Gas volumes and ideal gas

What you'll learn

How to use molar gas volume at room temperature and pressure.
How gas volumes connect to balanced chemical equations.
How to use the ideal gas equation, $p V = n R T$ , with correct units.
How to choose between “24.0 dm³ mol⁻¹” and the full ideal gas equation.

Start point: moles and equations

Before gas calculations, you need the idea of amount of substance.

Definition

Mole

One mole is an amount of substance containing the Avogadro constant of particles. In calculations, amount is written as $n$ and measured in mol.

You already know two key amount equations:

\frac{m}{M}

where $m$ is mass in g and $M$ is molar mass in g mol⁻¹.

n = c V

where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³.

A balanced chemical equation gives mole ratios. For example:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This tells you 1 mol of Mg produces 1 mol of H₂.

Gas volume at room temperature and pressure

For many A-Level calculations, gases are assumed to be measured at room temperature and pressure, often shortened to RTP.

Definition

Molar gas volume at RTP

At RTP, 1 mol of any gas occupies 24.0 dm³. This is called the molar gas volume.

So, for a gas at RTP:

\frac{V}{24.0}

where $V$ is in dm³.

If the volume is in cm³, use 24000 cm³ mol⁻¹ instead, because 24.0 dm³ = 24000 cm³.

Example

Hydrogen volume from a reacting mass

Magnesium reacts with excess hydrochloric acid:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the volume of H₂ produced at RTP when 0.120 g of Mg reacts completely. Use $Ar(Mg)=24.3A_\text{r}(\text{Mg}) = 24.3$ .

Calculate the amount of Mg: $moln(\text{Mg}) = \frac{0.120}{24.3} = 0.00494\text{ mol}$ .
Use the balanced equation. The ratio Mg : H₂ is 1 : 1, so $moln(\text{H}_2) = 0.00494\text{ mol}$ .
Convert moles of H₂ into volume at RTP: $\times 24.0 = 0.1185\text{ dm}^3$ .
Convert to cm³: $cm30.1185\text{ dm}^3 = 119\text{ cm}^3$ to 3 significant figures.

Common Mistake

Using 24.0 with the wrong units

The value 24.0 is in dm³ mol⁻¹, not cm³ mol⁻¹. If your volume is in cm³, use 24000 cm³ mol⁻¹ or convert cm³ to dm³ first.

Why gas volume ratios work

At the same temperature and pressure, equal volumes of gases contain equal numbers of particles. This is Avogadro’s law.

Key Idea

Gas volume ratios

For gases measured at the same temperature and pressure, the volume ratio is the same as the mole ratio in the balanced equation.

This means you can sometimes work directly with gas volumes without converting to mol first.

Example

Reacting gas volumes

Nitrogen and hydrogen react to form ammonia:

N₂(g) + 3H₂(g) → 2NH₃(g)

60 cm³ of N₂ is mixed with 150 cm³ of H₂ at the same temperature and pressure. Find the maximum volume of NH₃ formed and the gas left in excess.

Compare the ratio N₂ : H₂. The equation needs 1 volume of N₂ for every 3 volumes of H₂, so 60 cm³ of N₂ would need 180 cm³ of H₂.
Only 150 cm³ of H₂ is available, so H₂ is the limiting reactant.
Use the ratio H₂ : NH₃ = 3 : 2. The volume of NH₃ formed is $150 \times \frac{2}{3} = 100\text{ cm}^3$ .
Work out N₂ used. Since N₂ : H₂ = 1 : 3, $cm3150\text{ cm}^3$ of H₂ reacts with $cm350\text{ cm}^3$ of N₂, leaving 10 cm³ of N₂ unreacted.

The ideal gas equation

The simple 24.0 dm³ mol⁻¹ method only works for gases at RTP. If the pressure or temperature is different, use the ideal gas equation.

Definition

Ideal gas equation

The ideal gas equation is $p V = n R T$ , where $p$ is pressure, $V$ is volume, $n$ is amount, $R$ is the gas constant and $T$ is temperature.

Diagram linking pV = nRT to pressure, volume, amount, gas constant, temperature and common unit conversions

For Edexcel A-Level calculations, use:

$p$ in Pa
$V$ in m³
$n$ in mol
$T$ in K
$8.31\text{ J mol}^{-1}\text{ K}^{-1}$

To convert temperature from °C to K:

T(K)=T(∘C)+273T(\text{K}) = T(^{\circ}\text{C}) + 273

Definition

Ideal gas

An ideal gas is a model gas whose particles have negligible volume and no intermolecular forces. Real gases behave most ideally at low pressure and high temperature.

Example

Using the ideal gas equation

Calculate the volume occupied by 0.0500 mol of O₂ at 120 kPa and 27 °C.

Convert to the units needed for $p V = n R T$ : $120000\text{ Pa}$ and $300\text{ K}$ .
Rearrange the equation to make volume the subject: $\frac{nRT}{p}$ .
Substitute the values: $\frac{0.0500 \times 8.31 \times 300}{120000} = 0.00104\text{ m}^3$ .
Convert m³ to dm³: $dm30.00104\text{ m}^3 = 1.04\text{ dm}^3$ .

Common Mistake

Using Celsius in the ideal gas equation

Temperature in $p V = n R T$ must be in K, not °C. A temperature of 25 °C should be used as 298 K.

Finding molar mass from gas data

The ideal gas equation can also be used to find the molar mass of a gas or volatile liquid.

The overall route is:

Use $p V = n R T$ to find $n$ .
Use $\frac{m}{n}$ to find molar mass.

Example

Molar mass from gas data

A 0.204 g sample of a volatile liquid is vaporised. The vapour occupies 96.0 cm³ at 101 kPa and 373 K. Calculate its molar mass.

Convert pressure and volume into SI units: $101000\text{ Pa}$ and $96.0\text{ cm}^3 = 9.60 \times 10^{-5}\text{ m}^3$ .
Rearrange the ideal gas equation: $\frac{pV}{RT}$ .
Substitute the values: $\frac{101000 \times 9.60 \times 10^{-5}}{8.31 \times 373} = 0.00313\text{ mol}$ .
Calculate molar mass: $\frac{0.204}{0.00313} = 65.2\text{ g mol}^{-1}$ .

Choosing the right method

Key Idea

Which equation should you use?

Use 24.0 dm³ mol⁻¹ only for gases at RTP. Use $p V = n R T$ when pressure and temperature are given or are not RTP.

A good decision route is:

If the question says at RTP, use 24.0 dm³ mol⁻¹.
If the question gives pressure and temperature, use $p V = n R T$ .
If all gases are at the same temperature and pressure, you can use gas volume ratios directly from the balanced equation.

Tip

Sanity check

At RTP, 1 mol of gas should have a volume close to 24 dm³. If your answer is 0.024 dm³ or 24000 dm³ for 1 mol, a unit conversion has gone wrong.

Practical gas measurements

In experiments, gas volume is often measured using a gas syringe or an inverted measuring cylinder over water. A gas syringe is usually more accurate because it reduces loss of gas and avoids dissolving the gas in water.

To improve reliability:

check the apparatus is airtight before starting
record temperature and pressure if using $p V = n R T$
read the gas syringe at eye level
repeat the experiment and calculate a mean

Common Mistake

Gas collected over water

If a gas is collected over water and the vapour pressure of water is supplied, subtract it before using the ideal gas equation: $vapourp_{\text{gas}} = p_{\text{total}} - p_{\text{water vapour}}$ . If no vapour pressure data is given, do not invent a correction.

Exam technique

In the exam

Start with the balanced equation, then decide whether you need mole ratios, gas volume ratios, or $p V = n R T$ .
Write units beside every value before substituting into an equation, especially Pa, kPa, cm³, dm³, m³ and K.
Check whether the gas is at RTP. If it is not, avoid using 24.0 dm³ mol⁻¹ unless the question specifically tells you to.

Self review

Check yourself

When can you use gas volume ratios directly from a balanced equation?
Why must temperature be converted to K before using $p V = n R T$ ?
A gas volume is given in cm³ for an ideal gas calculation. What conversion is needed before using SI units?

What you'll learn

How to use molar gas volume at room temperature and pressure.
How gas volumes connect to balanced chemical equations.
How to use the ideal gas equation, $p V = n R T$ , with correct units.
How to choose between “24.0 dm³ mol⁻¹” and the full ideal gas equation.

Start point: moles and equations

Before gas calculations, you need the idea of amount of substance.

Definition

Mole

One mole is an amount of substance containing the Avogadro constant of particles. In calculations, amount is written as $n$ and measured in mol.

You already know two key amount equations:

\frac{m}{M}

where $m$ is mass in g and $M$ is molar mass in g mol⁻¹.

n = c V

where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³.

A balanced chemical equation gives mole ratios. For example:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This tells you 1 mol of Mg produces 1 mol of H₂.

Gas volume at room temperature and pressure

For many A-Level calculations, gases are assumed to be measured at room temperature and pressure, often shortened to RTP.

Definition

Molar gas volume at RTP

At RTP, 1 mol of any gas occupies 24.0 dm³. This is called the molar gas volume.

So, for a gas at RTP:

\frac{V}{24.0}

where $V$ is in dm³.

If the volume is in cm³, use 24000 cm³ mol⁻¹ instead, because 24.0 dm³ = 24000 cm³.

Example

Hydrogen volume from a reacting mass

Magnesium reacts with excess hydrochloric acid:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the volume of H₂ produced at RTP when 0.120 g of Mg reacts completely. Use $Ar(Mg)=24.3A_\text{r}(\text{Mg}) = 24.3$ .

Calculate the amount of Mg: $moln(\text{Mg}) = \frac{0.120}{24.3} = 0.00494\text{ mol}$ .
Use the balanced equation. The ratio Mg : H₂ is 1 : 1, so $moln(\text{H}_2) = 0.00494\text{ mol}$ .
Convert moles of H₂ into volume at RTP: $\times 24.0 = 0.1185\text{ dm}^3$ .
Convert to cm³: $cm30.1185\text{ dm}^3 = 119\text{ cm}^3$ to 3 significant figures.

Common Mistake

Using 24.0 with the wrong units

The value 24.0 is in dm³ mol⁻¹, not cm³ mol⁻¹. If your volume is in cm³, use 24000 cm³ mol⁻¹ or convert cm³ to dm³ first.

Why gas volume ratios work

At the same temperature and pressure, equal volumes of gases contain equal numbers of particles. This is Avogadro’s law.

Key Idea

Gas volume ratios

For gases measured at the same temperature and pressure, the volume ratio is the same as the mole ratio in the balanced equation.

This means you can sometimes work directly with gas volumes without converting to mol first.

Example

Reacting gas volumes

Nitrogen and hydrogen react to form ammonia:

N₂(g) + 3H₂(g) → 2NH₃(g)

60 cm³ of N₂ is mixed with 150 cm³ of H₂ at the same temperature and pressure. Find the maximum volume of NH₃ formed and the gas left in excess.

Compare the ratio N₂ : H₂. The equation needs 1 volume of N₂ for every 3 volumes of H₂, so 60 cm³ of N₂ would need 180 cm³ of H₂.
Only 150 cm³ of H₂ is available, so H₂ is the limiting reactant.
Use the ratio H₂ : NH₃ = 3 : 2. The volume of NH₃ formed is $150 \times \frac{2}{3} = 100\text{ cm}^3$ .
Work out N₂ used. Since N₂ : H₂ = 1 : 3, $cm3150\text{ cm}^3$ of H₂ reacts with $cm350\text{ cm}^3$ of N₂, leaving 10 cm³ of N₂ unreacted.

The ideal gas equation

The simple 24.0 dm³ mol⁻¹ method only works for gases at RTP. If the pressure or temperature is different, use the ideal gas equation.

Definition

Ideal gas equation

The ideal gas equation is $p V = n R T$ , where $p$ is pressure, $V$ is volume, $n$ is amount, $R$ is the gas constant and $T$ is temperature.

Diagram linking pV = nRT to pressure, volume, amount, gas constant, temperature and common unit conversions

For Edexcel A-Level calculations, use:

$p$ in Pa
$V$ in m³
$n$ in mol
$T$ in K
$8.31\text{ J mol}^{-1}\text{ K}^{-1}$

To convert temperature from °C to K:

T(K)=T(∘C)+273T(\text{K}) = T(^{\circ}\text{C}) + 273

Definition

Ideal gas

An ideal gas is a model gas whose particles have negligible volume and no intermolecular forces. Real gases behave most ideally at low pressure and high temperature.

Example

Using the ideal gas equation

Calculate the volume occupied by 0.0500 mol of O₂ at 120 kPa and 27 °C.

Convert to the units needed for $p V = n R T$ : $120000\text{ Pa}$ and $300\text{ K}$ .
Rearrange the equation to make volume the subject: $\frac{nRT}{p}$ .
Substitute the values: $\frac{0.0500 \times 8.31 \times 300}{120000} = 0.00104\text{ m}^3$ .
Convert m³ to dm³: $dm30.00104\text{ m}^3 = 1.04\text{ dm}^3$ .

Common Mistake

Using Celsius in the ideal gas equation

Temperature in $p V = n R T$ must be in K, not °C. A temperature of 25 °C should be used as 298 K.

Finding molar mass from gas data

The ideal gas equation can also be used to find the molar mass of a gas or volatile liquid.

The overall route is:

Use $p V = n R T$ to find $n$ .
Use $\frac{m}{n}$ to find molar mass.

Example

Molar mass from gas data

A 0.204 g sample of a volatile liquid is vaporised. The vapour occupies 96.0 cm³ at 101 kPa and 373 K. Calculate its molar mass.

Convert pressure and volume into SI units: $101000\text{ Pa}$ and $96.0\text{ cm}^3 = 9.60 \times 10^{-5}\text{ m}^3$ .
Rearrange the ideal gas equation: $\frac{pV}{RT}$ .
Substitute the values: $\frac{101000 \times 9.60 \times 10^{-5}}{8.31 \times 373} = 0.00313\text{ mol}$ .
Calculate molar mass: $\frac{0.204}{0.00313} = 65.2\text{ g mol}^{-1}$ .

Choosing the right method

Key Idea

Which equation should you use?

Use 24.0 dm³ mol⁻¹ only for gases at RTP. Use $p V = n R T$ when pressure and temperature are given or are not RTP.

A good decision route is:

If the question says at RTP, use 24.0 dm³ mol⁻¹.
If the question gives pressure and temperature, use $p V = n R T$ .
If all gases are at the same temperature and pressure, you can use gas volume ratios directly from the balanced equation.

Tip

Sanity check

At RTP, 1 mol of gas should have a volume close to 24 dm³. If your answer is 0.024 dm³ or 24000 dm³ for 1 mol, a unit conversion has gone wrong.

Practical gas measurements

To improve reliability:

check the apparatus is airtight before starting
record temperature and pressure if using $p V = n R T$
read the gas syringe at eye level
repeat the experiment and calculate a mean

Common Mistake

Gas collected over water

Exam technique

In the exam

Start with the balanced equation, then decide whether you need mole ratios, gas volume ratios, or $p V = n R T$ .
Write units beside every value before substituting into an equation, especially Pa, kPa, cm³, dm³, m³ and K.
Check whether the gas is at RTP. If it is not, avoid using 24.0 dm³ mol⁻¹ unless the question specifically tells you to.

Self review

Check yourself

When can you use gas volume ratios directly from a balanced equation?
Why must temperature be converted to K before using $p V = n R T$ ?
A gas volume is given in cm³ for an ideal gas calculation. What conversion is needed before using SI units?

What you'll learn

Start point: moles and equations

Gas volume at room temperature and pressure

Why gas volume ratios work

The ideal gas equation

Finding molar mass from gas data

Choosing the right method

Practical gas measurements

Review Gas volumes and ideal gas by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gas volumes and ideal gas

What you'll learn

Start point: moles and equations

Gas volume at room temperature and pressure

Why gas volume ratios work

The ideal gas equation

Finding molar mass from gas data

Choosing the right method

Practical gas measurements

Review Gas volumes and ideal gas by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gas volumes and ideal gas

What you'll learn

Start point: moles and equations

Gas volume at room temperature and pressure

Why gas volume ratios work

The ideal gas equation

Finding molar mass from gas data

Choosing the right method

Practical gas measurements

Review Gas volumes and ideal gas by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gas volumes and ideal gas Revision Guide

Gas volumes and ideal gas

What you'll learn

Start point: moles and equations

Gas volume at room temperature and pressure

Why gas volume ratios work

The ideal gas equation

Finding molar mass from gas data

Choosing the right method

Practical gas measurements

Review Gas volumes and ideal gas by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gas volumes and ideal gas Revision Guide