Thermodynamic Systems

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, Wby (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:
  • Wby = 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: Wby = Fx = (pA)x

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

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

• Expansion: Δ V > 0 Rightarrow Wby > 0 and Wₒₙ < 0
• Compression: Δ V < 0 Rightarrow Wby < 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

Example 1: Work done by and on a gasCore

A gas expands against constant external pressure p = 2.0 × 10⁵ Pa from Vᵢ = 3.0 × 10⁻³ m³ to Vf = 5.0 × 10⁻³ m³. Find Wby and Wₒₙ.

Show Answer

Δ V = 2.0 × 10⁻³ m³; Wby = pΔ V = (2.0 × 10⁵)(2.0 × 10⁻³) = 4.0 × 10² J; Wₒₙ = -pΔ V = -4.0 × 10² J

Example 2: Link to the first law (syllabus sign convention)Core

In Example 1, Q = + 600 J of heat is supplied to the gas. Find Δ U using Δ U = Q + Wₒₙ.

Show Answer

Δ U = Q + Wₒₙ = 600 + (-400) = 200 J

Example 3: Zeroth law (concept)Core

System A is in thermal equilibrium with thermometer C. System B is also in thermal equilibrium with thermometer C. What can you conclude about A and B?

Show Answer

By the zeroth law, A is in thermal equilibrium with B (so they have the same temperature).

Example 4: Compression: work done by vs onCore

A gas is compressed against constant external pressure p = 1.2 × 10⁵ Pa from Vᵢ = 4.0 × 10⁻³ m³ to Vf = 2.5 × 10⁻³ m³.

Find Wby and Wₒₙ.

Show Answer

Δ V = Vf - Vᵢ = -1.5 × 10⁻³ m³ Wby = pΔ V = (1.2 × 10⁵)(-1.5 × 10⁻³) = -1.8 × 10² J Wₒₙ = -pΔ V = + 1.8 × 10² J

Example 5: Identify system vs surroundingsCore

In a piston-cylinder setup, the system is “the gas in the cylinder”.

Give two examples of what counts as the surroundings in this setup.

Show Answer

Examples include:

• the piston and cylinder walls
• the external atmosphere applying pressure on the piston
• any heater or thermal reservoir in contact with the cylinder

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: Pacdotm³ = (N m⁻²)cdotm³ = 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 Wby = pₑₓₜΔ V.

8. Practice, Quiz and Next Step

Close your notes and use Thermodynamic Systems in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: An unfamiliar data set or physical system requires you to apply Thermodynamic Systems while stating the model, regime and assumptions.

1. Retrieve: define thermodynamic systems in your own words, including units, sign or conditions where relevant.
2. Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
3. Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.

Check the response before looking back

• The model, regime, coordinates and assumptions are explicit.
• The derivation or multi-step reasoning is visible rather than implied.
• The conclusion is tested against units, data quality and a limiting case.
• A practical control, uncertainty or model limitation is evaluated where applicable.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.