2.1.3(f) - The ideal gas equation

What the equation connects

The ideal gas equation is:

Each symbol has a precise meaning:

For this course, the value of $R$ is supplied on the data sheet. Use:

The ideal-gas model treats gas particles as having negligible volume and negligible intermolecular attractions. It is generally a better approximation at low pressure and high temperature, away from condensation. The equation treats the gas as ideal under the conditions in the question. You do not need to derive the equation from kinetic theory here. The assessed skill is using the relationship correctly with SI units.

SI units first

With $R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$, use pressure in Pa, volume in m³ and temperature in K as the standard SI route. Mixing an unconverted pressure or volume with otherwise SI values produces a scale error; using Celsius produces a temperature error. Other dimensionally consistent units can work—for example, kPa × dm³ also equals J—but you must keep the complete unit set compatible. These examples use the SI route expected in the specification.

The temperature must be absolute: 0 °C is about 273 K, not zero on the scale used by this equation. Use absolute pressure rather than a gauge pressure if the question distinguishes them.

Use these conversions often:

For example, $75.0\ \text{cm}^3$ is:

And $31^\circ\text{C}$ is:

Worked example: converting a data set

A gas sample is measured at $105\ \text{kPa}$, $250\ \text{cm}^3$ and $27^\circ\text{C}$.

Convert each value before using $pV = nRT$:

The amount $n$ is already in mol if it is given as a number of moles. If the question gives mass instead, find $n$ from the mole calculation already taught, then use the ideal gas equation.

Calculating amount of gas

Many ideal-gas questions ask for the amount of gas, $n$. Start from:

Divide both sides by $RT$:

The units provide a check: $\mathrm{Pa\,m^3}=\mathrm{J}$, so $pV/(RT)$ has unit mol. A volume of a few tens of cm³ near room conditions corresponds to only a few thousandths of a mole, not many moles.

Keep both R and T in the denominator. On a calculator, use brackets around the full product RT; dividing by R and then multiplying by T gives a different and incorrect result.

Worked example: finding $n$

A sample of gas has volume $48.0\ \text{cm}^3$ at $25^\circ\text{C}$ and $100\ \text{kPa}$. Calculate the amount of gas.

First convert:

Now substitute into the rearranged equation:

To three significant figures:

The answer is small, which is reasonable because $48.0\ \text{cm}^3$ is a small gas volume.

Rearranging for other variables

The equation can be rearranged for any one unknown. Do the algebra before substitution where possible, because it reduces calculator mistakes.

From:

You can make these forms:

Worked example 1: finding volume

A sample contains $0.0200\ \text{mol}$ of gas at $40^\circ\text{C}$ and $105\ \text{kPa}$. Calculate the volume in $\text{dm}^3$.

Convert first:

Use:

The question asks for $\text{dm}^3$, so convert back at the end:

Worked example 2: finding pressure

A sample contains $0.0200\ \text{mol}$ of gas in a volume of $650\ \text{cm}^3$ at $25^\circ\text{C}$. Calculate the pressure in kPa.

Convert:

Use:

The question asks for kPa:

Worked example 3: finding temperature

A sample contains 0.0200 mol of gas in 600 cm³ at 83.14 kPa. Temperature is the unknown, so use $T=pV/(nR)$. With $p=83140$ Pa and $V=6.00\times10^{-4}$ m³:

The equation already gives kelvin; do not add 273 to this result. It is about 27 °C, a reasonable near-room temperature.

Quick Recap

Convert pressure to Pa, volume to m³ and temperature to K before substituting into $pV=nRT$. Convert the calculated result into the requested unit only at the end.