PhysicsCore26 min read

Heat and Thermodynamics

Gas laws, kinetic theory and the laws that limit every engine

This topic appears in:

1st Year Physics · ch.6A Level Physics (9702) · ch.15A Level Physics (9702) · ch.16
01

The gas laws

Definition

Absolute zero — 0 K, or −273.15 °C. The temperature at which particles have the minimum possible energy and the pressure of an ideal gas would fall to zero.

Three experimental relationships describe how a fixed mass of gas behaves, each holding one quantity constant while two others vary.

Boyle's law: at constant temperature, pressure is inversely proportional to volume. Halve the volume and the pressure doubles, because the same particles strike a smaller area more often.

Charles's law: at constant pressure, volume is directly proportional to absolute temperature. Heat a gas and it expands, because the particles move faster and push the walls out until the pressure is back to what it was.

The pressure law: at constant volume, pressure is directly proportional to absolute temperature. This is why an aerosol can bursts in a fire.

The word absolute is essential. These proportionalities only hold in kelvin. Doubling from 20 °C to 40 °C does not double anything; doubling from 293 K to 586 K does.

p₁V₁ / T₁ = p₂V₂ / T₂pV = nRTT(K) = θ(°C) + 273ALWAYS in kelvin — this is the single most common source of wrong answers in the topic
p
pressurePa
V
volumem³
T
absolute temperatureK
n
number of molesmol
R
gas constant, 8.31J mol⁻¹ K⁻¹
Worked example 15 marks

A sealed cylinder holds gas at 1.0 × 10⁵ Pa and 27 °C in a volume of 0.020 m³. It is compressed to 0.008 m³ and warms to 87 °C. Find the new pressure.

  1. Convert both temperatures: 27 °C = 300 K, 87 °C = 360 K.Do this first, every time. Using Celsius here gives an answer roughly three times too small.
  2. Use p₁V₁/T₁ = p₂V₂/T₂.
  3. (1.0 × 10⁵ × 0.020) / 300 = (p₂ × 0.008) / 360.
  4. 6.67 = p₂ × 2.22 × 10⁻⁵.
  5. p₂ = 3.0 × 10⁵ Pa.Volume down by 2.5 and temperature up by 1.2 — a threefold rise is exactly right.

3.0 × 10⁵ Pa

02

Kinetic theory: what temperature really is

The gas laws are experimental facts. Kinetic theory explains them, by treating a gas as a very large number of tiny particles in constant random motion.

The model makes a few assumptions: the particles are far apart compared with their own size, they exert no forces on each other except during collisions, all collisions are perfectly elastic, and their motion is random. A real gas at ordinary pressures matches this closely.

Pressure then has a mechanical explanation. Each particle striking a wall exerts a tiny impulse, and the pressure is the total force from all those collisions divided by the wall area. Squeeze the gas and the collisions become more frequent, so the pressure rises — Boyle's law, explained.

Temperature also gets a meaning. The absolute temperature is proportional to the average kinetic energy of the particles. Heat the gas and they move faster, hit harder and more often, and the pressure rises — the pressure law, explained.

This is also why absolute zero is a genuine floor rather than an arbitrary point: it is where the kinetic energy is as low as it can be.

p V = ⅓ N m c̄²½ m c̄² = (3/2) k Tmean kinetic energy per particle is proportional to absolute temperature
N
number of particles
c̄²
mean square speedm² s⁻²
k
Boltzmann constant, 1.38 × 10⁻²³J K⁻¹

The flat sections are melting and boiling. Energy still flows in but the temperature holds steady, because it is going into separating particles rather than speeding them up — and temperature only measures the speeding up.

03

The first law of thermodynamics

Definition

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. It is conservation of energy, applied to heat.

Internal energy is the total energy of all the particles in a system — kinetic and potential. There are exactly two ways to change it: supply heat, or do work.

Both routes are equivalent. Rubbing your hands warms them by doing work; holding them near a fire warms them by heat transfer. The internal energy does not record which method was used.

Signs matter and are where marks are lost. Q is positive when heat goes into the system; W is positive when work is done on the system, which happens when a gas is compressed. A gas that expands does work on its surroundings, so W is negative and the gas cools unless heat is supplied.

That is why a spray can gets cold in use, and why compressing air in a bicycle pump makes the barrel warm.

ΔU = Q + WW = −p ΔV(gas expanding at constant pressure)Q into the system is positive; W is positive when work is done ON the gas
ΔU
change in internal energyJ
Q
heat suppliedJ
W
work done on the gasJ

An isothermal change is not a change of zero energy

At constant temperature ΔU = 0, so Q = −W. Heat still flows and work is still done — they simply cancel. Writing "nothing happens because the temperature is constant" throws away the whole question.

04

The second law, and why engines waste energy

The first law says energy is conserved. It does not say which direction things go, and left to itself it would permit a cup of tea to grow hotter while the room cooled. The second law supplies the missing direction: heat flows spontaneously from hot to cold and never the other way.

A consequence is that no heat engine can be perfectly efficient. An engine takes heat from a hot source, converts some to work, and must dump the rest into a cold sink. That waste is not bad engineering — it is required by the law.

The best possible efficiency depends only on the two temperatures, and it is reached only by an ideal reversible engine. A real engine does worse because of friction, turbulence and heat leaking away.

This is why a car engine is around 25-30% efficient and a modern power station around 40%. It is also why raising the temperature of the source is the main way to improve either.

A refrigerator is a heat engine run backwards: work is done on it to move heat from cold to hot, which is exactly what the second law forbids happening on its own. That is why a fridge needs a power supply and why the back of it is warm.

efficiency = W / Q_hmaximum efficiency = 1 − T_c / T_htemperatures in kelvin; only an ideal reversible engine reaches this
T_h
source temperatureK
T_c
sink temperatureK
Q_h
heat taken from the sourceJ

Key points

  1. Gas law calculations are in kelvin, always.
  2. Absolute temperature is proportional to the mean kinetic energy of the particles.
  3. ΔU = Q + W — heat in is positive, work done on the gas is positive.
  4. An expanding gas does work and cools unless heat is supplied.
  5. No engine can be 100% efficient; the maximum is 1 − T_c/T_h.

Practice questions

6 questions · 25 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State Boyle's law and the conditions under which it applies.
Model answer

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume. Both the mass and the temperature must be constant.

Examiner tip. Both conditions — fixed mass AND constant temperature. Each is worth stating.

SQ2[2 marks]
Explain, in terms of particles, why the pressure of a gas rises when it is heated at constant volume.
Model answer

The particles gain kinetic energy and move faster, so they strike the walls harder and more frequently. With the same wall area, the force per unit area increases.

Examiner tip. Both "harder" and "more often" are needed — they are separate marks.

SQ3[2 marks]
Explain why a gas cools when it expands rapidly without heat entering it.
Model answer

The gas does work pushing back the surroundings, so W is negative. With Q = 0, ΔU is negative, the internal energy falls and the temperature drops.

Examiner tip. Quote the first law with the signs. This is why an aerosol can goes cold in your hand.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
A gas at 2.0 × 10⁵ Pa occupies 0.015 m³ at 17 °C. It is heated at constant pressure to 137 °C. Find the new volume.
Full working
  1. Convert: 17 °C = 290 K, 137 °C = 410 K[1]
  2. Constant pressure, so V₁/T₁ = V₂/T₂Charles's law[1]
  3. 0.015 / 290 = V₂ / 410[1]
  4. V₂ = 0.015 × 410 / 290[1]
  5. V₂ = 0.021 m³a 41% rise in kelvin gives a 41% rise in volume[1]

0.021 m³

Examiner tip. In Celsius the ratio 137/17 would suggest the volume grows eightfold. It does not. Convert first.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
A heat engine takes 6000 J from a source at 500 K and rejects heat to a sink at 300 K.
  1. Calculate the maximum possible efficiency. [2]
  2. Calculate the maximum work it could do, and the heat it must reject. [3]
  3. The real engine delivers 1800 J of work. Calculate its actual efficiency and explain the difference. [3]
Mark scheme
  1. Uses 1 − T_c/T_h with kelvin[1]
  2. 1 − 300/500 = 0.40, so 40%[1]
  3. Maximum work = 0.40 × 6000 = 2400 J[1]
  4. Heat rejected = 6000 − 2400[1]
  5. = 3600 Jthis cannot be avoided — the second law requires it[1]
  6. Actual efficiency = 1800/6000 = 0.30, or 30%[1]
  7. Lower than the maximum because the real engine is not reversible[1]
  8. Friction, turbulence and heat lost to the surroundings all reduce the work obtained[1]

(a) 40% (b) 2400 J of work, 3600 J rejected (c) 30%

Examiner tip. The rejected heat is not waste caused by poor design — even a perfect engine must dump 3600 J here. That is the second law, not an engineering failure.

Exam questions

1 · 6 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
A gas is compressed, and 250 J of work is done on it. At the same time it loses 100 J of heat to the surroundings.
  1. State the first law of thermodynamics. [2]
  2. Calculate the change in internal energy. [2]
  3. State whether the temperature rises or falls, with a reason. [2]
Mark scheme
  1. The increase in internal energy equals the heat supplied plus the work done on the system[1]
  2. ΔU = Q + W[1]
  3. W = +250 J (work done ON the gas), Q = −100 J (heat lost)the signs are the difficult part[1]
  4. ΔU = −100 + 250 = +150 J[1]
  5. The temperature rises[1]
  6. Internal energy has increased, and for a gas that means greater mean kinetic energy[1]

ΔU = +150 J, so the temperature rises

Examiner tip. Write the signs down before substituting. Heat lost is negative Q; work done on the gas is positive W.