The Mole and the Ideal Gas Equation

A2 · 10 min

Every gas, whether it is air in a tyre, helium in a balloon or steam in an engine, behaves in roughly the same simple way when it is not too dense or too cold: its pressure, volume and temperature are linked by one equation, $pV = nRT$ . This note introduces the mole (how physicists count particles), defines an ideal gas, and shows how to use the equation of state in both its forms. It is the starting point for kinetic theory and thermodynamics, and gas calculations appear in nearly every Paper 4.

Amount of substance and the mole

Gases are made of enormous numbers of molecules. Counting them one at a time is hopeless, so we count them in large bundles called moles.

Key result

Amount of substance is one of the seven SI base quantities. Its base unit is the mole (mol).
One mole of any substance is the amount containing a number of particles equal to the Avogadro constant $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ .

So $1\ \text{mol}$ of helium contains $6.02 \times 10^{23}$ helium atoms, and $1\ \text{mol}$ of nitrogen gas contains $6.02 \times 10^{23}$ $\text{N}_2$ molecules. If a sample contains $N$ particles,

N = nN_A

where $n$ is the amount of substance in moles.

The molar mass $M$ is the mass of one mole. Numerically it is the relative molecular mass in grams per mole: helium $4.0\ \text{g mol}^{-1}$ , nitrogen $28\ \text{g mol}^{-1}$ , oxygen $32\ \text{g mol}^{-1}$ . A sample of mass $m$ contains

n = \frac{m}{M}

and the mass of one molecule is $M/N_A$ .

Watch out

Molar mass in kilograms. In SI calculations the molar mass must be in $\text{kg mol}^{-1}$ : helium is $4.0 \times 10^{-3}\ \text{kg mol}^{-1}$ , not $4.0$ . Forgetting this gives answers wrong by a factor of $1000$ .

The gas laws and the ideal gas

Experiments on a fixed mass of gas show three simple relationships:

Law	Constant	Relationship
Boyle's law	temperature	$p \propto 1/V$ , so $pV = \text{constant}$
Charles's law	pressure	$V \propto T$
Pressure law	volume	$p \propto T$

In each case $T$ is the thermodynamic temperature in kelvin. Combining them, for a fixed amount of gas, $pV \propto T$ .

Definition

An ideal gas is a gas that obeys $pV \propto T$ at all pressures, volumes and temperatures, where $T$ is the thermodynamic temperature.

No real gas is exactly ideal, but real gases come very close at low pressure and at temperatures well above their boiling point, when the molecules are far apart and moving fast. Air at room temperature and atmospheric pressure is very nearly ideal.

The equation of state

The constant of proportionality in $pV \propto T$ is proportional to the amount of gas. Writing it per mole gives the equation of state.

Key result

pV = nRT

$p$ : pressure (Pa)
$V$ : volume ( $\text{m}^3$ )
$n$ : amount of substance (number of moles)
$R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$ : the molar gas constant
$T$ : thermodynamic temperature (K)

Writing it per molecule instead:

Key result

pV = NkT

where $N$ is the number of molecules and $k$ is the Boltzmann constant:

k = \frac{R}{N_A} = \frac{8.31}{6.02 \times 10^{23}} = 1.38 \times 10^{-23}\ \text{J K}^{-1}

The two forms are the same equation, because $nR = (N/N_A)R = Nk$ . Use $pV = nRT$ when you are given moles or masses, and $pV = NkT$ when you are dealing with numbers of molecules.

For a fixed amount of gas changing from one state to another, $nR$ is constant, so

\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}

This avoids needing $n$ at all.

Graphs of ideal gas behaviour

At constant temperature, $p = nRT/V$ : a graph of $p$ against $V$ is a $1/V$ curve called an isotherm. Higher temperatures give isotherms further from the origin.

y = 2/x y = 4/x y = 6/x

The three curves are isotherms for the same gas at three temperatures in the ratio $1:2:3$ (inner curve coldest). For a fixed amount of gas:

$p$ against $1/V$ at constant $T$ is a straight line through the origin.
$pV$ against $T$ (in K) is a straight line through the origin, gradient $nR$ .
$V$ against $\theta$ (in $^\circ$ C) at constant pressure is a straight line that, extrapolated, meets $V = 0$ at $\theta = -273\ ^\circ\text{C}$ .

Ideal gas calculations

Convert every temperature to kelvin: $T = \theta + 273$ .
Convert volumes to $\text{m}^3$ ( $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$ , $1\ \text{litre} = 10^{-3}\ \text{m}^3$ ) and pressures to Pa.
If the amount of gas is fixed and the state changes, use $p_1V_1/T_1 = p_2V_2/T_2$ .
If you need the amount of gas, use $pV = nRT$ (moles) or $pV = NkT$ (molecules).
Convert between mass, moles and molecules with $n = m/M$ and $N = nN_A$ .

Worked examples

Moles and molecules in a box

A box of volume $1.0\ \text{m}^3$ contains air at $1.01 \times 10^{5}\ \text{Pa}$ and $273\ \text{K}$ . Calculate the amount of air in moles and the number of molecules.

Solution

n = \frac{pV}{RT} = \frac{1.01 \times 10^{5} \times 1.0}{8.31 \times 273} = 44.5\ \text{mol}

N = nN_A = 44.5 \times 6.02 \times 10^{23} = 2.68 \times 10^{25}

Tyre pressure on a hot day

A car tyre contains air at $2.40 \times 10^{5}\ \text{Pa}$ at $15\ ^\circ\text{C}$ . After a long drive the air is at $40\ ^\circ\text{C}$ . Assuming the volume of the tyre is constant, calculate the new pressure.

Solution

Fixed amount and fixed volume, so $p/T$ is constant. Temperatures in kelvin: $288\ \text{K}$ and $313\ \text{K}$ .

p_2 = p_1\frac{T_2}{T_1} = 2.40 \times 10^{5} \times \frac{313}{288} = 2.61 \times 10^{5}\ \text{Pa}

Using $15$ and $40$ directly would give a pressure of $6.4 \times 10^{5}\ \text{Pa}$ , wildly wrong.

Filling balloons from a cylinder

A cylinder of volume $0.050\ \text{m}^3$ contains helium at $2.0 \times 10^{7}\ \text{Pa}$ and $300\ \text{K}$ . (a) Calculate the mass of helium ( $M = 4.0\ \text{g mol}^{-1}$ ). (b) How many balloons, each of volume $0.010\ \text{m}^3$ at $1.05 \times 10^{5}\ \text{Pa}$ and $300\ \text{K}$ , can be filled?

Solution

(a)

n = \frac{pV}{RT} = \frac{2.0 \times 10^{7} \times 0.050}{8.31 \times 300} = 401\ \text{mol}, \qquad m = nM = 401 \times 4.0 \times 10^{-3} = 1.6\ \text{kg}

(b) At constant temperature $pV$ is constant. At $1.05 \times 10^{5}\ \text{Pa}$ the helium would occupy

V = \frac{2.0 \times 10^{7} \times 0.050}{1.05 \times 10^{5}} = 9.52\ \text{m}^3

But the cylinder cannot empty below atmospheric pressure: $0.050\ \text{m}^3$ of helium at $1.05 \times 10^{5}\ \text{Pa}$ stays inside. Available volume $= 9.52 - 0.05 = 9.47\ \text{m}^3$ , enough for $947$ complete balloons.

A rising bubble

An air bubble is released from the bottom of a lake $20\ \text{m}$ deep, where the temperature is $7\ ^\circ\text{C}$ . At the surface the pressure is $1.0 \times 10^{5}\ \text{Pa}$ and the temperature is $17\ ^\circ\text{C}$ . The density of water is $1000\ \text{kg m}^{-3}$ . Calculate the ratio of the bubble's volume at the surface to its volume at the bottom.

Solution

Pressure at the bottom: $p_1 = 1.0 \times 10^{5} + \rho gh = 1.0 \times 10^{5} + 1000 \times 9.81 \times 20 = 2.96 \times 10^{5}\ \text{Pa}$ .

\frac{V_2}{V_1} = \frac{p_1}{p_2} \times \frac{T_2}{T_1} = \frac{2.96 \times 10^{5}}{1.0 \times 10^{5}} \times \frac{290}{280} = 3.1

Connecting two containers

Container A ( $2.0\ \text{litres}$ ) holds gas at $3.0 \times 10^{5}\ \text{Pa}$ ; container B ( $3.0\ \text{litres}$ ) holds the same gas at $1.0 \times 10^{5}\ \text{Pa}$ . Both are at the same temperature. A tap between them is opened and the temperature stays constant. Calculate the final pressure.

Solution

At constant temperature, the amount of gas is proportional to $pV$ , and the total amount is conserved:

p_{\text{final}}(V_A + V_B) = p_AV_A + p_BV_B

p_{\text{final}} = \frac{3.0 \times 10^{5} \times 2.0 + 1.0 \times 10^{5} \times 3.0}{5.0} = 1.8 \times 10^{5}\ \text{Pa}

Litres can be used here because the units cancel; with $pV = nRT$ they would have to be in $\text{m}^3$ .

Density of air

Show that the density of air ( $M = 29\ \text{g mol}^{-1}$ ) at $1.01 \times 10^{5}\ \text{Pa}$ and $20\ ^\circ\text{C}$ is about $1.2\ \text{kg m}^{-3}$ .

Solution

Density $\rho = m/V = nM/V$ and $n/V = p/(RT)$ :

\rho = \frac{pM}{RT} = \frac{1.01 \times 10^{5} \times 0.029}{8.31 \times 293} = 1.20\ \text{kg m}^{-3}

Investigating Boyle's law

Apparatus: a column of air trapped in a thick-walled glass tube above oil; a Bourdon pressure gauge connected to the oil reservoir; a foot pump to increase the pressure; a scale beside the tube to read the length of the air column.

Method: increase the pressure in steps with the pump, wait each time for the temperature of the air to return to room temperature (compressing it warms it), then read the pressure $p$ and the length $l$ of the air column. Since the tube has uniform cross-section, $V \propto l$ .

Analysis: plot $p$ against $1/l$ . A straight line through the origin shows $p \propto 1/V$ at constant temperature.

Variables: independent: pressure; dependent: volume (length); controlled: temperature and mass of gas (no leaks).

Errors and improvements: parallax when reading the meniscus (read at eye level); temperature rising on compression (wait before reading); leaks (check that the reading is steady). Safety: high pressures in glass; use a safety screen and do not exceed the rated pressure.

A similar arrangement with a flask of air in a water bath, connected to a pressure gauge, investigates the pressure law: plot $p$ against $\theta$ and extrapolate to $p = 0$ to estimate absolute zero.

Watch out

Using $^\circ$ C in the gas equation. $pV = nRT$ only works with $T$ in kelvin. Any answer involving gas temperatures that does not add $273$ is almost certainly wrong.

Watch out

Confusing $n$ and $N$ . $n$ is the number of moles; $N$ is the number of molecules. They go with $R$ and $k$ respectively.

Exam tip

"What is meant by an ideal gas?" (1 mark): a gas that obeys $pV \propto T$ (where $T$ is thermodynamic temperature) at all values of $p$ , $V$ and $T$ . "A gas that obeys the gas laws" is often not enough; include the relationship.
"State what is meant by the Avogadro constant": the number of particles in one mole of a substance.
Many questions combine the gas equation with kinetic theory: for example find $N$ with $pV = NkT$ , then use it to find the mean kinetic energy or rms speed.
Show unit conversions in your working ( $250\ \text{cm}^3 = 2.50 \times 10^{-4}\ \text{m}^3$ ). A wrong final answer with a clear conversion can still earn method marks.

Summary

Amount of substance is an SI base quantity; unit mol; one mole contains $N_A = 6.02 \times 10^{23}$ particles.
$N = nN_A$ ; $n = m/M$ with $M$ in $\text{kg mol}^{-1}$ .
An ideal gas obeys $pV \propto T$ , with $T$ in kelvin.
Equation of state: $pV = nRT$ (moles) or $pV = NkT$ (molecules); $k = R/N_A$ .
For a fixed amount: $p_1V_1/T_1 = p_2V_2/T_2$ .
Real gases are close to ideal at low pressure and high temperature.

Practice questions

Question

Calculate the number of molecules in $1.0\ \text{cm}^3$ of gas at $1.0 \times 10^{-8}\ \text{Pa}$ (a good laboratory vacuum) and $300\ \text{K}$ .
Calculate the volume of $1.0\ \text{mol}$ of an ideal gas at $1.01 \times 10^{5}\ \text{Pa}$ and $0\ ^\circ\text{C}$ .
A gas occupies $500\ \text{cm}^3$ at $27\ ^\circ\text{C}$ and $1.0 \times 10^{5}\ \text{Pa}$ . It is compressed to $200\ \text{cm}^3$ and its temperature rises to $87\ ^\circ\text{C}$ . Calculate the new pressure.
Calculate the mass of oxygen ( $M = 32\ \text{g mol}^{-1}$ ) in a $10\ \text{litre}$ cylinder at $1.5 \times 10^{7}\ \text{Pa}$ and $290\ \text{K}$ .
Show that $k = R/N_A$ by comparing the two forms of the equation of state.
Explain why real gases deviate from ideal behaviour at high pressures.
A sealed flask contains gas at $1.0 \times 10^{5}\ \text{Pa}$ and $20\ ^\circ\text{C}$ . The flask can withstand a pressure of $1.8 \times 10^{5}\ \text{Pa}$ . Calculate the maximum temperature in $^\circ$ C to which it can be heated.
A weather balloon contains $8.0\ \text{m}^3$ of helium at ground level, where the pressure is $1.0 \times 10^{5}\ \text{Pa}$ and the temperature is $15\ ^\circ\text{C}$ . It rises to a height where the pressure is $2.5 \times 10^{4}\ \text{Pa}$ and the temperature is $-50\ ^\circ\text{C}$ . (a) Calculate the new volume. (b) Calculate the number of helium atoms. (c) The balloon fabric can stretch to a maximum volume of $30\ \text{m}^3$ . Determine whether the balloon bursts at this height.

Answers

$N = pV/kT = 1.0 \times 10^{-8} \times 1.0 \times 10^{-6}/(1.38 \times 10^{-23} \times 300) = 2.4 \times 10^{6}$ molecules.
$V = nRT/p = 1.0 \times 8.31 \times 273/1.01 \times 10^{5} = 2.25 \times 10^{-2}\ \text{m}^3$ ( $22.5$ litres).
$p_2 = p_1V_1T_2/(V_2T_1) = 1.0 \times 10^{5} \times 500 \times 360/(200 \times 300) = 3.0 \times 10^{5}\ \text{Pa}$ .
$n = pV/RT = 1.5 \times 10^{7} \times 0.010/(8.31 \times 290) = 62.2\ \text{mol}$ ; $m = 62.2 \times 0.032 = 2.0\ \text{kg}$ .
$pV = nRT$ and $pV = NkT$ , so $nR = Nk$ . With $N = nN_A$ : $nR = nN_Ak$ , so $k = R/N_A$ .
At high pressure the molecules are close together: their own volume is no longer negligible compared with the volume of the container, and intermolecular attractive forces become significant.
$T_2 = T_1 \times p_2/p_1 = 293 \times 1.8 = 527\ \text{K} = 254\ ^\circ\text{C}$ .
(a) $V_2 = V_1 \times (p_1/p_2) \times (T_2/T_1) = 8.0 \times 4.0 \times 223/288 = 24.8\ \text{m}^3$ . (b) $N = pV/kT = 1.0 \times 10^{5} \times 8.0/(1.38 \times 10^{-23} \times 288) = 2.0 \times 10^{26}$ . (c) $24.8\ \text{m}^3 < 30\ \text{m}^3$ , so it does not burst at this height (assuming the helium is at the temperature of the surroundings and the pressure inside is the same as outside).

Amount of substance and the mole

The gas laws and the ideal gas

The equation of state

Graphs of ideal gas behaviour

Worked examples

Practice questions

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