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.

Spec mapping (AQA 7408 §3.6.2.3 — Kinetic energy and temperature): This lesson covers the relationship (1/2)m〈c²〉 = (3/2)kT linking average molecular translational kinetic energy to absolute temperature, the corresponding total internal-energy expression for a monatomic ideal gas U = (3/2)NkT, the calculation of root-mean-square speed c_rms = √(3kT/m), and a qualitative treatment of the Maxwell–Boltzmann distribution of molecular speeds and how it changes with temperature. (Refer to the official AQA specification document for exact wording.)

Synoptic links: This lesson is the synthesis of the previous two — equating kinetic theory (§3.6.2.3) pV = (1/3)Nm〈c²〉 with the ideal gas equation (§3.6.2.2) pV = NkT yields (1/2)m〈c²〉 = (3/2)kT. It defines the operational meaning of temperature in microscopic terms and is the bridge from gas behaviour to statistical mechanics. The Maxwell–Boltzmann distribution connects forward to undergraduate kinetic theory and to chemistry's activation-energy arguments (Arrhenius equation). Synoptic exam routes include diffusion rates (Graham's law: lighter molecules have higher c_rms at the same T), and the explanation of why hydrogen escaped Earth's atmosphere but argon did not (escape-velocity vs c_rms at the relevant altitude).

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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)

T = (4.65 × 10⁻²⁰) / (4.14 × 10⁻²³)

T = 1123 K ≈ 850 °C

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The Maxwell-Boltzmann Distribution

Not all molecules in a gas move at the same speed. At any given temperature, there is a continuous distribution of speeds, described by the Maxwell-Boltzmann distribution.

Key Features of the Distribution Curve

1. The curve starts at the origin — no molecules have zero speed.
2. The curve rises to a peak and then falls off gradually with a long tail at high speeds.
3. The distribution is asymmetric — it has a positive skew (tail extends to the right).
4. Three characteristic speeds can be identified:
   • Most probable speed (v_p) — at the peak of the curve
   • Mean speed (v̄) — slightly higher than v_p
   • r.m.s. speed (c_rms) — higher still

For an ideal gas: v_p < v̄ < c_rms

Effect of Temperature on the Distribution

When the temperature increases:

1. The peak shifts to the right (higher speeds) — the most probable speed increases.
2. The peak becomes lower — fewer molecules at the most probable speed.
3. The curve becomes broader — a wider spread of speeds.
4. The tail at high speeds extends further — more molecules have very high speeds.
5. The area under the curve remains the same — the total number of molecules has not changed.

Effect of Molecular Mass on the Distribution

At the same temperature, lighter molecules have a broader distribution shifted to higher speeds, while heavier molecules have a narrower distribution centred at lower speeds. This is because c_rms ∝ 1/√m.

Exam Tip: When sketching two Maxwell-Boltzmann curves at different temperatures on the same axes, you MUST show: (1) both start at the origin; (2) the higher-T curve has a lower, broader peak shifted to the right; (3) the areas under both curves are equal. Examiners check all three features. Labelling the axes as "number of molecules" (or "probability density") vs "molecular speed" is also important.

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Degrees of Freedom and Equipartition

Although not always explicitly required by AQA, it is useful to know that the ³⁄₂kT result applies specifically to the translational kinetic energy of molecules.

The equipartition theorem states that each degree of freedom contributes ½kT to the mean energy of a molecule.

A monatomic ideal gas (e.g. He, Ne, Ar) has 3 translational degrees of freedom: movement in x, y, and z directions. Total energy = 3 × ½kT = ³⁄₂kT.

A diatomic gas (e.g. N₂, O₂, H₂) has:

• 3 translational degrees of freedom
• 2 rotational degrees of freedom (about two axes perpendicular to the bond axis)
• Total energy at moderate temperatures = 5 × ½kT = ⁵⁄₂kT

At very high temperatures, vibrational modes also become active, adding 2 more degrees of freedom (one kinetic, one potential), giving a total of ⁷⁄₂kT.

Type of Gas	Degrees of Freedom	Mean Energy per Molecule
Monatomic (He, Ne, Ar)	3 (translational)	³⁄₂kT
Diatomic at moderate T (N₂, O₂)	5 (3 trans + 2 rot)	⁵⁄₂kT
Diatomic at high T	7 (3 trans + 2 rot + 2 vib)	⁷⁄₂kT

Exam Tip: AQA A-Level only explicitly requires ½m⟨c²⟩ = ³⁄₂kT for the mean translational kinetic energy. However, understanding degrees of freedom can help you explain why the specific heat capacities of different gases differ, which is a potential synoptic question.

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Common Misconceptions

1. "At a given temperature, heavier molecules have more kinetic energy." No — at the same temperature, all molecules have the same mean translational kinetic energy (³⁄₂kT). Heavier molecules are simply slower.

2. "Doubling the temperature doubles the r.m.s. speed." No — since c_rms ∝ √T, doubling T increases c_rms by √2 ≈ 1.41. You need to quadruple T to double c_rms.

3. "The Maxwell-Boltzmann peak represents the average speed." The peak is the most probable speed, which is less than the mean speed and less than the r.m.s. speed.

4. "Increasing temperature shifts the curve upward." The peak actually moves down and to the right. The curve gets broader and flatter, but the total area remains constant.

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Additional Worked Example — Comparing Three Gases at the Same Temperature

A sealed container holds equal numbers of three gas species at a temperature of 500 K: hydrogen molecules (H₂, m = 3.32 × 10⁻²⁷ kg), oxygen molecules (O₂, m = 5.32 × 10⁻²⁶ kg), and xenon atoms (Xe, m = 2.18 × 10⁻²⁵ kg). Calculate the mean translational kinetic energy and root-mean-square speed for each species, and comment on what would happen if the container had a tiny pinhole.

Step 1 — mean translational kinetic energy. This is determined by temperature alone, independent of mass: 〈KE〉 = (3/2)kT = (3/2)(1.38 × 10⁻²³)(500) = 1.035 × 10⁻²⁰ J per molecule.

All three species — H₂, O₂, Xe — have exactly the same mean translational KE. This is a counter-intuitive but rigorous consequence of the equipartition theorem.

Step 2 — root-mean-square speeds. c_rms = √(3kT/m). Substituting T = 500 K:

For H₂: c_rms = √[(3 × 1.38 × 10⁻²³ × 500)/(3.32 × 10⁻²⁷)] = √(6.235 × 10⁶) = 2497 m s⁻¹ (≈ 2.5 km/s).

For O₂: c_rms = √[(3 × 1.38 × 10⁻²³ × 500)/(5.32 × 10⁻²⁶)] = √(3.891 × 10⁵) = 624 m s⁻¹.

For Xe: c_rms = √[(3 × 1.38 × 10⁻²³ × 500)/(2.18 × 10⁻²⁵)] = √(9.495 × 10⁴) = 308 m s⁻¹.

Step 3 — effusion (pinhole leak). The rate at which a gas effuses through a pinhole is proportional to the mean speed of the molecules, which scales as c_rms. So for the same temperature: Rate_H₂ : Rate_O₂ : Rate_Xe = 2497 : 624 : 308 ≈ 8.1 : 2.0 : 1.0.

Hydrogen escapes about 4 times faster than oxygen and 8 times faster than xenon. This is Graham's law of effusion: rate ∝ 1/√M. Over time, the container would become depleted of light gases first, leaving an increasingly xenon-rich mixture inside. This is the principle behind isotope separation of uranium hexafluoride for nuclear reactors (UF₆ with U-235 effuses 0.43% faster than UF₆ with U-238 — slow but exploitable in cascades).