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Thermodynamic Systems

Key idea: Internal energy, thermal equilibrium (zeroth law), and work done by/on a gas (W = pΔV) for A Level Physics.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Relate microscopic energy, internal energy and thermal equilibrium.

1. Definitions (Must Know)

  • System: the part of the universe you study (e.g. a gas in a cylinder).
  • Surroundings: everything outside the system.
  • Internal energy, U (J): sum of microscopic kinetic and potential energies of the particles in the system.
  • Heating, Q (J): energy transferred due to a temperature difference.
  • Thermal equilibrium: no net energy transfer by heating between systems in thermal contact (they have the same temperature).
  • Zeroth law of thermodynamics: if A is in thermal equilibrium with C, and B is in thermal equilibrium with C, then A is in thermal equilibrium with B.
  • Work done by a gas, W_by (J): energy transferred from the gas when it expands.
  • Work done on a gas, Wₒₙ (J): energy transferred to the gas by the surroundings.

2. Key Ideas (What Earns Marks)

  • Heating flows from higher T to lower T until thermal equilibrium is reached.
  • For expansion against a constant external pressure p:
    • W_by = pΔ V
    • Wₒₙ = -pΔ V
  • Unit check: Pa·m³ = J.
  • Be explicit about sign conventions when you use the first law: First Law of Thermodynamics.

3. Detailed Explanations

A. Internal energy (microscopic picture)

Internal energy is the total energy stored in the system’s particles:

  • random microscopic kinetic energy (translational/rotational/vibrational), and
  • microscopic potential energy due to interactions between particles.

For an ideal gas, intermolecular forces are neglected except during collisions, so its internal energy is microscopic kinetic energy. For a fixed amount of ideal gas, internal energy therefore depends only on thermodynamic temperature.

B. Thermal equilibrium and the zeroth law

When two systems are placed in thermal contact, energy is transferred by heating from the hotter system to the cooler system until:

  • both systems have the same temperature, and
  • there is no net heating between them (thermal equilibrium).

The zeroth law is what makes temperature measurement meaningful: if a thermometer is in thermal equilibrium with a system, they share the same temperature.

C. Work done by/on a gas: W = pΔ V (constant external pressure)

Consider a gas in a cylinder with a movable piston of area A, expanding a small distance x against constant external pressure p.

External force on the piston is: F = pA

Work done by the gas is: W_by = Fx = (pA)x

But the volume change is Δ V = Ax, so: W_by = pΔ V

For work done on the gas: Wₒₙ = -W_by = -pΔ V

  • Expansion: Δ V > 0 ⇒ W_by > 0 and Wₒₙ < 0
  • Compression: Δ V < 0 ⇒ W_by < 0 and Wₒₙ > 0

4. Common Mistakes

  • Using W = pΔ V inside the first law without stating whether W is by or on the gas.
  • Using the gas pressure when the question gives external pressure (for A Level, use the stated constant external pressure).
  • Forgetting that Δ V is in m³ (not cm³).

5. Exam Tips

  • If you’re using Δ U = Q + W, check that W is work done on the system (not by the system).
  • A quick sign check for expansion:
    • gas loses energy by doing work, so Wₒₙ should be negative.

6. Worked Examples

Modelled example 1

Work done by and on a gas

Core

Problem

A gas expands against constant external pressure 2.0 × 10⁵ Pa from 3.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³. Find the work done by and on the gas.
Study the worked solution

  1. Find the signed volume change

    Method

    The volume change is + 2.0 × 10⁻³ m³.

    Reason

    Expansion means V_f-Vᵢ is positive.

    Working

    Δ V = (5.0-3.0) × 10⁻³ = 2.0 × 10⁻³ m³

  2. Calculate work by the gas

    Method

    The gas does + 4.0 × 10² J of work.

    Reason

    For expansion against constant external pressure, W_by = pΔ V.

    Working

    W_by = (2.0 × 10⁵)(2.0 × 10⁻³) = 4.0 × 10² J

  3. Change viewpoint

    Method

    The work done on the gas is -4.0 × 10² J.

    Reason

    Work by the gas transfers energy out of the system, so work on the gas has the opposite sign.

    Working

    Wₒₙ = -W_by = -4.0 × 10² J

Guided practice 2

Link to the first law (syllabus sign convention)

About 4 min

Problem

For the expansion above, Q = +600 J is supplied and Wₒₙ = -400 J. Find Δ U using Δ U = Q + Wₒₙ.

Try this before viewing the solution

Unit: J

Hints

Hint 1: keep one convention
In Δ U = Q + W, W is work done on the gas.
View solution step by step

  1. Assign transfer signs

    Method

    Q = +600 J and Wₒₙ = -400 J.

    Reason

    Heating supplies energy, while expansion transfers energy out through work by the gas.

    Working

    Q > 0, Wₒₙ < 0

  2. Apply the convention

    Method

    The internal energy increases by 200 J.

    Reason

    The energy supplied by heating exceeds the energy transferred out by work.

    Working

    Δ U = 600 + (-400) = +200 J

Common misconception 3

Zeroth law (concept)

Find and correct the mistake

Learner claim

System A and system B are each in thermal equilibrium with thermometer C. A learner concludes that A and B must have equal internal energy. Diagnose the claim.

Try this before viewing the solution

View solution step by step

  1. Apply the zeroth law

    Method

    A and B are in thermal equilibrium with each other and have the same temperature.

    Reason

    Both are separately in equilibrium with the same third system C.

    Working

    A∼ C and B∼ C ⇒ A∼ B

  2. Limit the conclusion

    Method

    Equal temperature does not require equal total internal energy.

    Reason

    Internal energy also depends on particle number, phase and microscopic interactions, whereas temperature tracks mean microscopic kinetic energy.

    Working

    same T⇏ same U

Examiner practice 4

Compression: work done by vs on

4 marks

Examination question

A gas is compressed against constant external pressure 1.2 × 10⁵ Pa from 4.0 × 10⁻³ m³ to 2.5 × 10⁻³ m³. Find W_by and Wₒₙ. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Find the volume change

    1 mark

    Method

    Δ V = -1.5 × 10⁻³ m³.

    Reason

    Compression makes final volume smaller than initial volume.

    Working

    Δ V = (2.5-4.0) × 10⁻³ = -1.5 × 10⁻³ m³

  2. Calculate work by

    1 mark

    Method

    W_by = -1.8 × 10² J.

    Reason

    The negative value shows that the gas is not transferring energy out by expansion.

    Working

    W_by = pΔ V = (1.2 × 10⁵)(-1.5 × 10⁻³) = -180 J

  3. Calculate work on

    1 mark

    Method

    Wₒₙ = +1.8 × 10² J.

    Reason

    The surroundings transfer energy into the gas during compression.

    Working

    Wₒₙ = -W_by = +180 J

  4. State the sign meaning

    1 mark

    Method

    Work on the gas is positive under the syllabus first-law convention.

    Reason

    This is the work term used in Δ U = Q + W.

    Working

    W = Wₒₙ > 0 for compression

Challenge 5

Identify system vs surroundings

Minimal support

Independent transfer

In a piston–cylinder setup, define the system as “the gas in the cylinder”. Give two examples of surroundings and state one possible energy transfer across the system boundary for each.

Try this before viewing the solution

Hints

Hint 1: use the chosen boundary
Everything outside the gas is surroundings; look for mechanical contact and thermal contact.
View solution step by step

  1. Identify mechanical surroundings

    Method

    The piston, cylinder walls or external atmosphere are surroundings.

    Reason

    They lie outside the defined gas system and can exert forces at its boundary.

    Working

    gas | piston and atmosphere

  2. Name mechanical transfer

    Method

    Motion of the piston can transfer energy by work between the gas and surroundings.

    Reason

    Boundary displacement against external pressure produces pΔ V work.

    Working

    moving boundary → work transfer

  3. Identify thermal surroundings

    Method

    A heater or thermal reservoir in contact with the cylinder is also surroundings.

    Reason

    It is outside the chosen gas system but can exchange energy across the wall.

    Working

    temperature difference → heating

7. Mind Stretchers

Mind stretcher 1: Explain why pΔ V has unit of energyExtension

Show Answer

pΔ V has unit Pa·m³.

Since 1 Pa = 1 N m⁻²: Pa · m³ = (N m⁻²) · m³ = N m = J

Mind stretcher 2: Why use external pressure for pΔ V?Extension

In work calculations for a gas in a piston, some students use the gas pressure inside the cylinder. Explain why (in many exam questions) you should use the constant external pressure given instead.

Show Answer

Work done is force × distance on the piston, and the resisting force is set by the external pressure (plus any piston weight), which is often specified as constant.

Using the stated constant external pressure matches the model in the question and gives W_by = pₑₓₜΔ V.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027