible compared with the time between collisions.
• The molecules obey Newtonian mechanics in the classical model.

3. Pressure from Molecular Collisions

Gas pressure results from the continuous transfer of momentum when molecules collide with the walls. For an ideal gas containing N molecules of mass m in volume V, kinetic theory gives:

P = (1/3)(Nm/V)c_rms²

Since ρ=Nm/V, this can also be written as P=(1/3)ρc_rms².

4. Relation Between Temperature and Molecular Kinetic Energy

Combining the kinetic-theory pressure equation with the ideal-gas equation PV=Nk_BT gives:

½m c_rms² = 3/2 k_BT.

Thus the average translational kinetic energy of one molecule of an ideal gas is 3/2 k_BT. Temperature therefore measures the average translational kinetic energy scale of ideal-gas molecules.

5. RMS Speed

The root-mean-square speed is defined by c_rms=√(⟨c²⟩). For an ideal gas:

c_rms=√(3k_BT/m)=√(3RT/M), where M is molar mass in kg mol⁻¹.

At the same temperature, lighter molecules have greater RMS speeds than heavier molecules.

6. Gas Laws and Molecular Interpretation

• Boyle’s law: at constant temperature, PV=constant.
• Charles’ law: at constant pressure, V∝T.
• Pressure law: at constant volume, P∝T.
• Avogadro’s law: at the same temperature and pressure, equal volumes of gases contain equal numbers of molecules.

These laws combine to form PV=nRT or, microscopically, PV=Nk_BT.

7. Boltzmann Constant

The Boltzmann constant connects the microscopic and macroscopic descriptions of temperature: k_B=R/N_A. Therefore PV=Nk_BT and PV=nRT are equivalent because N=nN_A.

8. Degrees of Freedom

The degrees of freedom of a molecule are the independent ways in which energy can be associated with its motion and, where appropriate, other modes. A monatomic gas has three translational degrees of freedom. A simple diatomic molecule at ordinary temperatures is commonly treated as having five active degrees of freedom: three translational and two rotational, while vibrational modes become important when sufficiently excited.

9. Law of Equipartition of Energy

In classical statistical mechanics, each quadratic degree of freedom contributes an average energy of ½k_BT per molecule at thermal equilibrium. If a molecule has f active quadratic degrees of freedom, its average energy is f/2 k_BT.

For n moles, the corresponding internal energy in the classical ideal-gas model is U=(f/2)nRT.

10. Specific Heat and Degrees of Freedom

For an ideal gas in the classical equipartition approximation:

C_V,m=(f/2)R

and using C_P,m−C_V,m=R,

C_P,m=((f+2)/2)R.

The ratio of heat capacities is γ=C_P/C_V=(f+2)/f.

11. Mean Free Path

The mean free path is the average distance travelled by a molecule between successive collisions. It depends on molecular size, number density and the conditions of the gas. In a simple hard-sphere model, increasing number density reduces the mean free path because collisions become more frequent.

12. Brownian Motion

Brownian motion is the irregular motion of small suspended particles caused by random collisions with molecules of the surrounding fluid. It provides experimental evidence for the molecular nature of matter.

13. Temperature Dependence of Molecular Speed

Since c_rms∝√T for a fixed gas, increasing the absolute temperature increases the typical molecular speed. Doubling Kelvin temperature increases RMS speed by a factor of √2, not 2.

Worked Numerical 1 — RMS Speed

Question: Calculate the RMS speed of nitrogen molecules at 300 K. Take R=8.314 J mol⁻¹ K⁻¹ and molar mass M=0.028 kg mol⁻¹.

c_rms=√(3RT/M)=√[3(8.314)(300)/0.028]≈517 m/s.

Worked Numerical 2 — Average Translational Energy

Question: Find the average translational kinetic energy of one ideal-gas molecule at 300 K. Take k_B=1.38×10⁻²³ J/K.

E=3/2 k_BT=1.5(1.38×10⁻²³)(300)=6.21×10⁻²¹ J.

Worked Numerical 3 — Pressure from RMS Speed

Question: A gas has density 1.2 kg/m³ and RMS molecular speed 500 m/s. Find its pressure using kinetic theory.

P=(1/3)ρc_rms²=(1/3)(1.2)(500)²=1.0×10⁵ Pa.

Worked Numerical 4 — Temperature from RMS Speed

Question: The RMS speed of a gas is 400 m/s. If its molar mass is 0.032 kg/mol, estimate its temperature. Take R=8.314 J mol⁻¹ K⁻¹.

From c_rms²=3RT/M, T=Mc_rms²/(3R).

T=(0.032)(400)²/[3(8.314)]≈205 K.

Worked Numerical 5 — Effect of Temperature on RMS Speed

Question: A gas has RMS speed 300 m/s at 300 K. What is its RMS speed at 1200 K, assuming the gas is unchanged?

c_rms∝√T. Therefore c₂/c₁=√(1200/300)=2, so c₂=600 m/s.

Worked Numerical 6 — Degrees of Freedom

Question: In the classical approximation, a gas molecule has five active degrees of freedom. Find its molar internal energy at 300 K. Take R=8.314 J mol⁻¹ K⁻¹.

U_m=(f/2)RT=(5/2)(8.314)(300)=6235.5 J/mol.

Worked Numerical 7 — Heat Capacities

Question: For an ideal gas with five active degrees of freedom, find its molar C_V and C_P.

C_V=(5/2)R≈20.8 J mol⁻¹ K⁻¹.