How to Distinguish Gas Amount from Gas Concentration

Imagine two samples of the same gas, both at 25°C and 100 kPa. One occupies 1.0 L, while the other occupies 3.0 L.

Which sample contains more gas? The 3.0 L sample contains three times as many moles. But here’s the interesting part: both samples have exactly the same gas concentration.

This is the difference between gas amount and gas concentration. Amount tells you how much gas is present in total. Concentration tells you how much gas occupies each unit of volume.

Confusing these two quantities is an easy way to lose marks in ideal gas calculations. Let’s work through how to distinguish them, calculate each one, and recognise when experimental corrections are needed.

01Gas amount versus gas concentration

Think of a room full of students.

The number of students in the room tells you the total population. The number of students per cubic metre tells you how crowded the room is.

A larger room could contain more students without being more crowded. Gas amount and gas concentration work in the same way.

The analogy has one limitation: gas particles move freely, and their concentration depends on pressure and temperature as well as volume.

Gas amount: how much gas is present?

The amount of gas is the total quantity of gas particles in a sample, measured in moles (mol).

Its symbol is \(n\).

For example:

• A container holding 0.50 mol of oxygen has an oxygen amount of 0.50 mol.
• A container holding 2.00 mol of oxygen has four times the oxygen amount.

The amount does not tell you how tightly packed the particles are.

Gas concentration: how much gas occupies a given volume?

Molar concentration measures the amount of gas per unit volume.

Its symbol is \(c\).

The equation is:

\[ c=\frac{n}{V} \]

Where:

• \(c\) = molar concentration (mol L\(^{-1}\))
• \(n\) = amount of gas (mol)
• \(V\) = volume of gas (L)

For example, if a container holds 0.50 mol of oxygen in 2.00 L:

\[ c=\frac{0.50}{2.00} \]

\[ c=0.250\text{ mol L}^{-1} \]

The gas concentration is 0.250 mol L\(^{-1}\).

Notice that the amount and concentration describe different things, even though they are mathematically connected.

Quick comparison

The last row is important.

If you double the volume of an ideal gas sample while keeping its temperature and pressure constant, the amount doubles, but its concentration stays the same.

02How the ideal gas law connects amount and concentration

You can calculate gas amount and concentration using the ideal gas equation:

\[ PV=nRT \]

Where:

• \(P\) = absolute pressure (kPa)
• \(V\) = gas volume (L)
• \(n\) = amount of gas (mol)
• \(R\) = universal gas constant
• \(T\) = absolute temperature (K)

Rearranging for the amount:

\[ \boxed{n=\frac{PV}{RT}} \]

To find concentration, start with:

\[ c=\frac{n}{V} \]

Substitute the ideal gas expression for \(n\):

\[ c=\frac{PV}{RTV} \]

Cancel the volume:

\[ \boxed{c=\frac{P}{RT}} \]

This is one of the most useful results in this topic.

It tells you that, for an ideal gas, concentration depends on pressure and temperature. You do not need to know the total volume to calculate it if pressure and temperature are already known.

Why does the volume cancel?

Imagine two identical samples of gas at the same temperature and pressure.

One occupies 1.0 L, and the other occupies 5.0 L.

The larger sample contains five times as many moles, but it also occupies five times the volume.

The ratio \(n/V\) is unchanged.

That is why the ideal gas concentration equation contains pressure and temperature, but not volume.

Gas constant units: which value should you use?

The gas constant \(R\) has different numerical values depending on the units used for pressure and volume.

For most HSC Chemistry calculations using pressure in kPa and volume in litres, use:

\[ R=8.314\text{ kPa L mol}^{-1}\text{ K}^{-1} \]

Remember three important rules:

1. Convert temperature from degrees Celsius to kelvin:

\[ T(\mathrm K)=T(^{\circ}\mathrm C)+273.15 \]

1. Use absolute pressure, not gauge pressure. If a pressure reading is given as gauge pressure, add atmospheric pressure to obtain absolute pressure.

2. Keep your units consistent with the value of \(R\).

A temperature of 25°C, for example, must be entered as 298.15 K, not 25.

A useful decision rule

When a question asks for:

• Amount of gas: use \(n=PV/RT\).
• Concentration of gas: use \(c=n/V\) or \(c=P/RT\).
• Amount and concentration: calculate \(n\) first, then divide by volume. Alternatively, calculate concentration directly using \(P/RT\).

There is no need to calculate the amount first if the question only asks for concentration and the gas behaves ideally.

03An important distinction when conditions change

The relationship between amount, concentration, pressure, and temperature becomes clearer when you consider what happens to a gas under different conditions.

Consider the third situation.

You have a sealed, rigid container. You heat the gas inside it.

A common prediction is that the concentration increases because the pressure increases.

But this is incorrect.

The container cannot expand, and no gas enters or leaves. The amount and volume both remain constant, so:

\[ c=\frac{n}{V}=\text{constant} \]

The pressure increases because the particles move faster and collide with the container walls more frequently and forcefully.

This gives us an important distinction:

An increase in gas pressure does not necessarily mean an increase in gas concentration.

Temperature must also be considered.

04Experimental measurements: why corrections matter

So far, we have assumed that gases behave ideally and that the pressure and volume measurements represent exactly the gas we want to study.

Real experiments are not always that convenient.

Consider an experiment in which a chemical reaction produces a gas. You want to measure how many moles of gas were produced.

A common method is to collect the gas over water.

Collecting gas over water

Imagine a reaction flask connected to a delivery tube. The other end of the tube sits underneath an inverted, water-filled measuring cylinder in a water trough.

As the reaction produces gas, it travels through the delivery tube and enters the measuring cylinder. The gas pushes water out, allowing its volume to be measured.

The apparatus provides a volume reading, but it does not directly measure the number of moles.

To calculate the amount, you also need the gas temperature and pressure.

The first assumption: is the collected gas dry?

Suppose you measure 250 mL of gas.

You know the atmospheric pressure, temperature, and volume. Your first instinct might be to substitute atmospheric pressure directly into the ideal gas equation.

That would be correct if the collected gas were dry and its pressure equalled atmospheric pressure.

However, the gas was collected over water.

Some water evaporates into the measuring cylinder. The collected sample therefore contains:

• The gas produced by the reaction.
• Water vapour.

The measured volume belongs to the entire gas mixture, not just the gas produced by the reaction.

Using atmospheric pressure as though the entire pressure came from the target gas would overestimate the amount of that gas.

The correction: account for water vapour

Dalton’s law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of its components.

For gas collected over water:

\[ P_{\mathrm{total}}= P_{\mathrm{gas}}+P_{\ce{H2O}} \]

When the water levels inside and outside the measuring cylinder are equal, the total gas pressure equals atmospheric pressure.

Therefore:

\[ P_{\mathrm{atm}}= P_{\mathrm{gas}}+P_{\ce{H2O}} \]

Rearranging:

\[ \boxed{P_{\mathrm{gas}}= P_{\mathrm{atm}}-P_{\ce{H2O}}} \]

You must subtract the water vapour pressure before calculating the amount of the dry gas.

The corrected amount is:

\[ n_{\mathrm{gas}}= \frac{P_{\mathrm{gas}}V}{RT} \]

And its concentration is:

\[ c_{\mathrm{gas}}= \frac{P_{\mathrm{gas}}}{RT} \]

The second equation works because the gas concentration is the amount of the target gas divided by the total volume occupied by the gas mixture.

What if the water levels are not equal?

If the water level inside the measuring cylinder is higher than the water level outside, the gas pressure inside is lower than atmospheric pressure.

If the water level inside is lower, the gas pressure is higher than atmospheric pressure.

In either case, a hydrostatic pressure correction is needed before applying Dalton’s law.

Do not automatically assume that the gas pressure equals atmospheric pressure.

Other experimental limitations

Water vapour is not the only possible source of error.

• Gas solubility: Some gases dissolve in water, reducing the measured gas volume.
• Leaks: Gas escaping from the apparatus causes the measured amount to be lower than the actual amount produced.
• Temperature changes: The gas must be measured at its actual temperature, not automatically at room temperature.
• Incorrect volume readings: The measuring cylinder must be read carefully, using the appropriate scale and meniscus position.

These are experimental measurement corrections. They are different from correcting the ideal gas model itself.

05Ideal gases versus real gases

The ideal gas equation assumes that gas particles:

1. Have negligible volume compared with the container.
2. Experience no significant intermolecular attractions or repulsions.

These assumptions allow us to use:

\[ PV=nRT \]

Real gases do not follow these assumptions perfectly.

At relatively low pressures and high temperatures, many gases behave approximately ideally.

At high pressures and low temperatures, deviations become more significant. Particle volume and intermolecular attractions can no longer be ignored.

For a more accurate description, chemists can use real-gas corrections. One way to represent the deviation is through the compressibility factor, \(Z\):

\[ PV=ZnRT \]

Where:

• \(Z=1\) represents ideal gas behaviour.
• \(Z\neq1\) represents deviation from ideal behaviour.

Rearranging gives:

\[ n=\frac{PV}{ZRT} \]

And:

\[ c=\frac{P}{ZRT} \]

The value of \(Z\) is determined experimentally or using a suitable real-gas model.

For ordinary HSC ideal gas calculations, use \(Z=1\) unless the question explicitly provides a correction or asks you to consider non-ideal behaviour. Detailed real-gas equations, such as the van der Waals equation, are extension material rather than routine ideal-gas calculations.

For further reading, see Purdue University’s explanation of deviations from the ideal gas law.

A quick comparison

The key point is that the distinction between amount and concentration does not disappear for real gases. What changes is the equation used to relate them to pressure and temperature.

06Worked examples

07Questions and solutions

The following questions progress from straightforward ideal gas calculations to more demanding situations involving mixtures, temperature changes, and multiple containers.

Question 1

A sealed 2.00 L container holds an ideal gas at an absolute pressure of 110 kPa and a temperature of 27.0°C.

Calculate:

(a) The amount of gas in moles.

(b) The molar concentration of the gas.

Solution 1

The container holds 0.0882 mol of gas at a concentration of 0.0441 mol L\(^{-1}\).

First, convert the temperature:

\[ T=27.0+273.15=300.15\text{ K} \]

Calculate the amount using:

\[ n=\frac{PV}{RT} \]

Substitute:

\[ n=\frac{110\times2.00} {8.314\times300.15} \]

\[ \boxed{n=0.0882\text{ mol}} \]

Now calculate the concentration:

\[ c=\frac{n}{V} \]

\[ c=\frac{0.08816}{2.00} \]

\[ \boxed{c=0.0441\text{ mol L}^{-1}} \]

The container holds 0.0882 mol in total, but each litre contains approximately 0.0441 mol.

Question 2

A student collects a gas over water using an inverted measuring cylinder.

The following measurements are recorded:

The water levels inside and outside the cylinder are equal.

Assuming ideal gas behaviour and negligible dissolution or leakage, calculate:

(a) The amount of dry gas collected.

(b) The concentration of dry gas.

(c) Explain why using atmospheric pressure directly in the ideal gas equation would produce an incorrect result.

Solution 2

The amount of dry gas is 0.0161 mol, and its concentration is 0.0403 mol L\(^{-1}\).

The important step is correcting the pressure before calculating the amount.

Step 1: Convert the volume and temperature.

\[ V=400\text{ mL}=0.400\text{ L} \]

\[ T=20.0+273.15=293.15\text{ K} \]

Step 2: Calculate the pressure of the dry gas.

Using Dalton’s law:

\[ P_{\mathrm{gas}}= P_{\mathrm{atm}}-P_{\ce{H2O}} \]

\[ P_{\mathrm{gas}}=100.5-2.34 \]

\[ P_{\mathrm{gas}}=98.16\text{ kPa} \]

Step 3: Calculate the amount of dry gas.

\[ n=\frac{PV}{RT} \]

\[ n=\frac{98.16\times0.400} {8.314\times293.15} \]

\[ \boxed{n=0.0161\text{ mol}} \]

Step 4: Calculate the concentration.

\[ c=\frac{n}{V} \]

\[ c=\frac{0.01611}{0.400} \]

\[ \boxed{c=0.0403\text{ mol L}^{-1}} \]

Step 5: Explain the experimental correction.

The collected gas contains water vapour as well as the dry gas being investigated.

Using atmospheric pressure directly would treat the water vapour as though it were part of the dry gas. This would overestimate both the dry gas amount and its concentration.

The correct pressure is the partial pressure of the dry gas.

Question 3

A sealed 2.00 L container holds an ideal mixture of oxygen and nitrogen at 298 K.

The total pressure inside the container is 120 kPa, while the partial pressure of oxygen is 24.0 kPa.

Calculate:

(a) The total amount of gas in the container.

(b) The amount of oxygen present.

(c) The oxygen concentration.

(d) The mole fraction of oxygen in the mixture.

Solution 3

The container holds 0.0969 mol of gas in total, including 0.0194 mol of oxygen. The oxygen concentration is 0.00969 mol L\(^{-1}\), and its mole fraction is 0.200.

This question tests whether you can distinguish the concentration of one gas from the concentration of an entire mixture.

Step 1: Calculate the total amount of gas.

Use the total pressure:

\[ n_{\mathrm{total}}=\frac{PV}{RT} \]

\[ n_{\mathrm{total}}= \frac{120\times2.00}{8.314\times298} \]

\[ \boxed{n_{\mathrm{total}}=0.0969\text{ mol}} \]

Step 2: Calculate the amount of oxygen.

Oxygen contributes only part of the total pressure.

Use its partial pressure:

\[ n_{\ce{O2}}= \frac{p_{\ce{O2}}V}{RT} \]

\[ n_{\ce{O2}}= \frac{24.0\times2.00}{8.314\times298} \]

\[ \boxed{n_{\ce{O2}}=0.0194\text{ mol}} \]

Step 3: Calculate the oxygen concentration.

\[ c_{\ce{O2}}=\frac{n_{\ce{O2}}}{V} \]

\[ c_{\ce{O2}}=\frac{0.01937}{2.00} \]

\[ \boxed{c_{\ce{O2}}=0.00969\text{ mol L}^{-1}} \]

You could also calculate this directly using oxygen’s partial pressure:

\[ c_{\ce{O2}}=\frac{p_{\ce{O2}}}{RT} \]

Step 4: Calculate the mole fraction of oxygen.

Mole fraction is the amount of one gas divided by the total amount of gas.

\[ x_{\ce{O2}}= \frac{n_{\ce{O2}}}{n_{\mathrm{total}}} \]

\[ x_{\ce{O2}}= \frac{0.01937}{0.09687} \]

\[ \boxed{x_{\ce{O2}}=0.200} \]

This means oxygen represents 20.0% of the mixture by mole.

The tempting mistake is to use total pressure when calculating oxygen concentration. That would give the total gas concentration, not the concentration of oxygen alone.

For an ideal gas mixture, always use the relevant partial pressure when calculating the concentration of an individual component.

Question 4

A rigid 2.00 L vessel contains an ideal gas mixture at 298 K and a total pressure of 140 kPa.

Gas X makes up 30.0% of the mixture by mole.

The sealed vessel is heated to 348 K. No gas enters or leaves, no reaction occurs, and the volume remains constant.

Calculate:

(a) The initial amount of gas X.

(b) The initial concentration of gas X.

(c) The final total pressure.

(d) The final concentration of gas X.

Explain why the pressure changes even though the concentration of gas X does not.

Solution 4

The amount of gas X is 0.0339 mol, and its concentration remains 0.0169 mol L\(^{-1}\) throughout the heating process. The final total pressure is approximately 163 kPa.

The crucial point is that the container is sealed and rigid.

Step 1: Calculate the initial partial pressure of gas X.

Gas X makes up 30.0% of the mixture by mole.

For an ideal gas mixture, its mole fraction equals its proportion of the total pressure:

\[ p_X=x_XP_{\mathrm{total}} \]

\[ p_X=0.300\times140 \]

\[ p_X=42.0\text{ kPa} \]

Step 2: Calculate the initial amount of gas X.

\[ n_X=\frac{p_XV}{RT} \]

\[ n_X= \frac{42.0\times2.00}{8.314\times298} \]

\[ \boxed{n_X=0.0339\text{ mol}} \]

Step 3: Calculate the initial concentration.

\[ c_X=\frac{n_X}{V} \]

\[ c_X=\frac{0.03388}{2.00} \]

\[ \boxed{c_X=0.0169\text{ mol L}^{-1}} \]

Step 4: Calculate the final total pressure.

The amount and volume of the entire gas mixture remain constant.

Therefore, use the pressure-temperature relationship:

\[ \frac{P_1}{T_1}=\frac{P_2}{T_2} \]

Rearrange:

\[ P_2=\frac{P_1T_2}{T_1} \]

Substitute:

\[ P_2=\frac{140\times348}{298} \]

\[ \boxed{P_2=163.5\text{ kPa}} \]

Step 5: Determine the final concentration of gas X.

No gas X enters or leaves the container. The volume also remains constant.

Therefore:

\[ c_X=\frac{n_X}{V} \]

Both quantities on the right-hand side are unchanged.

Consequently:

\[ \boxed{c_{X,\mathrm{final}}= 0.0169\text{ mol L}^{-1}} \]

The pressure increases because the gas particles move faster at the higher temperature. However, the amount of gas X per litre remains unchanged.

This is why pressure alone cannot tell you the concentration of a gas unless you also know its temperature.

Question 5

Two rigid bulbs contain the same ideal gas. They are connected by a closed valve.

Their initial conditions are:

The valve is opened, allowing the gas to mix.

The combined system is maintained at 300 K, and the total available volume is 3.00 L.

Assume:

• Both initial pressures are absolute pressures.
• Neither bulb leaks.
• No gas reacts or is lost.
• Both samples behave ideally.

Two students make different predictions about the final pressure.

• Student A claims the final pressure is 100 kPa, using a volume-weighted average of the initial pressures.
• Student B claims the final pressure is 210 kPa, by adding the initial pressures.

Determine:

(a) The initial amount of gas in each bulb.

(b) The total amount of gas after mixing.

(c) The final concentration.

(d) The final pressure.

(e) Explain why both students’ predictions are incorrect.

Solution 5

The final pressure is 80.0 kPa, and the final concentration is 0.0321 mol L\(^{-1}\). Each bulb initially contains approximately 0.0481 mol of gas, giving a combined amount of 0.0962 mol.

The important detail is that the two bulbs initially have different temperatures.

You cannot compare their pressures without considering temperature.

Step 1: Calculate the initial amount in bulb A.

\[ n_A=\frac{P_AV_A}{RT_A} \]

\[ n_A=\frac{120\times1.00} {8.314\times300} \]

\[ \boxed{n_A=0.0481\text{ mol}} \]

Step 2: Calculate the initial amount in bulb B.

\[ n_B=\frac{P_BV_B}{RT_B} \]

\[ n_B=\frac{90.0\times2.00} {8.314\times450} \]

\[ \boxed{n_B=0.0481\text{ mol}} \]

Notice something interesting.

Although bulb B has twice the volume and a lower pressure, it contains exactly the same amount of gas as bulb A.

Its higher temperature changes the relationship between pressure and concentration.

Step 3: Calculate the total amount after mixing.

Opening the valve allows the gases to mix, but no gas enters or leaves the combined system.

Therefore, the total amount is conserved:

\[ n_{\mathrm{total}}=n_A+n_B \]

\[ n_{\mathrm{total}}=0.0481+0.0481 \]

\[ \boxed{n_{\mathrm{total}}=0.0962\text{ mol}} \]

Step 4: Calculate the final concentration.

The combined volume is 3.00 L.

\[ c_{\mathrm{final}}= \frac{n_{\mathrm{total}}}{V_{\mathrm{total}}} \]

\[ c_{\mathrm{final}}= \frac{0.0962}{3.00} \]

\[ \boxed{c_{\mathrm{final}}=0.0321\text{ mol L}^{-1}} \]

This is the total molar concentration after mixing.

Step 5: Calculate the final pressure.

The combined gas is now at 300 K.

Use the ideal gas equation:

\[ P_{\mathrm{final}}= \frac{n_{\mathrm{total}}RT_{\mathrm{final}}} {V_{\mathrm{total}}} \]

Substitute:

\[ P_{\mathrm{final}}= \frac{0.0962\times8.314\times300}{3.00} \]

\[ \boxed{P_{\mathrm{final}}=80.0\text{ kPa}} \]

Step 6: Explain why both predictions were wrong.

Student A used a volume-weighted average of the initial pressures:

\[ P_{\mathrm{average}}= \frac{120\times1+90.0\times2}{3} \]

This gives 100 kPa.

That approach would work if both initial samples were at the same temperature and the final temperature were also unchanged.

However, the initial temperatures are different. The amount of gas depends on \(PV/T\), not simply \(PV\).

Student B added the initial pressures to obtain 210 kPa.

This is also incorrect because pressures in separate containers do not add simply because the containers are connected. The final pressure depends on the total amount of gas, the combined volume, and the final temperature.

The correct sequence is to calculate the initial amounts separately, conserve the total amount, and then calculate the final concentration and pressure.

This is the deeper principle: gas amount is conserved during mixing, but concentration and pressure depend on the final conditions.

07What to remember in an HSC question

When you encounter a question involving gas amount or concentration, use the following approach.

Before calculating, check:

• Have you converted temperature into kelvin?
• Are you using absolute pressure?
• Are your pressure and volume units consistent with \(R\)?
• Are you calculating the total gas or one component of a mixture?
• Does the experimental setup require a pressure correction?
• Are ideal gas assumptions appropriate?

One final distinction is worth keeping in mind.

Gas concentration is not the same as gas density. Concentration measures moles per unit volume, while density measures mass per unit volume. They are connected through molar mass:

\[ \rho=cM \]

where \(\rho\) is density and \(M\) is molar mass.

You can explore this relationship further in How to Calculate Gas Density Using the Ideal Gas Law.

Once you can distinguish amount from concentration, you’re also in a much better position to tackle gas stoichiometry. The next step is using balanced chemical equations to calculate the amount of gas produced or consumed, then converting that amount into a volume under specified conditions. See Gas Volumes in Chemical Reactions for HSC Chemistry.

Course note: This resource is written for the Chemistry Stage 6 Syllabus (2017), which remains in use during 2026-2027. The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028, with the first HSC examination under that syllabus in 2029. See NESA’s syllabus implementation information.