# What are the units used for the ideal gas law?

Dec 25, 2013

The equation for the Ideal Gas Law is:

#### Explanation:

$P V = n R T$

On the whole, this is an easy equation to remember and use.

The problems lie almost entirely in the units.

SI units

Pressure, $P$

Pressure is measured in pascals ($\text{Pa}$) — sometimes expressed as newtons per square metre ($\text{N·m"^"-2}$). These mean exactly the same thing.

Be careful if you are given pressures in kilopascals ($\text{kPa}$). For example, $\text{150 kPa = 150 000 Pa}$. You must make that conversion before you use the ideal gas law.

The bar is “almost” an SI unit.

$\text{1 bar = 100 kPa = 100 000 Pa}$

Volume, $\text{V}$

This is one place for you to go wrong when you use the Ideal Gas Law.

That's because the SI base unit of volume is the cubic metre (${\text{m}}^{3}$) — not ${\text{cm}}^{3}$ or ${\text{dm}}^{3}$ or $\text{L}$.

$\text{1 m"^3 = "1000 dm"^3 = "1000 L" = 10^6 "cm"^3 = 10^6 "mL}$

Thus, if you are inserting values of volume into the equation, you first have to convert them into cubic metres.

Number of moles, $n$

This is easy, of course — the units are $\text{mol}$.

The gas constant, $R$

You will usually be given the value for $R$ if you need it.

The SI value for $R$ is

$R = \text{8.314 Pa·m"^3·"K"^"-1""mol"^"-1" = "8.314 J·K"^"-1""mol"^"-1" = "8.314 kPa·dm"^3·"K"^"-1""mol"^"-1" =" 8.314 kPa·L·K"^"-1""mol"^"-1}$

The temperature, $T$

The temperature has to be in kelvins.

Don't forget to add 273.15 if you are given a Celsius temperature.

ALWAYS make sure that the units you use for $R$ match the units for $P , V , n$, and $T$.

Non-SI units

The major difference will be that the pressure is given in atmospheres or millimetres of mercury or bars or millibars, and volume may be in litres or millilitres.

If you have to convert from other pressure measurements:

$\text{760 mm Hg = 1 atm = 101 325 Pa = 101.325 kPa}$

The most common combination is to have pressure in atmospheres or bars and the volume in litres.

For these combinations, the most convenient values of $R$ are

$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1" = "0.083 14 bar·L·K"^"-1""mol"^"-1}$.

Again, ALWAYS make sure that the units you use for $R$ match the units for $P , V , n$, and $T$.