Gas Volumes

When Experiment and Theory Collided

Combining Volumes

Gay-Lussac’s Evidence and Dalton’s Problem

The French chemist Gay-Lussac carried out extensive research into the behaviour of gases, and what he found was surprisingly simple. Through careful experiments, he showed that gases react in neat, predictable ratios of volumes. This was not random. It was consistent every time.

For example, when hydrogen reacts with oxygen, the volume of hydrogen needed is always double the volume of oxygen. In other words, if you have a certain volume of oxygen, you will always need twice that volume of hydrogen for the reaction to work. Similarly, when hydrogen reacts with chlorine, Gay-Lussac observed that they combine in equal volumes and produce double the volume of hydrogen chloride gas. So, if 100 cm³ of hydrogen reacts, it will need 100 cm³ of chlorine, and the result will be 200 cm³ of hydrogen chloride. Clean, simple ratios.

At first, this seemed like a strong and reliable pattern. Gay-Lussac was known for being extremely careful with his experimental work, so his results were trusted. However, they were not accepted straight away. The main issue came from John Dalton, whose atomic theory was already shaping how scientists thought about chemical reactions.

Dalton believed that these results did not fit with his model of how atoms combine. According to his thinking, the only way Gay-Lussac’s observations could be true is if equal volumes of gases contained the same number of atoms. For instance, 100 cm³ of hydrogen would need to contain exactly the same number of atoms as 100 cm³ of chlorine.

Joseph Louis Gay-Lussac

The Scientist Who Made Gas Volumes Measurable

Joseph Louis Gay-Lussac was not someone who stayed within limits. Born in France, he grew up during a period of major political and scientific change, and from early on showed a strong drive to push boundaries.

One of the most extreme examples of this was in 1804, when he ascended in a hydrogen balloon to over 21,000 feet, setting a world record at the time. This was not just for show. He used the journey to study the atmosphere, showing that his approach to science was hands-on and bold.

By his early thirties, Gay-Lussac had already secured major academic positions, becoming Professor of Physics at the Sorbonne and Professor of Chemistry at the École Polytechnique. He built a reputation for both energy and precision, contributing across multiple areas of chemistry rather than focusing on just one.

His most important scientific work came from studying gases. He established the laws of combining volumes, showing that gases react in simple, predictable ratios. This helped move chemistry away from vague ideas and toward measurable patterns.

He also developed industrial methods, including improvements in the production of sulphuric acid, linking scientific theory directly to real-world application. Outside the lab, he held government roles, including overseeing a gunpowder factory and working as chief assayer at the mint.

Gay-Lussac represents a scientist who combined theory, experimentation, and real-world impact. He did not just study chemistry. He tested it at altitude, applied it to industry, and helped define the patterns that students still learn today.

Dalton’s Problem

Why the Volume Evidence Did Not Fit

If that were the case, Dalton reasoned that each hydrogen atom would combine with a chlorine atom one by one until all atoms were used up. But this leads to a problem. Based on his understanding, the reaction would be:

H + Cl → HCl

This means one hydrogen atom reacts with one chlorine atom to form one molecule of hydrogen chloride. So if you start with equal numbers of hydrogen and chlorine atoms, you should end up with the same number of hydrogen chloride molecules. And if the number of particles stays the same, the volume should also stay the same.

So Dalton concluded that the final volume of hydrogen chloride should be 100 cm³, not 200 cm³.

This is where the conflict sits. Gay-Lussac’s experiments clearly showed that the volume doubles, but Dalton’s model predicted that it should stay the same. At the time, Dalton trusted his theory over the experimental pattern, which meant Gay-Lussac’s findings were not fully accepted.

What is important here is not just who was right, but what this reveals about science. Even strong experimental evidence can be questioned if it does not fit existing models. Progress often happens when those models are challenged, adjusted, or replaced.

In this case, resolving the disagreement required a deeper understanding of what particles in gases actually are, something that would later be clarified through Avogadro’s ideas about molecules and gas volumes.

Avogadro’s Law

The Breakthrough: Molecules, Not Single Atoms

Particle Theory

How Avogadro Resolved the Contradiction

At the time, the idea that there could be a direct relationship between the volume of a gas and the number of particles it contains seemed unlikely. It did not quite fit with how scientists were thinking about matter. However, there was a growing acceptance that it might actually be possible if, in gases, the particles were so far apart that their individual sizes did not really matter.

This is an important moment to recognise a recurring pattern in science. As new evidence appears, understanding does not always catch up immediately. Even when strong data suggests something new, it can be difficult to abandon or adapt a theory that already seems well supported. Scientists do not just switch ideas instantly. They test, question, and often resist until a clearer explanation emerges.

The breakthrough came from Avogadro. He proposed that gases like hydrogen and chlorine do not exist as single atoms, but as pairs of atoms bonded together. In other words, they exist as molecules. This simple shift in thinking completely resolves the earlier contradiction.

The reaction can now be written as:

H₂(g) + Cl₂(g) → 2HCl(g)

Once you view the particles correctly, the volume relationship suddenly makes sense. One volume of hydrogen reacts with one volume of chlorine to produce two volumes of hydrogen chloride. The doubling of volume is no longer a problem. It is exactly what you would expect based on the number of molecules formed.

This idea also led to a much wider realisation. Many gases do not exist as single atoms, but as diatomic molecules. Hydrogen, oxygen, nitrogen, and chlorine are all examples of this pattern. Recognising this was a major step forward in understanding how substances behave at the particle level.

Avogadro’s law also highlights the importance of conditions. Gases expand and contract easily depending on temperature and pressure, so comparisons of volume only make sense if these conditions are kept constant. Without controlling temperature and pressure, the relationships between volumes would break down.

Despite the strength of Avogadro’s explanation, his ideas were not accepted immediately. Dalton’s influence at the time was significant, and many scientists were reluctant to move away from his model. As a result, Avogadro’s work was overlooked during his lifetime. In fact, he died before the scientific community fully recognised that his explanation was correct.

This is a useful reminder for you as a student. Scientific progress is not always linear. Even accurate ideas can take time to be accepted, especially when they challenge established thinking.

Practice Question

Hydrocarbon Combustion and Gas Volumes

A hydrocarbon gas, C_xH_y, burns in oxygen to form carbon dioxide and water.

In an experiment, 40 cm³ of C_xH_y and 500 cm³ of oxygen are mixed. After reaction and cooling, the remaining gas volume is 420 cm³.

This gas contains carbon dioxide and unreacted oxygen. It is shaken with sodium hydroxide solution, which absorbs the carbon dioxide. The remaining gas volume is 260 cm³.

(a) Write a balanced equation for the reaction between carbon dioxide and sodium hydroxide solution. Include state symbols.

(b) What volume of carbon dioxide is formed in the reaction?

(c) What volume of oxygen is used in reacting with C_xH_y?

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Model Answer

Gas Volume Analysis

(a)

Carbon dioxide reacts with sodium hydroxide solution to form sodium carbonate and water.

CO₂(g) + 2NaOH(aq) → Na₂CO₃(aq) + H₂O(l)

(b)

Before absorption by sodium hydroxide, the gas volume is 420 cm³.

After absorption, the remaining gas volume is 260 cm³.

The decrease in volume is the volume of carbon dioxide absorbed:

420 − 260 = 160 cm³

Therefore, 160 cm³ of carbon dioxide is formed.

(c)

The gas remaining after sodium hydroxide treatment is unreacted oxygen.

Unreacted oxygen = 260 cm³.

Initial oxygen = 500 cm³.

Oxygen used:

500 − 260 = 240 cm³

Therefore, 240 cm³ of oxygen reacted with C_xH_y.

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Practice Question

Determining the Formula of a Hydrocarbon

A hydrocarbon gas, C_xH_y, burns completely in oxygen to form carbon dioxide and water.

From the previous parts, 160 cm³ of carbon dioxide is formed and 240 cm³ of oxygen is used when 40 cm³ of C_xH_y is burned.

(d) Complete the equation by providing the balancing number for the oxygen molecules.

(e) Use the gas volumes to determine the value of x and then y.

(f) Write the formula of C_xH_y.

(g) Use the balanced equation to determine the mass of water produced when 40 cm³ of C_xH_y is burned.

(h) The density of water is 1 g cm⁻³. What volume of water is obtained?

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Model Answer

Formula and Water Produced

(d)

The general combustion equation is:

C_xH_y + (x + y/4)O₂ → xCO₂ + (y/2)H₂O

Therefore, the balancing number for oxygen is x + y/4.

(e)

40 cm³ of hydrocarbon produces 160 cm³ of CO₂.

Ratio C_xH_y : CO₂ = 40 : 160 = 1 : 4

Therefore, x = 4.

40 cm³ of hydrocarbon uses 240 cm³ of oxygen.

Ratio C_xH_y : O₂ = 40 : 240 = 1 : 6

Therefore, x + y/4 = 6.

Since x = 4, then 4 + y/4 = 6.

y/4 = 2, so y = 8.

(f)

The formula of the hydrocarbon is:

C₄H₈

Balanced equation:

C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

(g)

At room temperature and pressure, 1 mol of gas occupies 24 000 cm³.

Moles of C₄H₈ burned:

40 ÷ 24 000 = 1.67 × 10⁻³ mol

From the equation, 1 mol of C₄H₈ produces 4 mol of H₂O.

Moles of water:

4 × 1.67 × 10⁻³ = 6.67 × 10⁻³ mol

Mass of water:

6.67 × 10⁻³ × 18 = 0.12 g

(h)

Density = 1 g cm⁻³, so 1 g of water has a volume of 1 cm³.

Therefore, 0.12 g of water has a volume of:

0.12 cm³

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Author

Luke Edwards-Stuart

Chemistry teacher, curriculum specialist and educational leader. Luke runs the free student resource website a-levelchemistry.co.uk, supporting students with high-quality Chemistry content.

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