The key idea
Thermal physics connects energy transfers in heating and changes of state with an ideal-gas model. Unit 5 calculations use absolute temperature and distinguish a sample's internal energy from its temperature. The first-law example is an additional energy-accounting extension.
Understand it · 01
Distinguish temperature from internal energy
Temperature determines the direction of net energy transfer when bodies are in thermal contact: energy transfers spontaneously from the higher-temperature body to the lower-temperature body. Internal energy is the sum of microscopic random kinetic and potential energies within a system. A large quantity of cool material can have more internal energy than a small hot sample. Temperature therefore cannot be used alone as a measure of the total energy stored in a substance.
The specific heat capacity relation Q = mcΔT applies when energy changes temperature without a change of state and when c is suitably constant over the interval. During a phase change, energy can change molecular arrangements without increasing temperature; use Q = mL for the appropriate specific latent heat. In a calorimetry experiment, some supplied energy may warm the container or escape to the surroundings, so equating all electrical input with sample energy is an assumption to assess.
Understand it · 02
Connect pressure to collisions and temperature
Gas pressure results from molecules changing momentum when they collide with container walls. A faster molecular motion means greater average momentum transfer and more frequent collisions. The ideal-gas model treats molecules as having negligible volume compared with the container, negligible intermolecular forces except during collisions, and elastic collisions. It works best for sufficiently dilute gases away from conditions where condensation or strong intermolecular interactions become important.
The macroscopic relationship is pV = nRT, equivalently pV = NkT. Temperature must be in kelvin and pressure must be absolute rather than gauge pressure. For a fixed amount of gas, pV/T stays constant between equilibrium states. At constant volume, raising absolute temperature increases pressure; at constant temperature, reducing volume increases pressure. The mean translational kinetic energy per molecule is 3kT/2, linking the macroscopic temperature to a microscopic average.
Understand it · 03
Keep averages and experimental controls distinct
Molecules in a gas have a range of speeds. Root-mean-square speed is the square root of the mean of their squared speeds; it is not the speed of every molecule or the ordinary average speed. For molecules of mass m, ½m〈c²〉 = 3kT/2. Raising kelvin temperature by a factor of four doubles the rms speed. Different gases at the same temperature have equal mean translational kinetic energy, but lighter molecules have a greater rms speed.
To test Boyle’s law, vary the volume of a fixed gas sample while keeping temperature constant and measure the corresponding absolute pressure. Compress slowly or allow time for thermal equilibration because rapid compression can heat the gas. Plot pressure against inverse volume to test the predicted straight line. Leaks change the amount of gas and undermine the fixed-n assumption; a pressure zero offset can distort the graph even if the volume measurements are repeatable.
Key definitions
- Ideal gas
- A gas that obeys pV = nRT at all pressures and temperatures — assuming negligible molecular volume and no intermolecular forces except during collisions.
- Internal energy
- The sum of microscopic random kinetic and potential energies within a system. For a fixed amount of ideal gas, intermolecular potential energy is neglected and internal energy depends on temperature.
- First law of thermodynamics
- The increase in internal energy of a system equals the heat supplied to it plus the work done on it: ΔU = q + W.
- Absolute zero
- 0 K (−273.15 °C): the temperature at which molecules have minimum internal energy.
Equations, units & conditions
Heating
ΔE = mcΔT
No phase change; c approximately constant over the interval.
Change of state
ΔE = LΔm
Use the specific latent heat for the phase change.
Equation of state
pV = NkT = nRT
N is molecule count; n is amount in moles; T is kelvin.
Kinetic theory
pV = ⅓ N m <c²>
Mean KE per molecule
E = 3/2 kT
Units: J
First law — extension
ΔU = q + W
W is work done on the system. This is additional context, not a separately listed Unit 5 equation.
Worked examples, step by step
Extension: first-law energy accounting
A gas expands at constant pressure of 1.0 × 10⁵ Pa from 2.0 × 10⁻³ m³ to 3.5 × 10⁻³ m³ while 500 J of heat is supplied. Find the change in internal energy.
Follow the method
- Work done BY the gas = pΔV = 1.0 × 10⁵ × 1.5 × 10⁻³ = 150 J.
- Work done ON the gas: W = −150 J (it expands).
- ΔU = q + W = 500 + (−150).
ΔU = +350 J
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Pressure at constant volume
A sealed rigid vessel contains gas at 100 kPa and 27 °C. Find the pressure at 87 °C using the ideal-gas model.
Show solution and self-check: question 1
- Convert temperatures: T₁ = 300 K and T₂ = 360 K, using 273 for this precision.
- At fixed volume and amount, p₂ = p₁T₂/T₁ = 100 × 360/300.
120 kPa.
Self-check · one mark per point
- Use absolute temperatures.
- Apply p/T = constant.
- Obtain 120 kPa.
Build · 3 suggested marks
2. Heating a sample
A 0.40 kg sample with specific heat capacity 900 J kg⁻¹ K⁻¹ warms from 20 °C to 65 °C. Find the energy gained, assuming no phase change.
Show solution and self-check: question 2
- Temperature change = 65 − 20 = 45 K.
- Q = mcΔT = 0.40 × 900 × 45.
1.62 × 10⁴ J.
Self-check · one mark per point
- Use the temperature difference of 45 K.
- Apply Q = mcΔT.
- Give the energy in joules.
Common mistakes & how to avoid them
- Celsius in gas equations. Every temperature in pV = nRT and E = 3/2 kT must be in kelvin. Using 25 instead of 298 is the most common zero on the topic.
- Sign errors in the first law. In ΔU = q + W, W is work done ON the gas: compression is positive, expansion is negative. State the sign convention given in the question; do not assume this extension equation is supplied on the IAL data sheet.
- Saying molecules have zero energy at 0 K — the syllabus wording is minimum internal energy.
- Mean square speed handled as if it were the square of the mean speed. <c²> is averaged after squaring; its square root (the r.m.s. speed) is what E = ½m<c²> connects to temperature.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
How is temperature related to molecular kinetic energy?
For an ideal gas, mean translational kinetic energy per molecule is 3kT/2. Temperature must be in kelvin, and the equation describes an average over many molecules rather than assigning the same speed or energy to every molecule.
At a given temperature, molecules continually collide and exchange energy, producing a distribution of speeds. A light molecule needs a greater speed than a heavy molecule to have the same kinetic energy. Raising temperature increases the average squared speed, while individual molecules can still move slower or faster than the average. The word translational matters: it concerns motion through space and should not automatically be treated as every possible microscopic energy contribution of a complex molecule. Use the particle count if converting an average per molecule into a total translational energy.
What are the assumptions of the kinetic theory of gases?
The ideal-gas model assumes particles have negligible volume compared with the container, negligible intermolecular forces between collisions, and elastic collisions. Their motion is random, and collision durations are short compared with the time between collisions.
These assumptions make pressure calculable from momentum transfer to the walls without tracking attractive forces or energy losses in every collision. The model is more successful for dilute gases away from condensation conditions. At high density, molecular size and interactions can no longer be ignored as easily. Random motion does not mean that every direction has a preferred net flow; in equilibrium the average velocity can be zero while average speed is positive. Distinguish the model from a claim that real molecules have literally no size or never interact outside impacts.
Why does the internal energy of an ideal gas depend only on temperature?
For a fixed amount of ideal gas, internal energy depends on temperature because the model neglects intermolecular potential-energy changes. Changing volume alone at unchanged temperature therefore does not change internal energy in that model.
The phrase fixed amount is essential: twice as many molecules can have twice the total energy at the same temperature. The relation between internal energy and temperature also depends on which molecular degrees of freedom are relevant, so the mean translational energy formula is not always the complete internal energy of every ideal gas. In school thermal reasoning, focus on whether temperature changes and on the stated particle model. A real gas can depart from this result because interactions store potential energy that depends on separation, especially near a phase change.
Why can Celsius be used for a temperature change but not in a gas-law ratio?
Kelvin is required in gas-law ratios because pressure and molecular kinetic energy are proportional to absolute temperature. Celsius has an offset, so doubling a Celsius reading does not represent doubling thermal motion or absolute temperature.
A Celsius temperature change has the same numerical size as a kelvin change, which is why either can describe ΔT in a heat-capacity calculation. Absolute temperatures are different: convert before using pV/T or rms-speed relationships. For example, raising a sample from 20 °C to 40 °C changes it from about 293 K to 313 K, not to twice its temperature. Use absolute pressure too; a gauge reading is measured relative to atmospheric pressure and may need that pressure added before the ideal-gas equation applies.
A rigid gas vessel
A fixed gas sample in a rigid vessel is at 150 kPa and 20 °C. Find its pressure at 80 °C using 273 for the kelvin conversion.
- T₁ = 293 K and T₂ = 353 K.
- At fixed volume and amount, p₂ = p₁T₂/T₁.
- p₂ = 150 × 353/293.
Approximately 181 kPa.
Which graph should I plot to test Boyle’s law?
For a fixed gas amount at constant temperature, pV is constant. A pressure–volume graph is a decreasing curve, while pressure plotted against 1/V should be a straight line through the origin for the ideal model.
The transformation separates a physical inverse relationship from an apparently curved trend. Label the inverse-volume axis with the correct reciprocal units, and use absolute pressure. Rapid compression can heat the gas, so allow time to return to the controlled temperature before each reading. A leak changes the number of molecules and invalidates the fixed-amount condition. A straight line with an unexpected intercept may indicate a pressure offset or an unaccounted dead volume in the apparatus rather than proving a new gas law. Evaluate the actual measurement model alongside the graph.
Why can heating continue without the temperature rising during a phase change?
During a phase change at fixed pressure, supplied energy can change molecular arrangement and potential energy without increasing temperature. Use Q = mL for that stage, and Q = mcΔT for temperature changes within one phase.
A heating curve can therefore contain sloping regions separated by a temperature plateau. The flat region does not mean that the heater transfers no energy: the energy is being used in changing state. If a sample must first warm and then melt, calculate the two contributions separately and add them. In a real experiment, some input warms the container or leaks to the surroundings, so dividing total electrical energy by melted mass without corrections can overestimate latent heat. Use a matched control or another justified method to assess those additional transfers.
Warm and then melt
A 0.10 kg solid warms by 20 K with c = 800 J kg⁻¹ K⁻¹, then melts completely with L = 2.0 × 10⁵ J kg⁻¹. Ignore losses. Find total energy.
- Heating energy = 0.10 × 800 × 20 = 1600 J.
- Melting energy = 0.10 × 2.0 × 10⁵ = 20,000 J.
- Add the two stages.
21,600 J, approximately 2.2 × 10⁴ J.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Use the thermodynamic (kelvin) scale: T/K = θ/°C + 273.15
- Use the equation of state pV = nRT (and pV = NkT)
- Recall the kinetic theory result pV = ⅓Nm<c²> and relate mean kinetic energy to temperature
- Define internal energy as the sum of random kinetic and potential energies of the molecules
- Use specific heat capacity and specific latent heat: ΔE = mcΔT and ΔE = LΔm; the first-law example below is an extension
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Temperature and ideal gases. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/temperature-and-ideal-gases
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