Introduction to Gas Volumes

In this chapter, we are going to look at how we measure the amount of "stuff" in a gas. Unlike solids or liquids, where we usually think about mass, with gases, we almost always talk about volume. The amazing thing about gases is that, under the same conditions, the identity of the gas (whether it's Oxygen, Nitrogen, or Carbon Dioxide) doesn't actually change how much space it takes up! This makes our calculations much simpler once you know the rules.

This topic is a key part of the Amount of substance and calculations section and links closely to your work on Formulae, equations and the mole. We will also see how these ideas apply to real-world contexts like The Ozone Story (OZ), where we look at the gases that make up our atmosphere.

1. Molar Gas Volume at RTP

At Room Temperature and Pressure (RTP), which is roughly \(25^\circ\text{C}\) and \(101\text{ kPa}\), one mole of any gas occupies the same volume. According to your OCR Salters Data Sheet, this value is:

Molar Gas Volume (\(V_m\)) = \(24.0\text{ dm}^3\text{ mol}^{-1}\)

Calculating Moles from Volume

To find the number of moles (\(n\)) in a certain volume of gas at RTP, we use this simple formula:

\(n = \frac{V}{24.0}\)

Where:

  • \(n\) = number of moles (\(\text{mol}\))
  • \(V\) = volume of the gas in \(\text{dm}^3\)

Top Tip: If your volume is given in \(\text{cm}^3\), you must convert it to \(\text{dm}^3\) first by dividing by \(1000\). Alternatively, use the value \(24000\text{ cm}^3\text{ mol}^{-1}\) in your calculation.

Quick Example:

How many moles are in \(480\text{ cm}^3\) of Argon gas at RTP?

1. Convert \(\text{cm}^3\) to \(\text{dm}^3\): \(480 / 1000 = 0.48\text{ dm}^3\)

2. Use the formula: \(n = \frac{0.48}{24.0} = 0.02\text{ mol}\)

Key Takeaway: At RTP, you don't need the molar mass of the gas to find the moles—just the volume!

2. Reacting Gas Volumes

Because equal volumes of gases contain equal numbers of moles (at the same temperature and pressure), we can use the molar ratios in a balanced equation to work out gas volumes directly.

Consider the combustion of methane:

\(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\)

The ratio of \(\text{CH}_4\) to \(\text{O}_2\) is \(1:2\). This means:

  • \(1\text{ mole}\) of \(\text{CH}_4\) reacts with \(2\text{ moles}\) of \(\text{O}_2\)
  • Therefore: \(10\text{ cm}^3\) of \(\text{CH}_4\) reacts with \(20\text{ cm}^3\) of \(\text{O}_2\)

Don't worry if this seems tricky at first! Just remember that for gases, the "big numbers" in the equation tell you the volume ratio as well as the mole ratio.

3. The Ideal Gas Equation

What if we aren't at Room Temperature and Pressure? If the gas is heated or under high pressure, we use the Ideal Gas Equation. This is found on your Data Sheet:

\(PV = nRT\)

The Variables:

  • \(P\) (Pressure): Measured in Pascals (\(\text{Pa}\)). (Note: \(1\text{ kPa} = 1000\text{ Pa}\)).
  • \(V\) (Volume): Measured in \(\text{m}^3\). This is the most common mistake!
  • \(n\) (Number of moles): Measured in \(\text{mol}\).
  • \(R\) (Gas Constant): Always \(8.314\text{ J mol}^{-1}\text{ K}^{-1}\) (on your Data Sheet).
  • \(T\) (Temperature): Measured in Kelvin (\(\text{K}\)).

The "Unit Trap" (How to avoid common errors):

Most students find this section hard because of the units. Here is a quick guide to staying safe:

  • Temperature: To get Kelvin, add \(273\) to the Celsius temperature. (\(T(K) = T(^\circ\text{C}) + 273\)).
  • Volume: To get from \(\text{dm}^3\) to \(\text{m}^3\), divide by \(1000\). To get from \(\text{cm}^3\) to \(\text{m}^3\), divide by \(1,000,000\).
  • Pressure: If the question says \(\text{kPa}\), multiply by \(1000\) to get \(\text{Pa}\).

Key Takeaway: Always convert your units before plugging them into \(PV = nRT\).

4. Composition by Volume

In The Ozone Story (OZ), we study the atmosphere. The "composition by volume" of a gas mixture is simply the percentage of the total volume that a specific gas occupies.

Formula:

\(\% \text{ by volume} = \frac{\text{Volume of specific gas}}{\text{Total volume of gas mixture}} \times 100\)

Because of the rule that volume is proportional to moles, the percentage by volume is the same as the percentage by moles.

Example: The Atmosphere

The Earth's atmosphere is roughly \(78\%\) Nitrogen and \(21\%\) Oxygen by volume. This means in any \(100\text{ cm}^3\) sample of air, there are \(78\text{ cm}^3\) of Nitrogen and \(21\text{ cm}^3\) of Oxygen.

Did you know? We often measure trace gases in the atmosphere, like Ozone, in parts per million (ppm). This is another way of expressing composition: \(1\text{ ppm}\) means \(1\text{ cm}^3\) of a gas in every \(1,000,000\text{ cm}^3\) of the total mixture.

Summary Checklist

  • Can you recall that \(1\text{ mole}\) of gas occupies \(24.0\text{ dm}^3\) at RTP?
  • Do you know how to convert \(^\circ\text{C}\) to \(\text{K}\) and \(\text{dm}^3\) to \(\text{m}^3\)?
  • Can you rearrange \(PV = nRT\) to find a missing variable?
  • Do you understand that the ratio of volumes in a reaction is the same as the ratio of moles in the equation (for gases)?

Next up: If you are comfortable with gas moles, you might want to review Atom economy and percentage yield to see how we calculate the efficiency of these reactions.