Ideal gases (A-level only)

Welcome to the thermal physics of gases! You already know that gases are made of particles zooming around randomly. In this section, we zoom out to look at the "big picture" (macroscopic) properties of a gas: pressure, volume, and temperature.

Here is what you'll learn:

How the three experimental gas laws link pressure, volume, and temperature.
How to extrapolate experimental graphs to find absolute zero.
How to use the ideal gas equations ( $p V = n R T$ and $p V = N k T$ ) to find the state of a gas.
How to calculate the work done when a gas expands or is compressed.

The Gas Laws and Absolute Zero

Before we can combine everything into one master equation, we need to look at the three historical gas laws. These laws show how two variables interact while the other variables (and the mass of the gas) are kept completely constant.

You will investigate two of these in Required Practical 8.

Boyle's Law (Constant Temperature) For a fixed mass of gas at a constant temperature, the pressure $p$ is inversely proportional to the volume $V$ . If you halve the volume, you double the pressure.

\propto \frac{1}{V} \quad \Rightarrow \quad pV = \text{constant}

Charles's Law (Constant Pressure) For a fixed mass of gas at a constant pressure, the volume $V$ is directly proportional to its absolute temperature $T$ .

\propto T \quad \Rightarrow \quad \frac{V}{T} = \text{constant}

The Pressure Law (Constant Volume) For a fixed mass of gas at a constant volume, the pressure $p$ is directly proportional to its absolute temperature $T$ .

\propto T \quad \Rightarrow \quad \frac{p}{T} = \text{constant}

Ideal gas law graphs

If you look at the third graph above, you'll see something fascinating. If you plot volume (or pressure) against temperature in degrees Celsius, you get a straight line that doesn't pass through the origin. However, if you extrapolate the line backwards, it crosses the temperature axis at exactly $\text{ }^\circ\text{C}$ .

Definition

Absolute Zero

Absolute zero ( $\text{ K}$ or $\text{ }^\circ\text{C}$ ) is the lowest possible temperature. At this theoretical temperature, an ideal gas would have zero volume and zero pressure, and its particles would have zero kinetic energy.

Common Mistake

Forgetting to use Kelvin

Charles's Law and the Pressure Law are only direct proportionalities if you measure temperature in Kelvin. Always add $273$ to your Celsius temperature before doing any gas calculations. An increase of $\text{ }^\circ\text{C}$ is exactly the same as an increase of $\text{ K}$ , but the zero points are different!

Moles, Molecules, and Mass

To describe the amount of gas we have, we can either count the exact number of individual molecules, or we can count "batches" of molecules called moles.

Number of molecules ( $N$ ): The literal count of individual particles in the container.
Number of moles ( $n$ ): The number of batches. One batch (one mole) contains exactly $6.02 \times 10^{23}$ particles. This number is the Avogadro constant ( $N_A$ ).

To convert between them, you simply multiply the number of moles by the Avogadro constant:

N = n N_A

We also need to be careful with mass. In physics, you will encounter two very similar-sounding terms that mean very different things:

Molecular mass ( $m$ ): The tiny mass of one single molecule in kilograms (e.g. roughly $\times 10^{-26} \text{ kg}$ for an oxygen molecule).
Molar mass ( $M$ ): The mass of one whole mole of the substance.

Tip

Finding the mass of one molecule

If an exam question gives you the molar mass $M$ (the mass of one mole) and asks for the mass of a single molecule $m$ , just divide the molar mass by the number of molecules in a mole:

\frac{M}{N_A}

The Ideal Gas Equation

By combining Boyle's, Charles's, and the Pressure law, we get a single relationship: $\propto T$ . To turn this into an equals sign, we need a constant of proportionality.

Depending on whether we want to work with moles or molecules, there are two versions of the ideal gas equation. You need to be completely comfortable using both.

Version 1: The Macroscopic Equation (Moles)

If you are dealing with the number of moles ( $n$ ), use the molar gas constant ( $R$ ), which is $\text{ J K}^{-1} \text{ mol}^{-1}$ .

p V = n R T

Version 2: The Microscopic Equation (Molecules)

If you are dealing with the number of individual molecules ( $N$ ), use the Boltzmann constant ( $k$ ), which is $1.38 \times 10^{-23} \text{ J K}^{-1}$ .

p V = N k T

Key Idea

The constants are linked!

The Boltzmann constant $k$ is just the molar gas constant $R$ divided by the Avogadro constant $N_A$ . In other words, $\frac{R}{N_A}$ . You can think of $k$ as the gas constant for a single molecule, while $R$ is the gas constant for a whole mole.

Example

Calculating state variables using the ideal gas equation

A sealed canister contains $\text{ moles}$ of an ideal gas at a temperature of $\text{ }^\circ\text{C}$ and a pressure of $\text{ kPa}$ . Calculate the volume of the gas. ( $\text{ J K}^{-1} \text{ mol}^{-1}$ )

First, identify your known variables and convert everything to standard SI units ( $Pa\text{Pa}$ , $m3\text{m}^3$ , $K\text{K}$ ). $\text{ mol}$ $\text{ K}$ $\times 10^3 \text{ Pa}$
State the correct ideal gas equation. Because we are given moles, we use the macroscopic version.

p V = n R T

Rearrange the equation to make volume ( $V$ ) the subject.

\frac{nRT}{p}

Substitute the values into the equation and calculate the final answer.

\frac{0.45 \times 8.31 \times 295}{120 \times 10^3}

\times 10^{-3} \text{ m}^3

Work Done by a Gas

When a gas is heated, it might expand. If it expands against a resisting pressure (like the air pushing down on a moving piston), it has to do mechanical work to push that boundary outward.

For a gas expanding at a constant pressure, the work done can be calculated using:

\Delta V

Where:

$W$ is the work done in joules ( $J\text{J}$ )
$p$ is the constant pressure in pascals ( $Pa\text{Pa}$ )
$ΔV\Delta V$ is the change in volume in cubic metres ( $m3\text{m}^3$ )

Why does this work? Work done is force multiplied by distance ( $\Delta x$ ). We know that force is pressure multiplied by area ( $F = p A$ ). So, $\Delta x$ . Since the cross-sectional area multiplied by the distance moved is exactly the change in volume ( $ΔV=AΔx\Delta V = A \Delta x$ ), the formula simplifies to $\Delta V$ .

Pressure-volume (p-V) graph showing work done

If the pressure isn't constant, you can still find the work done by looking at a $p$ - $V$ graph. The area under a pressure-volume graph is always equal to the work done.

Example

Calculating work done during expansion

A cylinder with a movable piston contains a gas at a constant pressure of $1.5 \times 10^5 \text{ Pa}$ . The gas is heated and expands, pushing the piston outwards. Its volume increases from $3.0 \times 10^{-3} \text{ m}^3$ to $7.5 \times 10^{-3} \text{ m}^3$ . Calculate the work done by the gas.

First, calculate the change in volume ( $ΔV\Delta V$ ).

\Delta V = V_{\text{final}} - V_{\text{initial}}

\Delta V = (7.5 \times 10^{-3}) - (3.0 \times 10^{-3}) = 4.5 \times 10^{-3} \text{ m}^3

State the formula for work done at constant pressure.

\Delta V

Substitute the values into the equation.

\times 10^5) \times (4.5 \times 10^{-3})

\text{ J}

Common Mistake

Expansion vs Compression

If a gas expands (volume increases), the gas does work on its surroundings. If a gas is compressed (volume decreases), work is done on the gas by the surroundings. Make sure you read the physical situation carefully!

Exam technique

In the exam

Always hunt for non-SI units. AQA examiners love to give pressure in $kPa\text{kPa}$ or $MPa\text{MPa}$ , volume in $cm3\text{cm}^3$ or $dm3\text{dm}^3$ , and temperature in $∘C\text{ }^\circ\text{C}$ . Convert everything to $Pa\text{Pa}$ , $m3\text{m}^3$ , and $K\text{K}$ before touching your calculator.
Remember your volume conversions. To convert $cm3\text{cm}^3$ to $m3\text{m}^3$ , you must multiply by $10^{-6}$ (because there are $100 \times 100 \times 100$ cubic centimetres in a cubic metre).
Choose the right constant. If the question gives you the number of molecules ( $N$ ), use $k$ . If it gives you the number of moles ( $n$ ), use $R$ .
Use ratios for "before and after" questions. If a fixed mass of gas changes state (e.g. moves from state 1 to state 2), you can often save time by using $p1V1T1=p2V2T2\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$ rather than finding $n$ or $N$ in an intermediate step.

Self review

Check yourself

What is the difference between molar mass and molecular mass?
What temperature scale must always be used in the ideal gas equations, and how do you convert to it from Celsius?
How does the Boltzmann constant $k$ relate to the molar gas constant $R$ ?
How can you determine the work done by a gas from a pressure-volume graph?

Ideal gases (A-level only)

Here is what you'll learn:

How the three experimental gas laws link pressure, volume, and temperature.
How to extrapolate experimental graphs to find absolute zero.
How to use the ideal gas equations ( $p V = n R T$ and $p V = N k T$ ) to find the state of a gas.
How to calculate the work done when a gas expands or is compressed.

The Gas Laws and Absolute Zero

You will investigate two of these in Required Practical 8.

Boyle's Law (Constant Temperature) For a fixed mass of gas at a constant temperature, the pressure $p$ is inversely proportional to the volume $V$ . If you halve the volume, you double the pressure.

\propto \frac{1}{V} \quad \Rightarrow \quad pV = \text{constant}

Charles's Law (Constant Pressure) For a fixed mass of gas at a constant pressure, the volume $V$ is directly proportional to its absolute temperature $T$ .

\propto T \quad \Rightarrow \quad \frac{V}{T} = \text{constant}

The Pressure Law (Constant Volume) For a fixed mass of gas at a constant volume, the pressure $p$ is directly proportional to its absolute temperature $T$ .

\propto T \quad \Rightarrow \quad \frac{p}{T} = \text{constant}

Ideal gas law graphs

Definition

Absolute Zero

Common Mistake

Forgetting to use Kelvin

Moles, Molecules, and Mass

To describe the amount of gas we have, we can either count the exact number of individual molecules, or we can count "batches" of molecules called moles.

Number of molecules ( $N$ ): The literal count of individual particles in the container.
Number of moles ( $n$ ): The number of batches. One batch (one mole) contains exactly $6.02 \times 10^{23}$ particles. This number is the Avogadro constant ( $N_A$ ).

To convert between them, you simply multiply the number of moles by the Avogadro constant:

N = n N_A

We also need to be careful with mass. In physics, you will encounter two very similar-sounding terms that mean very different things:

Molecular mass ( $m$ ): The tiny mass of one single molecule in kilograms (e.g. roughly $\times 10^{-26} \text{ kg}$ for an oxygen molecule).
Molar mass ( $M$ ): The mass of one whole mole of the substance.

Tip

Finding the mass of one molecule

If an exam question gives you the molar mass $M$ (the mass of one mole) and asks for the mass of a single molecule $m$ , just divide the molar mass by the number of molecules in a mole:

\frac{M}{N_A}

The Ideal Gas Equation

By combining Boyle's, Charles's, and the Pressure law, we get a single relationship: $\propto T$ . To turn this into an equals sign, we need a constant of proportionality.

Depending on whether we want to work with moles or molecules, there are two versions of the ideal gas equation. You need to be completely comfortable using both.

Version 1: The Macroscopic Equation (Moles)

If you are dealing with the number of moles ( $n$ ), use the molar gas constant ( $R$ ), which is $\text{ J K}^{-1} \text{ mol}^{-1}$ .

p V = n R T

Version 2: The Microscopic Equation (Molecules)

If you are dealing with the number of individual molecules ( $N$ ), use the Boltzmann constant ( $k$ ), which is $1.38 \times 10^{-23} \text{ J K}^{-1}$ .

p V = N k T

Key Idea

The constants are linked!

Example

Calculating state variables using the ideal gas equation

First, identify your known variables and convert everything to standard SI units ( $Pa\text{Pa}$ , $m3\text{m}^3$ , $K\text{K}$ ). $\text{ mol}$ $\text{ K}$ $\times 10^3 \text{ Pa}$
State the correct ideal gas equation. Because we are given moles, we use the macroscopic version.

p V = n R T

Rearrange the equation to make volume ( $V$ ) the subject.

\frac{nRT}{p}

Substitute the values into the equation and calculate the final answer.

\frac{0.45 \times 8.31 \times 295}{120 \times 10^3}

\times 10^{-3} \text{ m}^3

Work Done by a Gas

When a gas is heated, it might expand. If it expands against a resisting pressure (like the air pushing down on a moving piston), it has to do mechanical work to push that boundary outward.

For a gas expanding at a constant pressure, the work done can be calculated using:

\Delta V

Where:

$W$ is the work done in joules ( $J\text{J}$ )
$p$ is the constant pressure in pascals ( $Pa\text{Pa}$ )
$ΔV\Delta V$ is the change in volume in cubic metres ( $m3\text{m}^3$ )

Pressure-volume (p-V) graph showing work done

If the pressure isn't constant, you can still find the work done by looking at a $p$ - $V$ graph. The area under a pressure-volume graph is always equal to the work done.

Example

Calculating work done during expansion

First, calculate the change in volume ( $ΔV\Delta V$ ).

\Delta V = V_{\text{final}} - V_{\text{initial}}

\Delta V = (7.5 \times 10^{-3}) - (3.0 \times 10^{-3}) = 4.5 \times 10^{-3} \text{ m}^3

State the formula for work done at constant pressure.

\Delta V

Substitute the values into the equation.

\times 10^5) \times (4.5 \times 10^{-3})

\text{ J}

Common Mistake

Expansion vs Compression

Exam technique

In the exam

Always hunt for non-SI units. AQA examiners love to give pressure in $kPa\text{kPa}$ or $MPa\text{MPa}$ , volume in $cm3\text{cm}^3$ or $dm3\text{dm}^3$ , and temperature in $∘C\text{ }^\circ\text{C}$ . Convert everything to $Pa\text{Pa}$ , $m3\text{m}^3$ , and $K\text{K}$ before touching your calculator.
Remember your volume conversions. To convert $cm3\text{cm}^3$ to $m3\text{m}^3$ , you must multiply by $10^{-6}$ (because there are $100 \times 100 \times 100$ cubic centimetres in a cubic metre).
Choose the right constant. If the question gives you the number of molecules ( $N$ ), use $k$ . If it gives you the number of moles ( $n$ ), use $R$ .
Use ratios for "before and after" questions. If a fixed mass of gas changes state (e.g. moves from state 1 to state 2), you can often save time by using $p1V1T1=p2V2T2\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$ rather than finding $n$ or $N$ in an intermediate step.

Self review

Check yourself

What is the difference between molar mass and molecular mass?
What temperature scale must always be used in the ideal gas equations, and how do you convert to it from Celsius?
How does the Boltzmann constant $k$ relate to the molar gas constant $R$ ?
How can you determine the work done by a gas from a pressure-volume graph?

Ideal gases (A-level only)

The Gas Laws and Absolute Zero

Moles, Molecules, and Mass

The Ideal Gas Equation

Version 1: The Macroscopic Equation (Moles)

Version 2: The Microscopic Equation (Molecules)

Work Done by a Gas

Ideal gases (A-level only) Revision Guide

Ideal gases (A-level only)

The Gas Laws and Absolute Zero

Moles, Molecules, and Mass

The Ideal Gas Equation

Version 1: The Macroscopic Equation (Moles)

Version 2: The Microscopic Equation (Molecules)

Work Done by a Gas

Ideal gases (A-level only) Revision Guide