The ideal gas law

In a nutshell

For a constant temperature, the pressure of an ideal gas is inversely proportional to its volume. For a constant volume, the pressure of an ideal gas is directly proportional to its temperature. For a constant pressure, the volume of an ideal gas is directly proportional to its temperature. These relationships can be combined to find equations of state for an ideal gas.

Equations

description	equation
Initial and final state of a gas	$\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$
Ideal gas equations of state	$pV=nRT \\ pV=NkT$
Boltzmann constant conversion	$R=kN_A$

Constants

constant	symbol	value
$Avogadro\ constant$	$N_A$	$6.02\times10^{23}\ mol^{-1}$
$molar\ gas\ constant$	$R$	$8.31\ J\ K^{-1}\ mol^{-1}$
$Boltzmann\ constant$	$k$	$1.38\times10^{-23}\ J\ K^{-1}$

Variable definitions

quantity name	symbol	derived units	si base units
$pressure$	$p$	$Pa$	$kg \ m^{-1} \ s^{-2}$
$volume$	$V$	$m^3$	$m^3$
$temperature$	$T$	$K$	$K$
$number\ of\ atoms$	$N$
$number\ of\ moles$	$n$	$mol$	$mol$

Gas laws

The relationship between the pressure, volume and temperature of an ideal gas can be described by a set of gas laws.

Pressure and volume

For a constant mass and temperature (isothermal), the pressure exerted by an ideal gas is inversely proportional to its volume:

$p\propto\dfrac{1}{V}$

Physics; Thermal physics; KS5 Year 12; The ideal gas law

This means that for an ideal gas $pV=constant$ . This relationship can be seen through a $p$ - $V$ graph:

Physics; Thermal physics; KS5 Year 12; The ideal gas law

Pressure and temperature

For a constant mass and volume (isochoric), the pressure exerted by an ideal gas is directly proportional to its temperature:

$p\propto T$

Physics; Thermal physics; KS5 Year 12; The ideal gas law

This means that for an ideal gas $\dfrac{p}{T}=constant$ . This relationship can be seen through a $p$ - $T$ graph:

Physics; Thermal physics; KS5 Year 12; The ideal gas law

Volume and temperature

For a constant mass and pressure (isobaric), the volume of an ideal gas is directly proportional to its temperature:

$V\propto T$

This means that for an ideal gas $\dfrac{V}{T}=constant$ . This relationship can be seen through a $V$ - $T$ graph:

Physics; Thermal physics; KS5 Year 12; The ideal gas law

Combining the laws

When the laws above are combined, one finds that for an ideal gas:

$\dfrac{pV}{T}=constant$

This quantity will remain the same if the conditions change from an initial state $1$ to a final state $2$ :

$\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$

Equations of state

The constant from the combined relationship for $1\ mol$ of an ideal gas is known as the molar gas constant $R$ , with a value of $8.31\ J\ K^{-1}\ mol^{-1}$ . One can use it to find what is known as the equation of state of an ideal gas for $n$ moles:

$pV=nRT$

The molar gas constant $R$ is related to the Avogadro constant and another constant through the equation:

$R=kN_A$

This Boltzmann constant $k$ has a value of $1.38\times10^{-23}\ J\ K^{-1}$ and can be substituted into the equation of state to give the version for an $N$ number of particles:

$pV=NkT$

Note: when working with the equations of state the temperature must always be in $K$ !

Example

Calculate the pressure exerted by $50\ mol$ of an ideal gas on a container with a volume of $2\ m^3$ that is kept at a temperature of $20\degree C$ .

Firstly, write down all the known values:

$n=50\ mol\newline V=2\ m^3\newline T=20\degree C$ 

Convert the temperature into $K$ :

$T=20\degree C=293\ K$ 

Next, write down the correct equation of state and rearrange it to find pressure:

$pV=nRT\newline p=\dfrac{nRT}{V}$ 

Substitute all the values into the equation and calculate the pressure:

$p=\dfrac{50\times8.31\times293}{2}=60.9\ kPa$ 

The pressure exerted on the container by the gas is $\underline{60.9\ kPa}$ .