First law of thermodynamics (A-level only) - Revision Guide

Thermodynamic system diagram

Unpacking the terms and sign conventions

The single biggest place students lose marks in this topic is using the wrong positive or negative sign. Let's look at exactly what AQA expects.

1. Heating ( $Q$ )

$Q$ is the thermal energy transferred due to a temperature difference.

Positive ( $+ Q$ ): Heat is transferred into the gas (e.g., you put a Bunsen burner under the cylinder).
Negative ( $- Q$ ): Heat is transferred out of the gas (e.g., the cylinder is placed in an ice bath).

2. Internal Energy ( $ΔU\Delta U$ )

Internal energy is the sum of the randomly distributed kinetic and potential energies of the gas molecules. For an ideal gas, it is directly proportional to its absolute temperature.

Positive ( $+ΔU+\Delta U$ ): The internal energy increases. This means the gas is getting hotter.
Negative ( $−ΔU-\Delta U$ ): The internal energy decreases. This means the gas is getting cooler.

3. Work Done ( $W$ )

Work is done when a force moves through a distance. For a gas, this happens when the volume changes (the piston moves).

Positive ( $+ W$ ): Work is done by the gas. The gas pushes the piston outwards. It is expanding.
Negative ( $- W$ ): Work is done on the gas. An external force pushes the piston inwards. The gas is being compressed.

Key Idea

AQA Sign Convention Summary

Remember the standard AQA perspective: You are the gas.

If someone gives you heat, $Q$ is positive.
If your internal temperature goes up, $ΔU\Delta U$ is positive.
If you push out and do work on the world, $W$ is positive.

Common Mistake

Physics vs Chemistry Equations

If you study A-Level Chemistry, you might have learned the first law as $ΔU=Q+W\Delta U = Q + W$ . Do not use this in your Physics exam!

In chemistry, $W$ is usually defined as work done on the system. In AQA Physics, $W$ is strictly defined as work done by the system. Always stick to $\Delta U + W$ for your physics papers.

Analogy

The Bank Account Analogy

Think of the gas as your bank account:

$Q$ is your salary (money coming in).
$W$ is the bills you pay (money going out to do useful things).
$ΔU\Delta U$ is the change in your savings balance.

If your salary is £2000 ( $Q = + 2000$ ) and you pay £1500 in bills ( $W = + 1500$ ), your savings balance increases by £500 ( $ΔU=+500\Delta U = +500$ ). Equation: $2000 = 500 + 1500$ . It balances perfectly!

Applying the First Law

Let's see how this works in a standard exam-style calculation.

Example

Example 1: Expanding gas

A gas is heated by a burner, absorbing $\text{ J}$ of thermal energy. As it is heated, it expands, pushing a piston outwards and doing $\text{ J}$ of work against the atmosphere. Calculate the change in internal energy of the gas and state what happens to its temperature.

Assign signs to the given values: The gas absorbs heat, so $\text{ J}$ . The gas expands and does work, so $\text{ J}$ .
State the First Law equation: $\Delta U + W$
Substitute and rearrange: $\Delta U + 300$ $ΔU=850−300\Delta U = 850 - 300$ $J\Delta U = +550 \text{ J}$
Interpret the result: The change in internal energy is positive ( $\text{ J}$ ), which means the temperature of the gas increases.

Dealing with Compressions and Cooling

Things get a little trickier when the gas is being squashed or cooled down. You must be very careful with negative signs in your algebra.

Example

Example 2: Compressing a cooling gas

A mechanic rapidly compresses the gas in a cylinder, doing $\text{ J}$ of work on the gas. During this process, $\text{ J}$ of thermal energy is lost to the cold surroundings. Calculate the change in internal energy.

Assign signs to the given values: Work is done on the gas (it is compressed), so $W$ is negative: $\text{ J}$ . Thermal energy is lost to the surroundings, so $Q$ is negative: $\text{ J}$ .
State the First Law equation: $\Delta U + W$
Substitute the values carefully: $\Delta U + (-420)$
Rearrange for $ΔU\Delta U$ : $ΔU=−150−(−420)\Delta U = -150 - (-420)$ $ΔU=−150+420\Delta U = -150 + 420$ $J\Delta U = +270 \text{ J}$
Interpret the result: Even though the gas lost heat, the huge amount of work done on it caused its overall internal energy to increase. Its temperature goes up!

Special Cases

Sometimes, the exam question describes a scenario where one of the three terms is zero.

"Constant volume" (Isochoric): If the volume cannot change, the piston cannot move. No work is done, so $W = 0$ . The equation becomes $\Delta U$ . All heat added goes directly into raising the temperature.
"Thermally insulated" / "No heat transfer" (Adiabatic): If the cylinder is perfectly insulated, or the process happens so fast there is no time for heat to enter or leave, then $Q = 0$ . The equation becomes $\Delta U + W$ , meaning $ΔU=−W\Delta U = -W$ . If the gas does work (expands), it must use its own internal energy to do so, causing it to cool down rapidly.

Hint

Connecting topics

In upcoming topics, you will learn how to calculate the work done ( $W$ ) yourself using the equation $\Delta V$ (pressure $×\times$ change in volume), and how to calculate $ΔU\Delta U$ using the specific heat capacity of a gas. You will then plug those values right back into $\Delta U + W$ .

Exam technique

In the exam

Read carefully for direction words. Highlight words like "expands", "compressed", "absorbs", "loses", "heats up", and "cools".
Write out the variables with signs first. Before you do any math, write down $\text{...}$ , $\text{...}$ , $ΔU=...\Delta U = \text{...}$ with a $+$ or $-$ next to every number.
Trust the algebra. When rearranging an equation with negative numbers, take it one step at a time to avoid dropping a minus sign.
Check your final answer makes physical sense. If you squeezed a gas really hard (work done on it) and it didn't lose much heat, the temperature must go up. Check that your final $ΔU\Delta U$ is positive.

Self review

Check yourself

What does the symbol $Q$ represent in the first law of thermodynamics?
If a gas is compressed by an external force, is $W$ positive or negative in AQA Physics?
A gas is kept at a constant volume while being heated. Which term in the equation $\Delta U + W$ must be exactly zero?
If $ΔU\Delta U$ is negative at the end of your calculation, what does that tell you about the temperature of the ideal gas?

Thermodynamic system diagram

Unpacking the terms and sign conventions

The single biggest place students lose marks in this topic is using the wrong positive or negative sign. Let's look at exactly what AQA expects.

1. Heating ( $Q$ )

$Q$ is the thermal energy transferred due to a temperature difference.

Positive ( $+ Q$ ): Heat is transferred into the gas (e.g., you put a Bunsen burner under the cylinder).
Negative ( $- Q$ ): Heat is transferred out of the gas (e.g., the cylinder is placed in an ice bath).

2. Internal Energy ( $ΔU\Delta U$ )

Internal energy is the sum of the randomly distributed kinetic and potential energies of the gas molecules. For an ideal gas, it is directly proportional to its absolute temperature.

Positive ( $+ΔU+\Delta U$ ): The internal energy increases. This means the gas is getting hotter.
Negative ( $−ΔU-\Delta U$ ): The internal energy decreases. This means the gas is getting cooler.

3. Work Done ( $W$ )

Work is done when a force moves through a distance. For a gas, this happens when the volume changes (the piston moves).

Positive ( $+ W$ ): Work is done by the gas. The gas pushes the piston outwards. It is expanding.
Negative ( $- W$ ): Work is done on the gas. An external force pushes the piston inwards. The gas is being compressed.

Key Idea

AQA Sign Convention Summary

Remember the standard AQA perspective: You are the gas.

If someone gives you heat, $Q$ is positive.
If your internal temperature goes up, $ΔU\Delta U$ is positive.
If you push out and do work on the world, $W$ is positive.

Common Mistake

Physics vs Chemistry Equations

If you study A-Level Chemistry, you might have learned the first law as $ΔU=Q+W\Delta U = Q + W$ . Do not use this in your Physics exam!

In chemistry, $W$ is usually defined as work done on the system. In AQA Physics, $W$ is strictly defined as work done by the system. Always stick to $\Delta U + W$ for your physics papers.

Analogy

The Bank Account Analogy

Think of the gas as your bank account:

$Q$ is your salary (money coming in).
$W$ is the bills you pay (money going out to do useful things).
$ΔU\Delta U$ is the change in your savings balance.

Applying the First Law

Let's see how this works in a standard exam-style calculation.

Example

Example 1: Expanding gas

Assign signs to the given values: The gas absorbs heat, so $\text{ J}$ . The gas expands and does work, so $\text{ J}$ .
State the First Law equation: $\Delta U + W$
Substitute and rearrange: $\Delta U + 300$ $ΔU=850−300\Delta U = 850 - 300$ $J\Delta U = +550 \text{ J}$
Interpret the result: The change in internal energy is positive ( $\text{ J}$ ), which means the temperature of the gas increases.

Dealing with Compressions and Cooling

Things get a little trickier when the gas is being squashed or cooled down. You must be very careful with negative signs in your algebra.

Example

Example 2: Compressing a cooling gas

Assign signs to the given values: Work is done on the gas (it is compressed), so $W$ is negative: $\text{ J}$ . Thermal energy is lost to the surroundings, so $Q$ is negative: $\text{ J}$ .
State the First Law equation: $\Delta U + W$
Substitute the values carefully: $\Delta U + (-420)$
Rearrange for $ΔU\Delta U$ : $ΔU=−150−(−420)\Delta U = -150 - (-420)$ $ΔU=−150+420\Delta U = -150 + 420$ $J\Delta U = +270 \text{ J}$
Interpret the result: Even though the gas lost heat, the huge amount of work done on it caused its overall internal energy to increase. Its temperature goes up!

Special Cases

Sometimes, the exam question describes a scenario where one of the three terms is zero.

"Constant volume" (Isochoric): If the volume cannot change, the piston cannot move. No work is done, so $W = 0$ . The equation becomes $\Delta U$ . All heat added goes directly into raising the temperature.
"Thermally insulated" / "No heat transfer" (Adiabatic): If the cylinder is perfectly insulated, or the process happens so fast there is no time for heat to enter or leave, then $Q = 0$ . The equation becomes $\Delta U + W$ , meaning $ΔU=−W\Delta U = -W$ . If the gas does work (expands), it must use its own internal energy to do so, causing it to cool down rapidly.

Hint

Connecting topics

Exam technique

In the exam

Read carefully for direction words. Highlight words like "expands", "compressed", "absorbs", "loses", "heats up", and "cools".
Write out the variables with signs first. Before you do any math, write down $\text{...}$ , $\text{...}$ , $ΔU=...\Delta U = \text{...}$ with a $+$ or $-$ next to every number.
Trust the algebra. When rearranging an equation with negative numbers, take it one step at a time to avoid dropping a minus sign.
Check your final answer makes physical sense. If you squeezed a gas really hard (work done on it) and it didn't lose much heat, the temperature must go up. Check that your final $ΔU\Delta U$ is positive.

Self review

Check yourself

What does the symbol $Q$ represent in the first law of thermodynamics?
If a gas is compressed by an external force, is $W$ positive or negative in AQA Physics?
A gas is kept at a constant volume while being heated. Which term in the equation $\Delta U + W$ must be exactly zero?
If $ΔU\Delta U$ is negative at the end of your calculation, what does that tell you about the temperature of the ideal gas?

Unpacking the terms and sign conventions

1. Heating (QQQ)

2. Internal Energy (ΔU\Delta UΔU)

3. Work Done (WWW)

Applying the First Law

Dealing with Compressions and Cooling

Special Cases

First law of thermodynamics (A-level only) Revision Guide

Unpacking the terms and sign conventions

1. Heating (QQQ)

2. Internal Energy (ΔU\Delta UΔU)

3. Work Done (WWW)

Applying the First Law

Dealing with Compressions and Cooling

Special Cases

First law of thermodynamics (A-level only) Revision Guide

1. Heating ( $Q$ )

2. Internal Energy ( $ΔU\Delta U$ )

3. Work Done ( $W$ )

1. Heating ( $Q$ )

2. Internal Energy ( $ΔU\Delta U$ )

3. Work Done ( $W$ )