Kinetic Energy and Temperature

This lesson establishes the fundamental link between the kinetic energy of gas molecules and the thermodynamic temperature. It also introduces the Maxwell-Boltzmann distribution of molecular speeds, which describes how molecular speeds are distributed in a gas at a given temperature.

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Deriving the Kinetic Energy–Temperature Relationship

By combining the kinetic theory result with the ideal gas equation, we can derive the relationship between kinetic energy and temperature.

From kinetic theory: pV = ⅓Nm⟨c²⟩

From the ideal gas equation: pV = NkT

Setting these equal:

⅓Nm⟨c²⟩ = NkT

Dividing both sides by N:

⅓m⟨c²⟩ = kT

Multiplying both sides by ³⁄₂:

½m⟨c²⟩ = ³⁄₂kT

This is one of the most important equations in thermal physics.

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Interpreting the Equation

The left-hand side, ½m⟨c²⟩, is the mean translational kinetic energy of a single molecule. The equation tells us that:

1. The mean kinetic energy of a molecule is directly proportional to the absolute temperature.
2. At a given temperature, all ideal gas molecules have the same mean kinetic energy, regardless of their mass. Heavier molecules move more slowly; lighter molecules move faster.
3. Doubling the absolute temperature doubles the mean kinetic energy.
4. At absolute zero (T = 0 K), the mean kinetic energy is zero — molecules have no thermal motion.

Key Definition: Temperature is a measure of the average translational kinetic energy of the molecules in a substance.

Dependence of r.m.s. Speed on Temperature

From ½m⟨c²⟩ = ³⁄₂kT:

⟨c²⟩ = 3kT/m

c_rms = √(3kT/m)

Since c_rms ∝ √T:

• Doubling the temperature increases c_rms by a factor of √2 ≈ 1.41.
• To double c_rms, you must quadruple the temperature.

If temperature...	Then mean KE...	Then c_rms...
Doubles (×2)	Doubles (×2)	Increases by ×√2
Triples (×3)	Triples (×3)	Increases by ×√3
Quadruples (×4)	Quadruples (×4)	Doubles (×2)
Halves (×½)	Halves (×½)	Decreases by ×1/√2

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Worked Examples

Worked Example 1 — Mean Kinetic Energy

Question: Calculate the mean kinetic energy of a gas molecule at 20 °C. (k = 1.38 × 10⁻²³ J K⁻¹)

Solution:

T = 20 + 273 = 293 K

E_k = ³⁄₂kT = 1.5 × 1.38 × 10⁻²³ × 293

E_k = 1.5 × 4.043 × 10⁻²¹

E_k = 6.07 × 10⁻²¹ J

This is a tiny amount of energy for a single molecule, but multiplied by Avogadro's number of molecules it becomes significant.

Worked Example 2 — Comparing Speeds of Different Gases

Question: At the same temperature, calculate the ratio of the r.m.s. speed of hydrogen molecules (H₂, M = 2.0 × 10⁻³ kg mol⁻¹) to that of oxygen molecules (O₂, M = 32 × 10⁻³ kg mol⁻¹).

Solution:

c_rms = √(3kT/m), and since at the same temperature, 3kT is the same for both:

c_rms(H₂) / c_rms(O₂) = √(m_O₂ / m_H₂)

The molecular masses are in the ratio M_O₂/M_H₂ = 32/2 = 16, so:

c_rms(H₂) / c_rms(O₂) = √16 = 4

Hydrogen molecules move 4 times faster than oxygen molecules at the same temperature. This explains why hydrogen escapes from the Earth's atmosphere — many hydrogen molecules have speeds exceeding the escape velocity.

Worked Example 3 — Total Kinetic Energy of a Gas

Question: Calculate the total translational kinetic energy of the molecules in 1.0 mol of an ideal gas at 300 K.

Solution:

Total KE = N × ³⁄₂kT = nNₐ × ³⁄₂kT = n × ³⁄₂NₐkT = n × ³⁄₂RT

Total KE = 1.0 × 1.5 × 8.31 × 300 = 3740 J ≈ 3.7 kJ

Note: The total translational KE of an ideal gas depends only on n and T, not on the type of gas.

Worked Example 4 — Temperature Required for a Given Speed

Question: At what temperature would nitrogen molecules (m = 4.65 × 10⁻²⁶ kg) have an r.m.s. speed of 1000 m s⁻¹?

Solution:

From c_rms = √(3kT/m), squaring:

c_rms² = 3kT/m

T = mc_rms² / (3k) = (4.65 × 10⁻²⁶ × 1000²) / (3 × 1.38 × 10⁻²³)