First Law of Thermodynamics

The first law of thermodynamics is one of the most fundamental principles in all of physics. It is essentially a statement of the conservation of energy applied to thermal systems: energy cannot be created or destroyed, only transferred from one form to another. In this lesson, we apply it to gases and examine how energy transfers occur in different thermodynamic processes.

The First Law: Statement and Equation

The Edexcel convention for the first law is:

ΔU = Q + W

where:

ΔU = change in internal energy of the system (J)
Q = energy transferred to the system by heating (J) — positive when heat flows into the system
W = work done on the system (J) — positive when work is done on the system (compression)

Some textbooks use the convention ΔU = Q − W (where W is work done by the system). The Edexcel specification uses the convention above, where W is the work done on the gas, so be careful with signs.

In words: the increase in internal energy equals the heat added to the system plus the work done on the system.

Work Done by or on a Gas

When a gas expands against an external pressure, it does work on the surroundings. When a gas is compressed, work is done on the gas by the surroundings.

For a gas expanding at constant pressure:

W = pΔV

where:

W = work done by the gas (J)
p = pressure (Pa)
ΔV = change in volume (m³) = V_final − V_initial

Note: If the gas expands (ΔV > 0), it does positive work on the surroundings. In the Edexcel convention (W = work done on the gas), this means W is negative (work is done by the gas, not on it). If the gas is compressed (ΔV < 0), W on the gas is positive.

Work from a p-V Diagram

On a pressure-volume (p-V) diagram, the work done by the gas equals the area under the curve. This is true whether the process is at constant pressure or varying pressure.

For constant-pressure processes, the area is a simple rectangle: W = pΔV.
For varying-pressure processes (e.g., isothermal expansion), the area must be calculated from the shape of the curve.
For cyclic processes, the net work done is the area enclosed by the loop on the p-V diagram.

The Four Thermodynamic Processes

graph TD
    A["Thermodynamic\nProcesses"] --> B["Isothermal\n(constant T)"]
    A --> C["Adiabatic\n(Q = 0)"]
    A --> D["Isobaric\n(constant p)"]
    A --> E["Isochoric\n(constant V)"]
    B --> B1["ΔU = 0\nQ = −W"]
    C --> C1["ΔU = W\nT changes"]
    D --> D1["W = pΔV\nΔU = Q − pΔV"]
    E --> E1["W = 0\nΔU = Q"]

1. Isothermal Process (Constant Temperature)

Temperature is constant, so ΔU = 0 (for an ideal gas, internal energy depends only on temperature).
From ΔU = Q + W: 0 = Q + W, so Q = −W.
The heat added to the gas equals the work done by the gas (or equivalently, the work done by the gas equals the heat it absorbs).
On a p-V diagram: a curve (hyperbola, since pV = constant from Boyle's law).
Example: slow expansion of a gas in thermal contact with a heat reservoir.

2. Adiabatic Process (No Heat Transfer)

No heat enters or leaves the system: Q = 0.
From ΔU = Q + W: ΔU = W.
If the gas is compressed adiabatically (W > 0, work done on gas), internal energy increases and the temperature rises.
If the gas expands adiabatically (gas does work, W on gas < 0), internal energy decreases and the temperature falls.
On a p-V diagram: a curve steeper than an isothermal curve.
Example: rapid compression in a bicycle pump (gas heats up), rapid expansion of gas escaping from a pressurised container (gas cools).

3. Isobaric Process (Constant Pressure)

Pressure is constant.
Work done by gas: W_by = pΔV.
If the gas is heated (Q > 0), some energy goes into work (expansion) and the rest increases internal energy: ΔU = Q − pΔV (using work done by gas).
On a p-V diagram: a horizontal line.
Example: heating a gas in a cylinder with a freely moving piston at atmospheric pressure.

4. Isochoric (Isovolumetric) Process (Constant Volume)

Volume is constant, so ΔV = 0 and no work is done: W = 0.
From ΔU = Q + W: ΔU = Q.
All the heat supplied goes directly into increasing the internal energy (and hence the temperature).
On a p-V diagram: a vertical line.
Example: heating a gas in a rigid, sealed container.

Summary Table of Processes

Process	Constant	ΔU	Q	W (on gas)	p-V diagram
Isothermal	T	0	−W	−pΔV	Hyperbola (pV = const)
Adiabatic	— (Q = 0)	W	0	ΔU	Steep curve
Isobaric	p	Q + W	Q	−pΔV	Horizontal line
Isochoric	V	Q	ΔU	0	Vertical line

Worked Example 1: Isobaric Expansion

A gas in a cylinder expands at a constant pressure of 2.0 × 10⁵ Pa. The volume increases from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. During this expansion, 600 J of heat is supplied to the gas. Calculate: (a) the work done by the gas, (b) the change in internal energy.

(a) Work done by gas = pΔV = 2.0 × 10⁵ × (3.0 × 10⁻³ − 1.0 × 10⁻³) = 2.0 × 10⁵ × 2.0 × 10⁻³ = 400 J

(b) Using Edexcel convention: ΔU = Q + W, where W = work done ON gas = −400 J (since gas did work)

ΔU = 600 + (−400) = 200 J

The internal energy increases by 200 J. Of the 600 J of heat supplied, 400 J went into doing work (pushing back the surroundings) and 200 J went into increasing the internal energy of the gas.

Worked Example 2: Adiabatic Compression

A gas is compressed adiabatically. The work done on the gas is 150 J. What is the change in internal energy and what happens to the temperature?

Adiabatic means Q = 0.

ΔU = Q + W = 0 + 150 = 150 J

The internal energy increases by 150 J, so the temperature of the gas rises. This is why a bicycle pump gets hot when you pump quickly — the compression is approximately adiabatic.

Worked Example 3: Isothermal Expansion

During an isothermal expansion of an ideal gas, 250 J of work is done by the gas. How much heat is absorbed by the gas?

Isothermal means ΔU = 0 (for an ideal gas).

ΔU = Q + W → 0 = Q + (−250) → Q = 250 J

The gas absorbs 250 J of heat from the surroundings, and all of this energy is used to do the 250 J of work done during the expansion. No internal energy changes.

Worked Example 4: Complete First Law Problem

An ideal monatomic gas (n = 2.0 mol) is heated from 300 K to 500 K at constant pressure of 1.0 × 10⁵ Pa. Calculate: (a) the change in internal energy, (b) the work done by the gas, (c) the heat supplied.

(a) For a monatomic ideal gas: ΔU = (3/2)nRΔT = 1.5 × 2.0 × 8.31 × 200 = 4986 J ≈ 5.0 kJ

(b) At constant pressure, the gas expands. Using pV = nRT: V₁ = nRT₁/p = (2.0 × 8.31 × 300)/(1.0 × 10⁵) = 0.04986 m³ V₂ = nRT₂/p = (2.0 × 8.31 × 500)/(1.0 × 10⁵) = 0.08310 m³

Work done by gas = pΔV = 1.0 × 10⁵ × (0.08310 − 0.04986) = 1.0 × 10⁵ × 0.03324 = 3324 J ≈ 3.3 kJ

Alternatively: W_by = nRΔT = 2.0 × 8.31 × 200 = 3324 J (this shortcut works for isobaric processes of ideal gases).

Check: of the 8.3 kJ supplied, 3.3 kJ went into work (expanding against atmosphere) and 5.0 kJ went into internal energy. The ratio Q/ΔU = 8.3/5.0 = 5/3 — this is the ratio of C_p to C_v for a monatomic ideal gas.

Worked Example 5: Cyclic Process

A gas undergoes a cyclic process ABCA on a p-V diagram:

A → B: isobaric expansion at 3.0 × 10⁵ Pa from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³
B → C: isochoric (constant volume) pressure drop to 1.0 × 10⁵ Pa
C → A: isobaric compression at 1.0 × 10⁵ Pa back to 1.0 × 10⁻³ m³

(a) Calculate the work done by the gas in each stage.

A → B: W = pΔV = 3.0 × 10⁵ × 2.0 × 10⁻³ = 600 J (expansion, gas does work) B → C: ΔV = 0, so W = 0 J C → A: W = pΔV = 1.0 × 10⁵ × (−2.0 × 10⁻³) = −200 J (compression, work done on gas)

(b) What is the net work done by the gas per cycle?

Net W = 600 + 0 + (−200) = 400 J

This equals the area of the rectangle enclosed on the p-V diagram: Area = (3.0 − 1.0) × 10⁻³ × (3.0 − 1.0) × 10⁵ = 2.0 × 10⁻³ × 2.0 × 10⁵ = 400 J ✓

(c) What is the change in internal energy over one complete cycle?

The gas returns to its initial state, so ΔU = 0 for the complete cycle. By the first law: Q_net = −W_on = +400 J. The gas absorbs a net 400 J of heat per cycle and converts it to 400 J of net work — this is the principle of a heat engine.

p-V Diagrams: Key Features

Feature	What it represents
Area under any curve (expansion)	Work done by the gas
Area under any curve (compression)	Work done on the gas
Area enclosed by a complete cycle	Net work per cycle
Clockwise cycle	Heat engine (net work by gas)
Anticlockwise cycle	Heat pump/refrigerator (net work on gas)
Isothermal curve (hyperbola)	Temperature constant; ΔU = 0 for ideal gas
Adiabatic curve (steeper than isothermal)	No heat transfer; temperature changes
Horizontal line	Isobaric process (constant p)
Vertical line	Isochoric process (constant V, no work)

Gibbs Free Energy: A Brief Introduction

While the first law tells us whether a process is energetically possible, in chemistry and materials science, spontaneity is governed by the Gibbs free energy:

ΔG = ΔH − TΔS

where ΔH is the enthalpy change, T is absolute temperature, and ΔS is the entropy change. A process is spontaneous at constant temperature and pressure when ΔG < 0.

For an ideal gas expanding isothermally from V₁ to V₂:

ΔG = nRT ln(V₁/V₂) (which is negative for expansion, confirming spontaneity)

This links thermodynamics to the direction of natural processes — something the first law alone cannot determine. While detailed Gibbs free energy calculations are beyond the Edexcel A-Level physics syllabus, the concept illustrates that energy conservation (first law) is necessary but not sufficient to predict whether a process will actually occur.

Common Exam Mistakes

Mistake	How to avoid it
Getting the sign of W wrong	Clarify: does the question mean W on gas or W by gas?
Saying ΔU = 0 for adiabatic process	ΔU = 0 is isothermal; adiabatic means Q = 0, so ΔU = W
Forgetting work in isobaric heating	Not all heat goes to ΔU — some does work: Q = ΔU + pΔV
Saying work = 0 for all constant-T processes	Work is zero only for constant-V; in isothermal expansion, W ≠ 0
Confusing the area under the curve with the area of the enclosed cycle	Under = work in one process; enclosed = net work per cycle

Summary

First law: ΔU = Q + W (Edexcel convention: W = work done on gas).
Work done by gas at constant pressure: W = pΔV.
Isothermal: ΔU = 0, so Q = −W (heat absorbed = work done by gas).
Adiabatic: Q = 0, so ΔU = W (temperature changes when gas does/has work done on it).
Isobaric: constant pressure, both Q and W contribute to ΔU.
Isochoric: ΔV = 0, W = 0, so ΔU = Q.
On a p-V diagram, the area under the curve is the work done; the enclosed area of a cycle is the net work.
For a complete cycle: ΔU = 0, so net Q = net W.

A-Level Deep Dive: First Law of Thermodynamics

Spec mapping

Edexcel 9PH0 specification Topic 9 — Thermodynamics covers the first law of thermodynamics expressed as ΔU = Q + W, where ΔU is the change in internal energy, Q is the heat transferred to the system, and W is the work done on the system; learners must apply the law to isothermal, adiabatic, isobaric and isochoric processes, and use it in conjunction with the ideal-gas equation pV = nRT (refer to the official specification document for exact wording). The first law is examined primarily on Paper 2 (Advanced Physics II) and is highly synoptic: it ties Topic 5 (Conservation of energy) to Topic 14 (Gravitational fields, where adiabatic compression appears in stellar models). The Edexcel formula booklet provides ΔU = Q + W and W = pΔV but does not state the sign convention — candidates must memorise it and apply it consistently.

Worked example with full mark scheme

Question (8 marks):

A fixed mass of ideal gas is contained in a cylinder fitted with a frictionless piston. The gas is initially at pressure 1.0 × 10⁵ Pa and volume 4.0 × 10⁻³ m³.

(a) The gas is compressed isothermally to a volume of 1.0 × 10⁻³ m³. The work done on the gas is 555 J. State, with reasoning, the values of ΔU and Q for this process. (4)

(b) The gas is then allowed to expand adiabatically back to its original volume. Explain qualitatively how the temperature, internal energy and pressure of the gas change during this expansion, using the first law. (4)

Solution with mark scheme:

(a) Step 1 — apply the definition of an isothermal process.

For an isothermal change, temperature T is constant. For an ideal gas, internal energy U depends only on T (since U = (3/2)nRT for a monatomic ideal gas, or proportional to T more generally). Therefore ΔU = 0.

M1 — stating that internal energy of an ideal gas depends only on temperature.

A1 — concluding ΔU = 0 J.

Step 2 — apply the first law.

ΔU = Q + W, with W = +555 J (work done on the gas is positive in the Edexcel convention). Substituting ΔU = 0:

$0 = Q + 555 \implies Q = -555 \text{ J}$

M1 — correct substitution into ΔU = Q + W with the right sign for W.

A1 — Q = −555 J, with the explicit interpretation that 555 J of heat is transferred out of the gas to the surroundings (because Q is negative).

(b) Step 1 — characterise an adiabatic process.

In an adiabatic change, no heat is transferred between system and surroundings, so Q = 0.

B1 — stating Q = 0 for an adiabatic process.

Step 2 — apply the first law to expansion.

ΔU = Q + W. The gas expands, so the gas does work on the surroundings; therefore work done on the gas is negative (W < 0). With Q = 0:

$\Delta U = 0 + W < 0$

M1 — recognising that W is negative for an expansion under the Edexcel sign convention.

Step 3 — link ΔU to temperature.

Since ΔU < 0, the temperature falls (internal energy of an ideal gas is proportional to T). With T falling and V increasing, by pV = nRT pressure must also fall (and faster than for the equivalent isothermal expansion, because both T and V act to reduce p).

M1 — temperature falls, internal energy decreases.

A1 — pressure falls, with a coherent explanation tying the result to the first law and the gas equation.

Total: 8 marks (M4 A3 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A cylinder of ideal gas at 300 K is compressed isobarically at 1.5 × 10⁵ Pa from 2.0 × 10⁻³ m³ to 1.2 × 10⁻³ m³. During the compression, 320 J of heat is removed from the gas.

(a) Calculate the work done on the gas. (2)

(b) Hence calculate the change in internal energy of the gas, stating whether it increases or decreases. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2) — applying W = −pΔV with the correct Edexcel sign convention (work on the gas is positive when ΔV is negative). ΔV = 1.2 × 10⁻³ − 2.0 × 10⁻³ = −0.8 × 10⁻³ m³.
A1 (AO2) — W = −(1.5 × 10⁵)(−0.8 × 10⁻³) = +120 J.

(b)

M1 (AO2) — Q = −320 J (heat removed from gas).
M1 (AO2) — ΔU = Q + W = −320 + 120 = −200 J.
A1 (AO3) — internal energy decreases by 200 J; temperature falls.

(c)

B1 (AO1) — gas behaves as an ideal gas / pressure remains constant throughout / no friction in piston / no heat exchange with the cylinder walls (any one).