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

This study guide follows the IB Physics guide, first assessment 2025, and covers the first half of topic B.4 (Thermodynamics): the first law, work done by a gas, internal energy and the four gas processes. B.4 is entirely additional higher level (8 hours), so everything here is HL only. Entropy, the second law, heat engines and the Carnot limit are taught in a separate unit.

Use it with the revision notes and the practice set. The course page is IB DP Physics and the printable list is the IB DP Physics checklist. To find weak spots first, try the free 10-minute diagnostics. Thermal energy and the molecular model are in molecular model, temperature and internal energy.

What this unit covers

B.4 understanding	What you must be able to do	Level
First law, Q = ΔU + W	Apply energy conservation to a closed system	HL only
Work, W = PΔV	Find work from pressure and volume change, including varying pressure	HL only
Internal energy	Use ΔU = 3/2 N kB ΔT = 3/2 nRΔT	HL only
Gas processes	Recognise isovolumetric, isobaric, isothermal and adiabatic changes	HL only
Adiabatic law	Use PV^(5/3) = constant for a monatomic ideal gas	HL only
System types	Distinguish closed from isolated systems	HL only

The guide limits quantitative problems to monatomic ideal gases, and says pressure may vary during a process.

The first law of thermodynamics

The first law is conservation of energy applied to a closed system. Energy enters or leaves a gas in two ways: as heat across the boundary because of a temperature difference, and as work when the boundary moves. Any energy supplied that is not used to do work raises the internal energy.

Q = ΔU + W

The guide asks for the Clausius sign convention:

• Q is the net heat given to the gas. Heat in is positive, heat out is negative.
• W is the net work the gas does on its surroundings. Work done by the gas (expansion) is positive. Work done on the gas (compression) is negative.
• ΔU is positive when internal energy rises.

Read the equation as a sentence: heat supplied goes partly into raising internal energy and partly into work done by the gas.

A closed system allows no mass in or out but allows energy to move as heat or work. An isolated system allows neither mass nor energy across its boundary. For an isolated system Q = 0 and W = 0, so ΔU = 0.

Worked example 1

A gas absorbs 640 J of heat while it does 250 J of work on a piston. Find ΔU.

Q = +640 J and W = +250 J (done by the gas). ΔU = Q − W = 640 − 250 = 390 J. The internal energy rises.

Worked example 2

A gas is compressed so that 180 J of work is done on it, and it loses 70 J as heat. Find ΔU.

W = −180 J (work done on the gas). Q = −70 J (heat out). ΔU = Q − W = −70 − (−180) = +110 J.

Work done by a gas

When a gas pushes a piston of area A through a small distance, the force is PA and the work is PAΔx, which is PΔV. For a change at constant pressure:

W = PΔV

ΔV is final volume minus initial volume, so expansion gives W > 0 and compression gives W < 0. Use pressure in Pa and volume in m³ so that the work comes out in joules.

Varying pressure and P-V areas

If pressure changes during the process, W = PΔV can only be applied to thin slices. Adding the slices gives the rule you use in practice: the work done by the gas equals the area under the process line on a P-V graph.

• Expansion (volume increasing, moving right): the area is positive work done by the gas.
• Compression (moving left): the work is negative.
• An isovolumetric line is vertical, so the area under it is zero.
• For a straight line, split the area into a rectangle and a triangle, or use the average pressure multiplied by ΔV.

Worked example 3

A gas expands so that its pressure falls linearly from 4.0 × 10⁵ Pa at 1.0 × 10⁻³ m³ to 2.0 × 10⁵ Pa at 3.0 × 10⁻³ m³. Find the work done by the gas.

The area under the line is a trapezoid. Average pressure = (4.0 + 2.0)/2 × 10⁵ = 3.0 × 10⁵ Pa. ΔV = 3.0 × 10⁻³ − 1.0 × 10⁻³ = 2.0 × 10⁻³ m³. W = 3.0 × 10⁵ × 2.0 × 10⁻³ = 600 J.

Internal energy of a monatomic ideal gas

In an ideal gas there are no intermolecular forces, so internal energy is the random kinetic energy of the particles. For a monatomic gas each atom has translational kinetic energy only, and the guide gives the change as:

ΔU = 3/2 N kB ΔT = 3/2 nRΔT

Here N is the number of particles, kB the Boltzmann constant, n the amount in moles and R the molar gas constant. The two forms are linked by N kB = nR. Temperature must be in kelvin, though a temperature change is the same in kelvin and degrees Celsius.

Two consequences matter. First, ΔU depends only on the temperature change, not on how the change happened. Second, because PV = nRT, you can write ΔU = 3/2 (P₂V₂ − P₁V₁). This is useful when you are given a P-V graph and no temperatures.

Worked example 4

A sample contains 1.2 × 10²³ monatomic atoms. Its temperature rises by 50 K. Find ΔU (kB = 1.38 × 10⁻²³ J K⁻¹).

ΔU = 3/2 × 1.2 × 10²³ × 1.38 × 10⁻²³ × 50 = 124 J (3 s.f.).

The four gas processes

Each process holds one variable fixed. Apply the first law to each.

Process	Fixed	W	ΔU	Q
Isovolumetric	V	0 (no boundary movement)	3/2 nRΔT	equals ΔU
Isobaric	P	PΔV	3/2 nRΔT	ΔU + PΔV
Isothermal	T	area under the curve	0	equals W
Adiabatic	no heat transfer	−ΔU	3/2 nRΔT	0

In an isothermal change ΔU = 0 because the temperature does not change. The gas must absorb heat to do work, or lose heat when work is done on it. At fixed T, P ∝ 1/V, so pressure falls as volume rises. The work is read from the area under the curve; no formula is needed for it here.

In an adiabatic change no heat crosses the boundary, either because the system is insulated or because the change is too fast for heat to move. All work comes from internal energy, so an expanding gas cools and a compressed gas warms.

Worked example 5 (isobaric)

2.0 mol of a monatomic ideal gas expands at 1.5 × 10⁵ Pa from 0.030 m³ to 0.045 m³ (R = 8.31 J mol⁻¹ K⁻¹). Find W, ΔT, ΔU and Q.

• W = PΔV = 1.5 × 10⁵ × 0.015 = 2 250 J.
• Initial temperature: T₁ = PV/nR = (1.5 × 10⁵ × 0.030)/(2.0 × 8.31) = 270.8 K. Final: T₂ = (1.5 × 10⁵ × 0.045)/(2.0 × 8.31) = 406.1 K. So ΔT = 135 K.
• ΔU = 3/2 × 2.0 × 8.31 × 135.4 = 3 380 J (3 s.f.). Check: 3/2 PΔV = 3/2 × 2 250 = 3 375 J.
• Q = ΔU + W = 3 375 + 2 250 = 5 630 J (3 s.f.).

Worked example 6 (isovolumetric)

0.50 mol of monatomic gas in a rigid container is warmed from 300 K to 420 K. W = 0 because V is fixed. ΔU = 3/2 × 0.50 × 8.31 × 120 = 748 J, so Q = 748 J.

Worked example 7 (non-constant pressure, using the graph only)

On the straight line in worked example 3, find ΔU and Q.

P₁V₁ = 4.0 × 10⁵ × 1.0 × 10⁻³ = 400 J. P₂V₂ = 2.0 × 10⁵ × 3.0 × 10⁻³ = 600 J. ΔU = 3/2 × (600 − 400) = 300 J. With W = 600 J: Q = 300 + 600 = 900 J.

Adiabatic changes in a monatomic ideal gas

For an adiabatic change in a monatomic ideal gas the guide gives:

PV^(5/3) = constant

so P₁V₁^(5/3) = P₂V₂^(5/3). Combining with PV = nRT shows that T V^(2/3) is also constant, so the temperature falls on expansion and rises on compression. Use the first form for pressure and the ideal gas law for temperature.

On a P-V graph an adiabatic curve is steeper than an isothermal curve through the same point, because the temperature changes as well as the volume.

Worked example 8

A monatomic gas at 1.0 × 10⁵ Pa and 2.0 × 10⁻³ m³ is compressed adiabatically to 0.50 × 10⁻³ m³. Find the final pressure and the work done by the gas.

• Volume ratio V₁/V₂ = 4. P₂ = P₁ × 4^(5/3) = 1.0 × 10⁵ × 10.08 = 1.01 × 10⁶ Pa.
• ΔU = 3/2 (P₂V₂ − P₁V₁) = 3/2 × (504 − 200) = 456 J.
• Q = 0, so W = −ΔU = −456 J. The negative sign means work is done on the gas, which is why its internal energy rises.

Common errors

• Putting W with the wrong sign. Under the Clausius convention, compression gives negative W.
• Using Celsius temperatures in ΔU = 3/2 nRΔT. Use ΔT in K, or any temperature in K when you need T itself.
• Using volume in litres or cm³ with pressure in Pa. Convert to m³ first.
• Saying ΔU = 0 for an adiabatic change. It is Q that is zero. ΔU = 0 belongs to isothermal changes and isolated systems.
• Using PV^(5/3) = constant for an isothermal change, or PV = constant for an adiabatic one.
• Using 3/2 on a diatomic gas. The guide limits quantitative work to monatomic ideal gases.
• Reading work from the wrong area. It is the area between the line and the volume axis, not the area to the pressure axis.

Next steps

Condense this unit with the revision notes, then test yourself on the practice set.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This unit covers part of topic B.4 (Thermodynamics), which is additional higher level only.