IB DP Physics: First law of thermodynamics, gas processes and work done by a gas (DP Physics) – Revision Notes

Full explanations and extra worked examples are in the study guide. These notes follow the IB Physics guide, first assessment 2025, topic B.4 (first half), which is additional higher level only. After revising, use the practice set, the IB DP Physics course page, the printable checklist and the free 10-minute diagnostics.

Key ideas

• The first law is conservation of energy for a closed system: Q = ΔU + W.
• Closed system: no mass transfer, but energy can cross as heat or work.
• Isolated system: neither mass nor energy crosses, so Q = 0, W = 0 and ΔU = 0.
• Quantitative problems use monatomic ideal gases only.

Sign convention (Clausius)

Quantity	Positive	Negative
Q	heat supplied to the gas	heat removed from the gas
W	the gas pushes the piston out (expansion)	the piston pushes the gas in (compression)
ΔU	temperature rises	temperature falls

Formulas

Result	Equation	Notes
First law	Q = ΔU + W	Rearrange to ΔU = Q − W
Work at constant pressure	W = PΔV	P in Pa, V in m³
Work, varying pressure	area under line on P-V graph	right = positive, left = negative
Internal energy change	ΔU = 3/2 N kB ΔT = 3/2 nRΔT	monatomic ideal gas, ΔT in K
Using PV	ΔU = 3/2 (P₂V₂ − P₁V₁)	from PV = nRT
Adiabatic	PV^(5/3) = constant	monatomic ideal gas
Link	N kB = nR	N = n × Avogadro constant

Processes at a glance

Process	Fixed	W	ΔU	Q
Isovolumetric	V	0	3/2 nRΔT	= ΔU
Isobaric	P	PΔV	3/2 nRΔT	ΔU + PΔV
Isothermal	T	area	0	= W
Adiabatic	Q = 0	−ΔU	3/2 nRΔT	0

Particle reasoning to quote:

• Isothermal expansion: same mean kinetic energy, but particles hit a larger wall area less often per unit time, so pressure falls.
• Adiabatic compression: the piston does work on the gas, nothing escapes as heat, random kinetic energy rises, so temperature rises.
• Adiabatic expansion: the gas does work at the expense of its own internal energy, so it cools.

Method boxes

First-law problem

1. Draw or list: is the gas expanding or compressing?
2. Assign signs: W positive if expanding, Q positive if heat enters.
3. Find W (PΔV or area), then ΔU (from ΔT, or from PV values).
4. Use Q = ΔU + W for whatever is left.

Adiabatic problem

1. Write Q = 0, so ΔU = −W.
2. Find the missing pressure or volume from P₁V₁^(5/3) = P₂V₂^(5/3).
3. Find ΔU from 3/2 (P₂V₂ − P₁V₁), or from ΔT if n is known.
4. State W = −ΔU with its sign and meaning.

P-V graph problem

1. Identify each leg: vertical (W = 0), horizontal (W = PΔV), or sloping.
2. Split areas into rectangles and triangles.
3. Get ΔU between end points from P and V alone.
4. Remember ΔU depends only on end states, whereas Q and W depend on the path.

Worked reminders

Reminder 1 (isobaric). 3.0 mol at 2.0 × 10⁵ Pa expand by 0.012 m³. W = 2.0 × 10⁵ × 0.012 = 2 400 J. Since PΔV = nRΔT here, ΔU = 3/2 × 2 400 = 3 600 J, and Q = 6 000 J. For an isobaric change of monatomic gas, Q is five halves of PΔV.

Reminder 2 (isothermal). A gas absorbs 330 J during an isothermal expansion. ΔU = 0, so W = 330 J.

Reminder 3 (adiabatic). A monatomic gas is expanded adiabatically to twice its volume. The pressure falls by a factor 2^(5/3) = 3.17, which is more than the factor 2 for an isothermal change at the same volume ratio.

Reminder 4 (compression). 90 J of work is done on a gas that loses 40 J of heat. W = −90 J, Q = −40 J, so ΔU = −40 + 90 = +50 J.

Must-know distinctions

• Adiabatic is not isothermal. Adiabatic: Q = 0 and T changes. Isothermal: ΔU = 0 and heat flows.
• Closed is not isolated. A closed system can exchange energy.
• W by the gas is not W on the gas. The guide’s convention is work done by the system.
• ΔU is a state change; Q and W depend on the process.

Definitions to quote

• Internal energy of an ideal gas: the total random kinetic energy of its particles. There is no potential energy term because there are no intermolecular forces.
• Closed system: matter cannot cross the boundary, but energy can cross it in either direction, as heat or as work.
• Isolated system: the boundary lets through neither matter nor energy.
• Isovolumetric, isobaric, isothermal: volume, pressure or temperature is kept fixed.
• Adiabatic: no thermal energy is transferred to or from the gas.

Reading a P-V diagram

• Moving right on the diagram means expansion and positive work. Moving left means compression and negative work.
• A horizontal line is isobaric. A vertical line is isovolumetric. A hyperbola-like curve through points with equal PV is isothermal.
• Two points with the same value of PV are at the same temperature, so ΔU between them is zero, whatever path joins them.
• A higher value of PV means a higher temperature. If PV rises, ΔU is positive; if it falls, ΔU is negative.
• An adiabatic curve falls more steeply than an isothermal curve through the same starting point.

Reminder 5 (sloping line). A gas goes from 2.0 × 10⁵ Pa at 1.0 × 10⁻³ m³ to 6.0 × 10⁵ Pa at 3.0 × 10⁻³ m³ along a straight line. The area under the line is average pressure times ΔV: 4.0 × 10⁵ × 2.0 × 10⁻³ = 800 J. P₁V₁ = 200 J and P₂V₂ = 1 800 J, so ΔU = 3/2 × 1 600 = 2 400 J. Then Q = 2 400 + 800 = 3 200 J.

Command words and what they ask for

• State a sign or a value: give it with a unit, no explanation.
• Calculate: show the substitution, then the answer with a unit and sensible significant figures.
• Explain: link cause to effect, for example no heat transfer, so the work done changes the internal energy, so the temperature changes.
• Sketch a process on a P-V graph: label axes and mark the direction with an arrow.

Quick self-test

1. A gas takes in 500 J of heat and does 320 J of work. Find ΔU.
2. 75 J of work is done on a gas with no heat transfer. Find ΔU and say whether T rises or falls.
3. A gas at 2.5 × 10⁵ Pa expands from 0.80 × 10⁻³ m³ to 3.4 × 10⁻³ m³. Find W.
4. Why is W zero in an isovolumetric process?
5. Find ΔU for 0.40 mol of monatomic gas warmed by 60 K (R = 8.31 J mol⁻¹ K⁻¹).
6. State ΔU for an isothermal change and what that implies for Q and W.
7. A gas at 1.0 × 10⁵ Pa is compressed adiabatically to one eighth of its volume. Find the final pressure.
8. On a P-V graph, a gas moves from (1.0 × 10⁻³ m³, 3.0 × 10⁵ Pa) to (4.0 × 10⁻³ m³, 3.0 × 10⁵ Pa). Find W and ΔU.
9. Give the Q, W and ΔU of an isolated system.
10. Explain, using particles, why a gas cools when it expands adiabatically.

Answers

1. ΔU = Q − W = 500 − 320 = 180 J.
2. W = −75 J, Q = 0, so ΔU = +75 J. Temperature rises.
3. W = 2.5 × 10⁵ × 2.6 × 10⁻³ = 650 J.
4. The boundary does not move, so ΔV = 0 and W = PΔV = 0.
5. ΔU = 3/2 × 0.40 × 8.31 × 60 = 299 J.
6. ΔU = 0 because T is constant, so Q = W.
7. P₂ = 1.0 × 10⁵ × 8^(5/3) = 1.0 × 10⁵ × 32 = 3.2 × 10⁶ Pa.
8. W = 3.0 × 10⁵ × 3.0 × 10⁻³ = 900 J. P₁V₁ = 300 J, P₂V₂ = 1 200 J, ΔU = 3/2 × 900 = 1 350 J.
9. Q = 0, W = 0, ΔU = 0.
10. The gas does work on its surroundings while no heat enters, so the particles lose random kinetic energy and the mean kinetic energy, hence temperature, falls.

Where marks are usually lost

• Wrong sign on W for compression, then a wrong ΔU.
• Using Celsius temperatures where T itself is needed.
• Leaving volume in cm³ or litres when pressure is in Pa.
• Writing ΔU = 0 for an adiabatic process.
• Forgetting that a vertical line on a P-V graph has no work, while ΔU still changes.
• Using PV^(5/3) for a process that is not adiabatic.
• Giving an answer to the wrong number of significant figures, or without a unit.
• Explaining temperature changes without referring to work done and heat transferred.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. Topic B.4 (Thermodynamics) is additional higher level only.