Ch 3 · 16 Thermodynamics · A Level Physics 9702
Topic 3 of 3 in Thermal Physics — A Level Physics 9702 — 39 lessons.
Introduction
Engines, refrigerators and even your own body all convert energy between heat and work. The first law of thermodynamics is conservation of energy written for thermal systems: the internal energy of a gas changes only if you heat it or do work on it. (Syllabus topic 16: sections 16.1–16.2.)
Key ideas — 16.1 Internal energy
The internal energy U of a system is the sum of a random distribution of kinetic and potential energies of its molecules:
🧑🏫 Teacher's explanation
Imagine a crowded dance floor. Each person is moving around randomly (kinetic energy), and there are also invisible elastic bands connecting people to their neighbors (potential energy). The total energy of the entire crowd is the sum of everyone's individual movement and the energy stored in those elastic connections. This total sum is the 'internal energy' of the dance floor.
📖 Definitions
🔍 Worked example — Easy
A gas is compressed at a constant pressure of 1.5 × 10^5 Pa. The volume decreases from 0.50 m^3 to 0.20 m^3. Calculate the work done on the gas.
🔍 Worked example — Standard
A system undergoes a process where 250 J of heat is added to the system, and the system performs 120 J of work on its surroundings. Calculate the change in internal energy Δ U.
🔍 Worked example — Exam-style
An ideal gas is heated at constant volume. The temperature increases from 300 K to 350 K. If the internal energy of the gas was 1200 J at 300 K, what is the new internal energy? (Assume C_v = 20 J/K for this specific gas).
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — 16.2 (a) Work done by or on a gas
When a gas changes volume by Δ V at constant pressure p:
🧑🏫 Teacher's explanation
Think of a gas as a collection of tiny, frantic bumper cars. Each car has kinetic energy (speed) and potential energy (the 'springiness' of the space between them). The sum of all these energies is the internal energy of the gas. When the temperature rises, the cars move faster, so the internal energy increases.
📖 Definitions
🔍 Worked example — Easy
A gas at a constant pressure of 1.5 × 10^5 Pa expands from a volume of 2.0 × 10^-3 m^3 to 5.0 × 10^-3 m^3. Calculate the work done by the gas.
🔍 Worked example — Standard
A gas is compressed at a constant pressure of 2.0 × 10^5 Pa. The volume decreases from 4.0 × 10^-3 m^3 to 1.0 × 10^-3 m^3. Calculate the work done ON the gas.
🔍 Worked example — Exam-style
A system undergoes a process where 400 J of heat is added to the gas (q = +400 J). During this process, the gas expands at a constant pressure of 1.0 × 10^5 Pa, doing 150 J of work on the surroundings. Calculate the change in internal energy Δ U using Δ U = q + W, where W is the work done ON the gas.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — 16.2 (b) The first law of thermodynamics
Δ U = q + W
🧑🏫 Teacher's explanation
Think of a bank account for energy. The 'Internal Energy' (Δ U) is your current balance. 'Heating' (q) is like a cash deposit into your account. 'Work' (W) is a bit trickier: if someone pushes money into your account (compression), it's a deposit (+); if you spend money to buy something (expansion), it's a withdrawal (-).
📖 Definitions
🔍 Worked example — Easy
A gas is heated by 500 J at constant volume. Calculate the change in internal energy Δ U.
🔍 Worked example — Standard
A gas expands from 0.5 m^3 to 0.8 m^3 at a constant pressure of 1.2× 10^5 Pa. Calculate the work done by the gas.
🔍 Worked example — Exam-style
A gas is compressed at constant temperature. 200 J of work is done on the gas. How much heat is transferred to the gas?
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Example 1. A gas at a constant pressure of 1.0 × 10^5 Pa expands from 0.020 m³ to 0.030 m³ while 2500 J of heat is supplied. Find the change in internal energy.
🔍 Worked example — Easy
A gas at a constant pressure of 1.2 × 10^5 Pa expands from 0.050 m^3 to 0.080 m^3. Calculate the work done by the gas.
🔍 Worked example — Standard
A gas is compressed at a constant pressure of 2.0 × 10^5 Pa from 0.010 m^3 to 0.004 m^3. During this process, 500 J of heat is transferred to the gas. Find the change in internal energy Δ U.
🔍 Worked example — Exam-style
A system undergoes a process where 1500 J of heat is removed from the gas. Simultaneously, the gas performs 400 J of work on its surroundings. Calculate the change in internal energy Δ U.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) The sum of a random distribution of the kinetic and potential energies of the molecules in a system. (2) An ideal gas has no intermolecular forces, so no potential energy; its internal energy is the molecules' kinetic energy, which depends on temperature. (3) Δ U = q + W: increase in internal energy = energy transferred to the system by heating + work done on the system. (4) Δ U = 300 + (-300) = 0: no change in internal energy, so (for an ideal gas) the temperature stays the same. (5) Δ V = 2.5 × 10^-3 m³; W = 2.0 × 10^5 × 2.5 × 10^-3 = 500 J.