Gas volumes and the ideal gas equation

What you’ll learn

• Relate gas volumes to reacting mole ratios.
• Use the ideal gas equation with SI units.
• Evaluate measurements of gas amount and molar volume.

Understand the idea

At the same temperature and pressure, equal volumes of gases contain equal amounts. Consequently, a balanced equation gives volume ratios as well as mole ratios for its gaseous substances. A quoted molar volume belongs to specified conditions; it is not a universal constant.

The ideal gas equation connects pressure, volume, amount and absolute temperature. With $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$, use pressure in pascals and volume in cubic metres. Add 273 to a Celsius temperature for the usual examination approximation to kelvin, unless greater precision is required.

A molar-volume determination compares measured gas volume with an amount calculated from a limiting reagent. Leakage lowers the collected volume; trapped air can raise it. Measure temperature and pressure, consider gas dissolved in the liquid, and account for water vapour if the problem supplies the information. Repeat measurements test consistency but do not remove a systematic gas leak.

Key formulae

$$V=nV_m$$

$$pV=nRT$$

$$T/\mathrm K=\theta/{}^\circ\mathrm C+273$$

$$1\ \mathrm{dm^3}=10^{-3}\ \mathrm{m^3},\qquad1\ \mathrm{cm^3}=10^{-6}\ \mathrm{m^3}$$

Worked example

A gas occupies $2.00\ \mathrm{dm^3}$ at $100\ \mathrm{kPa}$ and $300\ \mathrm K$. Calculate its amount using $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

1. Convert $p=100000$ Pa and $V=0.00200\ \mathrm{m^3}$.
2. Rearrange to $n=pV/(RT)$.
3. Substitute: $n=100000(0.00200)/(8.31\times300)=0.0802246\ldots$ mol.

Answer: $0.0802$ mol to three significant figures.