The First Law of Thermodynamics Revision Notes | A Level AQA Physics

1. Why Energy Accounting Matters in Thermodynamics

Every engine, refrigerator, and power station is governed by one ruthlessly enforced rule: energy cannot be created or destroyed. The first law of thermodynamics is that rule applied to heat and work — and it tells you exactly where every joule goes.

Picture a gas trapped inside a cylinder with a movable piston. You supply heat to the gas. Two things can happen with that energy: the gas can get hotter (its internal energy rises), or it can push the piston outward and do mechanical work on the surroundings. The first law says those two destinations must account for every joule you put in — no exceptions.

2. The First Law: The Equation and Its Sign Convention

The AQA specification states the first law as $Q = \Delta U + W$, where each symbol carries a precise meaning and a sign convention you must get right before attempting any calculation.

The sign convention is the part that trips students up most often. Think of the system as a bank account. $Q$ is money paid in — positive means heat flows into the system, negative means heat flows out. $W$ is money spent — positive means the system does work on the surroundings (spends energy), negative means the surroundings do work on the system (energy is deposited back in). $\Delta U$ is the resulting change in the account balance.

Rearranging the equation gives you two useful forms: $\Delta U = Q - W$ (internal energy rises when heat in exceeds work out) and $W = Q - \Delta U$ (work done by the system equals heat in minus the rise in internal energy). All three forms are equivalent — choose whichever has the unknown on the left.

3. Internal Energy: What It Actually Means

Internal energy $U$ is the sum of the random kinetic energies and potential energies of all the molecules in the system. For an ideal gas, intermolecular potential energy is zero, so internal energy depends only on the kinetic energy of the molecules — which means it depends only on temperature.

This is a crucial point for calculations: if the temperature of an ideal gas does not change, $\Delta U = 0$, regardless of what else is happening. That constraint unlocks a powerful simplification in isothermal processes, which you will see shortly.

4. Work Done by a Gas: Connecting $W$ to Pressure and Volume

Before applying the first law quantitatively, you need a way to calculate $W$. When a gas expands at constant pressure $p$ and its volume changes by $\Delta V$, the work done by the gas is given by $W = p\Delta V$. This comes directly from the definition of work as force times distance, applied to a piston of area $A$ moving a distance $d$: $W = F \times d = pA \times d = p\Delta V$.

When pressure is not constant, $W$ is the area under a pressure-volume ($p$-$V$) graph. For AQA A Level, you are expected to read this area from a graph or use $W = p\Delta V$ when pressure is stated as constant. Always check whether the question specifies constant pressure before using this formula.

5. Applying the First Law to Specific Thermodynamic Processes

The real power of the first law emerges when you apply it to processes with a specific constraint — constant temperature, constant volume, constant pressure, or no heat exchange at all. Each constraint simplifies the equation in a different way, and AQA examiners test whether you can identify which simplification applies.

The adiabatic case deserves particular attention. When a gas expands adiabatically ($Q = 0$), it does positive work on the surroundings ($W > 0$), so $\Delta U = -W < 0$: internal energy falls and the gas cools. This is why air cools as it rises and expands in the atmosphere, and why a bicycle pump gets warm when you compress air rapidly — the compression is approximately adiabatic, so work done on the gas raises its internal energy and temperature.

6. Worked Examples: Quantitative Application

Exam questions on the first law almost always give you two of the three quantities and ask for the third. The method is always the same: identify the process, write the appropriate form of $Q = \Delta U + W$, substitute, and check the sign of your answer makes physical sense.

Worked example: First Law Calculation: Gas Expanding at Constant Pressure

A fixed mass of gas is heated at constant pressure. It absorbs 1200 J of heat and expands from $2.0 \times 10^{-3}$ m$^3$ to $5.0 \times 10^{-3}$ m$^3$ against a constant pressure of $1.0 \times 10^5$ Pa. Calculate (a) the work done by the gas and (b) the increase in internal energy.

1. $Q = +1200$ J (heat into system) $p = 1.0 \times 10^5$ Pa $V_i = 2.0 \times 10^{-3}$ m$^3$, $V_f = 5.0 \times 10^{-3}$ m$^3$ — Label each quantity with its sign. Heat enters the system so $Q$ is positive. The gas expands so we expect $W$ to be positive.
2. $\Delta V = V_f - V_i = 5.0 \times 10^{-3} - 2.0 \times 10^{-3} = 3.0 \times 10^{-3}$ m$^3$ — Always subtract initial from final. A positive result confirms expansion. [M1]
3. $W = p\Delta V = 1.0 \times 10^5 \times 3.0 \times 10^{-3}$ $W = 1.0 \times 10^5 \times 3.0 \times 10^{-3} = 300$ J — Expanding gas does positive work on surroundings. Check: $10^5 \times 10^{-3} = 10^2 = 100$, then $100 \times 3.0 = 300$ J. [M1][A1]
4. $Q = \Delta U + W$ $1200 = \Delta U + 300$ $\Delta U = 1200 - 300 = 900$ J — Rearrange to $\Delta U = Q - W$. The internal energy increases by 900 J — the gas gets hotter. [M1][A1]
5. Energy in: $Q = 1200$ J Energy out as work: $W = 300$ J Energy stored: $\Delta U = 900$ J $300 + 900 = 1200$ J ✓ — The three quantities sum correctly. Energy is conserved, confirming the calculation is consistent.

Answer: (a) Work done by the gas: $W = 300$ J. (b) Increase in internal energy: $\Delta U = 900$ J.

Worked example: First Law Calculation: Adiabatic Compression

A gas is compressed adiabatically. The surroundings do 450 J of work on the gas. Calculate the change in internal energy of the gas and state whether the temperature of the gas increases or decreases.

1. Adiabatic: $Q = 0$ Work done on the gas by surroundings $= 450$ J — Adiabatic means no heat exchange with surroundings. The surroundings do work on the gas, so from the gas's perspective, $W$ is negative.
2. $W = -450$ J — In the AQA convention, $W$ is work done by the system. Work done on the system is the negative of this: $W = -450$ J. This is the most common sign error in adiabatic questions.
3. $Q = \Delta U + W$ $0 = \Delta U + (-450)$ $\Delta U = +450$ J — With $Q = 0$: $\Delta U = -W = -(-450) = +450$ J. Internal energy increases. [M1][A1]
4. $\Delta U = +450$ J > 0, so temperature increases. — For an ideal gas, internal energy is proportional to temperature. A rise in $U$ means a rise in $T$. Compressing a gas adiabatically always raises its temperature.

Answer: $\Delta U = +450$ J. The internal energy increases, so the temperature of the gas rises.

7. Exam Technique and Common Mistakes

The first law is a short equation, but AQA mark schemes reveal that sign errors and failure to identify the process account for the majority of lost marks. The guidance below targets the specific errors that appear most frequently in examiner reports.

The first law of thermodynamics — $Q = \Delta U + W$ — is the quantitative statement of energy conservation for any thermodynamic process, and every AQA question on this topic reduces to correctly identifying $Q$, $\Delta U$, and $W$ with their signs, then solving for the unknown.