3.1.2.3 - The Ideal Gas Equation

What the equation means

The ideal gas equation is a single equation that connects pressure, volume, temperature, and moles for a gas:

The symbolic form is short, but each symbol must be interpreted with care.

In this equation:

• $p$ = pressure in pascals, Pa
• $V$ = volume in cubic metres, $\mathrm{m^{3}}$
• $n$ = amount of gas in moles, mol
• $R$ = gas constant
• $T$ = temperature in kelvin, K

One useful boundary to remember is that you do not need to memorise the value of $R$. If a calculation needs it, the value will be given.

This equation is especially useful because it gives a good model for many gases, especially at low pressure and at temperatures well above condensation, provided you use the correct units and substitute carefully. It is still an idealised model, so you should not treat it as exact for every gas under all conditions.

The SI unit conversions that matter

Most mistakes with the ideal gas equation happen before the calculation even begins. The algebra may be right, but if the values are not in SI units, the final answer will be wrong.

The essential unit conversions are:

• pressure: kPa to Pa by multiplying by 1000
• volume: $\mathrm{dm^{3}}$ to $\mathrm{m^{3}}$ by dividing by 1000
• volume: $\mathrm{cm^{3}}$ to $\mathrm{m^{3}}$ by multiplying by $1.0 \times 10^{-6}$
• temperature: °C to K by adding 273

These unit choices are not arbitrary. $R$ is usually given in $\mathrm{J}\,\mathrm{K^{-1}}\,\mathrm{mol^{-1}}$, and $1\,\mathrm{J} = 1\,\mathrm{Pa}\,\mathrm{m^{3}}$, so using Pa, $\mathrm{m^{3}}$, mol, and K keeps the units on both sides of $pV = nRT$ consistent.

Worked example:

A gas has pressure $95.0\,\mathrm{kPa}$, volume $250\,\mathrm{cm^{3}}$, and temperature $27 {}^\circ\mathrm{C}$.

• $p = 95.0\,\mathrm{kPa} = 95 000\,\mathrm{Pa}$
• $V = 250\,\mathrm{cm^{3}} = 250 \times 10^{-6}\,\mathrm{m^{3}} = 2.50 \times 10^{-4}\,\mathrm{m^{3}}$
• $T = 27 {}^\circ\mathrm{C} = 300\,\mathrm{K}$

The diagram below summarises the unit conversions you should check before substituting values into $pV = nRT$.

[DIAGRAM: asset_name: 1.2.3 - The Ideal Gas Equation - Diagram 1; asset_slug: 1.2.3 - The Ideal Gas Equation - Diagram 1; recommended_method: retained_png; description: Converting pressure, volume and temperature into the SI units required by the ideal gas equation.]

The best habit is to rewrite every value in SI units before rearranging the equation or reaching for a calculator.

Finding the number of moles

A very common use of the equation is to calculate the amount of gas present. To do that, rearrange the equation so that $n$ is the subject:

This is a good place to work methodically:

1. convert all values into SI units
2. write the rearranged equation
3. substitute with units
4. give the answer to a sensible number of significant figures

Worked example:

Calculate the number of moles of gas in $500\,\mathrm{cm^{3}}$ at $100 000\,\mathrm{Pa}$ and $25 {}^\circ\mathrm{C}$.

First convert the units:

• $V = 500\,\mathrm{cm^{3}} = 5.00 \times 10^{-4}\,\mathrm{m^{3}}$
• $T = 25 {}^\circ\mathrm{C} = 298\,\mathrm{K}$

Now substitute into the equation:

The step students most often miss is the volume conversion. $500\,\mathrm{cm^{3}}$ is not $0.500\,\mathrm{m^{3}}$; it is $5.00 \times 10^{-4}\,\mathrm{m^{3}}$.

This reflection matters because the chemistry is usually simple here, but one careless unit slip changes the whole answer.

Rearranging for volume or pressure

The same equation can be rearranged for other quantities. For example:

and

This means the ideal gas equation can be used to find:

• the volume occupied by a known amount of gas
• the pressure exerted by a known amount of gas
• the number of moles present in a measured sample

For a fixed amount of gas, if pressure stays the same, increasing temperature increases volume. If volume stays the same, increasing temperature increases pressure. The equation helps you calculate those changes rather than just describing them.

Worked example:

Find the volume of $0.250\,\mathrm{mol}$ of gas at $300\,\mathrm{K}$ and $1.00 \times 10^{5}\,\mathrm{Pa}$.

So the gas occupies $6.23 \times 10^{-3}\,\mathrm{m^{3}}$, which is $6.23\,\mathrm{dm^{3}}$.

If pressure is the unknown, rearrange to $p = nRT/V$ and check the volume is in $\mathrm{m^{3}}$ before substituting.

Worked example:

Find the pressure of $0.200\,\mathrm{mol}$ of gas at $300\,\mathrm{K}$ in $5.00\,\mathrm{dm^{3}}$.

First convert the volume:

• $V = 5.00\,\mathrm{dm^{3}} = 5.00 \times 10^{-3}\,\mathrm{m^{3}}$

Now substitute into the equation:

So the pressure is $9.97 \times 10^{4}\,\mathrm{Pa}$, which is about $1.00 \times 10^{5}\,\mathrm{Pa}$.

You can treat that as a checklist to run through before pressing $=$.

Using gas measurements to find a molar mass

A known mass of a volatile liquid can be vaporised in a suitable heated gas apparatus. Measure the mass that vaporised, the volume occupied by its vapour, its temperature and pressure. The sample must be fully vaporised, with no leaks or unaccounted air; the apparatus and heating method must suit the substance and its hazards. A hot-water bath may replace a flame for a flammable liquid.

Calculate $n=pV/RT$, then $M=m/n$ using mass in grams. For example, 0.250 g of vapour occupying $100.0\,\mathrm{cm^3}$ at 373 K and 100 kPa gives $n=0.003226\,\mathrm{mol}$, using $R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}$. Its molar mass is $77.5\,\mathrm{g\,mol^{-1}}$. A measured volume that includes trapped air would make the calculated amount too large and the inferred molar mass too small.

For a fixed amount of gas, the equation also gives $p_1V_1/T_1=p_2V_2/T_2$. The earlier proportional statements about volume or pressure assume this amount stays constant; a leaking container would break that assumption.

These calculations are usually short, but they reward disciplined method. If you choose the correct rearrangement and keep the units visible throughout, the result is usually very dependable.

For a volatile liquid, the gas volume is the increase in the syringe reading, and the mass delivered is the decrease in the delivery syringe's mass. Cap that syringe between measurements to limit evaporation. If the delivered mass is $0.400\ \mathrm{g}$ and each balance reading has uncertainty $\pm0.0005\ \mathrm{g}$, the mass difference has uncertainty $\pm0.0010\ \mathrm{g}$, or 0.25%. For $M=mRT/(pV)$, add percentage uncertainties in the measured factors to estimate the maximum percentage uncertainty in $M$. If these total 1.5% and $M=80.0\ \mathrm{g\,mol^{-1}}$, the estimated range is $80.0\pm1.2$, or 78.8–81.2 $\mathrm{g\,mol^{-1}}$. An uncertainty already stated for a difference must not be doubled again.