Thermal Physics

Thermal physics connects the macroscopic properties of gases (pressure, volume, temperature) with the microscopic behaviour of individual molecules. It also deals with how energy is stored and transferred within materials. This topic is central to A-Level Physics and appears in AQA Paper 2 and OCR Paper 2. A strong understanding of thermal physics requires fluency in both the mathematical relationships (ideal gas laws, kinetic theory) and the physical reasoning behind changes of state, energy transfer, and the molecular model of matter.

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Internal Energy

Key Definition: The internal energy of a system is the sum of the randomly distributed kinetic energies and potential energies of all its molecules.

• The kinetic energy component is associated with the random motion (translational, rotational, vibrational) of the molecules. In a monatomic ideal gas, only translational kinetic energy is present.
• The potential energy component is associated with the intermolecular forces between the molecules. In solids and liquids, the potential energy contribution is significant because molecules are close enough to interact strongly.

For an ideal gas, there are no intermolecular forces (except during collisions), so the internal energy is purely kinetic. Changing the temperature of an ideal gas changes only its kinetic energy.

When a substance changes state (e.g. melting or boiling), the internal energy increases even though the temperature remains constant. The additional energy goes into breaking or weakening intermolecular bonds (increasing potential energy), not into increasing kinetic energy. This is a critically important distinction that examiners test frequently.

Exam Tip: If asked what happens to internal energy during a change of state, always state explicitly that temperature remains constant, kinetic energy stays the same, and only the potential energy component of internal energy increases. Many candidates lose marks by failing to mention that kinetic energy does not change.

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Specific Heat Capacity

Key Definition: The specific heat capacity (c) of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 K (or 1 °C), without a change of state.

The equation relating energy, mass, and temperature change is:

E = mcΔθ

where E is the energy transferred (J), m is the mass (kg), c is the specific heat capacity (J kg⁻¹ K⁻¹), and Δθ is the temperature change (K or °C).

Some important specific heat capacity values:

Substance	Specific Heat Capacity (J kg⁻¹ K⁻¹)
Water	4200
Ice	2100
Aluminium	900
Copper	390
Lead	130

Water has a particularly high specific heat capacity, which is why it is used widely in central heating systems and why coastal regions experience milder temperature variations than inland areas.

Worked Example — Specific Heat Capacity Calculation

Question: A 2.0 kg aluminium block is heated from 20 °C to 85 °C using a 50 W heater that runs for 4 minutes. Calculate the specific heat capacity of aluminium, assuming no energy losses.

Solution:

Step 1: Calculate the energy supplied. E = Pt = 50 × (4 × 60) = 50 × 240 = 12 000 J

Step 2: Calculate the temperature change. Δθ = 85 − 20 = 65 K

Step 3: Rearrange E = mcΔθ for c. c = E / (mΔθ) = 12 000 / (2.0 × 65) = 12 000 / 130 = 92.3 J kg⁻¹ K⁻¹

Note: The true value is about 900 J kg⁻¹ K⁻¹. The significant discrepancy here would indicate substantial energy losses to the surroundings in a real experiment — a common discussion point in practical assessments.

Exam Tip: In practical questions about measuring specific heat capacity, always discuss sources of error. The main systematic error is energy loss to the surroundings, which causes the measured value of c to be too low (since not all the electrical energy goes into heating the block). Insulation around the block reduces this error.

Continuous-Flow Calorimetry

In the continuous-flow method (used to measure the specific heat capacity of a fluid), the fluid flows at a steady rate past a heater. Two measurements are taken at different flow rates but the same temperature change. By subtracting the two energy equations, the heat loss term cancels, giving a more accurate result for c.

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Specific Latent Heat

Key Definition: The specific latent heat (L) of a substance is the energy required to change the state of 1 kg of the substance at constant temperature.

E = mL

There are two types:

• Specific latent heat of fusion (Lf) — energy per kg for melting or freezing (e.g. ice: 3.34 × 10⁵ J kg⁻¹)
• Specific latent heat of vaporisation (Lv) — energy per kg for boiling or condensing (e.g. water: 2.26 × 10⁶ J kg⁻¹)

stateDiagram-v2
    [*] --> Solid
    Solid --> Liquid : Melting (energy = mLf)
    Liquid --> Solid : Freezing (energy released = mLf)
    Liquid --> Gas : Boiling/Evaporation (energy = mLv)
    Gas --> Liquid : Condensation (energy released = mLv)
    Solid --> Gas : Sublimation
    Gas --> Solid : Deposition
    note right of Solid : T constant during<br/>state changes.<br/>KE unchanged,<br/>PE increases.

The latent heat of vaporisation is always greater than the latent heat of fusion because the molecules must be completely separated from one another (overcoming all remaining intermolecular forces), which requires more energy than merely loosening the regular arrangement in a solid.

Heating and Cooling Curves

A heating curve (temperature-time graph) for a substance heated at a constant rate shows a characteristic shape:

1. Rising section (solid phase) — temperature increases linearly. The gradient depends on the specific heat capacity of the solid.
2. First flat section (melting) — temperature remains constant at the melting point. Energy supplied is used to break bonds (latent heat of fusion). The flat section lasts until all the solid has melted.
3. Rising section (liquid phase) — temperature increases again, but typically with a different gradient because the liquid has a different specific heat capacity.
4. Second flat section (boiling) — temperature remains constant at the boiling point. Energy is used to overcome remaining intermolecular forces (latent heat of vaporisation). This flat section is longer than the first because Lv > Lf.
5. Rising section (gas phase) — temperature increases once more.

A cooling curve shows the reverse process: the temperature drops, then plateaus at the boiling point (condensing), drops again, plateaus at the melting point (freezing), and continues to drop.

Described diagram — Heating/cooling curve (temperature-time graph): The horizontal axis is time and the vertical axis is temperature. Starting from the bottom left, the curve rises with a steady gradient (solid being heated). It then flattens into a horizontal plateau at the melting point — temperature stays constant while the solid melts. The curve rises again with a different gradient (liquid being heated). A second, longer horizontal plateau appears at the boiling point — temperature remains constant while the liquid vaporises. Finally, the curve rises once more (gas being heated). A cooling curve is the mirror image: the temperature falls, plateaus at the boiling point during condensation, falls again, plateaus at the melting point during freezing, and continues to fall.

Exam Tip: When interpreting a heating or cooling curve, the flat sections indicate a change of state. During these sections, the substance exists as a mixture of two phases. If asked to identify the state at a particular time, check whether the temperature is rising (single phase) or constant (two phases coexisting).

Worked Example — Heating Curve Calculation

Question: A 0.50 kg block of ice at −10 °C is heated until it becomes water at 30 °C. Calculate the total energy required. (c_ice = 2100 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹, Lf = 3.34 × 10⁵ J kg⁻¹)

Solution:

Step 1: Heat the ice from −10 °C to 0 °C. E₁ = mcΔθ = 0.50 × 2100 × 10 = 10 500 J

Step 2: Melt the ice at 0 °C. E₂ = mLf = 0.50 × 3.34 × 10⁵ = 167 000 J

Step 3: Heat the water from 0 °C to 30 °C. E₃ = mcΔθ = 0.50 × 4200 × 30 = 63 000 J

Step 4: Total energy = E₁ + E₂ + E₃ = 10 500 + 167 000 + 63 000 = 240 500 J ≈ 241 kJ

Note how the energy required for the change of state (167 kJ) dominates the total — this is typical and explains why ice is so effective at cooling drinks.

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Absolute Zero and the Kelvin Scale

Key Definition: Absolute zero (0 K = −273.15 °C) is the lowest temperature theoretically possible. At absolute zero, the molecules have minimum internal energy — they have zero kinetic energy (no random thermal motion) but retain zero-point energy due to quantum mechanical effects.

The Kelvin scale is the thermodynamic temperature scale. A change of 1 K is identical in magnitude to a change of 1 °C, but the zero point differs:

T (K) = θ (°C) + 273.15

The gas laws only work when temperature is measured in kelvin. The significance of absolute zero is that:

• All molecular motion associated with thermal energy ceases.
• Pressure and volume of an ideal gas would both reach zero (extrapolating the gas laws).
• It represents the lower limit of the thermodynamic temperature scale and cannot be reached in practice (Third Law of Thermodynamics).

If you plot volume against temperature in °C for a fixed mass of gas at constant pressure, you obtain a straight line. Extrapolating this line backwards, it crosses the temperature axis at −273 °C, providing experimental evidence for absolute zero.

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Brownian Motion — Evidence for the Kinetic Theory

In 1827, Robert Brown observed that pollen grains suspended in water moved in a random, jerky fashion. This Brownian motion was later explained by Einstein in 1905 as evidence for the existence of atoms and molecules.

The Smoke Cell Experiment

The classic A-Level demonstration uses a smoke cell viewed under a microscope with strong illumination:

1. A glass cell is filled with smoke (tiny visible particles).
2. The cell is illuminated from the side and viewed through a microscope.
3. Bright specks of light (smoke particles) are seen moving in random, jerky, unpredictable paths.

The explanation is that the visible smoke particles are being bombarded from all sides by invisible air molecules. Because the bombardment is random, the net force on each smoke particle changes direction and magnitude constantly, producing the characteristic random walk.

This experiment provides strong evidence that:

• Air is made up of molecules in constant random motion.
• The molecules are too small to see directly but their effects are observable.
• Increasing the temperature increases the vigour of the motion (molecules move faster).

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The Ideal Gas Laws

An ideal gas is a theoretical model in which:

• Molecules are point particles with negligible volume compared to the volume of the container
• There are no intermolecular forces between molecules (except during collisions)
• Collisions between molecules and with the walls of the container are perfectly elastic (no kinetic energy is lost)
• The time of a collision is negligible compared to the time between collisions
• The molecules move in random directions with a distribution of speeds

Boyle's Law

At constant temperature, for a fixed mass of gas:

pV = constant (or p₁V₁ = p₂V₂)

Pressure is inversely proportional to volume. A graph of p against V gives a curve (hyperbola); a graph of p against 1/V gives a straight line through the origin.

Described diagram — p-V graph for Boyle's law: The horizontal axis is volume (V) and the vertical axis is pressure (p). The curve is a smooth rectangular hyperbola in the first quadrant — it starts high on the pressure axis at small volumes, curves steeply downward, and then levels off at low pressures for large volumes. The curve never touches either axis. Each point on the curve satisfies pV = constant. If several isotherms are plotted (one for each temperature), higher-temperature isotherms lie further from the origin because the product pV is larger.

Charles's Law

At constant pressure, for a fixed mass of gas:

V/T = constant (or V₁/T₁ = V₂/T₂)

Volume is directly proportional to absolute temperature. A graph of V against T (in kelvin) gives a straight line through the origin.

Pressure Law (Gay-Lussac's Law)

At constant volume, for a fixed mass of gas:

p/T = constant (or p₁/T₁ = p₂/T₂)

Pressure is directly proportional to absolute temperature.

The Ideal Gas Equation

Combining the three gas laws:

pV = nRT

where p is pressure (Pa), V is volume (m³), n is the number of moles, R is the molar gas constant (8.31 J mol⁻¹ K⁻¹), and T is the absolute temperature (K).

Alternatively, using the number of molecules N:

pV = NkT

where k is the Boltzmann constant (1.38 × 10⁻²³ J K⁻¹) and k = R/Nₐ, with Nₐ being the Avogadro constant (6.02 × 10²³ mol⁻¹).

The two forms are related by N = nNₐ.

Worked Example — Ideal Gas Law Calculation

Question: A cylinder contains 0.040 mol of an ideal gas at a pressure of 3.0 × 10⁵ Pa and a temperature of 450 K. Calculate the volume of the gas.

Solution:

Using pV = nRT:

V = nRT / p = (0.040 × 8.31 × 450) / (3.0 × 10⁵)

V = 149.58 / 300 000 = 4.99 × 10⁻⁴ m³

V ≈ 5.0 × 10⁻⁴ m³ (or 0.50 litres)

Exam Tip: Always convert temperature to kelvin and volume to m³ before substituting into pV = nRT. A common error is using cm³ or litres — remember that 1 litre = 1 × 10⁻³ m³ and 1 cm³ = 1 × 10⁻⁶ m³.

Real Gases vs Ideal Gases

Real gases deviate from ideal behaviour at:

• High pressures — molecules are forced close together, so their volume is no longer negligible and intermolecular forces become significant.
• Low temperatures — molecules move slowly enough for intermolecular attractions to influence their paths.

Real gases behave most like ideal gases at low pressure and high temperature, when the molecules are far apart and moving quickly.

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Molecular Kinetic Theory

The kinetic theory model derives the ideal gas equation from first principles by considering the motion of individual molecules in a container.

The Derivation — Key Assumptions

The derivation of pV = ⅓Nm⟨c²⟩ is based on the following explicit assumptions:

1. The gas contains a very large number of identical molecules.
2. The molecules are in constant random motion.
3. The volume of the molecules is negligible compared to the volume of the container.
4. Collisions between molecules and with the walls are perfectly elastic.
5. The duration of a collision is negligible compared to the time between collisions.
6. There are no intermolecular forces except during collisions.
7. The molecules exert pressure on the container walls by colliding with them and changing momentum.

The Derivation Outline

Consider a single molecule of mass m moving with velocity c in a cubic container of side length L:

1. The molecule travels to a wall, bounces back elastically, and returns — a round trip of distance 2L.
2. Its component of velocity perpendicular to the wall is cₓ. The change in momentum per collision with one wall is 2mcₓ.
3. The time between successive collisions with the same wall is 2L/cₓ.
4. Therefore the force from one molecule on one wall = rate of change of momentum = 2mcₓ / (2L/cₓ) = mcₓ²/L.
5. Summing over all N molecules and averaging: total force = Nm⟨cₓ²⟩/L.
6. Since the motion is random, ⟨cₓ²⟩ = ⟨c²⟩/3 (the mean square speed is shared equally among three dimensions).
7. Pressure = Force / Area = (Nm⟨c²⟩/3L) / L² = Nm⟨c²⟩ / (3L³) = Nm⟨c²⟩ / (3V).

flowchart TD
    A["Single molecule bounces<br/>between walls of cubic box"] --> B["Change in momentum per<br/>collision = 2mcx"]
    B --> C["Time between collisions<br/>with same wall = 2L/cx"]
    C --> D["Force from one molecule<br/>= mcx²/L"]
    D --> E["Sum over N molecules<br/>Total force = Nm⟨cx²⟩/L"]
    E --> F["Random motion:<br/>⟨cx²⟩ = ⟨c²⟩/3"]
    F --> G["Pressure = Force/Area<br/>= Nm⟨c²⟩/(3V)"]
    G --> H["pV = ⅓Nm⟨c²⟩"]

This gives:

pV = ⅓Nm⟨c²⟩

Root Mean Square Speed

Combining pV = NkT with pV = ⅓Nm⟨c²⟩ gives:

½m⟨c²⟩ = 3/2 kT

This shows that the mean kinetic energy of a molecule is directly proportional to the absolute temperature. The root mean square (r.m.s.) speed is:

c_rms = √⟨c²⟩ = √(3kT/m)

This shows that at a given temperature, lighter molecules move faster on average than heavier molecules. It also explains why hydrogen escapes from the Earth's atmosphere more readily than nitrogen or oxygen.

The Maxwell-Boltzmann Distribution

At any given temperature, not all molecules in a gas move at the same speed. The distribution of molecular speeds is described by the Maxwell-Boltzmann distribution:

• The distribution curve is asymmetric, with a tail extending to high speeds.
• The most probable speed is at the peak of the curve.
• The mean speed is slightly higher than the most probable speed.
• The r.m.s. speed is higher still.
• At higher temperatures, the distribution flattens and shifts to the right — the peak is lower and broader, but the area under the curve (total number of molecules) remains the same.

Described diagram — Maxwell-Boltzmann distribution at two temperatures: The horizontal axis is molecular speed and the vertical axis is the number of molecules (or probability density). Both curves start at the origin. The lower-temperature curve (T₁) has a tall, narrow peak at a lower speed. The higher-temperature curve (T₂) has a shorter, broader peak shifted to the right (higher speed). The right-hand tail of the T₂ curve extends further, showing more molecules with very high speeds. Crucially, the total area under both curves is the same, because the total number of molecules has not changed.

Exam Tip: When sketching Maxwell-Boltzmann distributions at two different temperatures on the same axes, ensure that: (1) the higher-temperature curve has a lower, broader peak shifted to the right; (2) both curves start at the origin; (3) the areas under both curves are equal (same number of molecules). Examiners specifically check all three of these features.

Pressure and Kinetic Energy

The pressure exerted by an ideal gas on the walls of its container arises from the collisions of molecules with the walls. Each collision transfers momentum to the wall, and the cumulative effect of billions of collisions per second produces a measurable, steady force.

From pV = ⅓Nm⟨c²⟩ and pV = NkT:

⅓m⟨c²⟩ = kT

This confirms that temperature is fundamentally a measure of the average translational kinetic energy of the molecules. Doubling the absolute temperature doubles the average kinetic energy and increases the r.m.s. speed by a factor of √2.

Quantity	Relationship to Temperature
Mean kinetic energy	Directly proportional to T
r.m.s. speed	Proportional to √T
Pressure (at constant V)	Directly proportional to T
Volume (at constant p)	Directly proportional to T