Kinetic Theory

Kinetic Theory of Gases

The kinetic theory explains the macroscopic properties of gases (pressure, temperature, volume) in terms of the microscopic behaviour of their constituent molecules.

Assumptions of Kinetic Theory

Gases consist of a very large number of molecules in random motion.
Molecules are point-like; volume of molecules is negligible compared to total volume.
Molecules exert no forces on each other except during elastic collisions.
Collisions are perfectly elastic (kinetic energy conserved).
Time of collision is negligible compared to time between collisions.
All directions of motion are equally probable.

Pressure of an Ideal Gas

The pressure exerted by an ideal gas on container walls arises from molecular collisions. Derived result:
P = (1/3) rho v_rms² = (1/3) (n m / V) v_rms²

where n = number of molecules, m = mass of one molecule, v_rms = root mean square speed.

Relation to ideal gas law: PV = nRT (n = number of moles, R = 8.314 J/mol/K)
Also PV = N k_B T (N = number of molecules, k_B = Boltzmann constant = 1.38 x 10^-23 J/K)
R = N_A k_B (N_A = Avogadro number = 6.022 x 10²³)

Kinetic Energy and Temperature

Average kinetic energy per molecule: (1/2) m v_rms² = (3/2) k_B T
Total internal energy of ideal gas: U = (3/2) N k_B T = (3/2) n R T (for monatomic gas)

This shows temperature is a measure of average kinetic energy of molecules.

Speed Distribution

Root mean square speed: v_rms = √(3RT/M) = √(3k_BT/m)
Mean speed: v_mean = √(8RT/(pi M))
Most probable speed: v_p = √(2RT/M)

Order: v_p < v_mean < v_rms

Law of Equipartition of Energy

Each degree of freedom contributes (1/2) k_B T of energy per molecule (average).
Monatomic gas: 3 translational DOF → U = (3/2) k_B T per molecule
Diatomic gas (rigid): 3 translational + 2 rotational DOF → U = (5/2) k_B T per molecule
Diatomic gas (with vibration): 7 DOF → U = (7/2) k_B T per molecule

This explains C_v: monatomic C_v = (3/2)R, diatomic C_v = (5/2)R.

Mean Free Path

Average distance a molecule travels between successive collisions:
lambda = 1 / (√(2) x pi x d² x n)
where d = diameter of molecule, n = number density (N/V).

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Example 1

Find average kinetic energy per molecule of oxygen at 27 C.
T = 27 + 273 = 300 K; KE = (3/2) k_B T = (3/2) x 1.38 x 10^-23 x 300 = 6.21 x 10^-21 J

Example 2

Find v_rms for hydrogen (M = 2 x 10^-3 kg/mol) at 300 K.
v_rms = √(3RT/M) = √(3 x 8.314 x 300 / (2 x 10^-3)) = √(3 x 8.314 x 300 / 0.002) = √(3742300) = 1934 m/s

Example 3

Calculate v_rms for oxygen (M = 32 x 10^-3 kg/mol) at 300 K.
v_rms = √(3 x 8.314 x 300 / 0.032) = √(233831) = 483 m/s

Example 4

How many degrees of freedom does a diatomic molecule have at room temperature?
At room temperature, vibrational modes are generally not excited; 3 translational + 2 rotational = 5 degrees of freedom.

Example 5

Find total internal energy of 1 mol of a monatomic ideal gas at 300 K.
U = (3/2) n R T = (3/2) x 1 x 8.314 x 300 = 3741.3 J

Example 6

At what temperature will the v_rms of oxygen equal v_rms of hydrogen at 300 K?
v_rms proportional to √(T/M). For equal speeds: T_O2/M_O2 = T_H2/M_H2
T_O2 = T_H2 x (M_O2/M_H2) = 300 x (32/2) = 300 x 16 = 4800 K

Example 7

Compare v_rms, v_mean, and v_p in terms of √(RT/M).
v_p = √(2RT/M); v_mean = √(8RT/piM) = 1.596 √(RT/M); v_rms = √(3RT/M) = 1.732 √(RT/M)
Order: v_p < v_mean < v_rms.

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Common mistakes

v_rms is NOT the same as the average speed — v_rms = √(3RT/M), mean speed = √(8RT/piM).
Degrees of freedom: at room temperature, diatomic molecules have only 5 (not 7; vibrational modes are frozen).
Mean free path decreases as number density increases — more molecules means more collisions.

Summary

Kinetic theory derives macroscopic gas properties from molecular motion. Temperature measures average molecular kinetic energy. The equipartition theorem distributes energy equally among degrees of freedom, explaining specific heats. Mean free path characterises average distance between collisions.