WORD EQUATION	SYMBOL EQUATION
$number \space of \space moles \, = \, \frac{volume}{molar \space gas \space volume}$	$n = \frac{V}{V_m}$
$pressure \space \times \space volume \, = \, number \space of \space moles \space \times \space gas \space constant \space \times \space temperature$	$pV=nRT$

Constants

constant	symbol	value
$molar \space gas \space volume$	$V_m$	$24 \, dm^3 \,mol^{-1}$ (at r.t.p) $22.4 \, dm^3 \, mol^{-1}$ (at s.t.p)
$gas \space constant$	$R$	$8.31 \, J \, K^{-1} \, mol^{-1}$

Note: r.t.p is room temperature and pressure ( $293 \, K, 101.3 \, kPa$ ), whereas s.t.p is standard temperature and pressure ( $273 \, K, 101.3 \, kPa$ ).

Variable units

quantity name	symbol	unit name	unit
$number \space of \space moles$	$n$	$mole$	$mol$
$volume$	$V$	$cubic \space decimeter \space or \newline cubic \, metre$	$dm^3 \space or \newline m^3$
$molar \space gas \space volume$	$V_m$	$cubic \space decimeter \space per \space mole$	$dm^3 \, mol^{-1}$
$pressure$	$p$	$pascal$	$Pa$
$gas \space constant$	$R$	$joule \space per \space kelvin \space per \space per \space mole$	$J \, K^{-1} \, mol^{-1}$
$temperature$	$T$	$kelvin$	$K$

Molar gas volume equation

The molar gas volume is known as the volume that one mole of gas occupies at a certain temperature and pressure.

The equation to work out the number of moles given the volume and molar gas volume is:

$number \space of \space moles \, = \, \frac{volume}{molar \space gas \space volume} \newline n = \frac{V}{V_m}$

This equation can be rearranged to calculate the volume of a gas:

$volume \, = \, number \space of \space moles \, \times \, molar \space gas \space volume \newline V \, = \, n \, \times \, V_m$

At room temperature and pressure the molar gas volume is $24 \, dm^3 \, mol^{-1}$ .

At standard temperature and pressure the molar gas volume is $22.4 \, dm^3 \, mol^{-1}$ .

Note: $cm^3$ can be converted to $dm^3$ by dividing by 1000.

Example

Work out the number of moles there are in $50 \, 000 \,cm^3$ of nitrogen at room temperature and pressure.

First convert $cm^3$ to $dm^3$ to work out the volume in $dm^3$ :

$V \, = \, \frac{50 \, 000 \, cm^3}{1000} \, = \, 50 \, dm^3$

Work out which value for the molar gas volume is needed based on the conditions stated in the question. At room temperature and pressure (r.tp):

$V_m \, = \, 24 \, dm^3 \, mol^{-1}$

Next, write down the equation you will use:

$n = \frac{V}{V_m}$

Substitute the values into the equation to find your answer:

$n \, = \, \frac{50 \, dm^3}{24 \, dm^3 \, mol^{-1}} \, = \, 2.08 \, mol$

There are $\underline{2.08 \, moles}$ in $50 \, 000 \, cm^3$  of nitrogen gas.

Measuring molar gas volume

The experiment below can be used to measure molar gas volume, using the measurement of the amount of gas produced in a reaction.

Chemistry; Formulae, equations and amounts of substances; KS5 Year 12; Calculations involving gases

Note: 1. Reaction mixture 2. Delivery tube 3. Gas collection 4. Gas syringe

procedure

1.	Measure a known volume of acid into a conical flask connected to a gas syringe.
2.	Add a known mass of salt into the conical flask filled with acid. Make sure to replace the bung and let the reaction progress until finished.
3.	Record the volume of gas produced in the gas syringe.
4.	Repeat steps 1-3 with a different mass of salt each time.
5.	Plot a graph of mass of salt ( $x$ -axis) against volume of gas produced ( $y$ -axis).
6.	Read the volume of gas produced at a certain mass of salt.
7.	Calculate the amount of salt using the mass and $M$ . $number \space of \space moles \, = \, \frac{mass}{Molar \ mass}$
8.	Use the balanced equation to work out the molar ratios of the salt and gaseous product. The number of moles of the gas product can be calculated using this.
9.	Use the number of moles of the gas product along with the volume to find the molar gas volume by rearranging the following equation. $number \space of \space moles \, = \frac{volume}{molar \space gas \space volume}$ Rearranged: $molar \space gas \space volume \, = \, \frac{volume}{number \space of \space moles}$

Mole calculations to calculate gas volumes

The mole equation can be used to calculate the volume of gas produced using the molar ratio of the reactants compared to the products.

procedure

Calculate the number of moles of the reactant mass using the following equation.

$number \space of \space moles \, = \, \frac{volume}{molar \space gas \space volume}$

Use the number of moles of the reactant, along with the molar ratios from the balanced symbol equation of the reaction to work out the number of moles of the gas that is produced.

Use the number of moles and the molar gas volume (depending on the reaction conditions) to calculate the volume of gas produced.

$volume \, = \, number \space of \space moles \space \times \space molar \space gas \space volume$

Volume calculations to calculate gas volumes

The volume of reactant gases can be used alongside the molar ratios within the balanced equation to calculate the volume of gas that is produced.

procedure

1.	Use the balanced symbol equation to work out the molar ratio of the gaseous reactant (with the known volume) compared to the gaseous product.
2.	Once the molar ratio is determined, work out the volume of gaseous product produced using the volume of gaseous reactant.

Ideal gas equation

The ideal gas equation is used to calculate the number of moles of a gas. This can be used to work out either the mass or molar mass of a substance depending on the information provided in the question.

$pressure \space \times \space volume \, = \, number \space of \space moles \space \times \space gas \space constant \space \times \space temperature \newline pV=nRT$

Note: The gas constant has a value of $8.31 \, J \, K^{-1} \, mol^{-1}$ .

This equation can be rearranged to make each variable the subject, so can therefore be used to calculate a number of different things.

Unit conversions

There are lots of different units in this equation. Use the following unit conversions to help you:

To convert from a kilo to a gram, multiply by 1000 ( $10^3$ ).
To convert from $cm^3$ to $m^3$ , multiply by $10^{-6}$ .
To convert from $dm^3$ to $m^3$ multiply by $10^{-3}$ (divide by 1000).
To convert from $^oC$ to $K$ add 273.

Example

At a temperature of $25^oC$ and pressure of $150 \, kPa$ , methane occupies a volume of $1760 \, cm^3$ . What is the mass of methane?

First convert all of the variables that have been given in the question:

$T \, (K) \, = \, 25 \, (^oC) \, + \, 273 \, = \, 298 \, K$

$P \, (Pa) \, = \, 150 \, (kPa) \, \times \, 10^3 \, = \, 150 \, 000 \, Pa$

$V \, (m^3) \, = \, 1760 \, (cm^3) \, \times 10^{-6} \, = \, 1760 \, \times \, 10^{-6} \, m^3$

You will need your gas constant with the correct unit:

$8.31 \, J \, K^{-1} \, mol^{-1}$

Write the equation you will use:

$pV=nRT$

Rearrange the equation to make the number of moles the subject (this can be used to calculate the mass of methane):

$n \, = \, \frac{pV}{RT}$ 

Substitute in your values into this equation:

$n \, = \, \frac{150 \, 000 \, Pa \, \times \, (1760 \, \times \, 10^{-6}) \, m^3}{8.31 \, J \, K^{-1} \, mol^{-1} \, \times \, 298 \, K} \, = \, 0.11 \, mol$

Write down the equation you will use to find the mass:

$m \, = \, n \, \times \, M$ 

Work out the $M_r$ of methane ( $CH_4$ ):

$M \, (CH_4) \, = \, (12 \, (A_r \, of \, C) \, \times \, 1) \, + \, (1 \, (A_r \, of \, H) \, \times \, 4) \, = \, 16 \ g/mol$

Substitute your values for the number of moles and relative molecular mass ( $M_r$ ) to find the mass of methane:

$m \, = \, 0.11 \, mol \, \times \, 16 \, = \, 1.76 \, g$

The mass of methane is $\underline{1.76 \, g}$ , given that there is $0.11 \, mol$ of methane.

Learn with Basics

Length:

Unit 1

Mole calculations and Avogadro's constant - Higher

Unit 2

Calculating gas volumes using moles - Higher

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Optional

Unit 3